Control Systems — Chapter 3: System Transfer Function and Responses

Asked 4 times
2081 Baisakh · 8 marks
2075 Chaitra · 8 marks
2068 Chaitra · 8 marks
2080 Baisakh · 8 marks

Find the overall transfer function C(s)/R(s) using the block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → G1 → node A → summing point S2 (+, −) → G2 → node B → summing point S3 (+) → C(s). From node A, G3 feeds forward to S3 (+). From node B, H1 feeds back to S1 (−). From C(s), H2 feeds back to S2 (−).]

Answer

Name the signals: $E$ = output of $S_1$ , $A$ = output of $G_1$ , $X$ = output of $S_2$ , $B = G_2X$ , and $C = B + G_3A$ . So $E = R - H_1B$ and $X = A - H_2C$ .

Step 1: Move the $H_1$ take-off point behind $G_2$

Since $B = G_2X$ , the feedback $H_1B$ can be taken from $X$ through $G_2H_1$ .

Step 2: Move the $G_3$ take-off point from $A$ to after $S_2$

Because $A = X + H_2C$ , the branch $G_3A$ equals $G_3X + G_3H_2C$ . So $G_3$ now starts at $X$ , and an extra path $G_3H_2$ goes from $C$ into $S_3$ (positive).

Step 3: Combine parallel blocks and the positive loop at $S_3$

$G_2$ and $G_3$ are in parallel from $X$ : $(G_2 + G_3)$ . The extra path forms a positive feedback loop $G_3H_2$ around $S_3$ :

\frac{C}{X} = \frac{G_2 + G_3}{1 - G_3H_2}

 R ─►[G1]─►(Σ)──► X ──►[ (G2+G3)/(1−G3H2) ]──┬──► C
            ▲ − ▲ −                          │
            │   └──────────[H2]◄─────────────┤
            └─[G1G2H1]◄── X                  │

Step 4: Move $S_1$ forward past $G_1$

$R$ becomes $G_1R$ and the $H_1$ feedback becomes $G_1G_2H_1$ from $X$ , so $X = G_1R - G_1G_2H_1X - H_2C$ . The loop on $X$ gives

\frac{X}{U} = \frac{1}{1 + G_1G_2H_1}, \qquad U = G_1R - H_2C

Step 5: Close the $H_2$ loop

Forward path from $U$ : $P = \dfrac{G_2 + G_3}{(1 + G_1G_2H_1)(1 - G_3H_2)}$ , feedback $H_2$ :

\frac{C}{G_1R} = \frac{P}{1 + PH_2} = \frac{G_2 + G_3}{(1 + G_1G_2H_1)(1 - G_3H_2) + H_2(G_2 + G_3)}

Expanding the denominator: $1 + G_1G_2H_1 - G_3H_2 - G_1G_2G_3H_1H_2 + G_2H_2 + G_3H_2$ ; the $G_3H_2$ terms cancel.

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_1G_2H_1 + G_2H_2 - G_1G_2G_3H_1H_2}

Check by Mason's rule

Forward paths $G_1G_2$ , $G_1G_3$ . Loops $-G_1G_2H_1$ , $-G_2H_2$ , and $+G_1G_3H_2G_2H_1$ (path $A \to G_3 \to C \to H_2 \to G_2 \to B \to H_1 \to G_1 \to A$ ). All loops touch each other and both paths, so $\Delta_1 = \Delta_2 = 1$ , giving the same result.

Asked 4 times
2067 Asar (old course) · 8 marks
2081 Bhadra · 8 marks
2080 Bhadra · 8 marks
2079 Bhadra · 6 marks

The open loop transfer function of a unity feedback system is given by G(s) = K/[s(1+sT)], where 'K' is the gain constant and 'T' is time constant. With the gain multiplied by a factor K1 the maximum overshoot of the system is increased from 25% to 50%. Determine K1.

Answer

The closed-loop system is second order; increasing the gain lowers the damping ratio and raises the overshoot.

Closed-loop characteristic equation

\frac{C(s)}{R(s)} = \frac{K}{Ts^2 + s + K} = \frac{K/T}{s^2 + \frac{1}{T}s + \frac{K}{T}}

Comparing with $s^2 + 2\zeta\omega_ns + \omega_n^2$ :

\omega_n = \sqrt{\frac{K}{T}}, \qquad 2\zeta\omega_n = \frac{1}{T} \;\Rightarrow\; \zeta = \frac{1}{2\sqrt{KT}}

So $\zeta \propto 1/\sqrt{K}$ (for fixed $T$ ).

Damping ratio from overshoot

M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} \;\Rightarrow\; \zeta = \frac{-\ln M_p}{\sqrt{\pi^2 + (\ln M_p)^2}}

For $M_p = 0.25$ : $\ln 0.25 = -1.3863$

\zeta_1 = \frac{1.3863}{\sqrt{9.8696 + 1.9218}} = \frac{1.3863}{3.4339} = 0.4037

For $M_p = 0.50$ : $\ln 0.5 = -0.6931$

\zeta_2 = \frac{0.6931}{\sqrt{9.8696 + 0.4805}} = \frac{0.6931}{3.2172} = 0.2155

Gain factor

With gain $K$ : $\zeta_1 = \dfrac{1}{2\sqrt{KT}}$ . With gain $K_1K$ : $\zeta_2 = \dfrac{1}{2\sqrt{K_1KT}}$ . Dividing:

\frac{\zeta_1}{\zeta_2} = \sqrt{K_1} \;\Rightarrow\; K_1 = \left(\frac{\zeta_1}{\zeta_2}\right)^2 = \left(\frac{0.4037}{0.2155}\right)^2 = (1.8738)^2

Answer: $K_1 \approx 3.51$ . The gain must be multiplied by about 3.51 to raise the peak overshoot from 25% to 50%.

Asked 2 times
2082 Baisakh · 8 marks
2080 Bhadra · 8 marks

For a control system shown in figure below, find the value of K & Kt so that the damping ratio (ζ) of system is 0.6 & settling time (ts) is 0.1 sec for the unit step response. [Figure: R(s) → summing point (+, −) → K → summing point (+, −) → 100/(1+0.2s) → 1/(20s) → C(s). Minor loop: output of 100/(1+0.2s) block fed back through Kt to the second summing point (−). Major loop: C(s) fed back with unity gain to the first summing point (−).]

Answer

Step 1: Reduce the minor (tachometer) loop

The minor loop has forward block $\dfrac{100}{1 + 0.2s}$ and feedback $K_t$ taken from its own output:

G_{in}(s) = \frac{\frac{100}{1 + 0.2s}}{1 + \frac{100K_t}{1 + 0.2s}} = \frac{100}{0.2s + 1 + 100K_t}

Step 2: Open-loop transfer function

G(s) = K\cdot\frac{100}{0.2s + 1 + 100K_t}\cdot\frac{1}{20s} = \frac{5K}{s(0.2s + 1 + 100K_t)} = \frac{25K}{s\left[s + 5(1 + 100K_t)\right]}

Step 3: Closed-loop characteristic equation (unity feedback)

s^2 + 5(1 + 100K_t)s + 25K = 0

Comparing with $s^2 + 2\zeta\omega_ns + \omega_n^2 = 0$ :

\omega_n^2 = 25K, \qquad 2\zeta\omega_n = 5(1 + 100K_t)

Step 4: Use the specifications

Settling time (2% criterion):

t_s = \frac{4}{\zeta\omega_n} = 0.1 \;\Rightarrow\; \zeta\omega_n = 40

\omega_n = \frac{40}{0.6} = 66.67\ \text{rad/s}

Step 5: Solve for $K$ and $K_t$

\begin{aligned} 25K &= \omega_n^2 = (66.67)^2 = 4444.4 \;\Rightarrow\; K = 177.8 \\ 5(1 + 100K_t) &= 2\zeta\omega_n = 80 \;\Rightarrow\; 1 + 100K_t = 16 \;\Rightarrow\; K_t = 0.15 \end{aligned}

Answer: $K \approx 177.8$ and $K_t = 0.15$ (using the 2% settling criterion).

If the 5% criterion $t_s = 3/(\zeta\omega_n)$ is used instead, $\zeta\omega_n = 30$ , $\omega_n = 50$ rad/s, giving $K = 100$ and $K_t = 0.11$ .

Note how the tachometer feedback $K_t$ sets the damping, while $K$ sets the natural frequency, so both specifications can be met independently.

Asked 2 times
2081 Bhadra · 8 marks
2068 Baisakh (old course) · 8 marks

From the given block diagram, find the transfer ratio C(s)/R(s) using block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → node A → G1 → summing point S2 (+) → G2 → summing point S3 (+, −) → node B → G3 → C(s). Feedforward from node A to S2 (+). H1 from node B fed back to S2 (−). Unity feedback from C(s) to S3 (−). H2 from C(s) fed back to S1 (−).]

Answer

Name the signals: $A = R - H_2C$ (output of $S_1$ ), $X$ = output of $S_2$ , $B$ = output of $S_3$ , $C = G_3B$ . Then $X = G_1A + A - H_1B$ and $B = G_2X - C$ .

Step 1: Combine $G_1$ and the feedforward path

$G_1$ and the unity feedforward from $A$ are in parallel into $S_2$ :

G_a = 1 + G_1

Step 2: Move the $H_1$ take-off point from $B$ to $C$

Since $B = C/G_3$ , the feedback block becomes $\dfrac{H_1}{G_3}$ , taken from $C$ .

Step 3: Reduce the unity loop around $G_3$

$G_3$ with unity negative feedback to $S_3$ :

\frac{C}{G_2X} = \frac{G_3}{1 + G_3}

So from $X$ to $C$ : $G_b = \dfrac{G_2G_3}{1 + G_3}$ .

Step 4: Reduce the $H_1/G_3$ loop

\frac{C}{X'} = \frac{G_b}{1 + G_b\frac{H_1}{G_3}} = \frac{\frac{G_2G_3}{1+G_3}}{1 + \frac{G_2H_1}{1+G_3}} = \frac{G_2G_3}{1 + G_3 + G_2H_1}

(here $X'$ is the signal $(1 + G_1)A$ entering $S_2$ )

Step 5: Series with $(1 + G_1)$ and close the $H_2$ loop

Forward path: $G_f = \dfrac{(1 + G_1)G_2G_3}{1 + G_3 + G_2H_1}$ , feedback $H_2$ :

\frac{C}{R} = \frac{G_f}{1 + G_fH_2}

 R +    ┌──────────────────────────┐
 ──►(Σ)─►│ (1+G1)G2G3/(1+G3+G2H1) ├──┬──► C
    ▲ -  └──────────────────────────┘  │
    │            ┌────┐                │
    └────────────┤ H2 │◄───────────────┘
                 └────┘

Answer:

\frac{C(s)}{R(s)} = \frac{G_2G_3(1 + G_1)}{1 + G_3 + G_2H_1 + G_1G_2G_3H_2 + G_2G_3H_2}

Check by Mason's rule

Forward paths: $P_1 = G_1G_2G_3$ , $P_2 = G_2G_3$ (via feedforward). Loops: $-G_3$ , $-G_2H_1$ , $-G_1G_2G_3H_2$ , $-G_2G_3H_2$ . All loops touch each other and both paths ( $\Delta_1 = \Delta_2 = 1$ ), giving the same answer.

Asked 2 times
2081 Baisakh · 8 marks
2066 Bhadra (old course) · 8 marks

For the given mechanical system, obtain the transfer function with angular displacement θ(t) as output and torque T(t) as input. Hence find J and D to give 20% overshoot and settling time of 2 second for step input of torque T(t). [Figure: fixed wall – torsional spring K = 5 N-m/rad – inertia J (torque T(t) applied, angle θ(t)) – viscous damper D – fixed wall]

Answer

Transfer function

Torque balance on $J$ (spring $K$ and damper $D$ both to the frame):

J\ddot\theta + D\dot\theta + K\theta = T(t)

\frac{\theta(s)}{T(s)} = \frac{1}{Js^2 + Ds + K} = \frac{1/J}{s^2 + \frac{D}{J}s + \frac{K}{J}}, \qquad K = 5\ \text{N-m/rad}

Comparing with the standard form: $\omega_n^2 = \dfrac{K}{J}$ , $2\zeta\omega_n = \dfrac{D}{J}$ .

Damping ratio for 20% overshoot

\zeta = \frac{-\ln(0.20)}{\sqrt{\pi^2 + \ln^2(0.20)}} = \frac{1.6094}{\sqrt{9.8696 + 2.5903}} = \frac{1.6094}{3.5299} = 0.456

Natural frequency from settling time (2% criterion)

t_s = \frac{4}{\zeta\omega_n} = 2 \;\Rightarrow\; \zeta\omega_n = 2

\omega_n = \frac{2}{0.456} = 4.386\ \text{rad/s}

Solve for $J$ and $D$

\begin{aligned} J &= \frac{K}{\omega_n^2} = \frac{5}{(4.386)^2} = \frac{5}{19.24} = 0.260\ \text{kg-m}^2 \\ D &= 2\zeta\omega_nJ = 2\times 2\times 0.260 = 1.04\ \text{N-m-s/rad} \end{aligned}

Answer: $J \approx 0.26\ \text{kg-m}^2$ and $D \approx 1.04\ \text{N-m-s/rad}$ .

With these values, $\dfrac{\theta(s)}{T(s)} = \dfrac{3.85}{s^2 + 4s + 19.24}$ (poles at $s = -2 \pm j3.90$ ), giving 20% overshoot and $t_s = 2$ s. The steady-state angle for a unit step torque is $1/K = 0.2$ rad.

Asked 2 times
2076 Asoj · 2+6 marks
2067 Asar (old course) · 8 marks

Develop signal flow graph for the block diagram model below and find transfer function using Mason's gain formula. [Figure: R → summing point S1 (+, −) → G1 → node A → summing point S2 (+, −) → G2 → node B → summing point S3 (+, +) → C. G3 from node A to S3 (+). H1 from node B fed back to S1 (−). H2 from C fed back to S2 (−).]

Answer

Signal flow graph

Nodes: $R$ , $e_1$ (after $S_1$ ), $A$ (after $G_1$ ), $e_2$ (after $S_2$ ), $B$ (after $G_2$ ), $C$ . A dummy output node $C$ is drawn with unity gain.

       1      G1      1       G2       1
 R ───►e1 ───►A ────►e2 ────► B ─────►C ──1──► C
       ▲      │       ▲       │        │
       │      └───────┼──G3───┼───────►│
       │              │       │        │
       │              └──────(−H2)─────┘
       └────────(−H1)─────────┘

Branches: $R\to e_1$ (1), $e_1\to A$ ( $G_1$ ), $A\to e_2$ (1), $e_2\to B$ ( $G_2$ ), $B\to C$ (1), $A\to C$ ( $G_3$ ), $B\to e_1$ ( $-H_1$ ), $C\to e_2$ ( $-H_2$ ).

Mason's gain formula

T = \frac{1}{\Delta}\sum_k P_k\Delta_k

Forward paths

$P_1 = G_1G_2$ ( $R\to e_1\to A\to e_2\to B\to C$ )
$P_2 = G_1G_3$ ( $R\to e_1\to A\to C$ )

Individual loops

$L_1 = -G_1G_2H_1$ ( $e_1\to A\to e_2\to B\to e_1$ )
$L_2 = -G_2H_2$ ( $e_2\to B\to C\to e_2$ )
$L_3 = +G_1G_3G_2H_1H_2$ ( $e_1\to A\to C\to e_2\to B\to e_1$ ; two negative branches give a positive gain)

Non-touching loops: none (all share node $e_2$ or $B$ ).

Determinant

\Delta = 1 - (L_1 + L_2 + L_3) = 1 + G_1G_2H_1 + G_2H_2 - G_1G_2G_3H_1H_2

Cofactors: both forward paths touch all loops, so $\Delta_1 = \Delta_2 = 1$ .

Transfer function

\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_1G_2H_1 + G_2H_2 - G_1G_2G_3H_1H_2}

This agrees with block diagram reduction of the same figure.

Asked 2 times
2075 Asoj · 8 marks
2081 Baisakh · 6 marks

Using Mason's gain formula, find the transfer function C(s)/R(s) of the fig given below. [Figure: R → summing point S1 (+, −) → summing point S2 (+, −) → G1 → summing point S3 (+, −) → node A → G2 → node B → G3 → summing point S4 (+, +) → C. G4 from node A to S4 (+). H2 from C fed back to S3 (−). H1 from node B fed back to S2 (−). Unity feedback from C to S1 (−).]

Answer

Name the nodes: $R$ , $x_1$ (after $S_1$ ), $x_2$ (after $S_2$ ), $x_3$ (after $G_1$ ), $A$ (after $S_3$ ), $B$ (after $G_2$ ), $x_4$ (after $G_3$ ), $C$ (after $S_4$ ).

Signal flow graph

     1     1     G1     1     G2     G3     1
 R ─►x1 ─►x2 ──►x3 ──►A ───►B ───►x4 ───►C
     ▲     ▲          ▲│     │            ▲│
     │     │          ││     │            ││
     │     └───(−H1)──┼┼─────┘            ││
     │                │└──────G4──────────┘│
     │                └───────(−H2)────────┤
     └──────────────────(−1)───────────────┘

Forward paths

$P_1 = G_1G_2G_3$ (through $G_2$ , $G_3$ )
$P_2 = G_1G_4$ (through $G_4$ )

Individual loops

Loop	Path	Gain
$L_1$	$x_2\to x_3\to A\to B\to x_2$	$-G_1G_2H_1$
$L_2$	$A\to B\to x_4\to C\to A$	$-G_2G_3H_2$
$L_3$	$A\to C$ (via $G_4$ ) $\to A$	$-G_4H_2$
$L_4$	$x_1\to\dots\to B\to x_4\to C\to x_1$	$-G_1G_2G_3$
$L_5$	$x_1\to x_2\to x_3\to A\to C\to x_1$ (via $G_4$ )	$-G_1G_4$

Non-touching loops: none; every loop passes through node $A$ .

Determinant

\Delta = 1 - \sum L_i = 1 + G_1G_2H_1 + G_2G_3H_2 + G_4H_2 + G_1G_2G_3 + G_1G_4

Cofactors

Both forward paths pass through $A$ , so they touch all loops: $\Delta_1 = \Delta_2 = 1$ .

Transfer function

\frac{C(s)}{R(s)} = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 + G_2G_3H_2 + G_4H_2 + G_1G_2G_3 + G_1G_4}

Asked 2 times
2074 Chaitra · 8 marks
2080 Bhadra · 8 marks

Determine the transfer function of the given system by reducing blocks. [Figure: R(s) → summing point S1 (+, −) → G1 → G3 → node A → summing point S2 (+) → C(s); G2 from node A also feeds S2 (sign printed unclear, take +). Feedback: C(s) → H1 → summing point S3 (+) → H2 → S1 (−). G4 takes the input R(s) (branch before S1) into S3 (−).]

Answer

Reading of the figure: the forward path is $G_1$ then $G_3$ to node $A$ ; from $A$ the signal reaches $S_2$ directly and also through $G_2$ (sign taken as $+$ ). The output $C$ goes through $H_1$ to $S_3$ (+), where $G_4R$ is subtracted; the result passes through $H_2$ to $S_1$ (−).

Step 1: Series and parallel blocks in the forward path

$G_1$ and $G_3$ are in series; the direct path and $G_2$ from $A$ are in parallel:

G(s) = G_1G_3(1 + G_2)

Step 2: Move summing point $S_3$ past $H_2$

The signal into $S_1$ from the feedback is $H_2(H_1C - G_4R) = H_1H_2C - G_4H_2R$ . So after moving $S_3$ beyond $H_2$ , the feedback path is $H_1H_2$ from $C$ , and a path $G_4H_2$ from $R$ enters $S_1$ with sign $(-)(-) = +$ .

Step 3: Combine the two input paths into $S_1$

$R$ reaches $S_1$ directly and through $G_4H_2$ (both +), a parallel combination:

E = (1 + G_4H_2)R - H_1H_2C

 R ─►[1+G4H2]─►(Σ)─►[G1G3(1+G2)]──┬──► C
               ▲ -                │
               └─────[H1H2]◄──────┘

Step 4: Close the feedback loop

\frac{C}{(1 + G_4H_2)R} = \frac{G_1G_3(1 + G_2)}{1 + G_1G_3(1 + G_2)H_1H_2}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_3(1 + G_2)(1 + G_4H_2)}{1 + G_1G_3H_1H_2 + G_1G_2G_3H_1H_2}

If the sign at the $G_2$ input of $S_2$ is negative, replace $(1 + G_2)$ by $(1 - G_2)$ throughout.

Asked 2 times
2074 Chaitra · 5 marks
2069 Chaitra · 6 marks

Consider a unity feedback control system with the closed loop transfer function C(s)/R(s) = (Ks + b)/(s² + as + b). Determine the open loop transfer function. Show that the steady state error in the unit ramp input response is given by ess = (a − K)/b.

Answer

Open-loop transfer function

For unity feedback, $T(s) = \dfrac{G(s)}{1 + G(s)}$ , so

G(s) = \frac{T(s)}{1 - T(s)}

1 - T(s) = \frac{s^2 + as + b - Ks - b}{s^2 + as + b} = \frac{s^2 + (a - K)s}{s^2 + as + b}

G(s) = \frac{Ks + b}{s^2 + (a - K)s} = \frac{Ks + b}{s\left[s + (a - K)\right]}

The open-loop system is type 1 (one pole at the origin).

Steady-state error for unit ramp

For a unit ramp $R(s) = 1/s^2$ , $e_{ss} = \dfrac{1}{K_v}$ where the velocity error constant is

K_v = \lim_{s\to 0} sG(s) = \lim_{s\to 0}\frac{Ks + b}{s + (a - K)} = \frac{b}{a - K}

Therefore

e_{ss} = \frac{1}{K_v} = \frac{a - K}{b}

Direct check from the error transfer function

E(s) = R(s)\left[1 - T(s)\right] = \frac{1}{s^2}\cdot\frac{s\left[s + (a - K)\right]}{s^2 + as + b}

e_{ss} = \lim_{s\to 0}sE(s) = \lim_{s\to 0}\frac{s + (a - K)}{s^2 + as + b} = \frac{a - K}{b}

Hence proved (valid when the closed loop is stable, i.e. $a > 0$ , $b > 0$ ). Choosing $K = a$ makes the ramp error zero, because the open loop then becomes type 2.

Asked 2 times
2074 Asoj · 8 marks
2081 Baisakh · 6 marks

The open loop transfer function of a unity feedback system is given by G(s) = 108/[s²(s+4)(s²+3s+12)]. Find the static error coefficients and steady state error of the system when subjected to an input given by r(t) = 2 + 5t + 8t².

Answer

The steady state error of a unity feedback system depends on the type of $G(s)$ and on the static error coefficients $K_p$ , $K_v$ and $K_a$ .

Type of the system

G(s) = \frac{108}{s^2(s+4)(s^2+3s+12)}

There are two poles at the origin, so the system is Type 2.

Static error coefficients

\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \lim_{s\to 0}\frac{108}{s(s+4)(s^2+3s+12)} = \infty \\ K_a &= \lim_{s\to 0} s^2G(s) = \frac{108}{(4)(12)} = \frac{108}{48} = 2.25 \end{aligned}

Error for each input term

For a unity feedback system:

Input	Error formula	Value here
Step $A$	$A/(1+K_p)$	$0$
Ramp $Bt$	$B/K_v$	$0$
Parabola $Ct^2/2$	$C/K_a$	finite

Input: $r(t) = 2 + 5t + 8t^2$ .

Step part $2$ : $e_1 = \dfrac{2}{1+\infty} = 0$
Ramp part $5t$ : $e_2 = \dfrac{5}{\infty} = 0$
Parabolic part $8t^2 = 16\cdot\dfrac{t^2}{2}$ , so $C = 16$ : $e_3 = \dfrac{16}{K_a} = \dfrac{16}{2.25} = 7.11$

By superposition:

e_{ss} = 0 + 0 + \frac{16}{2.25} = \frac{64}{9} = 7.11

Answer: $K_p = \infty$ , $K_v = \infty$ , $K_a = 2.25$ , and $e_{ss} = 7.11$ (units of the output).

Note on stability: the closed-loop characteristic equation is

s^5 + 7s^4 + 24s^3 + 48s^2 + 0\cdot s + 108 = 0

The $s$ term is missing, so by the R-H necessary condition the closed loop is actually unstable, and strictly a steady state does not exist. The value $7.11$ is the result the error-coefficient method gives, which is what is normally expected for this question.

Asked 2 times
2070 Chaitra · 6 marks
2078 Bhadra · 8 marks

Reduce the following block diagram model to obtain its overall transfer function. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → node A → G1 → summing point S3 (+, −) → G2 → node B → G3 → summing point S4 (+, +) → C(s). G4 from node A to S4 (+). H2 from node B fed back to S2 (−). H3 from C(s) fed back to S3 (−). H1 from C(s) fed back to S1 (−).]

Answer

Name the signals: $A$ is the output of $S_2$ (input of $G_1$ and of $G_4$ ), $B$ is the output of $G_2$ .

From the figure the signal equations are:

\begin{aligned} A &= R - H_1C - H_2B \\ B &= G_2(G_1A - H_3C) \\ C &= G_3B + G_4A \end{aligned}

(The two summers $S_1$ and $S_2$ are in cascade, so they can be merged into one summer at $A$ .)

Step 1: Move the take-off point of $H_2$ from $B$ past $G_3$ and $S_4$

From the last equation, $B = \dfrac{C - G_4A}{G_3}$ . So the signal $H_2B$ can be produced as

H_2B = \frac{H_2}{G_3}C - \frac{G_4H_2}{G_3}A

The $H_2$ path is replaced by a block $H_2/G_3$ from $C$ (negative) and a block $G_4H_2/G_3$ from $A$ back to $A$ (positive).

Step 2: Combine the feedback blocks from $C$ and remove the self loop at $A$

Feedback from $C$ to the summer: $H_1 + \dfrac{H_2}{G_3}$ (parallel blocks).

The positive self loop at $A$ gives the block $\dfrac{1}{1 - G_4H_2/G_3}$ .

Step 3: Reduce the forward part from $A$ to $C$

C = G_3G_2(G_1A - H_3C) + G_4A \;\Rightarrow\; \frac{C}{A} = \frac{G_1G_2G_3 + G_4}{1 + G_2G_3H_3}

This is the inner loop $G_2G_3$ with feedback $H_3$ , plus the parallel path $G_4$ .

Step 4: Close the main loop

Forward gain:

G_f = \frac{G_1G_2G_3 + G_4}{(1 + G_2G_3H_3)\left(1 - \frac{G_4H_2}{G_3}\right)}, \qquad H = H_1 + \frac{H_2}{G_3}

\frac{C}{R} = \frac{G_f}{1 + G_fH}

Multiplying out, the $G_4H_2/G_3$ terms cancel:

\begin{aligned} 1+G_fH &\propto (1+G_2G_3H_3)\left(1-\tfrac{G_4H_2}{G_3}\right) + (G_1G_2G_3+G_4)\left(H_1+\tfrac{H_2}{G_3}\right) \\ &= 1 + G_1G_2H_2 + G_2G_3H_3 + G_1G_2G_3H_1 + G_4H_1 - G_2G_4H_2H_3 \end{aligned}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_4}{1 + G_1G_2H_2 + G_2G_3H_3 + G_1G_2G_3H_1 + G_4H_1 - G_2G_4H_2H_3}

Check by Mason's formula: forward paths $P_1 = G_1G_2G_3$ , $P_2 = G_4$ ; loops $-G_1G_2H_2$ , $-G_2G_3H_3$ , $-G_1G_2G_3H_1$ , $-G_4H_1$ and $+G_2G_4H_2H_3$ (path $A \to G_4 \to C \to H_3 \to G_2 \to B \to H_2 \to A$ ). All loops touch each other and both paths, so $\Delta_1 = \Delta_2 = 1$ . This gives the same result.

2082 Baisakh · 8 marks

Obtain the overall transfer function C(s)/R(s) using block diagram reduction technique for the system represented by the block diagram as shown in the figure below. [Figure: R(s) enters summing point S1 (+); output of S1 → G1 → node A → G2 → node B → summing point S2 (+) → G4 → summing point S3 (+) → C(s). From node A, branch → G3 → added (+) at S3. From node B, H1 feeds back (+) to S1. From C(s), H2 feeds back (+) to S2.]

Answer

Name the signals: $E_1$ is the output of $S_1$ , $A$ the output of $G_1$ , $B$ the output of $G_2$ , $E_2$ the output of $S_2$ .

The signal equations from the figure are:

\begin{aligned} E_1 &= R + H_1B, \quad A = G_1E_1, \quad B = G_2A \\ E_2 &= B + H_2C, \quad C = G_4E_2 + G_3A \end{aligned}

Step 1: Reduce the first loop ( $G_1$ , $G_2$ with positive feedback $H_1$ )

The loop $G_1G_2$ with positive feedback $H_1$ gives

\frac{B}{R} = \frac{G_1G_2}{1 - G_1G_2H_1}

and, since $A = B/G_2$ ,

\frac{A}{R} = \frac{G_1}{1 - G_1G_2H_1}

        +------- G3 (from A) -------+
        |                           v
R -> [G1/(1-G1G2H1)] -A-> G2 -B-> (S2)+-> G4 -> (S3)+-> C
                                   ^                 |
                                   +------ H2 -------+ (+)

Step 2: Reduce the second loop ( $G_4$ with positive feedback $H_2$ )

$C$ is fed back through $H_2$ to $S_2$ with a plus sign, and $G_3A$ is added after $G_4$ :

C = G_4(B + H_2C) + G_3A \;\Rightarrow\; C(1 - G_4H_2) = G_4B + G_3A

Step 3: Substitute $A$ and $B$

\begin{aligned} C(1 - G_4H_2) &= \frac{G_1G_2G_4 + G_1G_3}{1 - G_1G_2H_1}\,R \\ \frac{C}{R} &= \frac{G_1(G_2G_4 + G_3)}{(1 - G_1G_2H_1)(1 - G_4H_2)} \end{aligned}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2G_4 + G_1G_3}{1 - G_1G_2H_1 - G_4H_2 + G_1G_2G_4H_1H_2}

Check (Mason): paths $G_1G_2G_4$ and $G_1G_3$ ; loops $L_1 = G_1G_2H_1$ , $L_2 = G_4H_2$ , which do not touch each other, so $\Delta = 1 - L_1 - L_2 + L_1L_2$ . Both paths touch both loops, so $\Delta_1=\Delta_2=1$ . Same result.

2081 Bhadra · 8 marks

A unity feedback system having feed forward transfer function G(s) = 16/[s(s+1)], determine the value of undamped natural frequency, damping ratio. If tachometer feedback is introduced, the feedback transfer function becomes (1+ks). What should be the value of 'k' to obtain damping ratio 0.6. Also calculate the percentage peak overshoot for unit step response before and after introduction of feedback.

Answer

A second order system $\omega_n^2/(s^2 + 2\zeta\omega_n s + \omega_n^2)$ has peak overshoot $M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}}$ . Tachometer (derivative output) feedback increases the $s$ coefficient, so it increases damping without changing $\omega_n$ .

Without tachometer feedback

\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{16}{s^2 + s + 16}

Comparing with $s^2 + 2\zeta\omega_n s + \omega_n^2$ :

\begin{aligned} \omega_n^2 &= 16 \;\Rightarrow\; \omega_n = 4\ \text{rad/s} \\ 2\zeta\omega_n &= 1 \;\Rightarrow\; \zeta = \frac{1}{2\times 4} = 0.125 \end{aligned}

M_p = e^{-\pi(0.125)/\sqrt{1-0.125^2}} = e^{-0.3958} = 0.6731 = 67.31\%

With tachometer feedback $H(s) = 1 + ks$

\frac{C(s)}{R(s)} = \frac{G}{1+GH} = \frac{16}{s(s+1) + 16(1+ks)} = \frac{16}{s^2 + (1+16k)s + 16}

$\omega_n$ is still $4$ rad/s. For $\zeta = 0.6$ :

\begin{aligned} 2\zeta\omega_n &= 1 + 16k \\ 2(0.6)(4) = 4.8 &= 1 + 16k \\ k &= \frac{3.8}{16} = 0.2375 \end{aligned}

Peak overshoot after adding feedback

M_p = e^{-\pi(0.6)/\sqrt{1-0.36}} = e^{-0.75\pi} = e^{-2.356} = 0.0948 = 9.48\%

Summary

Quantity	Before	After
$\omega_n$	4 rad/s	4 rad/s
$\zeta$	0.125	0.6
$M_p$	67.31%	9.48%

Answer: $\omega_n = 4$ rad/s, $\zeta = 0.125$ ; $k = 0.2375$ s; $M_p$ falls from 67.31% to 9.48%. Tachometer feedback greatly reduces overshoot without changing $\omega_n$ .

2080 Bhadra · 8 marks

Obtain the overall transfer function of given system by signal flow graph technique. [Figure: R(s) → summing point S1 (+, −) → G1 → summing point S2 (+, −) → summing point S3 (+, −) → node A. From node A: G2 → node B, and H1 (feedforward branch) going to summing point S4 (+). Node B → S4 (+) → G3 → C(s). From node B, H2 feeds back to S2 (−). From C(s), H3 feeds back to S3 (−). From C(s), unity feedback to S1 (−).]

Answer

In the signal flow graph method, each signal becomes a node and each block a branch; then Mason's gain formula gives the transfer function.

Signal flow graph

Nodes: $x_1$ = output of $S_1$ , $x_2$ = output of $G_1$ , $x_3$ = output of $S_2$ , $x_4$ = output of $S_3$ (node A), $x_5$ = node B, $x_6$ = output of $S_4$ , $C$ .

                         H1
                   +--------------+
                   |              v
R -1-> x1 -G1-> x2 -1-> x3 -1-> x4 -G2-> x5 -1-> x6 -G3-> C
       ^                ^       ^         |                |
       |                +-(-H2)-+---------+                |
       |                        +-------(-H3)--------------+
       +--------------------(-1)---------------------------+

Branches: $x_4\to x_6$ ( $H_1$ ), $x_5\to x_3$ ( $-H_2$ ), $C\to x_4$ ( $-H_3$ ), $C\to x_1$ ( $-1$ ); all others as labelled.

Forward paths

$P_1 = G_1G_2G_3$ (through $G_2$ )
$P_2 = G_1H_1G_3$ (through the $H_1$ branch)

Individual loops

$L_1 = -G_2H_2$ ( $x_3, x_4, x_5, x_3$ )
$L_2 = -G_2G_3H_3$ ( $x_4, x_5, x_6, C, x_4$ )
$L_3 = -G_3H_1H_3$ ( $x_4, x_6, C, x_4$ )
$L_4 = -G_1G_2G_3$ ( $x_1 \dots C, x_1$ )
$L_5 = -G_1G_3H_1$ ( $x_1, x_2, x_3, x_4, x_6, C, x_1$ )

Non-touching loops

$L_1$ touches $L_2$ and $L_3$ at $x_4$ , and all others pass through $x_3$ / $x_4$ . So there are no non-touching pairs.

\Delta = 1 + G_2H_2 + G_2G_3H_3 + G_3H_1H_3 + G_1G_2G_3 + G_1G_3H_1

Cofactors

Both forward paths pass through $x_3$ , $x_4$ and $C$ , so they touch every loop: $\Delta_1 = \Delta_2 = 1$ .

Mason's gain formula

T = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_3(G_2 + H_1)}{1 + G_2H_2 + G_2G_3H_3 + G_3H_1H_3 + G_1G_2G_3 + G_1G_3H_1}

2080 Bhadra · 4 marks

Discuss the effect of addition of pole and zero in a system.

Answer

Adding a pole or a zero changes the shape of the root locus and the time response. Poles slow the system down; zeros speed it up.

Addition of a pole

Open-loop pole (to $G(s)H(s)$ ): the root locus bends to the right, towards the imaginary axis. The system becomes less stable and the range of $K$ for stability shrinks.
Closed-loop response becomes slower: rise time and settling time increase.
Peak overshoot usually increases (lower effective damping).
A pole at the origin raises the system type, which reduces steady state error.
Closed-loop pole: adds a slow mode; overshoot falls but rise time rises (more sluggish response).

Addition of a zero

Open-loop zero: pulls the root locus to the left, away from the imaginary axis. Relative stability improves (this is the idea behind PD and lead compensation).
Closed-loop response becomes faster: rise time and peak time decrease.
Closed-loop zero: adds a derivative term $\frac{1}{a}\frac{dc}{dt}$ to the response, so overshoot increases, especially when the zero is close to the origin.
Bandwidth increases.

 jw         pole added            zero added
  |  locus bends right      locus bends left
  |      \                         /
--+-------x-----             -----o-------

Effect on	Adding pole	Adding zero
Root locus	shifts right	shifts left
Stability	decreases	increases
Rise time	increases	decreases
Bandwidth	decreases	increases

A pole or zero far from the dominant poles (more than about 5 times further left) has little effect.

2080 Baisakh · 8 marks

Obtain the overall transfer function of given system by block diagram reduction techniques. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → G1 → node A → summing point S3 (+, −) → G2 → node B → summing point S4 (+, +) → C(s). Feedforward from node A to S4 (+). H1 from node B fed back to S2 (−). H2 from C(s) fed back to S3 (−). Unity feedback from C(s) to S1 (−).]

Answer

Name the signals: $E$ = output of $S_2$ (input of $G_1$ ), $A$ = output of $G_1$ , $B$ = output of $G_2$ .

From the figure:

\begin{aligned} E &= R - C - H_1B, \quad A = G_1E \\ B &= G_2(A - H_2C), \quad C = B + A \end{aligned}

( $S_1$ and $S_2$ are in cascade, so they act as one summer.)

Step 1: Move the take-off point of $H_1$ from $B$ past summer $S_4$

Since $C = B + A$ , we have $B = C - A$ . So

H_1B = H_1C - H_1A = H_1C - G_1H_1E

The $H_1$ path becomes $H_1$ from $C$ (negative) plus a positive self loop $G_1H_1$ around $G_1$ .

Step 2: Combine the feedback from $C$ and remove the self loop

Feedback from $C$ : $1 + H_1$ (parallel). Loop $G_1$ with positive feedback $H_1$ : $\dfrac{G_1}{1 - G_1H_1}$ .

Step 3: Reduce from $A$ to $C$

C = G_2(A - H_2C) + A \;\Rightarrow\; \frac{C}{A} = \frac{1 + G_2}{1 + G_2H_2}

(Inner loop $G_2$ , $H_2$ plus the parallel unity feedforward path.)

R ->(+)-> [G1/(1-G1H1)] -> [(1+G2)/(1+G2H2)] --> C
     ^-                                      |
     +------------- (1 + H1) <---------------+

Step 4: Close the main loop

G_f = \frac{G_1(1+G_2)}{(1 - G_1H_1)(1 + G_2H_2)}, \qquad H = 1 + H_1

\begin{aligned} \frac{C}{R} &= \frac{G_f}{1+G_fH} = \frac{G_1(1+G_2)}{(1-G_1H_1)(1+G_2H_2) + G_1(1+G_2)(1+H_1)} \\ \text{Denominator} &= 1 + G_2H_2 - G_1H_1 - G_1G_2H_1H_2 + G_1 + G_1G_2 + G_1H_1 + G_1G_2H_1 \end{aligned}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1 + G_1G_2}{1 + G_1 + G_1G_2 + G_2H_2 + G_1G_2H_1 - G_1G_2H_1H_2}

Check (Mason): paths $G_1G_2$ , $G_1$ ; loops $-G_1G_2H_1$ , $-G_2H_2$ , $-G_1G_2$ , $-G_1$ , and $+G_1G_2H_1H_2$ (through the feedforward, $H_2$ , $G_2$ , $H_1$ ). All loops touch, $\Delta_1=\Delta_2=1$ . Same result.

2080 Baisakh · 8 marks

For a mechanical system with closed loop transfer function θ(s)/T(s) = 1/(as² + bs + c), where θ(s) is the output of step input T(t) = 10 Nm. Determine values of a, b and c if maximum overshoot is 6%, peak time (tp) = 1 sec, and ess = 0.5.

Answer

Compare the system with the standard second order form. The final value fixes $c$ ; the overshoot fixes $\zeta$ ; the peak time then fixes $\omega_n$ .

Interpretation: " $e_{ss} = 0.5$ " is taken as the steady state value of the output, $\theta(\infty) = 0.5$ rad, for the 10 Nm step (the usual reading of this problem, since $\theta$ and $T$ have different units).

\frac{\theta(s)}{T(s)} = \frac{1}{as^2+bs+c} = \frac{1/a}{s^2 + \frac{b}{a}s + \frac{c}{a}} \;\Rightarrow\; \omega_n^2 = \frac{c}{a},\quad 2\zeta\omega_n = \frac{b}{a}

Step 1: Find $c$ from the steady state

$T(s) = 10/s$ . By the final value theorem:

\theta(\infty) = \lim_{s\to 0} s\cdot\frac{10}{s}\cdot\frac{1}{as^2+bs+c} = \frac{10}{c} = 0.5 \;\Rightarrow\; c = 20\ \text{Nm/rad}

Step 2: Damping ratio from 6% overshoot

\zeta = \frac{-\ln M_p}{\sqrt{\pi^2 + (\ln M_p)^2}} = \frac{-\ln 0.06}{\sqrt{\pi^2 + (\ln 0.06)^2}} = \frac{2.8134}{\sqrt{9.8696 + 7.9153}} = 0.6671

Step 3: Natural frequency from $t_p = 1$ s

t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} \;\Rightarrow\; \omega_n = \frac{\pi}{1\times\sqrt{1 - 0.6671^2}} = \frac{3.1416}{0.7449} = 4.217\ \text{rad/s}

Step 4: Find $a$ and $b$

\begin{aligned} a &= \frac{c}{\omega_n^2} = \frac{20}{4.217^2} = \frac{20}{17.785} = 1.125\ \text{kg-m}^2 \\ b &= 2\zeta\omega_n a = 2(0.6671)(4.217)(1.125) = 6.328\ \text{Nm-s/rad} \end{aligned}

Answer: $a \approx 1.125$ kg-m², $b \approx 6.33$ Nm-s/rad, $c = 20$ Nm/rad.

(If instead $e_{ss} = 10 - \theta(\infty) = 0.5$ is used, then $c = 10/9.5 = 1.053$ , $a = 0.0592$ , $b = 0.333$ ; $\zeta$ and $\omega_n$ are unchanged.)

2079 Bhadra · 8 marks

From the given block diagram, draw the signal flow graph and find the transfer ratio C(s)/R(s) using Mason's gain formula. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → node A → G1 → summing point S3 (+, −) → G2 → node B → G3 → summing point S4 (+, +) → C(s). G4 feeds forward from node A to S4 (+). H2 from node B fed back to S2 (−). H3 from C(s) fed back to S3 (−). H1 from C(s) fed back to S1 (−).]

Answer

A signal flow graph has a node for each signal and a directed branch (with gain) for each block. Summing points become nodes; a minus sign at a summer goes into the branch gain.

Signal flow graph

Nodes: $x_1$ = output of $S_1$ , $x_2$ = output of $S_2$ (node A), $x_3$ = output of $S_3$ , $x_4$ = node B, $C$ = output.

                  G4
        +--------------------------+
        |                          v
R --1--> x1 --1--> x2 --G1--> x3 --G2--> x4 --G3--> C --1--> C
         ^          ^          ^          |         |
         |          +---(-H2)--+----------+         |
         |                     +--------(-H3)-------+
         +-----------------(-H1)--------------------+

Branches: $R\to x_1$ (1), $x_1\to x_2$ (1), $x_2\to x_3$ ( $G_1$ ), $x_3\to x_4$ ( $G_2$ ), $x_4\to C$ ( $G_3$ ), $x_2\to C$ ( $G_4$ ), $x_4\to x_2$ ( $-H_2$ ), $C\to x_3$ ( $-H_3$ ), $C\to x_1$ ( $-H_1$ ).

Mason's gain formula

T = \frac{C}{R} = \frac{\sum_k P_k\Delta_k}{\Delta}

Forward paths

$P_1 = G_1G_2G_3$ (R, x1, x2, x3, x4, C)
$P_2 = G_4$ (R, x1, x2, C)

Individual loops

$L_1 = -G_1G_2H_2$ (x2, x3, x4, x2)
$L_2 = -G_2G_3H_3$ (x3, x4, C, x3)
$L_3 = -G_1G_2G_3H_1$ (x1, x2, x3, x4, C, x1)
$L_4 = -G_4H_1$ (x1, x2, C, x1)
$L_5 = (G_4)(-H_3)(G_2)(-H_2) = +G_2G_4H_2H_3$ (x2, C, x3, x4, x2)

Non-touching loops: every pair of loops shares at least one node (all pass through x2/x3 or C), so there are none.

\Delta = 1 - (L_1+L_2+L_3+L_4+L_5) = 1 + G_1G_2H_2 + G_2G_3H_3 + G_1G_2G_3H_1 + G_4H_1 - G_2G_4H_2H_3

Cofactors: $P_1$ and $P_2$ both touch all loops (they pass through x2 and C), so $\Delta_1 = \Delta_2 = 1$ .

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_4}{1 + G_1G_2H_2 + G_2G_3H_3 + G_1G_2G_3H_1 + G_4H_1 - G_2G_4H_2H_3}

2079 Bhadra · 8 marks

Find the impulse response of the given circuit. [Figure: input ei; series capacitor 1 F, then shunt resistor 2 Ω across the line, then series resistor 1 Ω, then shunt capacitor 2 F; output eo taken across the 2 F capacitor]

Answer

The impulse response is the inverse Laplace transform of the transfer function, $h(t) = \mathcal{L}^{-1}\{E_o(s)/E_i(s)\}$ , because $\mathcal{L}\{\delta(t)\} = 1$ .

Circuit in the s-domain

Impedances: $C_1 = 1$ F $\to \dfrac{1}{s}$ , $R_1 = 2\ \Omega$ , $R_2 = 1\ \Omega$ , $C_2 = 2$ F $\to \dfrac{1}{2s}$ .

 ei --||-----+----/\/\/----+------ eo
     1/s     |     1 ohm   |
            2 ohm        1/(2s)
             |             |
 ----------- +-------------+------

Let $V_1$ be the voltage at the junction of $C_1$ and $R_1$ .

Node equations

At the output node (current through $R_2$ flows into $C_2$ ):

\frac{V_1 - E_o}{1} = 2sE_o \;\Rightarrow\; V_1 = (1 + 2s)E_o

At node $V_1$ (KCL):

(E_i - V_1)s = \frac{V_1}{2} + (V_1 - E_o)

sE_i = V_1\left(s + \tfrac{3}{2}\right) - E_o

Substituting $V_1$ :

\begin{aligned} sE_i &= E_o\left[(1+2s)\left(s+\tfrac{3}{2}\right) - 1\right] = E_o\left[2s^2 + 4s + \tfrac{1}{2}\right] \\ \frac{E_o(s)}{E_i(s)} &= \frac{2s}{4s^2 + 8s + 1} = \frac{0.5\,s}{s^2 + 2s + 0.25} \end{aligned}

Poles

s = -1 \pm \frac{\sqrt{3}}{2} \;\Rightarrow\; s_1 = -0.134,\quad s_2 = -1.866

Partial fractions

\frac{0.5s}{(s+0.134)(s+1.866)} = \frac{k_1}{s+0.134} + \frac{k_2}{s+1.866}

\begin{aligned} k_1 &= \frac{0.5(-0.134)}{-0.134+1.866} = \frac{-0.067}{1.732} = -0.0387 \\ k_2 &= \frac{0.5(-1.866)}{-1.866+0.134} = \frac{-0.933}{-1.732} = 0.5387 \end{aligned}

Impulse response

h(t) = 0.5387\,e^{-1.866t} - 0.0387\,e^{-0.134t}, \quad t \ge 0

Exact form:

h(t) = e^{-t}\left[\tfrac{1}{2}\cosh\tfrac{\sqrt{3}}{2}t - \tfrac{1}{\sqrt{3}}\sinh\tfrac{\sqrt{3}}{2}t\right]

Answer: $e_o(t) = 0.5387e^{-1.866t} - 0.0387e^{-0.134t}$ V for a unit impulse input. Check: $h(0) = 0.5$ , which matches $\lim_{s\to\infty} sT(s) = 2/4$ .

2079 Bhadra · 4 marks

The system below shows a potential tracking problem with the reference input r(t) and disturbance d(t). What value of K will limit the steady state component of c(t) due to d(t) to 2% of d(t)? [Figure: r(t) → summing point (+, −) → 5K/(s+10) → summing point (+, + with disturbance d(t)) → 6/(s+2) → c(t); unity feedback from c(t) to the first summing point (−)]

Answer

With $R = 0$ , the output due to disturbance is found from the loop with $G_2$ forward and $G_1$ in feedback; then the final value theorem gives its steady state.

Let $G_1 = \dfrac{5K}{s+10}$ and $G_2 = \dfrac{6}{s+2}$ .

Transfer function from $D$ to $C$ (set $R = 0$ )

\frac{C(s)}{D(s)} = \frac{G_2}{1 + G_1G_2} = \frac{\frac{6}{s+2}}{1 + \frac{30K}{(s+10)(s+2)}} = \frac{6(s+10)}{(s+10)(s+2) + 30K}

Steady state for a step disturbance $D(s) = d/s$

c_{ss} = \lim_{s\to 0} s\cdot\frac{d}{s}\cdot\frac{6(s+10)}{(s+10)(s+2)+30K} = \frac{60\,d}{20 + 30K} = \frac{3d}{1 + 1.5K}

Condition: $c_{ss} \le 2\%$ of $d$

\begin{aligned} \frac{3}{1 + 1.5K} &\le 0.02 \\ 1 + 1.5K &\ge 150 \\ K &\ge \frac{149}{1.5} = 99.33 \end{aligned}

Stability check: characteristic equation $s^2 + 12s + 20 + 30K = 0$ ; all coefficients positive for $K > 0$ , so the second order system is stable for this $K$ .

Answer: $K \ge 99.33$ (about $K = 100$ ) keeps the steady state output due to the disturbance within 2% of $d(t)$ .

2078 Bhadra · 8 marks

Determine the transfer function of the given system by block reduction technique. [Figure: R(s) → summing point S1 → G1 → node A → summing point S2 (+, −) → G2 → node B → summing point S3 (+) → C(s). G3 from node A to S3 (−). H2 from C(s) fed back to S2 (−). H1 from node B fed back to S1, and C(s) also fed back directly (unity) to S1 (signs at S1 not clearly printed; take as negative feedback)]

Answer

Assumption: both feedback signals at $S_1$ ( $H_1B$ and $C$ ) are negative, as stated.

Name the signals: $E$ = output of $S_1$ , $A$ = output of $G_1$ , $B$ = output of $G_2$ .

\begin{aligned} E &= R - H_1B - C, \quad A = G_1E \\ B &= G_2(A - H_2C), \quad C = B - G_3A \end{aligned}

Step 1: Move the take-off point of $H_1$ from $B$ past summer $S_3$

From $C = B - G_3A$ : $B = C + G_3A$ . So

H_1B = H_1C + G_3H_1A = H_1C + G_1G_3H_1E

The $H_1$ path becomes $H_1$ from $C$ plus a negative self loop $G_1G_3H_1$ around $G_1$ .

Step 2: Combine feedbacks and remove the self loop

Feedback from $C$ to $S_1$ : $1 + H_1$ (parallel blocks). $G_1$ with negative feedback $G_3H_1$ : $\dfrac{G_1}{1 + G_1G_3H_1}$ .

Step 3: Reduce from $A$ to $C$

C = G_2(A - H_2C) - G_3A \;\Rightarrow\; \frac{C}{A} = \frac{G_2 - G_3}{1 + G_2H_2}

R ->(+)-> [G1/(1+G1G3H1)] -> [(G2-G3)/(1+G2H2)] --> C
     ^-                                         |
     +--------------- (1 + H1) <----------------+

Step 4: Close the main loop

\frac{C}{R} = \frac{G_1(G_2 - G_3)}{(1 + G_1G_3H_1)(1 + G_2H_2) + G_1(G_2 - G_3)(1 + H_1)}

Expanding the denominator (the $G_1G_3H_1$ terms cancel):

1 + G_2H_2 + G_1G_3H_1 + G_1G_2G_3H_1H_2 + G_1G_2 - G_1G_3 + G_1G_2H_1 - G_1G_3H_1

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2 - G_1G_3}{1 + G_1G_2 - G_1G_3 + G_1G_2H_1 + G_2H_2 + G_1G_2G_3H_1H_2}

Check (Mason): paths $G_1G_2$ and $-G_1G_3$ ; loops $-G_1G_2H_1$ , $-G_2H_2$ , $-G_1G_2$ , $+G_1G_3$ , $-G_1G_2G_3H_1H_2$ ; all touch, $\Delta_1 = \Delta_2 = 1$ . Same result.

2078 Bhadra · 8 marks

The open loop transfer function of unity feedback system is given by G(s) = 108/[s²(s+4)(s²+3s+12)]. Find the static error coefficients and steady state error of the system when subjected to an input given by r(t) = 2 + 5t + 2t².

Answer

The steady state error of a unity feedback system depends on the type of $G(s)$ and on the static error coefficients $K_p$ , $K_v$ and $K_a$ .

Type of the system

G(s) = \frac{108}{s^2(s+4)(s^2+3s+12)}

There are two poles at the origin, so the system is Type 2.

Static error coefficients

\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \lim_{s\to 0}\frac{108}{s(s+4)(s^2+3s+12)} = \infty \\ K_a &= \lim_{s\to 0} s^2G(s) = \frac{108}{(4)(12)} = \frac{108}{48} = 2.25 \end{aligned}

Error for each input term

For a unity feedback system:

Input	Error formula	Value here
Step $A$	$A/(1+K_p)$	$0$
Ramp $Bt$	$B/K_v$	$0$
Parabola $Ct^2/2$	$C/K_a$	finite

Input: $r(t) = 2 + 5t + 2t^2$ .

Step part $2$ : $e_1 = \dfrac{2}{1+\infty} = 0$
Ramp part $5t$ : $e_2 = \dfrac{5}{\infty} = 0$
Parabolic part $2t^2 = 4\cdot\dfrac{t^2}{2}$ , so $C = 4$ : $e_3 = \dfrac{4}{K_a} = \dfrac{4}{2.25} = 1.78$

By superposition:

e_{ss} = 0 + 0 + \frac{4}{2.25} = \frac{16}{9} = 1.78

Answer: $K_p = \infty$ , $K_v = \infty$ , $K_a = 2.25$ , and $e_{ss} = 1.78$ (units of the output).

Note on stability: the closed-loop characteristic equation is

s^5 + 7s^4 + 24s^3 + 48s^2 + 0\cdot s + 108 = 0

The $s$ term is missing, so by the R-H necessary condition the closed loop is unstable and strictly no steady state exists. The value $1.78$ is the result the error-coefficient method gives, which is the answer normally expected here.

2078 Kartik · 8 marks

From the given block diagram, draw the signal flow graph and find the transfer ratio C(s)/R(s) using Mason gain's formula. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → G1 → node A → summing point S3 (+, +) → G2 → node B → G3 → summing point S4 (+, +) → C(s). G4 from node A to S4 (+). H1 from node B fed back to S2 (−). H2 from C(s) fed back to S3 (+). Unity feedback from C(s) to S1 (−).]

Answer

Each signal becomes a node and each block a branch; summer signs go into branch gains. Then Mason's gain formula is applied.

Signal flow graph

Nodes: $x_1$ = output of $S_1$ , $x_2$ = output of $S_2$ , $x_3$ = node A (output of $G_1$ ), $x_4$ = output of $S_3$ , $x_5$ = node B, $C$ .

                         G4
               +------------------------+
               |                        v
R -1-> x1 -1-> x2 -G1-> x3 -1-> x4 -G2-> x5 -G3-> C -1-> C
       ^       ^                ^        |        |
       |       +-----(-H1)------+--------+        |
       |                        +-----(+H2)-------+
       +-------------------(-1)-------------------+

Branches: $x_3\to C$ ( $G_4$ ), $x_5\to x_2$ ( $-H_1$ ), $C\to x_4$ ( $+H_2$ ), $C\to x_1$ ( $-1$ ).

Forward paths

$P_1 = G_1G_2G_3$
$P_2 = G_1G_4$

Individual loops

$L_1 = -G_1G_2H_1$ ( $x_2, x_3, x_4, x_5, x_2$ )
$L_2 = +G_2G_3H_2$ ( $x_4, x_5, C, x_4$ )
$L_3 = -G_1G_2G_3$ ( $x_1, \dots, C, x_1$ )
$L_4 = -G_1G_4$ ( $x_1, x_2, x_3, C, x_1$ )
$L_5 = G_1G_4(H_2)(G_2)(-H_1) = -G_1G_2G_4H_1H_2$ ( $x_2, x_3, C, x_4, x_5, x_2$ )

Non-touching loops

Every loop passes through $x_4/x_5$ or $C$ together with another loop, so no two loops are non-touching.

\Delta = 1 + G_1G_2H_1 - G_2G_3H_2 + G_1G_2G_3 + G_1G_4 + G_1G_2G_4H_1H_2

Cofactors

$P_1$ touches all loops; $P_2$ passes through $x_2$ , $x_3$ , $C$ , which lie on every loop. So $\Delta_1 = \Delta_2 = 1$ .

Mason's gain formula

\frac{C}{R} = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 - G_2G_3H_2 + G_1G_2G_3 + G_1G_4 + G_1G_2G_4H_1H_2}

2078 Kartik · 8 marks

For the system as in figure (a), the unit step response is as in figure (b), determine M, B and K. [Figure (a): mass M hanging from a fixed support through spring K, with a damper B from M to the fixed ground below; displacement x downward; input is a force on M. Figure (b): unit step response x(t) rises to a peak of 2.36 at t = 2 sec and settles to a final value of 2]

Answer

For a force input the mass-spring-damper is a standard second order system. The final value gives $K$ , the overshoot gives $\zeta$ , and the peak time gives $\omega_n$ .

Transfer function

Force balance on $M$ : $F = M\ddot{x} + B\dot{x} + Kx$ , so

\frac{X(s)}{F(s)} = \frac{1}{Ms^2 + Bs + K} = \frac{1/M}{s^2 + \frac{B}{M}s + \frac{K}{M}}

with $\omega_n^2 = K/M$ and $2\zeta\omega_n = B/M$ .

Step 1: $K$ from the final value

For a unit step force, $F(s) = 1/s$ :

x(\infty) = \lim_{s\to 0} s\cdot\frac{1}{s}\cdot\frac{1}{Ms^2+Bs+K} = \frac{1}{K} = 2 \;\Rightarrow\; K = 0.5\ \text{N/m}

Step 2: $\zeta$ from the overshoot

M_p = \frac{2.36 - 2}{2} = 0.18

\zeta = \frac{-\ln 0.18}{\sqrt{\pi^2 + (\ln 0.18)^2}} = \frac{1.7148}{\sqrt{9.8696 + 2.9405}} = \frac{1.7148}{3.5791} = 0.4791

Step 3: $\omega_n$ from the peak time

t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} = 2 \;\Rightarrow\; \omega_n = \frac{\pi}{2\sqrt{1 - 0.4791^2}} = \frac{3.1416}{2(0.8778)} = 1.790\ \text{rad/s}

Step 4: $M$ and $B$

\begin{aligned} M &= \frac{K}{\omega_n^2} = \frac{0.5}{1.790^2} = \frac{0.5}{3.2025} = 0.156\ \text{kg} \\ B &= 2\zeta\omega_n M = 2(0.4791)(1.790)(0.156) = 0.268\ \text{N-s/m} \end{aligned}

Answer: $M \approx 0.156$ kg, $B \approx 0.268$ N-s/m, $K = 0.5$ N/m.

2076 Chaitra · 8 marks

Reduce the following block diagram and find transfer function. [Figure: R → summing point S1 (+, −) → summing point S2 (+, −) → G1 → node A → summing point S3 (+, −) → G2 → node B → G3 → summing point S4 (+, +) → C. G4 from node A to S4 (+). H1 from node B fed back to S2 (−). H2 from C fed back to S3 (−). Unity feedback from C to S1 (−).]

Answer

Name the signals: $E$ = output of $S_2$ (input of $G_1$ ), $A$ = output of $G_1$ , $B$ = output of $G_2$ .

\begin{aligned} E &= R - C - H_1B, \quad A = G_1E \\ B &= G_2(A - H_2C), \quad C = G_3B + G_4A \end{aligned}

( $S_1$ and $S_2$ are in cascade and are treated as one summer.)

Step 1: Move the take-off point of $H_1$ from $B$ past $G_3$ and $S_4$

From $C = G_3B + G_4A$ : $B = \dfrac{C - G_4A}{G_3}$ . Hence

H_1B = \frac{H_1}{G_3}C - \frac{G_1G_4H_1}{G_3}E

So the $H_1$ path becomes a block $H_1/G_3$ from $C$ (negative) and a positive self loop $G_4H_1/G_3$ around $G_1$ .

Step 2: Combine the feedbacks and remove the self loop

Feedback from $C$ : $H = 1 + \dfrac{H_1}{G_3}$ . $G_1$ with positive feedback $G_4H_1/G_3$ : $\dfrac{G_1}{1 - G_1G_4H_1/G_3}$ .

Step 3: Reduce from $A$ to $C$

C = G_3G_2(A - H_2C) + G_4A \;\Rightarrow\; \frac{C}{A} = \frac{G_2G_3 + G_4}{1 + G_2G_3H_2}

R ->(+)-> [G1/(1-G1G4H1/G3)] -> [(G2G3+G4)/(1+G2G3H2)] -> C
     ^-                                               |
     +------------------ (1 + H1/G3) <----------------+

Step 4: Close the main loop

\frac{C}{R} = \frac{G_1(G_2G_3+G_4)}{\left(1 - \frac{G_1G_4H_1}{G_3}\right)(1 + G_2G_3H_2) + G_1(G_2G_3 + G_4)\left(1 + \frac{H_1}{G_3}\right)}

On expanding, the $G_1G_4H_1/G_3$ terms cancel.

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 + G_2G_3H_2 + G_1G_2G_3 + G_1G_4 - G_1G_2G_4H_1H_2}

Check (Mason): paths $G_1G_2G_3$ , $G_1G_4$ ; loops $-G_1G_2H_1$ , $-G_2G_3H_2$ , $-G_1G_2G_3$ , $-G_1G_4$ and $+G_1G_2G_4H_1H_2$ (via $G_4$ , $H_2$ , $G_2$ , $H_1$ ). All loops touch each other and both paths, so $\Delta_1 = \Delta_2 = 1$ . Same result.

2076 Chaitra · 8 marks

For a unity feedback system, the open loop transfer function is G(s) = 50/[s(s+2)]. With unit step input find maximum overshoot and settling time. Also determine static error coefficients and steady state error if the input to the system is r(t) = 2 + 4t + 6t², t ≥ 0.

Answer

Find $\zeta$ and $\omega_n$ from the closed-loop characteristic equation for the transient part. Then use the error coefficients (the system is Type 1) for the steady state part.

Closed-loop transfer function

\frac{C(s)}{R(s)} = \frac{50}{s^2 + 2s + 50}

\begin{aligned} \omega_n^2 &= 50 \;\Rightarrow\; \omega_n = 7.071\ \text{rad/s} \\ 2\zeta\omega_n &= 2 \;\Rightarrow\; \zeta = \frac{1}{7.071} = 0.1414 \end{aligned}

Maximum overshoot

M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} = e^{-\pi(0.1414)/0.9899} = e^{-0.4488} = 0.6384 = 63.84\%

Settling time (2% criterion)

t_s = \frac{4}{\zeta\omega_n} = \frac{4}{0.1414 \times 7.071} = \frac{4}{1} = 4\ \text{s}

(With the 5% criterion, $t_s = 3/\zeta\omega_n = 3$ s.)

Static error coefficients

$G(s)$ has one pole at the origin, so it is Type 1.

\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \frac{50}{2} = 25 \\ K_a &= \lim_{s\to 0} s^2G(s) = 0 \end{aligned}

Steady state error for $r(t) = 2 + 4t + 6t^2$

Write $6t^2 = 12\cdot\dfrac{t^2}{2}$ .

\begin{aligned} e_{ss} &= \frac{2}{1+K_p} + \frac{4}{K_v} + \frac{12}{K_a} \\ &= 0 + \frac{4}{25} + \frac{12}{0} = 0 + 0.16 + \infty = \infty \end{aligned}

Input part	Error
Step 2	0
Ramp 4t	0.16
Parabola 6t²	∞

Answer: $M_p = 63.84\%$ , $t_s = 4$ s (2%); $K_p = \infty$ , $K_v = 25$ , $K_a = 0$ ; $e_{ss} = \infty$ , because a Type 1 system cannot follow a parabolic input.

2076 Asoj · 6 marks

Determine values of a and b of the closed loop control system shown below, so that maximum overshoot for unit step input is 25% and the peak time is 2 sec. Assume that J = 1 kg-m². [Figure: R(s) → summing point (+, −) → summing point (+, −) → a/(Js) → node → 1/s → C(s); minor loop: the node after a/(Js) fed back through b to the second summing point (−); major loop: C(s) fed back with unity gain to the first summing point (−)]

Answer

Reduce the minor (velocity) feedback loop first, then compare the closed-loop characteristic equation with $s^2 + 2\zeta\omega_n s + \omega_n^2$ .

Step 1: Reduce the minor loop

The block $\dfrac{a}{Js}$ has feedback $b$ :

\frac{a/(Js)}{1 + ab/(Js)} = \frac{a}{Js + ab}

Open-loop transfer function (in series with $1/s$ ):

G(s) = \frac{a}{s(Js + ab)}

Step 2: Closed-loop transfer function (unity feedback, $J = 1$ )

\frac{C(s)}{R(s)} = \frac{a}{s^2 + ab\,s + a} \;\Rightarrow\; \omega_n^2 = a,\quad 2\zeta\omega_n = ab

Step 3: $\zeta$ from $M_p = 25\%$

\zeta = \frac{-\ln 0.25}{\sqrt{\pi^2 + (\ln 0.25)^2}} = \frac{1.3863}{\sqrt{9.8696 + 1.9218}} = \frac{1.3863}{3.4339} = 0.4037

Step 4: $\omega_n$ from $t_p = 2$ s

\omega_n = \frac{\pi}{t_p\sqrt{1-\zeta^2}} = \frac{3.1416}{2\sqrt{1 - 0.4037^2}} = \frac{3.1416}{2(0.9149)} = 1.717\ \text{rad/s}

Step 5: Find $a$ and $b$

\begin{aligned} a &= \omega_n^2 = 1.717^2 = 2.948 \\ b &= \frac{2\zeta\omega_n}{a} = \frac{2(0.4037)(1.717)}{2.948} = \frac{1.386}{2.948} = 0.470 \end{aligned}

Answer: $a \approx 2.95$ and $b \approx 0.47$ (with $J = 1$ kg-m²).

2075 Chaitra · 8 marks

A system has 40% overshoot and requires a settling time of 4 seconds when given a step input. Find peak time and rise time.

Answer

Overshoot fixes $\zeta$ . The settling time fixes $\zeta\omega_n$ (2% criterion, $t_s = 4/\zeta\omega_n$ ). Then $\omega_d$ gives $t_p$ and $t_r$ .

Step 1: Damping ratio from $M_p = 40\%$

\zeta = \frac{-\ln 0.40}{\sqrt{\pi^2 + (\ln 0.40)^2}} = \frac{0.9163}{\sqrt{9.8696 + 0.8396}} = \frac{0.9163}{3.2725} = 0.280

Step 2: Natural frequency from $t_s = 4$ s (2% criterion)

t_s = \frac{4}{\zeta\omega_n} = 4 \;\Rightarrow\; \zeta\omega_n = 1 \;\Rightarrow\; \omega_n = \frac{1}{0.280} = 3.571\ \text{rad/s}

Step 3: Damped frequency

\omega_d = \omega_n\sqrt{1-\zeta^2} = 3.571\sqrt{1 - 0.0784} = 3.571(0.960) = 3.429\ \text{rad/s}

Step 4: Peak time

t_p = \frac{\pi}{\omega_d} = \frac{3.1416}{3.429} = 0.916\ \text{s}

Step 5: Rise time (0 to 100%)

\theta = \cos^{-1}\zeta = \cos^{-1}(0.280) = 1.287\ \text{rad}

t_r = \frac{\pi - \theta}{\omega_d} = \frac{3.1416 - 1.287}{3.429} = \frac{1.8546}{3.429} = 0.541\ \text{s}

Answer: $t_p \approx 0.916$ s and $t_r \approx 0.541$ s ( $\zeta = 0.28$ , $\omega_n = 3.57$ rad/s).

(With the 5% criterion, $t_s = 3/\zeta\omega_n$ , you get $\omega_n = 2.679$ rad/s, $t_p = 1.222$ s and $t_r = 0.721$ s.)

2075 Asoj · 6 marks

Fig (ii) is step response of system as in fig (i); find K and P. [Fig (i): R(s) → summing point (+, −) → K/[s(s+2)] → C(s), with feedback path (1 + sP) from C(s) to the summing point. Fig (ii): unit step response c(t) with final value 1, peak overshoot 0.343 above the final value, occurring at t = 3 sec]

Answer

Find the closed-loop transfer function with feedback $(1 + sP)$ . Then use the measured overshoot and peak time to get $\zeta$ and $\omega_n$ .

Closed-loop transfer function

\frac{C(s)}{R(s)} = \frac{G}{1+GH} = \frac{\frac{K}{s(s+2)}}{1 + \frac{K(1+sP)}{s(s+2)}} = \frac{K}{s^2 + (2 + KP)s + K}

Its DC gain is 1, which agrees with the final value of 1 in Fig (ii). Comparing:

\omega_n^2 = K, \qquad 2\zeta\omega_n = 2 + KP

Step 1: $\zeta$ from $M_p = 0.343$

\zeta = \frac{-\ln 0.343}{\sqrt{\pi^2 + (\ln 0.343)^2}} = \frac{1.0700}{\sqrt{9.8696 + 1.1449}} = \frac{1.0700}{3.3188} = 0.3224

Step 2: $\omega_n$ from $t_p = 3$ s

\omega_n = \frac{\pi}{t_p\sqrt{1-\zeta^2}} = \frac{3.1416}{3\sqrt{1 - 0.1040}} = \frac{3.1416}{3(0.9466)} = 1.106\ \text{rad/s}

Step 3: $K$

K = \omega_n^2 = 1.106^2 = 1.224

Step 4: $P$

\begin{aligned} 2 + KP &= 2\zeta\omega_n = 2(0.3224)(1.106) = 0.713 \\ P &= \frac{0.713 - 2}{1.224} = -1.05\ \text{s} \end{aligned}

Answer: $K \approx 1.22$ and $P \approx -1.05$ s.

Remark: $P$ comes out negative. With these values of $t_p$ and $M_p$ , the response is less damped than $K/[s(s+2)]$ with unity feedback would give ( $\zeta = 2/(2\sqrt{1.224}) = 0.90$ ). So the $sP$ term must act as positive rate feedback. If the figure values in your paper differ, use the same four steps with them.

2075 Asoj · 4 marks

Discuss effect of addition of a zero to a system.

Answer

Adding a zero to a system, either in the open-loop $G(s)H(s)$ or in the closed-loop transfer function, generally makes the response faster and changes the stability.

Zero added to the open-loop transfer function

The root locus is pulled to the left, away from the imaginary axis.
Relative stability improves. A system that was unstable for high $K$ can become stable for all $K$ . Example: $K/[s^2(s+a)]$ is always unstable, but $K(s+b)/[s^2(s+a)]$ with $b < a$ is stable for all $K > 0$ .
Damping increases, so overshoot reduces. This is the basis of PD control and lead compensation.

Zero added to the closed-loop transfer function

If $T(s)$ becomes $T(s)(1 + s/z)$ , the new output is

c_{new}(t) = c(t) + \frac{1}{z}\frac{dc(t)}{dt}

The derivative term adds to the response while it is rising, so rise time and peak time decrease.
Peak overshoot increases. The closer the zero is to the origin (small $z$ ), the larger the overshoot.
Bandwidth increases, so the system lets more high-frequency noise through.
A zero far to the left (about 5 times further than the dominant poles) has little effect.

 c(t)       with zero (faster, more overshoot)
  |       _/\_
  |     /  _--\___________
  |    / /                original
  |   //
  +--------------------------- t

Parameter	Effect of adding zero
Rise time	decreases
Overshoot (CL zero)	increases
Root locus	shifts left
Stability	improves
Bandwidth	increases

2074 Chaitra · 4 marks

Suppose that the step response of a first order system is c(t) = 5(1 − e^(−t/5)). What are impulse and ramp responses?

Answer

For a linear time-invariant system, the impulse response is the derivative of the step response, and the ramp response is the integral of the step response.

Transfer function

C(s) = \mathcal{L}\{5(1 - e^{-t/5})\} = 5\left[\frac{1}{s} - \frac{1}{s + 0.2}\right] = \frac{1}{s(s+0.2)}

Since $R(s) = 1/s$ :

G(s) = \frac{1}{s + 0.2} = \frac{5}{5s + 1}

This is a first-order system with gain 5 and time constant $T = 5$ s.

Impulse response

c_\delta(t) = \frac{d}{dt}\left[5(1 - e^{-t/5})\right] = 5\cdot\frac{1}{5}e^{-t/5} = e^{-t/5},\quad t \ge 0

Check: $\mathcal{L}^{-1}\{1/(s+0.2)\} = e^{-0.2t}$ .

Ramp response (unit ramp)

\begin{aligned} c_r(t) &= \int_0^t 5(1 - e^{-\tau/5})\,d\tau \\ &= 5\left[\tau + 5e^{-\tau/5}\right]_0^t \\ &= 5t + 25e^{-t/5} - 25 \\ &= 5t - 25 + 25e^{-t/5},\quad t \ge 0 \end{aligned}

Check: $\dfrac{1}{s^2(s+0.2)} = \dfrac{5}{s^2} - \dfrac{25}{s} + \dfrac{25}{s+0.2}$ , which gives the same result.

Answer: impulse response $= e^{-t/5}$ ; ramp response $= 5t - 25 + 25e^{-t/5}$ for $t \ge 0$ .

2074 Asoj · 8 marks

Determine the overall transfer function C(s)/R(s) of the given system by block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → G1 → summing point S3 (+, −) → node A → G2 → node B → G3 → summing point S4 (+, +) → C(s). G4 from node A to S4 (+). H1 from node B fed back to S2 (−). H2 from C(s) fed back to S3 (−). Unity feedback from C(s) to S1 (−).]

Answer

Name the signals: $E$ = output of $S_2$ (input of $G_1$ ), $A$ = output of $S_3$ , $B$ = output of $G_2$ .

Step 1: Move the take-off point of $H_1$ from $B$ back to $A$

$B = G_2A$ , so the $H_1$ branch becomes $G_2H_1$ taken from $A$ .

Step 2: Combine the paths from $A$ to $C$

$G_2$ and $G_3$ are in series, and that series pair is in parallel with $G_4$ :

\frac{C}{A} = G_2G_3 + G_4

Step 3: Move the take-off point of $G_2H_1$ from $A$ forward to $C$

Since $A = \dfrac{C}{G_2G_3 + G_4}$ , the branch becomes $\dfrac{G_2H_1}{G_2G_3 + G_4}$ taken from $C$ .

R->(S1)->(S2)-> G1 ->(S3)-> [G2G3+G4] --+--> C
    ^-     ^-          ^-               |
    |      |           +----- H2 -------+
    |      +--- G2H1/(G2G3+G4) ---------+
    +------------- 1 -------------------+

Step 4: Reduce the inner loop with $H_2$

\frac{C}{G_1E} = \frac{G_2G_3 + G_4}{1 + H_2(G_2G_3 + G_4)}

So the forward gain from $S_2$ is

G_f = \frac{G_1(G_2G_3 + G_4)}{1 + G_2G_3H_2 + G_4H_2}

Step 5: Combine the two outer feedbacks and close the loop

$S_1$ and $S_2$ are cascaded summers, so the feedbacks from $C$ add in parallel:

H = 1 + \frac{G_2H_1}{G_2G_3 + G_4}

\frac{C}{R} = \frac{G_f}{1 + G_fH} = \frac{G_1(G_2G_3 + G_4)}{1 + G_2G_3H_2 + G_4H_2 + G_1(G_2G_3 + G_4) + G_1G_2H_1}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 + G_2G_3H_2 + G_4H_2 + G_1G_2G_3 + G_1G_4}

Check (Mason): paths $G_1G_2G_3$ , $G_1G_4$ ; loops $-G_1G_2H_1$ , $-G_2G_3H_2$ , $-G_4H_2$ , $-G_1G_2G_3$ , $-G_1G_4$ ; all loops touch, $\Delta_1 = \Delta_2 = 1$ . Same result.

2073 Shrawan · 8 marks

Determine the overall transfer function C(s)/R(s) of the given system by block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → node A → G1 → summing point S2 (+, −) → G2 → summing point S3 (+, −) → G3 → node B → summing point S4 (+, +) → C(s). G4 from node A to S4 (+). H2 from node B fed back to S2 (−). H1 from C(s) fed back to S3 (−). H3 from C(s) fed back to S1 (−).]

Answer

Name the signals: $A$ = output of $S_1$ , $B$ = output of $G_3$ . From the figure, $C = B + G_4A$ .

Step 1: Move the take-off point of $H_1$ from $C$ back to $B$

Since $C = B + G_4A$ :

H_1C = H_1B + G_4H_1A

So $H_1$ is now taken from $B$ (a loop around $G_3$ ), and an extra branch $G_4H_1$ goes from $A$ to $S_3$ with a minus sign.

Step 2: Move the extra input at $S_3$ back before $G_2$

A signal entering after $G_2$ can be moved before it by dividing by $G_2$ : the branch becomes $G_4H_1/G_2$ entering $S_2$ (minus).

Step 3: Reduce the loop $G_3$ , $H_1$

\frac{G_3}{1 + G_3H_1}

Step 4: Reduce the loop with $H_2$

Forward: $G_2\cdot\dfrac{G_3}{1+G_3H_1}$ , feedback $H_2$ :

\frac{B}{E_2} = \frac{G_2G_3}{1 + G_3H_1 + G_2G_3H_2}

where $E_2 = G_1A - \dfrac{G_4H_1}{G_2}A$ is the input of $G_2$ . Hence

\frac{B}{A} = \frac{G_3(G_1G_2 - G_4H_1)}{1 + G_3H_1 + G_2G_3H_2}

Step 5: Add the parallel path $G_4$

\begin{aligned} \frac{C}{A} &= \frac{G_3(G_1G_2 - G_4H_1) + G_4(1 + G_3H_1 + G_2G_3H_2)}{1 + G_3H_1 + G_2G_3H_2} \\ &= \frac{G_1G_2G_3 + G_4 + G_2G_3G_4H_2}{1 + G_3H_1 + G_2G_3H_2} \end{aligned}

R ->(S1)--A--> [ N / D' ] -----+---> C
     ^-                        |
     +--------- H3 ------------+
 N  = G1G2G3 + G4 + G2G3G4H2
 D' = 1 + G3H1 + G2G3H2

Step 6: Close the outer loop with $H_3$

\frac{C}{R} = \frac{N/D'}{1 + H_3N/D'} = \frac{N}{D' + H_3N}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_4 + G_2G_3G_4H_2}{1 + G_3H_1 + G_2G_3H_2 + G_1G_2G_3H_3 + G_4H_3 + G_2G_3G_4H_2H_3}

Check (Mason): paths $P_1 = G_1G_2G_3$ ( $\Delta_1 = 1$ ), $P_2 = G_4$ ( $\Delta_2 = 1 + G_2G_3H_2$ , since the $H_2$ loop does not touch it). Loops $-G_2G_3H_2$ , $-G_3H_1$ , $-G_1G_2G_3H_3$ , $-G_4H_3$ ; the pair ( $-G_2G_3H_2$ , $-G_4H_3$ ) is non-touching. Same result.

2073 Shrawan · 4 marks

A closed loop servo is represented by the differential equation d²y/dt² + 8 dy/dt = 64z, where 'y' is the displacement of the output shaft and 'u' is the displacement of the input shaft and z = u − y. Determine frequency of sustained oscillation, damping ratio and percentage maximum overshoot for unit step input.

Answer

Take the Laplace transform with zero initial conditions and compare with the standard second order form $\omega_n^2/(s^2 + 2\zeta\omega_n s + \omega_n^2)$ .

Transfer function

\frac{d^2y}{dt^2} + 8\frac{dy}{dt} = 64(u - y) \;\Rightarrow\; (s^2 + 8s + 64)Y(s) = 64U(s)

\frac{Y(s)}{U(s)} = \frac{64}{s^2 + 8s + 64}

Parameters

\begin{aligned} \omega_n^2 &= 64 \;\Rightarrow\; \omega_n = 8\ \text{rad/s} \\ 2\zeta\omega_n &= 8 \;\Rightarrow\; \zeta = \frac{8}{16} = 0.5 \end{aligned}

Frequency of oscillation

The step response oscillates at the damped frequency:

\omega_d = \omega_n\sqrt{1 - \zeta^2} = 8\sqrt{0.75} = 6.928\ \text{rad/s}

(If there were no damping, sustained oscillation would be at $\omega_n = 8$ rad/s.)

Maximum overshoot

M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} = e^{-\pi(0.5)/0.866} = e^{-1.814} = 0.1630 = 16.30\%

Answer: $\omega_d = 6.93$ rad/s ( $\omega_n = 8$ rad/s), $\zeta = 0.5$ , $M_p = 16.3\%$ .

2073 Shrawan · 4 marks

Draw the region in s-plane that satisfies following requirements: i) ζ > 0.707 ii) ts < 2 s.

Answer

Each specification puts a boundary on where the closed-loop poles $s = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2}$ may lie.

i) $\zeta > 0.707$

A pole makes angle $\theta$ with the negative real axis, where $\cos\theta = \zeta$ .

\zeta > 0.707 \;\Rightarrow\; \theta < \cos^{-1}(0.707) = 45^\circ

So the poles must lie inside the wedge between the two lines at $\pm 45^\circ$ to the negative real axis.

ii) $t_s < 2$ s

Using the 2% criterion, $t_s = \dfrac{4}{\zeta\omega_n} = \dfrac{4}{\sigma}$ :

\frac{4}{\sigma} < 2 \;\Rightarrow\; \sigma = \zeta\omega_n > 2

So the poles must lie to the left of the vertical line $\text{Re}(s) = -2$ .

Required region

Both conditions together: the shaded region (marked #), left of $\sigma = -2$ and inside the $\pm 45^\circ$ lines.

                          jw
  ##############\     :   |
  ################\   :   |
  ##################\ :   |
  ####################\   |
  ####################: \ |
----------------------+---+------ sigma
  ####################: / |
  ####################/   |
  ##################/ :   |
  ################/   :   |
  ##############/     :   |
                     -2   0
  # = allowed region; \ / = 45 deg lines

The region is bounded by the lines $\omega = \pm\sigma$ (45 degrees) and the line $\sigma = -2$ ; it extends to the left without limit. (With the 5% criterion, $t_s = 3/\sigma$ , the vertical line is at $\sigma = -1.5$ .)

2073 Shrawan · 8 marks

For a closed loop system given by the block diagram below: i) Can the system track a step reference input 'r' with zero steady state error? ii) Can the system reject a step disturbance 'w' with zero steady state error? iii) Compute the sensitivity of closed loop transfer function to change in the plant pole at '−2'. [Figure: R → summing point (+, −) → controller 160(s+4)/(s+30) → summing point (+, + with disturbance w) → plant 1/[s(s+2)] → y; unity negative feedback from y]

Answer

Controller $D(s) = \dfrac{160(s+4)}{s+30}$ , plant $P(s) = \dfrac{1}{s(s+2)}$ .

Stability first

Characteristic equation $1 + DP = 0$ :

s(s+2)(s+30) + 160(s+4) = s^3 + 32s^2 + 220s + 640 = 0

Routh array:

Row	Col 1	Col 2
$s^3$	1	220
$s^2$	32	640
$s^1$	$(32\cdot 220 - 640)/32 = 200$	0
$s^0$	640

No sign change, so the closed loop is stable and the final value theorem can be used.

i) Tracking a step reference

G(s) = DP = \frac{160(s+4)}{s(s+2)(s+30)}

There is one pole at the origin, so the system is Type 1: $K_p = \infty$ .

e_{ss} = \frac{1}{1 + K_p} = 0

Yes, the system tracks a step reference with zero steady state error.

ii) Rejecting a step disturbance $w$

With $r = 0$ :

\frac{Y(s)}{W(s)} = \frac{P}{1 + DP} = \frac{s+30}{s^3 + 32s^2 + 220s + 640}

For a unit step $W(s) = 1/s$ :

y_{ss} = \lim_{s\to 0} s\cdot\frac{1}{s}\cdot\frac{s+30}{s^3+32s^2+220s+640} = \frac{30}{640} = 0.0469

No. The disturbance enters after the controller, and the integrator is in the plant, not in the controller. So a steady error of $0.0469w$ (about 4.7% of the disturbance) remains.

iii) Sensitivity to the plant pole at $-2$

Let the pole be at $-a$ , with $a = 2$ : $P = \dfrac{1}{s(s+a)}$ , $T = \dfrac{DP}{1+DP}$ . By the chain rule:

S_a^T = S_P^T \cdot S_a^P

\begin{aligned} S_P^T &= \frac{1}{1 + DP} = \frac{s(s+2)(s+30)}{s^3 + 32s^2 + 220s + 640} \\ S_a^P &= \frac{\partial P}{\partial a}\cdot\frac{a}{P} = \frac{-1}{s(s+a)^2}\cdot a\,s(s+a) = \frac{-a}{s+a} = \frac{-2}{s+2} \end{aligned}

Answer:

S_a^T = \frac{-2s(s+30)}{s^3 + 32s^2 + 220s + 640}

At DC ( $s = 0$ ) the sensitivity is zero, so a change in the pole location has no effect on the steady state output. Its effect grows at higher frequencies.

2072 Chaitra · 8 marks

Find transfer function of the following system. [Figure: R(s) → node P → summing point S1 (+, −) → summing point S2 (+, +, −) → G1 → node A → G2 → node B → summing point S3 (+, +) → C(s). H4 from node P (input R) to S2 (+). H2 from node A to S3 (+). H1 from C(s) fed back to S2 (−). H3 from node B fed back to S1 (−).]

Answer

Name the signals: $E$ = output of $S_2$ (input of $G_1$ ), $A$ = output of $G_1$ , $B$ = output of $G_2$ .

Step 1: Combine the cascaded summers $S_1$ and $S_2$

$R$ enters $S_1$ directly and $S_2$ through $H_4$ . Since the two summers are in series:

E = R + H_4R - H_3B - H_1C = (1 + H_4)R - H_3B - H_1C

So the input becomes a block $(1 + H_4)$ in front of one summer.

Step 2: Move the take-off point of $H_3$ from $B$ back to $A$

$B = G_2A$ , so $H_3$ becomes $G_2H_3$ taken from $A$ .

Step 3: Reduce the loop $G_1$ , $G_2H_3$

\frac{A}{E'} = \frac{G_1}{1 + G_1G_2H_3}

where $E'$ is the summer output before this loop.

Step 4: Parallel paths from $A$ to $C$

$G_2$ and $H_2$ both go from $A$ to $S_3$ :

\frac{C}{A} = G_2 + H_2

R -> [1+H4] ->(+)-> [G1/(1+G1G2H3)] -> [G2+H2] --+-> C
               ^-                                |
               +------------- H1 ----------------+

Step 5: Close the loop with $H_1$

G_f = \frac{G_1(G_2 + H_2)}{1 + G_1G_2H_3}

\frac{C}{R} = (1 + H_4)\frac{G_f}{1 + G_fH_1} = \frac{(1 + H_4)G_1(G_2 + H_2)}{1 + G_1G_2H_3 + G_1G_2H_1 + G_1H_1H_2}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1(G_2 + H_2)(1 + H_4)}{1 + G_1G_2H_1 + G_1G_2H_3 + G_1H_1H_2}

Check (Mason): four forward paths $G_1G_2$ , $G_1H_2$ , $H_4G_1G_2$ , $H_4G_1H_2$ ; loops $-G_1G_2H_3$ , $-G_1G_2H_1$ , $-G_1H_2H_1$ , all touching each other and all paths. Same result.

2072 Chaitra · 6 marks

Open loop pole/zero plot of a unity feedback system is shown in figure below. Determine maximum overshoot and settling time for its step response. [Figure: s-plane with open-loop poles at s = −2 ± j1; no zeros]

Answer

The open-loop poles at $s = -2 \pm j1$ give

G(s) = \frac{K}{(s+2-j)(s+2+j)} = \frac{K}{s^2 + 4s + 5}

No gain is marked on the plot, so take $K = 1$ (and show the general result).

Closed-loop transfer function (unity feedback)

\frac{C(s)}{R(s)} = \frac{K}{s^2 + 4s + 5 + K}

Closed-loop poles: $s = -2 \pm j\sqrt{1 + K}$ , so $\sigma = \zeta\omega_n = 2$ for every $K$ .

For $K = 1$ :

\begin{aligned} \omega_n^2 &= 6 \;\Rightarrow\; \omega_n = 2.449\ \text{rad/s} \\ 2\zeta\omega_n &= 4 \;\Rightarrow\; \zeta = \frac{2}{2.449} = 0.8165 \end{aligned}

Maximum overshoot

M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} = e^{-\pi\sigma/\omega_d} = e^{-2\pi/\sqrt{2}} = e^{-4.443} = 0.0118 = 1.18\%

In general, $M_p = e^{-2\pi/\sqrt{1+K}}$ .

Settling time (2% criterion)

t_s = \frac{4}{\zeta\omega_n} = \frac{4}{2} = 2\ \text{s}

This is the same for any $K$ , because the real part of the closed-loop poles is always $-2$ .

Answer: with $K = 1$ : $M_p \approx 1.18\%$ and $t_s = 2$ s.

(If the plotted poles $-2 \pm j1$ are taken directly as the dominant poles, $\zeta = 2/\sqrt{5} = 0.894$ , $M_p = e^{-2\pi} = 0.19\%$ , and $t_s$ is still 2 s.)

2071 Chaitra · 8 marks

Following figure shows a mechanical vibratory system and the response when 10 lb of force is applied to the system. Determine the transfer function and value of M, D and K. The displacement x is measured from the equilibrium position. [Figure: mass M suspended from a fixed support by spring K, with damper D from M to the fixed ground below; a 10 lb step force acts on M; x(t) downward. Response: x(t) settles at 0.02 (ft), with peak overshoot 0.0093 above the final value occurring at t = 3 sec]

Answer

The system is a mass-spring-damper driven by a force. Find $K$ from the final value, $\zeta$ from the overshoot, and $\omega_n$ from the peak time.

Transfer function

M\ddot{x} + D\dot{x} + Kx = f(t) \;\Rightarrow\; \frac{X(s)}{F(s)} = \frac{1}{Ms^2 + Ds + K}

with $\omega_n^2 = K/M$ and $2\zeta\omega_n = D/M$ .

Step 1: $K$ from steady state

For $F(s) = 10/s$ :

x(\infty) = \lim_{s\to 0} s\cdot\frac{10}{s}\cdot\frac{1}{Ms^2+Ds+K} = \frac{10}{K} = 0.02 \;\Rightarrow\; K = 500\ \text{lb/ft}

Step 2: $\zeta$ from overshoot

M_p = \frac{0.0093}{0.02} = 0.465

\zeta = \frac{-\ln 0.465}{\sqrt{\pi^2 + (\ln 0.465)^2}} = \frac{0.7657}{\sqrt{9.8696 + 0.5863}} = \frac{0.7657}{3.2336} = 0.2368

Step 3: $\omega_n$ from $t_p = 3$ s

\omega_n = \frac{\pi}{t_p\sqrt{1-\zeta^2}} = \frac{3.1416}{3\sqrt{1 - 0.0561}} = \frac{3.1416}{3(0.9715)} = 1.078\ \text{rad/s}

Step 4: $M$ and $D$

\begin{aligned} M &= \frac{K}{\omega_n^2} = \frac{500}{1.1618} = 430.4\ \text{slug} \\ D &= 2\zeta\omega_n M = 2(0.2368)(1.078)(430.4) = 219.7\ \text{lb-s/ft} \end{aligned}

Transfer function

\frac{X(s)}{F(s)} = \frac{1}{430.4s^2 + 219.7s + 500}

Answer: $M \approx 430$ slug, $D \approx 220$ lb-s/ft, $K = 500$ lb/ft.

2071 Chaitra · 8 marks

Show that using the velocity feedback technique shown in figure below damping ratio and steady state error are both increased. [Figure: R(s) → summing point (+, −) → summing point (+, −) → ωn²/[s(s + 2ζωn)] → C(s); minor loop: output fed back through skd to the second summing point (−); major loop: unity feedback from output to the first summing point (−)]

Answer

Velocity (derivative output) feedback adds a term proportional to $\dot{c}$ to the feedback signal. This increases the coefficient of $s$ in the characteristic equation.

Without velocity feedback ( $k_d = 0$ )

G(s) = \frac{\omega_n^2}{s(s + 2\zeta\omega_n)}, \qquad \frac{C(s)}{R(s)} = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}

Damping ratio $= \zeta$ . Velocity error constant and ramp error:

K_v = \lim_{s\to 0}sG(s) = \frac{\omega_n}{2\zeta}, \qquad e_{ss} = \frac{1}{K_v} = \frac{2\zeta}{\omega_n}

With velocity feedback

The minor loop has forward gain $\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}$ and feedback $sk_d$ :

G'(s) = \frac{\frac{\omega_n^2}{s(s+2\zeta\omega_n)}}{1 + \frac{\omega_n^2 sk_d}{s(s+2\zeta\omega_n)}} = \frac{\omega_n^2}{s(s + 2\zeta\omega_n + k_d\omega_n^2)}

Closed loop (unity outer feedback):

\frac{C(s)}{R(s)} = \frac{\omega_n^2}{s^2 + (2\zeta\omega_n + k_d\omega_n^2)s + \omega_n^2}

New damping ratio

Comparing with $s^2 + 2\zeta'\omega_n s + \omega_n^2$ ( $\omega_n$ unchanged):

2\zeta'\omega_n = 2\zeta\omega_n + k_d\omega_n^2 \;\Rightarrow\; \zeta' = \zeta + \frac{k_d\omega_n}{2}

Since $k_d > 0$ , $\zeta' > \zeta$ . Damping increases, so overshoot falls.

New steady state error (unit ramp)

K_v' = \lim_{s\to 0}sG'(s) = \frac{\omega_n^2}{2\zeta\omega_n + k_d\omega_n^2} = \frac{\omega_n}{2\zeta + k_d\omega_n}

e_{ss}' = \frac{1}{K_v'} = \frac{2\zeta + k_d\omega_n}{\omega_n} = \frac{2\zeta}{\omega_n} + k_d

Since $k_d > 0$ , $e_{ss}' > e_{ss}$ . The steady state error to a ramp increases by $k_d$ . (The step error stays zero, as the system is still Type 1.)

Quantity	Without $k_d$	With $k_d$
$\omega_n$	$\omega_n$	$\omega_n$
Damping ratio	$\zeta$	$\zeta + k_d\omega_n/2$
$K_v$	$\omega_n/2\zeta$	$\omega_n/(2\zeta + k_d\omega_n)$
Ramp $e_{ss}$	$2\zeta/\omega_n$	$2\zeta/\omega_n + k_d$

So both the damping ratio and the steady state error increase. This error can be reduced again by raising the forward gain.

2071 Shrawan · 4 marks

For an open loop transfer function with unity feedback G(s) = ωn²/[s(s + 2ξωn)] where ξ < 1, derive an expression for output when unit step input is applied.

Answer

For $\xi < 1$ (underdamped) the unit step response is a damped sinusoid that settles at 1.

Closed-loop transfer function

\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{\omega_n^2}{s^2 + 2\xi\omega_n s + \omega_n^2}

For $\xi < 1$ the poles are complex:

s = -\xi\omega_n \pm j\omega_n\sqrt{1-\xi^2} = -\xi\omega_n \pm j\omega_d, \qquad \omega_d = \omega_n\sqrt{1-\xi^2}

Output for a unit step

With $R(s) = 1/s$ :

C(s) = \frac{\omega_n^2}{s(s^2 + 2\xi\omega_n s + \omega_n^2)}

Partial fractions:

\begin{aligned} C(s) &= \frac{1}{s} - \frac{s + 2\xi\omega_n}{s^2 + 2\xi\omega_n s + \omega_n^2} \\ &= \frac{1}{s} - \frac{s + \xi\omega_n}{(s + \xi\omega_n)^2 + \omega_d^2} - \frac{\xi\omega_n}{(s + \xi\omega_n)^2 + \omega_d^2} \end{aligned}

Using $\mathcal{L}^{-1}\left\{\frac{s+a}{(s+a)^2+\omega^2}\right\} = e^{-at}\cos\omega t$ and $\mathcal{L}^{-1}\left\{\frac{\omega}{(s+a)^2+\omega^2}\right\} = e^{-at}\sin\omega t$ :

\begin{aligned} c(t) &= 1 - e^{-\xi\omega_n t}\cos\omega_d t - \frac{\xi\omega_n}{\omega_d}e^{-\xi\omega_n t}\sin\omega_d t \\ &= 1 - e^{-\xi\omega_n t}\left[\cos\omega_d t + \frac{\xi}{\sqrt{1-\xi^2}}\sin\omega_d t\right] \end{aligned}

Put $\cos\theta = \xi$ and $\sin\theta = \sqrt{1-\xi^2}$ , so that $\cos\omega_d t\,\sin\theta + \sin\omega_d t\,\cos\theta = \sin(\omega_d t + \theta)$ :

c(t) = 1 - \frac{e^{-\xi\omega_n t}}{\sqrt{1-\xi^2}}\sin(\omega_d t + \theta), \qquad \theta = \tan^{-1}\frac{\sqrt{1-\xi^2}}{\xi}

The response is a sine wave at $\omega_d$ that decays inside the envelopes $1 \pm e^{-\xi\omega_n t}/\sqrt{1-\xi^2}$ and settles at the final value 1, so the steady state error is zero.

 c(t)
  |    _
  |   / \     _
 1|--/---\---/-\--------- final value
  | /     \_/
  |/
  +------------------------ t

2070 Chaitra (old course) · 8 marks

Determine the transfer function C(s)/R(s) for the following system. [Figure: R(s) → summing point S1 (+, −) → G1 → summing point S2 (+, −) → node A → G2 → node B → summing point S3 (+, +) → C(s). G3 from node A to S3 (+). H1 from node B fed back to S1 (−). Unity feedback from C(s) to S2 (−).]

Answer

Name the signals: $E$ = output of $S_1$ , $A$ = output of $S_2$ , $B$ = output of $G_2$ .

Step 1: Combine the parallel paths from $A$ to $C$

$G_2$ (through $B$ ) and $G_3$ both go from $A$ to $S_3$ :

\frac{C}{A} = G_2 + G_3

Step 2: Reduce the inner unity-feedback loop

$C$ is fed back to $S_2$ with unity gain:

\frac{C}{G_1E} = \frac{G_2 + G_3}{1 + G_2 + G_3}

Step 3: Move the take-off point of $H_1$ from $B$ to $C$

$B = G_2A$ and $A = \dfrac{C}{G_2 + G_3}$ , so

H_1B = \frac{G_2H_1}{G_2 + G_3}\,C

R ->(S1)-> G1 -> [(G2+G3)/(1+G2+G3)] --+--> C
     ^-                                |
     +----- G2H1/(G2+G3) <-------------+

Step 4: Close the outer loop

\begin{aligned} G_f &= \frac{G_1(G_2 + G_3)}{1 + G_2 + G_3}, \qquad H = \frac{G_2H_1}{G_2 + G_3} \\ \frac{C}{R} &= \frac{G_f}{1 + G_fH} = \frac{G_1(G_2 + G_3)}{1 + G_2 + G_3 + G_1G_2H_1} \end{aligned}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_2 + G_3 + G_1G_2H_1}

Check (Mason): paths $G_1G_2$ , $G_1G_3$ ; loops $-G_2$ , $-G_3$ , $-G_1G_2H_1$ , all touching each other and both paths. Same result.

2070 Chaitra (old course) · 8 marks

The unit step response of a linear control system is shown in figure below. Find the transfer function of a second order system to model the system. [Figure: unit step response c(t) with final value 1 and peak value 1.25 occurring at t = 0.01 sec]

Answer

The response has an overshoot, so model it as an underdamped second order system with unity DC gain (final value 1):

T(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}

Step 1: Read the response

Final value $= 1$ , so the DC gain is 1.
Peak value $= 1.25$ , so $M_p = \dfrac{1.25 - 1}{1} = 0.25$ (25%).
Peak time $t_p = 0.01$ s.

Step 2: Damping ratio

M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} \;\Rightarrow\; \zeta = \frac{-\ln M_p}{\sqrt{\pi^2 + (\ln M_p)^2}} = \frac{1.3863}{\sqrt{9.8696 + 1.9218}} = \frac{1.3863}{3.4339} = 0.4037

Step 3: Natural frequency

t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} \;\Rightarrow\; \omega_n = \frac{\pi}{0.01\sqrt{1 - 0.4037^2}} = \frac{3.1416}{0.01(0.9149)} = 343.4\ \text{rad/s}

Step 4: Coefficients

\begin{aligned} \omega_n^2 &= 343.4^2 = 1.179\times 10^5 \\ 2\zeta\omega_n &= 2(0.4037)(343.4) = 277.3 \end{aligned}

Answer:

T(s) = \frac{C(s)}{R(s)} = \frac{117914}{s^2 + 277.3s + 117914}

with $\zeta = 0.404$ and $\omega_n = 343.4$ rad/s. If the system is taken as unity feedback, the equivalent open-loop model is $G(s) = \dfrac{117914}{s(s + 277.3)}$ .

2070 Chaitra · 7 marks

For a second order system with G(s) = ωn²/[s(s + 2ξωn)] and H(s) = 1, find expression for maximum overshoot on its unit step response, where ωn is natural frequency of oscillation and ξ is damping ratio, at underdamped situation. [Figure: R(s) → summing point (+, −) → G(s) → C(s), feedback H(s)]

Answer

Maximum (peak) overshoot is the largest amount by which the response exceeds its final value, usually given as a percentage of the final value. To find it, derive $c(t)$ , find the time of the first peak, and substitute.

Closed-loop transfer function

\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{\omega_n^2}{s^2 + 2\xi\omega_n s + \omega_n^2}

For $\xi < 1$ the poles are complex:

s = -\xi\omega_n \pm j\omega_n\sqrt{1-\xi^2} = -\xi\omega_n \pm j\omega_d, \qquad \omega_d = \omega_n\sqrt{1-\xi^2}

Output for a unit step

With $R(s) = 1/s$ :

C(s) = \frac{\omega_n^2}{s(s^2 + 2\xi\omega_n s + \omega_n^2)}

Partial fractions:

\begin{aligned} C(s) &= \frac{1}{s} - \frac{s + 2\xi\omega_n}{s^2 + 2\xi\omega_n s + \omega_n^2} \\ &= \frac{1}{s} - \frac{s + \xi\omega_n}{(s + \xi\omega_n)^2 + \omega_d^2} - \frac{\xi\omega_n}{(s + \xi\omega_n)^2 + \omega_d^2} \end{aligned}

Using $\mathcal{L}^{-1}\left\{\frac{s+a}{(s+a)^2+\omega^2}\right\} = e^{-at}\cos\omega t$ and $\mathcal{L}^{-1}\left\{\frac{\omega}{(s+a)^2+\omega^2}\right\} = e^{-at}\sin\omega t$ :

\begin{aligned} c(t) &= 1 - e^{-\xi\omega_n t}\cos\omega_d t - \frac{\xi\omega_n}{\omega_d}e^{-\xi\omega_n t}\sin\omega_d t \\ &= 1 - e^{-\xi\omega_n t}\left[\cos\omega_d t + \frac{\xi}{\sqrt{1-\xi^2}}\sin\omega_d t\right] \end{aligned}

Put $\cos\theta = \xi$ and $\sin\theta = \sqrt{1-\xi^2}$ , so that $\cos\omega_d t\,\sin\theta + \sin\omega_d t\,\cos\theta = \sin(\omega_d t + \theta)$ :

c(t) = 1 - \frac{e^{-\xi\omega_n t}}{\sqrt{1-\xi^2}}\sin(\omega_d t + \theta), \qquad \theta = \tan^{-1}\frac{\sqrt{1-\xi^2}}{\xi}

Peak time

At a peak, $\dfrac{dc}{dt} = 0$ . Differentiating $c(t)$ :

\frac{dc}{dt} = \frac{\omega_n}{\sqrt{1-\xi^2}}e^{-\xi\omega_n t}\sin\omega_d t = 0

So $\sin\omega_d t = 0$ , which gives $\omega_d t = n\pi$ . The first peak (maximum overshoot) is at $n = 1$ :

t_p = \frac{\pi}{\omega_d} = \frac{\pi}{\omega_n\sqrt{1-\xi^2}}

Maximum overshoot

Substitute $t_p$ in $c(t)$ . Since $\omega_d t_p = \pi$ , $\sin(\pi + \theta) = -\sin\theta = -\sqrt{1-\xi^2}$ :

\begin{aligned} c(t_p) &= 1 - \frac{e^{-\xi\omega_n\pi/\omega_d}}{\sqrt{1-\xi^2}}\left(-\sqrt{1-\xi^2}\right) \\ &= 1 + e^{-\xi\pi/\sqrt{1-\xi^2}} \end{aligned}

With final value $c(\infty) = 1$ :

M_p = c(t_p) - c(\infty) = e^{-\xi\pi/\sqrt{1-\xi^2}}

\%M_p = 100\,e^{-\xi\pi/\sqrt{1-\xi^2}}

Remarks

$M_p$ depends only on $\xi$ , not on $\omega_n$ .
When $\xi$ increases, $M_p$ decreases: e.g. $\xi = 0.5$ gives 16.3%, $\xi = 0.707$ gives 4.3%, and $\xi \ge 1$ gives no overshoot.

2070 Chaitra · 6 marks

Find all static error constants for a unity feedback system with feedforward transfer function G(s) = 1000/[s(s+10)(s+100)]. Evaluate steady state error if system is excited with r(t) = 2 + t.

Answer

The steady state error of a unity feedback system follows from its static error constants. The system is Type 1 (one pole at the origin).

Stability check

$1 + G = 0$ gives $s^3 + 110s^2 + 1000s + 1000 = 0$ . Routh: $110 \times 1000 > 1000$ , so all first-column terms are positive and the system is stable.

Static error constants

\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \lim_{s\to 0}\frac{1000}{s(s+10)(s+100)} = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \frac{1000}{(10)(100)} = 1 \\ K_a &= \lim_{s\to 0} s^2G(s) = 0 \end{aligned}

Steady state error for $r(t) = 2 + t$

By superposition:

\begin{aligned} e_{ss} &= \frac{2}{1 + K_p} + \frac{1}{K_v} \\ &= \frac{2}{1 + \infty} + \frac{1}{1} = 0 + 1 = 1 \end{aligned}

Input part	Error
Step 2	0
Ramp t	1

Answer: $K_p = \infty$ , $K_v = 1\ \text{s}^{-1}$ , $K_a = 0$ ; $e_{ss} = 1$ .

2069 Chaitra · 6 marks

Convert the given block diagram to signal flow graph and determine the overall transfer function using Mason's Gain Formula. [Figure: R(s) → summing point S1 (+, −) → node A → G1(s) → node B → G2(s) → summing point S2 (+, −) → C(s). G3(s) from node A to S2 (−). H(s) from node B fed back to S1 (−).]

Answer

Signal flow graph

Nodes: $R$ , $A$ (output of $S_1$ ), $B$ (output of $G_1$ ), $C$ .

From the diagram: $A = R - HB$ , $B = G_1A$ , $C = G_2B - G_3A$ .

              -G3
        +----------------+
        |                v
R --1--> A --G1--> B --G2--> C
         ^         |
         +---(-H)--+

Branches: $R\to A$ (1), $A\to B$ ( $G_1$ ), $B\to C$ ( $G_2$ ), $A\to C$ ( $-G_3$ ), $B\to A$ ( $-H$ ).

Mason's gain formula

T = \frac{\sum P_k\Delta_k}{\Delta}

Forward paths

$P_1 = G_1G_2$ (R, A, B, C)
$P_2 = -G_3$ (R, A, C)

Loops

$L_1 = -G_1H$ (A, B, A)

There is only one loop, so no non-touching loops.

\Delta = 1 - L_1 = 1 + G_1H

Cofactors: both paths pass through node $A$ , which is on $L_1$ , so $\Delta_1 = \Delta_2 = 1$ .

Transfer function

T = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2 - G_3}{1 + G_1H}

2067 Asar (old course) · 8 marks

A step torque T(t) is applied in a system shown in figure below. Find the percent overshoot, settling time and peak time for output θ2(t). [Figure: torque T(t) applied to inertia 1 kg-m² (angle θ1(t)); inertia connected through viscous damper 1 N-m-s/rad to a point of angle θ2(t), which is connected through a torsional spring 1 N-m/rad to the fixed wall]

Answer

Write the torque equations at each angle, eliminate $\theta_1$ to get $\Theta_2(s)/T(s)$ , then compare with the standard second order form.

Equations of motion

At $\theta_1$ (inertia $J = 1$ , damper $D = 1$ between $\theta_1$ and $\theta_2$ ):

J\ddot{\theta}_1 + D(\dot{\theta}_1 - \dot{\theta}_2) = T \;\Rightarrow\; (s^2 + s)\Theta_1 - s\Theta_2 = T(s)

At $\theta_2$ (no inertia; damper and spring $K = 1$ to the wall):

D(\dot{\theta}_2 - \dot{\theta}_1) + K\theta_2 = 0 \;\Rightarrow\; -s\Theta_1 + (s + 1)\Theta_2 = 0

Transfer function

From the second equation, $\Theta_1 = \dfrac{s+1}{s}\Theta_2$ . Substituting in the first:

\begin{aligned} (s^2 + s)\frac{s+1}{s}\Theta_2 - s\Theta_2 &= T \\ \left[(s+1)^2 - s\right]\Theta_2 &= T \\ \frac{\Theta_2(s)}{T(s)} &= \frac{1}{s^2 + s + 1} \end{aligned}

Parameters

\omega_n^2 = 1 \Rightarrow \omega_n = 1\ \text{rad/s}, \qquad 2\zeta\omega_n = 1 \Rightarrow \zeta = 0.5

Percent overshoot

\%OS = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}} = 100\,e^{-\pi(0.5)/0.866} = 100\,e^{-1.814} = 16.3\%

Settling time (2% criterion)

t_s = \frac{4}{\zeta\omega_n} = \frac{4}{0.5 \times 1} = 8\ \text{s}

Peak time

t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} = \frac{3.1416}{0.866} = 3.63\ \text{s}

Answer: $\%OS = 16.3\%$ , $t_s = 8$ s, $t_p = 3.63$ s.

2067 Asar (old course) · 8 marks

The open loop transfer function of a unity feedback system is given by G(s) = 5/[s(s+2)(s²+2s+8)]. Find the static error coefficients and steady state error of the system when subjected to an input given by r(t) = 2 + 5t + 2t².

Answer

The steady state error of a unity feedback system follows from its static error coefficients. $G(s)$ has one pole at the origin, so the system is Type 1.

Stability check

$1 + G = 0$ : $s^4 + 4s^3 + 12s^2 + 16s + 5 = 0$ .

Row
$s^4$	1	12	5
$s^3$	4	16
$s^2$	8	5
$s^1$	13.5
$s^0$	5

No sign change, so the system is stable.

Static error coefficients

\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \frac{5}{(2)(8)} = 0.3125 \\ K_a &= \lim_{s\to 0} s^2G(s) = 0 \end{aligned}

Steady state error for $r(t) = 2 + 5t + 2t^2$

Write $2t^2 = 4\cdot\dfrac{t^2}{2}$ .

\begin{aligned} e_{ss} &= \frac{2}{1 + K_p} + \frac{5}{K_v} + \frac{4}{K_a} \\ &= 0 + \frac{5}{0.3125} + \frac{4}{0} = 0 + 16 + \infty = \infty \end{aligned}

Input part	Error
Step 2	0
Ramp 5t	16
Parabola 2t²	∞

Answer: $K_p = \infty$ , $K_v = 0.3125$ , $K_a = 0$ ; $e_{ss} = \infty$ . A Type 1 system cannot follow the parabolic part of the input.

2066 Bhadra (old course) · 8 marks

Reduce the block diagram of fig. 2(a) and find the overall transfer function C(s)/R(s). [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → G1 → summing point S3 (+, −) → node A → G2 → node B → summing point S4 (+, +) → C(s). G3 from node A to S4 (+). H1 from node B fed back to S2 (−). H2 from C(s) fed back to S3 (−). Unity feedback from C(s) to S1 (−).]

Answer

Name the signals: $E$ = output of $S_2$ (input of $G_1$ ), $A$ = output of $S_3$ , $B$ = output of $G_2$ .

Step 1: Combine the parallel paths from $A$ to $C$

$G_2$ and $G_3$ both go from $A$ to $S_4$ :

\frac{C}{A} = G_2 + G_3

Step 2: Move the take-off point of $H_1$ from $B$ to $C$

$B = G_2A$ and $A = \dfrac{C}{G_2 + G_3}$ , so the branch becomes $\dfrac{G_2H_1}{G_2 + G_3}$ taken from $C$ .

Step 3: Reduce the inner loop with $H_2$

\frac{C}{G_1E} = \frac{G_2 + G_3}{1 + H_2(G_2 + G_3)}

Step 4: Combine the outer feedbacks

$S_1$ and $S_2$ are cascaded, so the unity feedback and the moved $H_1$ branch add in parallel:

H = 1 + \frac{G_2H_1}{G_2 + G_3}

R ->(+)-> G1 -> [(G2+G3)/(1+G2H2+G3H2)] --+--> C
     ^-                                   |
     +------ 1 + G2H1/(G2+G3) <-----------+

Step 5: Close the loop

G_f = \frac{G_1(G_2 + G_3)}{1 + G_2H_2 + G_3H_2}

\frac{C}{R} = \frac{G_f}{1 + G_fH} = \frac{G_1(G_2 + G_3)}{1 + G_2H_2 + G_3H_2 + G_1(G_2 + G_3) + G_1G_2H_1}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_1G_2 + G_1G_3 + G_1G_2H_1 + G_2H_2 + G_3H_2}

Check (Mason): paths $G_1G_2$ , $G_1G_3$ ; loops $-G_1G_2$ , $-G_1G_3$ , $-G_1G_2H_1$ , $-G_2H_2$ , $-G_3H_2$ , all touching. Same result.

2066 Bhadra (old course) · 8 marks

The open loop transfer function of a unity negative feedback system is given by G(s) = 20/[s(0.5s+1)(s+2)]. Calculate the static error constants for this system. Also calculate the steady state error due to input r(t) = 10 + 5t.

Answer

The steady state error follows from the static error constants. $G(s)$ has one pole at the origin, so the system is Type 1.

Static error constants

G(s) = \frac{20}{s(0.5s + 1)(s + 2)}

\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \frac{20}{(1)(2)} = 10\ \text{s}^{-1} \\ K_a &= \lim_{s\to 0} s^2G(s) = 0 \end{aligned}

Steady state error for $r(t) = 10 + 5t$

\begin{aligned} e_{ss} &= \frac{10}{1 + K_p} + \frac{5}{K_v} \\ &= \frac{10}{1 + \infty} + \frac{5}{10} = 0 + 0.5 = 0.5 \end{aligned}

Answer: $K_p = \infty$ , $K_v = 10\ \text{s}^{-1}$ , $K_a = 0$ ; $e_{ss} = 0.5$ .

Note on stability: $G(s) = \dfrac{40}{s(s+2)^2}$ , so the characteristic equation is $s^3 + 4s^2 + 4s + 40 = 0$ . Routh's condition for a cubic needs $4 \times 4 > 40$ , which fails. So the closed loop as given is unstable, and the error value above (the usual expected answer) holds only in the formal sense. The gain would have to be below 16 (i.e. $K < 8$ in $K/[s(0.5s+1)(s+2)]$ ) for the result to apply.

2066 Bhadra (old course) · 8 marks

Evaluate the percentage overshoot and peak time for the unity feedback system with open loop transfer function G(s) = 5/[s(s+4)].

Answer

Find the closed-loop transfer function, read off $\omega_n$ and $\zeta$ , then use the standard second order formulas.

Closed-loop transfer function

\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{5}{s^2 + 4s + 5}

Parameters

\begin{aligned} \omega_n^2 &= 5 \;\Rightarrow\; \omega_n = \sqrt{5} = 2.236\ \text{rad/s} \\ 2\zeta\omega_n &= 4 \;\Rightarrow\; \zeta = \frac{4}{2(2.236)} = 0.894 \end{aligned}

Since $\zeta < 1$ , the response is underdamped.

\omega_d = \omega_n\sqrt{1 - \zeta^2} = 2.236\sqrt{1 - 0.8} = 2.236(0.4472) = 1\ \text{rad/s}

(The closed-loop poles are $s = -2 \pm j1$ .)

Percentage overshoot

\%M_p = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}} = 100\,e^{-\pi(0.894)/0.447} = 100\,e^{-2\pi} = 0.187\%

Peak time

t_p = \frac{\pi}{\omega_d} = \frac{\pi}{1} = 3.14\ \text{s}

Answer: $\%M_p \approx 0.19\%$ and $t_p = 3.14$ s. The system is heavily damped, so the overshoot is very small.

2066 Jestha (old course) · 8 marks

The open loop transfer function of a unity feedback system is given by G(s) = K/[s(s+1)(s²+2s+2)]. Calculate the static error constants for this system and find the range of K if static error is less than 0.5 for input r(t) = 10 + 50t.

Answer

$G(s)$ has one pole at the origin, so the system is Type 1: the step error is zero and the ramp error is $1/K_v$ .

Static error constants

\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \frac{K}{(1)(2)} = \frac{K}{2} \\ K_a &= \lim_{s\to 0} s^2G(s) = 0 \end{aligned}

Steady state error for $r(t) = 10 + 50t$

e_{ss} = \frac{10}{1 + K_p} + \frac{50}{K_v} = 0 + \frac{50}{K/2} = \frac{100}{K}

Condition $e_{ss} < 0.5$

\frac{100}{K} < 0.5 \;\Rightarrow\; K > 200

Stability check (needed for the result to be valid)

Characteristic equation: $s^4 + 3s^3 + 4s^2 + 2s + K = 0$ .

Row
$s^4$	1	4	$K$
$s^3$	3	2
$s^2$	$10/3$	$K$
$s^1$	$2 - 0.9K$
$s^0$	$K$

For stability: $2 - 0.9K > 0$ and $K > 0$ , so $0 < K < 2.22$ .

Answer: the error condition needs $K > 200$ , but the system is stable only for $0 < K < 2.22$ . The two ranges do not overlap, so no value of K meets $e_{ss} < 0.5$ with a stable system. A compensator (for example lag compensation) is needed to raise $K_v$ without causing instability.

2066 Jestha (old course) · 6 marks

Evaluate the transfer function Y(s)/R(s) for the system represented by following block diagram. [Figure: R(s) → summing point S1 (+, −) → node A → G1 → summing point S2 (+, −) → node B → G2 → summing point S3 (+, −) → node C → G3 → summing point S4 (+, +) → Y(s). G4 from node A to S4 (+). H1 from node B fed back to S1 (−). H2 from node C fed back to S2 (−). H3 from Y(s) fed back to S3 (−).]

Answer

Mason's gain formula is used (the block diagram reduction gives the same result). Name the signals: $A$ = output of $S_1$ , $B$ = output of $S_2$ , $N$ = output of $S_3$ (node C in the figure), $Y$ = output.

Signal flow graph

                       G4
        +--------------------------------+
        |                                v
R -1--> A --G1--> B --G2--> N --G3--> Y --1--> Y
        ^         |^        |^        |
        +--(-H1)--+|        ||        |
                   +-(-H2)--+|        |
                             +-(-H3)--+

Branches: $B\to A$ ( $-H_1$ ), $N\to B$ ( $-H_2$ ), $Y\to N$ ( $-H_3$ ), $A\to Y$ ( $G_4$ ).

Forward paths

$P_1 = G_1G_2G_3$ (A, B, N, Y)
$P_2 = G_4$ (A, Y)

Individual loops

$L_1 = -G_1H_1$ (A, B, A)
$L_2 = -G_2H_2$ (B, N, B)
$L_3 = -G_3H_3$ (N, Y, N)
$L_4 = G_4(-H_3)(-H_2)(-H_1) = -G_4H_1H_2H_3$ (A, Y, N, B, A)

Non-touching loops

$L_1$ (A, B) and $L_3$ (N, Y) do not touch: $L_1L_3 = G_1G_3H_1H_3$ . No three loops are mutually non-touching.

\begin{aligned} \Delta &= 1 - (L_1 + L_2 + L_3 + L_4) + L_1L_3 \\ &= 1 + G_1H_1 + G_2H_2 + G_3H_3 + G_4H_1H_2H_3 + G_1G_3H_1H_3 \end{aligned}

Cofactors

$P_1$ touches all loops: $\Delta_1 = 1$ .
$P_2$ (A, Y) does not touch $L_2$ (B, N): $\Delta_2 = 1 - L_2 = 1 + G_2H_2$ .

Transfer function

\frac{Y}{R} = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta}

Answer:

\frac{Y(s)}{R(s)} = \frac{G_1G_2G_3 + G_4(1 + G_2H_2)}{1 + G_1H_1 + G_2H_2 + G_3H_3 + G_1G_3H_1H_3 + G_4H_1H_2H_3}

2065 Shrawan (old course) · 8 marks

Find the unit step response of a unity feedback system whose open loop transfer function is given by G(s)H(s) = 4/[s(s+2)].

Answer

For unity feedback, the closed-loop transfer function is $T(s)=\frac{G(s)}{1+G(s)}$ . We find $C(s)$ for $R(s)=1/s$ and take the inverse Laplace transform.

Closed-loop transfer function

\begin{aligned} \frac{C(s)}{R(s)} &= \frac{\frac{4}{s(s+2)}}{1+\frac{4}{s(s+2)}} = \frac{4}{s^2+2s+4} \end{aligned}

Compare with the standard form $\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$ :

$\omega_n^2 = 4 \Rightarrow \omega_n = 2$ rad/s
$2\zeta\omega_n = 2 \Rightarrow \zeta = 0.5$ (underdamped, $0<\zeta<1$ )
$\omega_d = \omega_n\sqrt{1-\zeta^2} = 2\sqrt{0.75} = \sqrt{3} = 1.732$ rad/s
$\zeta\omega_n = 1$ , $\theta = \cos^{-1}\zeta = 60^\circ = \pi/3$ rad

Response by partial fractions

\begin{aligned} C(s) &= \frac{4}{s(s^2+2s+4)} = \frac{1}{s} - \frac{s+2}{s^2+2s+4} \\ &= \frac{1}{s} - \frac{(s+1)}{(s+1)^2+3} - \frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{(s+1)^2+3} \end{aligned}

Taking the inverse Laplace transform:

\begin{aligned} c(t) &= 1 - e^{-t}\left[\cos\sqrt{3}t + \frac{1}{\sqrt{3}}\sin\sqrt{3}t\right] \\ &= 1 - \frac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}}\sin(\omega_d t+\theta) \\ &= 1 - 1.1547\,e^{-t}\sin\left(1.732t + 60^\circ\right), \quad t\ge 0 \end{aligned}

Check: at $t=0$ , $c(0)=1-1.1547\sin 60^\circ = 1-1 = 0$ ; as $t\to\infty$ , $c(t)\to 1$ , so the steady-state error is zero (type-1 system with step input).

Time-domain specifications (for completeness)

Quantity	Formula	Value
Rise time $t_r$	$\frac{\pi-\theta}{\omega_d}$	$\frac{2.094}{1.732}=1.209$ s
Peak time $t_p$	$\frac{\pi}{\omega_d}$	$1.814$ s
Peak overshoot $M_p$	$e^{-\zeta\pi/\sqrt{1-\zeta^2}}$	$0.163$ = 16.3%
Settling time $t_s$ (2%)	$\frac{4}{\zeta\omega_n}$	$4$ s

 c(t)
 1.163 |      _
   1.0 |----/---\_____________
       |   /       ''
       |  /
       | /
     0 |/_____________________ t
          tp=1.81 s

Answer: $c(t) = 1 - 1.1547\,e^{-t}\sin(1.732t+60^\circ)$ ; the response is underdamped ( $\zeta=0.5$ , $\omega_n=2$ rad/s) with 16.3% overshoot and settles to 1.

2065 Shrawan (old course) · 8 marks

Derive the overall transfer function Y(s)/R(s) for the system shown below using block reduction technique. [Figure: R(s) → node P → summing point S1 (+, −) → G1 → summing point S2 (+, −, −) → G2 → node B → G3 → node C → summing point S3 (+, +) → Y(s). G4 from node P (input R) to S3. H1 from node B fed back to both S1 (−) and S2 (−). H2 from node C fed back to S2 (−).]

Answer

Result: $\dfrac{Y(s)}{R(s)} = \dfrac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} + G_4$

The diagram is reduced step by step using block diagram algebra rules (series, parallel, feedback, moving take-off points).

Step 1: Separate the parallel feed-forward path

$G_4$ takes the input $R$ directly to the output summing point $S_3$ (both inputs $+$ ). So the system is two parallel paths:

\frac{Y(s)}{R(s)} = \frac{C(s)}{R(s)} + G_4

where $C(s)$ is the output of $G_3$ (node C). We now reduce the main path from $R$ to $C$ .

Step 2: Move the take-off point of $H_2$ from C to B

Node C is after $G_3$ . Moving a take-off point before a block multiplies the branch by that block. So the $H_2$ feedback becomes $G_3H_2$ taken from node B.

Now two negative feedbacks go from B to $S_2$ : $H_1$ and $G_3H_2$ . They are in parallel:

H_{eq} = H_1 + G_3H_2

Step 3: Reduce the inner loop around $G_2$

\frac{B}{E_2} = \frac{G_2}{1+G_2(H_1+G_3H_2)} = \frac{G_2}{1+G_2H_1+G_2G_3H_2}

Step 4: Series with $G_1$ , then the outer loop through $H_1$ to $S_1$

Forward path from $S_1$ to B: $\frac{G_1G_2}{1+G_2H_1+G_2G_3H_2}$ , with negative feedback $H_1$ :

\begin{aligned} \frac{B}{R} &= \frac{\frac{G_1G_2}{1+G_2H_1+G_2G_3H_2}}{1+\frac{G_1G_2H_1}{1+G_2H_1+G_2G_3H_2}} \\ &= \frac{G_1G_2}{1+G_2H_1+G_2G_3H_2+G_1G_2H_1} \end{aligned}

Step 5: Series with $G_3$ and add $G_4$

\frac{C}{R} = \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2}

        +------------ G4 -------------+
        |                             | +
 R ---->+--> [ G1G2G3 / Delta ] ----->(+)---> Y
                                        +
 Delta = 1 + G1G2H1 + G2H1 + G2G3H2

\frac{Y(s)}{R(s)} = \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} + G_4

or, over a common denominator,

\frac{Y(s)}{R(s)} = \frac{G_1G_2G_3 + G_4\left(1+G_1G_2H_1+G_2H_1+G_2G_3H_2\right)}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2}

Check (Mason's rule): loops $-G_1G_2H_1$ , $-G_2H_1$ , $-G_2G_3H_2$ all touch each other, so $\Delta = 1+G_1G_2H_1+G_2H_1+G_2G_3H_2$ . Path $G_1G_2G_3$ touches all loops ( $\Delta_1=1$ ); path $G_4$ touches none ( $\Delta_2=\Delta$ ). This gives the same result.

2065 Shrawan (old course) · 6 marks

The open loop transfer function of a unity feedback system is given by G(s)H(s) = 4/[s(s²+4s+4)]. Find the static error constants and calculate error due to input r(t) = 4t + 1.

Answer

Static error constants are the limits $K_p=\lim_{s\to0}G(s)H(s)$ , $K_v=\lim_{s\to0}sG(s)H(s)$ and $K_a=\lim_{s\to0}s^2G(s)H(s)$ . They give the steady-state error for step, ramp and parabolic inputs.

G(s)H(s) = \frac{4}{s(s^2+4s+4)} = \frac{4}{s(s+2)^2}

There is one pole at the origin, so the system is type 1.

Stability check (needed before using final value theorem): characteristic equation $s^3+4s^2+4s+4=0$ . All coefficients are positive and $4\times4=16>1\times4=4$ , so by Routh's condition for a cubic the closed loop is stable.

Static error constants

\begin{aligned} K_p &= \lim_{s\to0}\frac{4}{s(s+2)^2} = \infty \\ K_v &= \lim_{s\to0}s\cdot\frac{4}{s(s+2)^2} = \frac{4}{4} = 1\ \text{s}^{-1} \\ K_a &= \lim_{s\to0}s^2\cdot\frac{4}{s(s+2)^2} = 0 \end{aligned}

Steady-state error for $r(t)=4t+1$

By superposition, split the input into a step and a ramp:

Step part: $r_1(t)=1\cdot u(t)$ , $A=1$ : $e_{ss1} = \frac{A}{1+K_p} = \frac{1}{1+\infty} = 0$
Ramp part: $r_2(t)=4t$ , $A=4$ : $e_{ss2} = \frac{A}{K_v} = \frac{4}{1} = 4$

e_{ss} = e_{ss1} + e_{ss2} = 0 + 4 = 4

Input	Error constant	Error
Step $1$	$K_p=\infty$	0
Ramp $4t$	$K_v=1$	4
Total		4

Answer: $K_p=\infty$ , $K_v=1\ \text{s}^{-1}$ , $K_a=0$ ; steady-state error for $r(t)=4t+1$ is $e_{ss}=4$ (units of the output).

2081 Bhadra · 8 marks

Reduce the blocks and find transfer function for the following model. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → G1 → summing point S3 (+, −) → G2 → node B → G3 → C(s). H1 from node B fed back to S2 (−). H2 from C(s) fed back to S3 (−). Unity feedback from C(s) to S1 (−).]

Answer

Result: $\dfrac{C(s)}{R(s)} = \dfrac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3}$

The diagram has three nested negative feedback loops. We remove them from the inside out.

R -->(S1)-->(S2)--> G1 -->(S3)--> G2 --+--> G3 --+--> C
      - ^    - ^           - ^         B |        |
        |      +---- H1 ---|-----------+        |
        |                  +-------- H2 --------+
        +------------------ 1 -------------------+

Step 1: Move the take-off point of $H_2$ from C to B

C $= G_3 \times$ B. Moving a take-off point before the block $G_3$ means the feedback branch becomes $G_3H_2$ from node B to $S_3$ .

Step 2: Reduce the innermost loop ( $G_2$ with feedback $G_3H_2$ )

\frac{B}{E_3} = \frac{G_2}{1+G_2G_3H_2}

Step 3: Series with $G_1$ , then the $H_1$ loop

Forward gain from $S_2$ to B is $\frac{G_1G_2}{1+G_2G_3H_2}$ , feedback $H_1$ (negative):

\begin{aligned} \frac{B}{E_2} &= \frac{\frac{G_1G_2}{1+G_2G_3H_2}}{1+\frac{G_1G_2H_1}{1+G_2G_3H_2}} = \frac{G_1G_2}{1+G_2G_3H_2+G_1G_2H_1} \end{aligned}

Step 4: Series with $G_3$

\frac{C}{E_1} = \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2}

Step 5: Unity negative feedback loop

\begin{aligned} \frac{C}{R} &= \frac{\frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2}}{1+\frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2}} \\ &= \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3} \end{aligned}

Check (Mason's rule): one forward path $P_1=G_1G_2G_3$ ; loops $-G_1G_2H_1$ , $-G_2G_3H_2$ , $-G_1G_2G_3$ , all touching each other and the path. So $\Delta = 1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3$ and $\Delta_1=1$ , giving the same answer.

Answer: $\dfrac{C(s)}{R(s)} = \dfrac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3}$

2081 Baisakh · 4 marks

For a second order system given by the transfer function G(s) = 25/(s² + 8s + 25), find the rise time, peak time, settling time and maximum overshoot.

Answer

Compare with the standard second-order form $\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$ :

$\omega_n^2=25 \Rightarrow \omega_n = 5$ rad/s
$2\zeta\omega_n = 8 \Rightarrow \zeta = 0.8$ (underdamped)
$\omega_d = \omega_n\sqrt{1-\zeta^2} = 5\times0.6 = 3$ rad/s
$\theta = \cos^{-1}(0.8) = 36.87^\circ = 0.6435$ rad

\begin{aligned} t_r &= \frac{\pi-\theta}{\omega_d} = \frac{3.1416-0.6435}{3} = 0.833\ \text{s} \\ t_p &= \frac{\pi}{\omega_d} = \frac{3.1416}{3} = 1.047\ \text{s} \\ t_s &= \frac{4}{\zeta\omega_n} = \frac{4}{0.8\times5} = 1\ \text{s}\quad(2\%\ \text{criterion}) \\ M_p &= e^{-\zeta\pi/\sqrt{1-\zeta^2}} = e^{-0.8\pi/0.6} = e^{-4.189} = 0.0152 \end{aligned}

With the 5% criterion, $t_s = \frac{3}{\zeta\omega_n} = 0.75$ s.

Answer: $t_r = 0.833$ s, $t_p = 1.047$ s, $t_s = 1$ s (2%), $M_p = 1.52\%$ .

2080 Baisakh · 6 marks

What is the importance of error coefficients in design of control system? For a unity feedback system having G(s) = k(s+2)/[s²(s²+7s+12)], determine (i) Type and order of the system (ii) Error coefficients and (iii) Steady state error for parabolic input r(t) = 2.5t².

Answer

Importance of error coefficients

Static error coefficients ( $K_p$ , $K_v$ , $K_a$ ) measure how well a closed-loop system follows standard test inputs in steady state. Their role in design:

They give the steady-state error directly: step $e_{ss}=\frac{A}{1+K_p}$ , ramp $e_{ss}=\frac{A}{K_v}$ , parabola $e_{ss}=\frac{A}{K_a}$ .
They act as design specifications: e.g. " $K_v \ge 20$ " fixes the minimum gain needed for a tracking system.
They show the effect of the type of the system: adding an integrator (higher type) makes the error constant for that input infinite, so the error becomes zero.
They show the trade-off between accuracy and stability: increasing gain increases $K_p, K_v, K_a$ (better accuracy) but usually reduces relative stability. Lag/PI compensators are designed mainly to raise error constants without disturbing the transient response.

(i) Type and order

G(s) = \frac{k(s+2)}{s^2(s^2+7s+12)} = \frac{k(s+2)}{s^2(s+3)(s+4)}

Two poles at the origin: Type 2.
Highest power of $s$ in the denominator is $s^4$ : Order 4.

(ii) Error coefficients

\begin{aligned} K_p &= \lim_{s\to0}G(s) = \infty \\ K_v &= \lim_{s\to0}sG(s) = \infty \\ K_a &= \lim_{s\to0}s^2G(s) = \frac{k\times2}{12} = \frac{k}{6} \end{aligned}

(iii) Steady-state error for $r(t)=2.5t^2$

Standard parabolic input is $r(t)=\frac{A t^2}{2}$ , so $\frac{A}{2}=2.5 \Rightarrow A=5$ .

e_{ss} = \frac{A}{K_a} = \frac{5}{k/6} = \frac{30}{k}

Answer: Type 2, order 4; $K_p=\infty$ , $K_v=\infty$ , $K_a = k/6$ ; $e_{ss} = 30/k$ .

Note on stability: the error formula is valid only if the closed loop is stable. The characteristic equation is $s^4+7s^3+12s^2+ks+2k=0$ . Its Routh array has first column $1,\ 7,\ \frac{84-k}{7},\ \frac{k(k+14)}{k-84},\ 2k$ , which always has a sign change for $k>0$ . So this particular closed loop is unstable for every positive $k$ ; $30/k$ is the theoretical value that would apply once the system is stabilised (e.g. by a lead compensator).

2079 Bhadra · 8 marks

Determine the transfer function C(s)/R(s) of the block diagram given below, using block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → node A → G1 → summing point S3 (+, +) → G2 → node B → G3 → summing point S4 (+, +) → C(s). G4 from node A to S4 (+). H1 from node B fed back to S2 (−). H2 from C(s) fed back to S3 (+). Unity feedback from C(s) to S1 (−).]

Answer

Result: $\dfrac{C(s)}{R(s)} = \dfrac{G_1G_2G_3+G_4}{1+G_1G_2H_1-G_2G_3H_2+G_2G_4H_1H_2+G_1G_2G_3+G_4}$

The feedback $H_2$ from the output is taken after summing point $S_4$ , which is fed by both $G_3$ and $G_4$ . We first shift this take-off point behind $S_4$ , then reduce the loops.

R->(S1)->(S2)--A--> G1 ->(S3)-> G2 --B--> G3 ->(S4)--+--> C
   -^     -^    |        +^            |       +^    |
    |      +----|--- H1 -|-------------+        |    |
    |           +--------|------- G4 -----------+    |
    |                    +--------- H2 --------------+
    +------------------------- 1 --------------------+

Step 1: Split the $H_2$ feedback

$C = G_3B + G_4A$ , so $H_2C = G_3H_2\,B + G_4H_2\,A$ . The $H_2$ branch is replaced by two branches into $S_3$ (both $+$ ):

$G_3H_2$ from node B (a positive feedback loop around $G_2$ )
$G_4H_2$ from node A (in parallel with $G_1$ )

Step 2: Combine the parallel blocks from A to $S_3$

G_1 + G_4H_2

Step 3: Positive loop around $G_2$

\frac{B}{E_3} = \frac{G_2}{1-G_2G_3H_2} \quad\Rightarrow\quad \frac{B}{A} = \frac{G_2(G_1+G_4H_2)}{1-G_2G_3H_2}

Step 4: Negative loop through $H_1$ (B back to $S_2$ )

\frac{A}{E_1} = \frac{1}{1+H_1\frac{G_2(G_1+G_4H_2)}{1-G_2G_3H_2}} = \frac{1-G_2G_3H_2}{1-G_2G_3H_2+G_1G_2H_1+G_2G_4H_1H_2}

Step 5: Output from A

\begin{aligned} \frac{C}{A} &= G_3\frac{B}{A} + G_4 = \frac{G_2G_3(G_1+G_4H_2) + G_4(1-G_2G_3H_2)}{1-G_2G_3H_2} \\ &= \frac{G_1G_2G_3+G_4}{1-G_2G_3H_2} \end{aligned}

Hence

\frac{C}{E_1} = \frac{G_1G_2G_3+G_4}{1+G_1G_2H_1-G_2G_3H_2+G_2G_4H_1H_2}

Step 6: Unity negative feedback

\frac{C}{R} = \frac{G_1G_2G_3+G_4}{1+G_1G_2H_1-G_2G_3H_2+G_2G_4H_1H_2+G_1G_2G_3+G_4}

Check (Mason's rule): forward paths $G_1G_2G_3$ and $G_4$ ; loops $-G_1G_2H_1$ , $+G_2G_3H_2$ , $-G_1G_2G_3$ , $-G_4$ , $-G_2G_4H_1H_2$ (A → $G_4$ → C → $H_2$ → $G_2$ → B → $H_1$ → A). All loops touch each other and both paths, so $\Delta_1=\Delta_2=1$ and the same result follows.

2078 Bhadra · 4 marks

A servomechanism as shown in block diagram below is designed to keep a radar antenna pointed at a flying aeroplane. If the aeroplane is flying with a velocity of 600 km/hr, at a range of 2 km and the maximum tracking error is to be within 0.1°, determine the required velocity error coefficient. [Figure: R(s) → summing point (+, −) → Kv/[s(1+sT)] → C(s), unity negative feedback]

Answer

For a type-1 system with a ramp input of slope $\omega$ (constant angular velocity), the steady-state error is $e_{ss} = \frac{\omega}{K_v}$ . So the required $K_v = \frac{\omega}{e_{ss,\max}}$ .

Angular velocity of the line of sight

The antenna must turn at the angular rate of the aeroplane as seen from the radar:

\begin{aligned} v &= 600\ \text{km/h} = \frac{600\times1000}{3600} = 166.67\ \text{m/s} \\ \omega &= \frac{v}{r} = \frac{166.67}{2000} = 0.08333\ \text{rad/s} \\ &= 0.08333\times\frac{180}{\pi} = 4.775^\circ/\text{s} \end{aligned}

So the input to the servo is a ramp $\theta_r(t) = 4.775\,t$ degrees.

Velocity error coefficient

For $G(s)=\frac{K_v}{s(1+sT)}$ (type 1):

\lim_{s\to0}sG(s) = K_v

e_{ss} = \frac{\omega}{K_v} \le 0.1^\circ \quad\Rightarrow\quad K_v \ge \frac{4.775}{0.1} = 47.75\ \text{s}^{-1}

Answer: Required velocity error coefficient $K_v \ge 47.75\ \text{s}^{-1}$ (about $48\ \text{s}^{-1}$ ).

2078 Bhadra · 4 marks

A system has 25% overshoot and settling time of 6 seconds for a unit step input. Determine the transfer function and calculate peak time. Assume ess as 2%.

Answer

Assume a standard second-order system $\frac{C(s)}{R(s)} = \frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$ . Find $\zeta$ from overshoot and $\omega_n$ from settling time.

Damping ratio from $M_p = 0.25$

\begin{aligned} M_p &= e^{-\zeta\pi/\sqrt{1-\zeta^2}} = 0.25 \\ \frac{\zeta\pi}{\sqrt{1-\zeta^2}} &= -\ln 0.25 = 1.3863 \\ \zeta &= \frac{1.3863}{\sqrt{\pi^2+1.3863^2}} = \frac{1.3863}{3.4339} = 0.4037 \end{aligned}

Natural frequency from $t_s = 6$ s (2% band)

\begin{aligned} t_s &= \frac{4}{\zeta\omega_n} = 6 \Rightarrow \zeta\omega_n = 0.6667 \\ \omega_n &= \frac{0.6667}{0.4037} = 1.651\ \text{rad/s} \end{aligned}

Transfer function

$\omega_n^2 = 2.727$ , $2\zeta\omega_n = 1.333$ :

\frac{C(s)}{R(s)} = \frac{2.727}{s^2 + 1.333s + 2.727}

Peak time

\begin{aligned} \omega_d &= \omega_n\sqrt{1-\zeta^2} = 1.651\sqrt{1-0.4037^2} = 1.511\ \text{rad/s} \\ t_p &= \frac{\pi}{\omega_d} = \frac{3.1416}{1.511} = 2.079\ \text{s} \end{aligned}

Answer: $\zeta = 0.404$ , $\omega_n = 1.651$ rad/s, $T(s) = \dfrac{2.727}{s^2+1.333s+2.727}$ , peak time $t_p = 2.08$ s.

2078 Kartik · 8 marks

Find overall transfer function of the following diagram using block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → node A → G1 → summing point S2 (+, −) → G2 → node B → summing point S3 (+, +) → C(s). G3 from node A (before G1) to S3 (+). H2 from C(s) fed back to S2 (−). H1 from node B fed back to S1 (−).]

Answer

Result: $\dfrac{C(s)}{R(s)} = \dfrac{G_1G_2+G_3}{1+G_2H_2+G_1G_2H_1-G_2G_3H_1H_2}$

The feedback $H_2$ is taken from C, after summing point $S_3$ where $G_3$ joins. We shift this take-off point back over $S_3$ , then reduce.

R ->(S1)--A--> G1 ->(S2)-> G2 --B-->(S3)--+--> C
    -^    |         -^          |     +^   |
     |    |          +-----H2---|------|---+
     +----|------ H1 -----------+      |
          +----------- G3 -------------+

Step 1: Move the $H_2$ take-off point before $S_3$

$C = B + G_3A$ , so $H_2C = H_2B + G_3H_2A$ . The single $H_2$ branch is replaced by:

$H_2$ from node B to $S_2$ (negative)
$G_3H_2$ from node A to $S_2$ (negative)

Step 2: Combine blocks from A into $S_2$

$G_1$ (+) and $G_3H_2$ (−) both start at A and end at $S_2$ , so they are in parallel: $G_1 - G_3H_2$ .

Step 3: Inner loop around $G_2$

\frac{B}{E_2} = \frac{G_2}{1+G_2H_2} \qquad \frac{B}{A} = \frac{G_2(G_1-G_3H_2)}{1+G_2H_2}

Step 4: Loop through $H_1$ back to $S_1$

\frac{A}{R} = \frac{1}{1+\frac{G_2H_1(G_1-G_3H_2)}{1+G_2H_2}} = \frac{1+G_2H_2}{1+G_2H_2+G_1G_2H_1-G_2G_3H_1H_2}

Step 5: Output

\begin{aligned} \frac{C}{A} &= \frac{B}{A} + G_3 = \frac{G_1G_2 - G_2G_3H_2 + G_3 + G_2G_3H_2}{1+G_2H_2} = \frac{G_1G_2+G_3}{1+G_2H_2} \\ \frac{C}{R} &= \frac{C}{A}\cdot\frac{A}{R} = \frac{G_1G_2+G_3}{1+G_2H_2+G_1G_2H_1-G_2G_3H_1H_2} \end{aligned}

Check (Mason's rule): paths $G_1G_2$ , $G_3$ ; loops $-G_1G_2H_1$ , $-G_2H_2$ , and $+G_2G_3H_1H_2$ (A → $G_3$ → C → $-H_2$ → $G_2$ → B → $-H_1$ → A). All loops touch both paths, so $\Delta = 1+G_1G_2H_1+G_2H_2-G_2G_3H_1H_2$ , matching the result.

2078 Kartik · 8 marks

The system below in figure (a), when subjected to unit step input, gives the output response as shown in figure (b). Determine the value of K and T. [Figure (a): R(s) → summing point (+, −) → K/[s(1+sT)] → C(s), unity negative feedback. Figure (b): unit step response with final value 1, peak overshoot 0.254 above the final value at t = 3 sec]

Answer

From the response curve, peak overshoot $M_p = 0.254$ (25.4%) and peak time $t_p = 3$ s. These give $\zeta$ and $\omega_n$ , which are then matched with the closed-loop transfer function.

Closed-loop transfer function

\frac{C(s)}{R(s)} = \frac{\frac{K}{s(1+sT)}}{1+\frac{K}{s(1+sT)}} = \frac{K}{Ts^2+s+K} = \frac{K/T}{s^2+\frac{1}{T}s+\frac{K}{T}}

Comparing with $\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$ :

\omega_n^2 = \frac{K}{T}, \qquad 2\zeta\omega_n = \frac{1}{T}

Damping ratio

\begin{aligned} e^{-\zeta\pi/\sqrt{1-\zeta^2}} &= 0.254 \\ \frac{\zeta\pi}{\sqrt{1-\zeta^2}} &= -\ln 0.254 = 1.3704 \\ \zeta &= \frac{1.3704}{\sqrt{\pi^2+1.3704^2}} = 0.400 \end{aligned}

Natural frequency

\begin{aligned} t_p &= \frac{\pi}{\omega_d} = 3 \Rightarrow \omega_d = 1.0472\ \text{rad/s} \\ \omega_n &= \frac{\omega_d}{\sqrt{1-\zeta^2}} = \frac{1.0472}{\sqrt{1-0.16}} = 1.1425\ \text{rad/s} \end{aligned}

K and T

\begin{aligned} T &= \frac{1}{2\zeta\omega_n} = \frac{1}{2\times0.4\times1.1425} = 1.0946\ \text{s} \\ K &= \omega_n^2 T = 1.3053\times1.0946 = 1.429 \end{aligned}

Answer: $\zeta = 0.4$ , $\omega_n = 1.142$ rad/s, giving $T \approx 1.095$ s and $K \approx 1.43$ .

2076 Chaitra · 8 marks

Determine the transfer function C/R for the block diagram below by signal flow graph (SFG) technique. [Figure: R → node P → summing point S1 (+, ±) → G1 → summing point S2 (+, −, −) → G2 → node B → G3 → node C → summing point S3 (+, +) → C. G4 from node P (input R) to S3 (+). H1 from node B fed back to both S1 (sign unclear in scan) and S2 (−). H2 from node C fed back to S2 (−).]

Answer

Result (taking the sign of $H_1$ at $S_1$ as negative): $\dfrac{C}{R} = \dfrac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} + G_4$

The block diagram is converted to a signal flow graph (SFG) and Mason's gain formula is applied:

T = \frac{1}{\Delta}\sum_k P_k\Delta_k

Assumption: the sign of $H_1$ at $S_1$ is unclear in the scan. It is taken as negative, as in the usual version of this question.

Signal flow graph

Nodes: $R$ , $x_1$ (output of $S_1$ ), $x_2$ (output of $S_2$ ), $x_3$ (= B, output of $G_2$ ), $x_4$ (= node C, output of $G_3$ ), $C$ .

  +------------------ G4 -------------------+
  |                                         v
  R -1-> x1 -G1-> x2 -G2-> x3 -G3-> x4 -1-> C
         ^        ^  ^     |  |     |
         |        |  +-H1-+  |     |
         +--------|---(-H1)---+     |
                  +-----(-H2)-------+

Branches: $x_3 \to x_1$ gain $-H_1$ , $x_3 \to x_2$ gain $-H_1$ , $x_4 \to x_2$ gain $-H_2$ , $R \to C$ gain $G_4$ .

Forward paths

Path	Route	Gain
$P_1$	$R\to x_1\to x_2\to x_3\to x_4\to C$	$G_1G_2G_3$
$P_2$	$R\to C$	$G_4$

Individual loops

Loop	Route	Gain
$L_1$	$x_1\to x_2\to x_3\to x_1$	$-G_1G_2H_1$
$L_2$	$x_2\to x_3\to x_2$	$-G_2H_1$
$L_3$	$x_2\to x_3\to x_4\to x_2$	$-G_2G_3H_2$

All three loops share the branch $G_2$ , so there are no non-touching loop pairs.

Determinants

\begin{aligned} \Delta &= 1-(L_1+L_2+L_3) = 1+G_1G_2H_1+G_2H_1+G_2G_3H_2 \\ \Delta_1 &= 1 \quad (P_1 \text{ touches all loops}) \\ \Delta_2 &= \Delta \quad (P_2 \text{ touches no loop}) \end{aligned}

Transfer function

\begin{aligned} \frac{C}{R} &= \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta} \\ &= \frac{G_1G_2G_3 + G_4(1+G_1G_2H_1+G_2H_1+G_2G_3H_2)}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} \\ &= \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} + G_4 \end{aligned}

If $H_1$ enters $S_1$ with a positive sign instead, only $L_1$ changes to $+G_1G_2H_1$ , giving $\frac{C}{R} = \frac{G_1G_2G_3}{1-G_1G_2H_1+G_2H_1+G_2G_3H_2}+G_4$ .

Past questions and answers

Step 1: Move the H1H_1H1​ take-off point behind G2G_2G2​

Step 2: Move the G3G_3G3​ take-off point from AAA to after S2S_2S2​

Step 3: Combine parallel blocks and the positive loop at S3S_3S3​

Step 4: Move S1S_1S1​ forward past G1G_1G1​

Step 5: Close the H2H_2H2​ loop

Check by Mason's rule

2.The open loop transfer function of a unity feedback system is given by G(s) = K/[s(1+sT)], where 'K' is the gain constant and 'T' is time constant. With the gain multiplied by a factor K1 the maximum overshoot of the system is increased from 25% to 50%. Determine K1.

Closed-loop characteristic equation

Damping ratio from overshoot

Gain factor

Step 1: Reduce the minor (tachometer) loop

Step 2: Open-loop transfer function

Step 3: Closed-loop characteristic equation (unity feedback)

Step 4: Use the specifications

Step 5: Solve for KKK and KtK_tKt​

Step 1: Combine G1G_1G1​ and the feedforward path

Step 2: Move the H1H_1H1​ take-off point from BBB to CCC

Step 3: Reduce the unity loop around G3G_3G3​

Step 4: Reduce the H1/G3H_1/G_3H1​/G3​ loop

Step 5: Series with (1+G1)(1 + G_1)(1+G1​) and close the H2H_2H2​ loop

Check by Mason's rule

Transfer function

Damping ratio for 20% overshoot

Natural frequency from settling time (2% criterion)

Solve for JJJ and DDD

Signal flow graph

Mason's gain formula

Transfer function

Signal flow graph

Forward paths

Individual loops

Determinant

Cofactors

Transfer function

Step 1: Series and parallel blocks in the forward path

Step 2: Move summing point S3S_3S3​ past H2H_2H2​

Step 3: Combine the two input paths into S1S_1S1​

Step 4: Close the feedback loop

9.Consider a unity feedback control system with the closed loop transfer function C(s)/R(s) = (Ks + b)/(s² + as + b). Determine the open loop transfer function. Show that the steady state error in the unit ramp input response is given by ess = (a − K)/b.

Open-loop transfer function

Steady-state error for unit ramp

Direct check from the error transfer function

10.The open loop transfer function of a unity feedback system is given by G(s) = 108/[s²(s+4)(s²+3s+12)]. Find the static error coefficients and steady state error of the system when subjected to an input given by r(t) = 2 + 5t + 8t².

Type of the system

Static error coefficients

Error for each input term

Step 1: Move the take-off point of H2H_2H2​ from BBB past G3G_3G3​ and S4S_4S4​

Step 2: Combine the feedback blocks from CCC and remove the self loop at AAA

Step 3: Reduce the forward part from AAA to CCC

Step 4: Close the main loop

Step 1: Reduce the first loop (G1G_1G1​, G2G_2G2​ with positive feedback H1H_1H1​)

Step 2: Reduce the second loop (G4G_4G4​ with positive feedback H2H_2H2​)

Step 3: Substitute AAA and BBB

Without tachometer feedback

With tachometer feedback H(s)=1+ksH(s) = 1 + ksH(s)=1+ks

Peak overshoot after adding feedback

Summary

Signal flow graph

Forward paths

Individual loops

Non-touching loops

Cofactors

Mason's gain formula

15.Discuss the effect of addition of pole and zero in a system.

Addition of a pole

Addition of a zero

Step 1: Move the take-off point of H1H_1H1​ from BBB past summer S4S_4S4​

Step 2: Combine the feedback from CCC and remove the self loop

Step 3: Reduce from AAA to CCC

Step 4: Close the main loop

17.For a mechanical system with closed loop transfer function θ(s)/T(s) = 1/(as² + bs + c), where θ(s) is the output of step input T(t) = 10 Nm. Determine values of a, b and c if maximum overshoot is 6%, peak time (tp) = 1 sec, and ess = 0.5.

Step 1: Find ccc from the steady state

Step 2: Damping ratio from 6% overshoot

Step 3: Natural frequency from tp=1t_p = 1tp​=1 s

Step 4: Find aaa and bbb

Signal flow graph

Mason's gain formula

19.Find the impulse response of the given circuit. [Figure: input ei; series capacitor 1 F, then shunt resistor 2 Ω across the line, then series resistor 1 Ω, then shunt capacitor 2 F; output eo taken across the 2 F capacitor]

Circuit in the s-domain

Step 1: Move the $H_1$ take-off point behind $G_2$

Step 2: Move the $G_3$ take-off point from $A$ to after $S_2$

Step 3: Combine parallel blocks and the positive loop at $S_3$

Step 4: Move $S_1$ forward past $G_1$

Step 5: Close the $H_2$ loop

The open loop transfer function of a unity feedback system is given by G(s) = K/[s(1+sT)], where 'K' is the gain constant and 'T' is time constant. With the gain multiplied by a factor K1 the maximum overshoot of the system is increased from 25% to 50%. Determine K1.

Step 5: Solve for $K$ and $K_t$

Step 1: Combine $G_1$ and the feedforward path

Step 2: Move the $H_1$ take-off point from $B$ to $C$

Step 3: Reduce the unity loop around $G_3$

Step 4: Reduce the $H_1/G_3$ loop

Step 5: Series with $(1 + G_1)$ and close the $H_2$ loop

Solve for $J$ and $D$

Step 2: Move summing point $S_3$ past $H_2$

Step 3: Combine the two input paths into $S_1$

Consider a unity feedback control system with the closed loop transfer function C(s)/R(s) = (Ks + b)/(s² + as + b). Determine the open loop transfer function. Show that the steady state error in the unit ramp input response is given by ess = (a − K)/b.

The open loop transfer function of a unity feedback system is given by G(s) = 108/[s²(s+4)(s²+3s+12)]. Find the static error coefficients and steady state error of the system when subjected to an input given by r(t) = 2 + 5t + 8t².

Step 1: Move the take-off point of $H_2$ from $B$ past $G_3$ and $S_4$

Step 2: Combine the feedback blocks from $C$ and remove the self loop at $A$

Step 3: Reduce the forward part from $A$ to $C$

Step 1: Reduce the first loop ( $G_1$ , $G_2$ with positive feedback $H_1$ )

Step 2: Reduce the second loop ( $G_4$ with positive feedback $H_2$ )

Step 3: Substitute $A$ and $B$

With tachometer feedback $H(s) = 1 + ks$

Discuss the effect of addition of pole and zero in a system.

Step 1: Move the take-off point of $H_1$ from $B$ past summer $S_4$

Step 2: Combine the feedback from $C$ and remove the self loop

Step 3: Reduce from $A$ to $C$

For a mechanical system with closed loop transfer function θ(s)/T(s) = 1/(as² + bs + c), where θ(s) is the output of step input T(t) = 10 Nm. Determine values of a, b and c if maximum overshoot is 6%, peak time (tp) = 1 sec, and ess = 0.5.

Step 1: Find $c$ from the steady state

Step 3: Natural frequency from $t_p = 1$ s

Step 4: Find $a$ and $b$

Find the impulse response of the given circuit. [Figure: input ei; series capacitor 1 F, then shunt resistor 2 Ω across the line, then series resistor 1 Ω, then shunt capacitor 2 F; output eo taken across the 2 F capacitor]

Transfer function from $D$ to $C$ (set $R = 0$ )

Steady state for a step disturbance $D(s) = d/s$

Condition: $c_{ss} \le 2\%$ of $d$

Step 1: Move the take-off point of $H_1$ from $B$ past summer $S_3$

Step 3: Reduce from $A$ to $C$

The open loop transfer function of unity feedback system is given by G(s) = 108/[s²(s+4)(s²+3s+12)]. Find the static error coefficients and steady state error of the system when subjected to an input given by r(t) = 2 + 5t + 2t².

Step 1: $K$ from the final value

Step 2: $\zeta$ from the overshoot

Step 3: $\omega_n$ from the peak time

Step 4: $M$ and $B$

Step 1: Move the take-off point of $H_1$ from $B$ past $G_3$ and $S_4$

Step 3: Reduce from $A$ to $C$

For a unity feedback system, the open loop transfer function is G(s) = 50/[s(s+2)]. With unit step input find maximum overshoot and settling time. Also determine static error coefficients and steady state error if the input to the system is r(t) = 2 + 4t + 6t², t ≥ 0.

Steady state error for $r(t) = 2 + 4t + 6t^2$

Step 2: Closed-loop transfer function (unity feedback, $J = 1$ )

Step 3: $\zeta$ from $M_p = 25\%$

Step 4: $\omega_n$ from $t_p = 2$ s

Step 5: Find $a$ and $b$

A system has 40% overshoot and requires a settling time of 4 seconds when given a step input. Find peak time and rise time.

Step 1: Damping ratio from $M_p = 40\%$

Step 2: Natural frequency from $t_s = 4$ s (2% criterion)