Digital Control System — Chapter 4: Design and Compensation of Discrete Time Control System

Asked 6 times
2082 Chaitra (new course) · 5 marks
2082 Chaitra · 5 marks
2082 Kartik · 5 marks
2078 Chaitra · 8 marks
2074 Bhadra · 8 marks
2071 Magh · 8 marks

Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²) (equivalently GD(z)=(2z²+2.2z+0.2)/(z²+0.4z-0.12)). Realize this filter in the ladder scheme.

Answer

In ladder programming, $G(z)$ is written as a continued fraction in $z$ :

G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2}}}}

Each $1/(B_iz)$ becomes a delay $z^{-1}/B_i$ , and each $A_i$ becomes a constant gain.

Write the filter in positive powers of $z$ :

G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.12z^{-2}} = \frac{2z^2 + 2.2z + 0.2}{z^2 + 0.4z - 0.12}

Step 1: Constant term $A_0$

G(z) = 2 + \frac{(2z^2+2.2z+0.2) - 2(z^2+0.4z-0.12)}{z^2+0.4z-0.12} = 2 + \frac{1.4z + 0.44}{z^2 + 0.4z - 0.12}

So $A_0 = 2$ .

Step 2: $B_1$

\frac{z^2+0.4z-0.12}{1.4z+0.44} = \frac{1}{1.4}z + \frac{0.0857z - 0.12}{1.4z+0.44}

So $B_1 = 1/1.4 = 0.7143$ . The remainder is $0.4z - 0.3143z = 0.0857z$ (exactly $3z/35$ ).

Step 3: $A_1$

\frac{1.4z+0.44}{0.0857z-0.12} = \frac{1.4}{0.0857} + \frac{0.44 + 16.333(0.12)}{0.0857z - 0.12} = 16.333 + \frac{2.4}{0.0857z-0.12}

So $A_1 = 49/3 = 16.333$ .

Step 4: $B_2$ and $A_2$

\frac{0.0857z - 0.12}{2.4} = 0.0357z - 0.05 = B_2z + \frac{1}{A_2}

So $B_2 = 1/28 = 0.0357$ and $A_2 = -1/0.05 = -20$ .

G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{16.333 + \cfrac{1}{0.0357z + \cfrac{1}{-20}}}}

$A_0$	$B_1$	$A_1$	$B_2$	$A_2$
2	0.7143	16.333	0.0357	−20

Ladder block diagram

X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         |
      |                           |
      +<--E2--[1/A2]<-------------+

Signal equations (these define the ladder diagram; $o$ marks a take-off point):

\begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}Y_2 \\ Y &= A_0X + Y_1 \end{aligned}

With the values: $1/B_1 = 1.4$ , $1/A_1 = 0.0612$ , $1/B_2 = 28$ , $1/A_2 = -0.05$ .

Simulating these equations gives the same impulse response as the original difference equation $y(k) = -0.4y(k-1) + 0.12y(k-2) + 2x(k) + 2.2x(k-1) + 0.2x(k-2)$ . The realization needs only 2 delays.

Asked 3 times
2082 Chaitra · 12 marks
2076 Bhadra · 12 marks
2073 Bhadra · 12 marks

Design a digital controller such that the compensated system should have damping ratio ζ=0.5 and settling time tₛ=2 s using root locus approach. [Figure: unity-feedback loop: R(s) → summing junction → E(s) → sampler (T = 0.2) → digital controller → ZOH → 1/s(s+2) → C(s)]

Answer

Step 1: Desired closed-loop poles

From the specifications $\zeta = 0.5$ and $t_s = 2$ s (2% criterion, $t_s = 4/\zeta\omega_n$ ):

\begin{aligned} \zeta\omega_n &= \frac{4}{t_s} = 2, \qquad \omega_n = 4\ \text{rad/s} \\ \omega_d &= \omega_n\sqrt{1-\zeta^2} = 3.464\ \text{rad/s} \end{aligned}

With $T = 0.2$ s, the samples per cycle are $\omega_s/\omega_d = 31.42/3.464 \approx 9$ .

\begin{aligned} |z| &= e^{-\zeta\omega_nT} = e^{-0.4} = 0.6703 \\ \angle z &= \omega_dT = 0.6928\ \text{rad} = 39.69^\circ \\ z_{1,2} &= 0.6703\angle \pm 39.69^\circ = 0.5158 \pm j0.4281 \end{aligned}

Step 2: Plant pulse transfer function

\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+2)}\right] = \frac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} \\ &= \frac{0.01758(z + 0.8753)}{(z-1)(z-0.6703)} \end{aligned}

Step 3: Angle deficiency at $z_1 = 0.5158 + j0.4281$

Factor	Angle at $z_1$
zero at $-0.8753$	$+17.11^\circ$
pole at $1$	$-138.52^\circ$
pole at $0.6703$	$-109.85^\circ$
Total $\angle G(z_1)$	$-231.26^\circ$

The angle condition requires $-180^\circ$ , so the controller must add $+51.26^\circ$ of phase lead.

Step 4: Lead controller

Take $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$ . Put the zero on the plant pole, $\alpha = 0.6703$ , to cancel it. This zero adds $+109.85^\circ$ . The pole must then contribute $109.85^\circ - 51.26^\circ = 58.59^\circ$ :

\beta = 0.5158 - \frac{0.4281}{\tan 58.59^\circ} = 0.2543

Step 5: Gain from the magnitude condition

After cancellation, the open-loop function is $\dfrac{0.01758K(z+0.8753)}{(z-1)(z-0.2543)}$ , and $|G_DG| = 1$ at $z_1$ :

K = \frac{|z_1-1|\,|z_1-0.2543|}{0.01758\,|z_1+0.8753|} = \frac{0.6464 \times 0.5017}{0.01758 \times 1.4555} = 12.67

G_D(z) = 12.67\,\frac{z - 0.6703}{z - 0.2543}

Step 6: Check

\begin{aligned} G_D(z)G(z) &= \frac{0.2228(z + 0.8753)}{(z-1)(z-0.2543)} \\ \text{Characteristic eq.:}\quad z^2 &- 1.0316z + 0.4493 = 0 \end{aligned}

The roots are $z = 0.5158 \pm j0.4281$ , the desired poles. ✔ This is a second-order closed loop, so the poles are exactly dominant.

Static velocity error constant:

K_v = \lim_{z\to1}\frac{(1-z^{-1})G_DG}{T} = \frac{1}{0.2}\cdot\frac{0.2228 \times 1.8753}{0.7457} = 2.80\ \text{s}^{-1}

Answer: $G_D(z) = 12.67\dfrac{z-0.6703}{z-0.2543}$ , a lead compensator, with $K_v = 2.80\ \text{s}^{-1}$ .

Asked 3 times
2081 Chaitra · 12 marks
2074 Bhadra · 12 marks
2082 Kartik · 12 marks

Design a digital PI controller such that the dominant closed loop poles have damping ratio ζ=0.5, sampling period T = 1 and ωd/ωₛ=1/10 (10 samples per cycle of sinusoidal oscillation) and dead time of 2 sec. Also find Kᵥ and eₛₛ in response to unit ramp input. [Figure: unity-feedback loop: sampler → GD(z) → ZOH (1-e⁻ᵀˢ)/s → e⁻²ˢ/(s+1) → output]

Answer

Step 1: Plant pulse transfer function ( $T = 1$ s)

The dead time of 2 s is exactly $2T$ , so it becomes $z^{-2}$ :

\begin{aligned} G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+1)}\right] = z^{-2}\frac{(1-e^{-1})z^{-1}}{1-e^{-1}z^{-1}} \\ &= \frac{0.6321}{z^2(z - 0.3679)} \end{aligned}

Step 2: Desired dominant poles

With 10 samples per cycle, $\omega_dT = 2\pi/10 = 36^\circ$ , and

\begin{aligned} |z| &= \exp\left(-\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{\omega_d}{\omega_s}\right) = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac{1}{10}\right) = 0.6958 \\ z_1 &= 0.6958\angle 36^\circ = 0.5629 + j0.4090 \end{aligned}

Step 3: PI controller and angle condition

G_D(z) = K_P + \frac{K_I}{1-z^{-1}} = K\frac{z-\alpha}{z-1}, \qquad K = K_P + K_I,\; \alpha = \frac{K_P}{K_P+K_I}

Angles at $z_1$ :

Factor	Angle
double pole at origin ( $z^2$ )	$-2(36^\circ) = -72^\circ$
pole at $0.3679$	$-64.51^\circ$
controller pole at $1$	$-136.91^\circ$
Sum of pole angles	$-273.41^\circ$

For $-180^\circ$ , the controller zero must contribute $+93.41^\circ$ :

\alpha = 0.5629 - \frac{0.4090}{\tan 93.41^\circ} = 0.5629 + 0.0244 = 0.5873

Step 4: Gain $K$

K = \frac{|z_1|^2\,|z_1-0.3679|\,|z_1-1|}{0.6321\,|z_1-0.5873|} = \frac{0.4841 \times 0.4531 \times 0.5986}{0.6321 \times 0.4097} = 0.5070

G_D(z) = 0.5070\,\frac{z - 0.5873}{z - 1}

K_P = K\alpha = 0.2977, \qquad K_I = K(1-\alpha) = 0.2093

Step 5: $K_v$ and steady-state error for a ramp

\begin{aligned} K_v &= \lim_{z\to1}\frac{(1-z^{-1})G_D(z)G(z)}{T} = \frac{K(1-\alpha)}{T}\cdot\frac{0.6321}{1-0.3679} = 0.2093 \times 1 = 0.2093\ \text{s}^{-1} \\ e_{ss} &= \frac{1}{K_v} = 4.78 \end{aligned}

Check: the closed-loop characteristic equation $z^3(z-1) - 0.3679z^2(z-1) + 0.3205(z-0.5873) = 0$ has roots $0.5629 \pm j0.4090$ , $0.7562$ and $-0.5141$ . The desired pair is obtained. The real pole at 0.7562 is close to the PI zero at 0.5873 and partly cancelled by it, but it makes the response a little slower than the pair alone suggests.

Answer: $G_D(z) = 0.5070\dfrac{z-0.5873}{z-1}$ (that is, $K_P = 0.2977$ , $K_I = 0.2093$ ), with $K_v = 0.2093\ \text{s}^{-1}$ and $e_{ss} = 4.78$ for a unit ramp.

Asked 2 times
2082 Chaitra · 3 marks
2077 Chaitra · 4 marks

Realize the given digital controller by series programming: G(z)=4(z-1)(z²+1.2z+1)/((z+0.1)(z²-0.3z+0.8)).

Answer

In series (cascade) programming, $G(z)$ is factored into first- and second-order sections. Each section is realized by standard programming, and the sections are connected in cascade.

Step 1: Factor into sections (divide each factor by the highest power of $z$ )

G(z) = 4\cdot\underbrace{\frac{1 - z^{-1}}{1 + 0.1z^{-1}}}_{G_1(z)}\cdot\underbrace{\frac{1 + 1.2z^{-1} + z^{-2}}{1 - 0.3z^{-1} + 0.8z^{-2}}}_{G_2(z)}

Step 2: Difference equations

Section 1: $G_1 = W_1/X$ with

$w_1(k) = x(k) - 0.1w_1(k-1)$
$y_1(k) = w_1(k) - w_1(k-1)$

Section 2: input $y_1$ , intermediate $w_2$

$w_2(k) = y_1(k) + 0.3w_2(k-1) - 0.8w_2(k-2)$
$y(k) = 4\big[w_2(k) + 1.2w_2(k-1) + w_2(k-2)\big]$

Step 3: Block diagram (two standard-programmed sections in cascade)

Section 1, $G_1(z)$ :

x(k)-->(S)------------o--[1]-------->(S)--> y1(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w1(k-1)       |
        +<[-0.1]------o--[-1]---------+

Section 2, $G_2(z)$ , followed by the gain 4:

y1(k)-->(S)------------o--[1]-------->(S)->[4]-> y(k)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-1)       |
        (S)<[0.3]------o--[1.2]------>(S)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-2)       |
         +<[-0.8]------o--[1]----------+

The realization uses only $1 + 2 = 3$ delay elements, the minimum for a third-order filter. The overall gain 4 can be placed at either end.

Why series programming

A coefficient error in one section moves only that section's poles and zeros. The structure is less sensitive to coefficient quantization than direct programming of the whole third-order polynomial.
The complex poles $z = 0.15 \pm j0.882$ (with $|z| = 0.894$ ) are kept together in one real-coefficient second-order section.

Asked 2 times
2081 Chaitra · 5 marks
2078 Chaitra · 4 marks

Consider the digital filter defined by parallel programming: Y(z)/X(z)=4(z-1)(z²+1.2z+1)/((z+0.1)(z²-0.3z+0.8)).

Answer

In parallel programming, $G(z)$ is expanded into partial fractions in $z^{-1}$ . Each term is realized separately (first- or second-order), and the outputs are added.

Step 1: Write in powers of $z^{-1}$

\frac{Y(z)}{X(z)} = \frac{4(1-z^{-1})(1+1.2z^{-1}+z^{-2})}{(1+0.1z^{-1})(1-0.3z^{-1}+0.8z^{-2})}

The numerator and denominator have equal degree. So there is a constant term, plus a first-order term for the real pole and a second-order term for the complex pair, $z = 0.15 \pm j0.882$ .

Step 2: Partial fractions (let $w = z^{-1}$ )

G = C_0 + \frac{C_1}{1+0.1w} + \frac{D_0 + D_1w}{1 - 0.3w + 0.8w^2}

$C_0 = \lim_{w\to\infty} G = \dfrac{4(-1)(1)}{(0.1)(0.8)} = -50$
$C_1 = \dfrac{4(1-w)(1+1.2w+w^2)}{1-0.3w+0.8w^2}\Big|_{w=-10} = \dfrac{4(11)(89)}{84} = 46.619$
$D_0$ and $D_1$ : at $w = 0$ , $4 = -50 + 46.619 + D_0$ , so $D_0 = 7.381$ . At $w = 1$ (that is, $z = 1$ , where $G = 0$ ): $0 = -50 + \dfrac{46.619}{1.1} + \dfrac{7.381 + D_1}{1.5}$ , so $D_1 = 4.0476$ .

\frac{Y(z)}{X(z)} = -50 + \frac{46.619}{1 + 0.1z^{-1}} + \frac{7.381 + 4.0476z^{-1}}{1 - 0.3z^{-1} + 0.8z^{-2}}

Check: At $z = 2$ , both forms give the same value. At $z \to \infty$ : $-50 + 46.619 + 7.381 = 4$ . ✔

Step 3: Difference equations of the branches

$y_1(k) = -50x(k)$
$w_2(k) = x(k) - 0.1w_2(k-1)$ , and $y_2(k) = 46.619\,w_2(k)$
$w_3(k) = x(k) + 0.3w_3(k-1) - 0.8w_3(k-2)$ , and $y_3(k) = 7.381w_3(k) + 4.0476w_3(k-1)$
$y(k) = y_1(k) + y_2(k) + y_3(k)$

Step 4: Block diagram

Overall structure:

        +-------------[-50]-------------+ y1
        |                               v
x(k)-o--+-->[ 46.619/(1+0.1z^-1) ]--->(S)--> y(k)
        |                       y2      ^
        +-->[ second-order branch ]-----+ y3

Branch 2:

x(k)-->(S)------------o--[1]-------->(S)->[46.619]-> y2(k)
        ^             |
        |          [z^-1]
        |             | w2(k-1)
        +<[-0.1]------o

Branch 3:

x(k)-->(S)------------o--[7.381]---->(S)--> y3(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w3(k-1)       |
       (S)<[0.3]------o--[4.0476]-----+
        ^             |
        |          [z^-1]
        |             | w3(k-2)
        +<[-0.8]------o

Total delays: $1 + 2 = 3$ , the minimum for a third-order controller. A coefficient error in one branch affects only that branch's pole(s).

Asked 2 times
2071 Magh · 10 marks
2070 Magh · 12 marks

Design a digital controller such that the compensated system should have damping ratio ζ=0.5 and settling time = 2 seconds of the following system using root locus approach. [Figure: unity-feedback loop: R(s) → summing junction → E(s) → sampler (period T) → digital controller G₁*(s) → ZOH → 1/s(s+1) → C(s)]

Answer

The sampling period is not given, so we choose it first.

Step 1: Desired poles and choice of $T$

From $\zeta = 0.5$ and $t_s = 4/(\zeta\omega_n) = 2$ s:

\zeta\omega_n = 2, \quad \omega_n = 4\ \text{rad/s}, \quad \omega_d = 4\sqrt{1-0.25} = 3.464\ \text{rad/s}

Choose $T = 0.2$ s. Then $\omega_s = 2\pi/T = 31.4$ rad/s, giving about 9 samples per damped cycle. This is fast enough (the usual rule is at least 8–10), and the ZOH adds little phase lag.

\begin{aligned} |z| &= e^{-\zeta\omega_nT} = e^{-0.4} = 0.6703 \\ \angle z &= \omega_dT = 0.6928\ \text{rad} = 39.69^\circ \\ z_{1,2} &= 0.5158 \pm j0.4281 \end{aligned}

Step 2: Plant with ZOH

\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})} \\ &= \frac{0.01873(z + 0.9355)}{(z-1)(z-0.8187)} \end{aligned}

Step 3: Angle condition at $z_1$

Factor	Angle
zero at $-0.9355$	$+16.44^\circ$
pole at $1$	$-138.52^\circ$
pole at $0.8187$	$-125.28^\circ$
Total	$-247.36^\circ$

The angle deficiency is $247.36^\circ - 180^\circ = 67.36^\circ$ , so a lead controller is needed.

Step 4: Lead controller $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$

Cancel the slow plant pole with the zero: $\alpha = 0.8187$ , which adds $+125.28^\circ$ . The controller pole must give $125.28^\circ - 67.36^\circ = 57.92^\circ$ :

\beta = 0.5158 - \frac{0.4281}{\tan 57.92^\circ} = 0.2474

Step 5: Gain

K = \frac{|z_1-1|\,|z_1-0.2474|}{0.01873\,|z_1+0.9355|} = \frac{0.6464 \times 0.5053}{0.01873 \times 1.5131} = 11.52

G_D(z) = 11.52\,\frac{z - 0.8187}{z - 0.2474}

Step 6: Check

The open loop is $\dfrac{0.2158(z+0.9355)}{(z-1)(z-0.2474)}$ . The characteristic equation is $z^2 - 1.0316z + 0.4493 = 0$ , whose roots are $0.5158 \pm j0.4281$ . ✔

K_v = \frac{1}{T}\lim_{z\to1}(1-z^{-1})G_DG = \frac{0.2158 \times 1.9355}{0.2 \times 0.7526} = 2.78\ \text{s}^{-1}

Answer (for $T = 0.2$ s): $G_D(z) = 11.52\dfrac{z-0.8187}{z-0.2474}$ . The closed-loop poles are at $0.5158 \pm j0.4281$ ( $\zeta = 0.5$ , $t_s = 2$ s), and $K_v = 2.78\ \text{s}^{-1}$ . A different $T$ changes the numbers but not the method.

Asked 2 times
2080 Chaitra · 12 marks
2075 Bhadra · 12 marks

Design a digital proportional-plus-derivative controller for the plant as shown in figure below. It is desired that the damping ratio ζ of the dominant closed loop poles be 0.5 and the undamped natural frequency be 4 rad/sec. The sampling period is 0.1 sec. [Figure: R(z) → summing junction → sampler (T) → digital PD controller → ZOH (1-e⁻ᵀˢ)/s → 1/s² → C(z); feedback path H(s) (unity in the 2080 paper)]

Answer

Step 1: Plant with ZOH ( $T = 0.1$ s)

G(z) = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^3}\right] = (1-z^{-1})\frac{T^2z(z+1)}{2(z-1)^3} = \frac{T^2(z+1)}{2(z-1)^2} = \frac{0.005(z+1)}{(z-1)^2}

Step 2: Desired closed-loop poles

With $\zeta = 0.5$ and $\omega_n = 4$ :

\begin{aligned} \zeta\omega_n &= 2, \qquad \omega_d = 4\sqrt{0.75} = 3.464\ \text{rad/s} \\ |z| &= e^{-\zeta\omega_nT} = e^{-0.2} = 0.8187 \\ \angle z &= \omega_dT = 0.3464\ \text{rad} = 19.85^\circ \\ z_1 &= 0.7701 + j0.2780 \end{aligned}

Step 3: Digital PD controller

G_D(z) = K_P + K_D(1-z^{-1}) = (K_P+K_D)\frac{z - \alpha}{z} = K\frac{z-\alpha}{z}, \qquad \alpha = \frac{K_D}{K_P+K_D}

Step 4: Angle condition at $z_1$

Factor	Angle
plant zero at $-1$	$+8.92^\circ$
double plant pole at $1$	$-2(129.59^\circ) = -259.19^\circ$
controller pole at $0$	$-19.85^\circ$
Sum	$-270.11^\circ$

For a total of $-180^\circ$ , the controller zero must give $+90.11^\circ$ :

\alpha = 0.7701 - \frac{0.2780}{\tan 90.11^\circ} = 0.7706

The zero lies almost directly below the desired pole.

Step 5: Magnitude condition

K = \frac{|z_1|\,|z_1-1|^2}{0.005\,|z_1+1|\,|z_1-0.7706|} = \frac{0.8187 \times 0.3607^2}{0.005 \times 1.7918 \times 0.2780} = 42.78

G_D(z) = 42.78\,\frac{z - 0.7706}{z} = 42.78\,(1 - 0.7706z^{-1})

Step 6: PD gains

K_P + K_D = 42.78, \qquad K_D = K\alpha = 32.97, \qquad K_P = K(1-\alpha) = 9.81

Step 7: Check

1 + G_DG = 0 \;\Rightarrow\; z^3 - 1.7861z^2 + 1.0491z - 0.1648 = 0

The roots are $z = 0.7701 \pm j0.2780$ (the desired pair) and $z = 0.2459$ . The third pole is much closer to the origin, so it decays fast, and the complex pair is dominant. ✔

Answer: $G_D(z) = 42.78\dfrac{z-0.7706}{z}$ , i.e. $K_P = 9.81$ and $K_D = 32.97$ . The difference equation is $u(k) = 42.78e(k) - 32.97e(k-1)$ .

2080 Chaitra · 8 marks

Realize the given digital filter by ladder programming: G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.1z⁻²).

Answer

In positive powers of $z$ :

G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.1z^{-2}} = \frac{2z^2 + 2.2z + 0.2}{z^2 + 0.4z - 0.1}

Ladder programming expands $G(z)$ as:

G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2}}}}

Step 1: $A_0$

G(z) = 2 + \frac{(2z^2+2.2z+0.2) - 2(z^2+0.4z-0.1)}{z^2+0.4z-0.1} = 2 + \frac{1.4z + 0.4}{z^2 + 0.4z - 0.1}

Step 2: $B_1$

\frac{z^2 + 0.4z - 0.1}{1.4z + 0.4} = 0.7143z + \frac{0.1143z - 0.1}{1.4z + 0.4}

The remainder is $0.4z - 0.7143(0.4)z = 0.1143z$ (exactly $4z/35$ ). So $B_1 = 0.7143$ .

Step 3: $A_1$

\frac{1.4z + 0.4}{0.1143z - 0.1} = 12.25 + \frac{0.4 + 12.25(0.1)}{0.1143z - 0.1} = 12.25 + \frac{1.625}{0.1143z - 0.1}

So $A_1 = 12.25$ .

Step 4: $B_2$ and $A_2$

\frac{0.1143z - 0.1}{1.625} = 0.07033z - 0.06154 = B_2z + \frac{1}{A_2}

So $B_2 = 0.07033$ and $A_2 = -16.25$ .

G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{12.25 + \cfrac{1}{0.07033z + \cfrac{1}{-16.25}}}}

$A_0$	$B_1$	$A_1$	$B_2$	$A_2$
2	0.7143	12.25	0.07033	−16.25

Ladder block diagram

X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         |
      |                           |
      +<--E2--[1/A2]<-------------+

Signal equations of the ladder ( $o$ = take-off point, (S) = summer):

\begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}Y_2 \\ Y &= A_0X + Y_1 \end{aligned}

Numerical gains: $1/B_1 = 1.4$ , $1/A_1 = 0.08163$ , $1/B_2 = 14.22$ , $1/A_2 = -0.06154$ .

Substituting back reproduces $G(z)$ exactly, so the realization is correct. It uses two delay elements.

2076 Bhadra · 2+6 marks

Higher order pulse transfer functions are generally realized by decomposing it in to several lower order pulse transfer function why? Consider a digital filter shown below. Realize this filter in ladder scheme. G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²)

Answer

Why higher-order pulse transfer functions are decomposed

A high-order $G(z)$ realized directly (direct or standard programming) is very sensitive to coefficient errors. The controller coefficients are stored with finite word length. A small rounding error in one coefficient of a high-order denominator can move all the poles a lot, even outside the unit circle, and make the filter unstable. So the transfer function is split into first- and second-order blocks (series, parallel or ladder programming):

A coefficient error then affects only the one or two poles of that block.
The structure is less sensitive to quantization, and the computations are better conditioned.
Low-order blocks are easier to program, test and scale.

Ladder realization

Write the filter in positive powers of $z$ :

G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.12z^{-2}} = \frac{2z^2 + 2.2z + 0.2}{z^2 + 0.4z - 0.12}

In ladder programming, $G(z)$ is expanded as a continued fraction:

G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2}}}}

Each $1/(B_iz)$ is realized by a delay $z^{-1}/B_i$ , and each $A_i$ by a constant.

Step 1: $A_0$ .

G(z) = 2 + \frac{1.4z + 0.44}{z^2 + 0.4z - 0.12}

So $A_0 = 2$ .

Step 2: $B_1$ .

\frac{z^2 + 0.4z - 0.12}{1.4z + 0.44} = 0.7143z + \frac{0.0857z - 0.12}{1.4z + 0.44}

So $B_1 = 1/1.4 = 0.7143$ .

Step 3: $A_1$ .

\frac{1.4z + 0.44}{0.0857z - 0.12} = 16.333 + \frac{2.4}{0.0857z - 0.12}

So $A_1 = 49/3 = 16.333$ .

Step 4: $B_2$ and $A_2$ .

\frac{0.0857z - 0.12}{2.4} = 0.0357z - 0.05

So $B_2 = 1/28 = 0.0357$ and $A_2 = -1/0.05 = -20$ .

G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{16.333 + \cfrac{1}{0.0357z - 0.05}}}

$A_0$	$B_1$	$A_1$	$B_2$	$A_2$
2	0.7143	16.333	0.0357	−20

Ladder diagram:

X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         |
      |                           |
      +<--E2--[1/A2]<-------------+

Signal equations of the ladder ( $o$ = take-off point, (S) = summer):

\begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}Y_2 \\ Y &= A_0X + Y_1 \end{aligned}

With the values: $1/B_1 = 1.4$ , $1/A_1 = 0.06122$ , $1/B_2 = 28$ , $1/A_2 = -0.05$ . Only two delays are needed.

The ladder needs only two delays. Because of its continued-fraction structure, it has low sensitivity to coefficient rounding.

2073 Magh · 6 marks

Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²). Realize this filter in the parallel scheme.

Answer

In parallel programming, $G(z)$ is expanded into partial fractions in $z^{-1}$ , and each term is realized as a separate branch.

Step 1: Factor the denominator

1 + 0.4z^{-1} - 0.12z^{-2} = (1 + 0.6z^{-1})(1 - 0.2z^{-1})

The poles are $z = -0.6$ and $z = 0.2$ .

Step 2: Partial fractions

The numerator and denominator have the same degree in $z^{-1}$ , so

G(z) = C_0 + \frac{C_1}{1 - 0.2z^{-1}} + \frac{C_2}{1 + 0.6z^{-1}}

Let $w = z^{-1}$ .

\begin{aligned} C_0 &= \lim_{w\to\infty}\frac{2 + 2.2w + 0.2w^2}{1 + 0.4w - 0.12w^2} = \frac{0.2}{-0.12} = -1.6667 \\ C_1 &= \frac{2 + 2.2w + 0.2w^2}{1 + 0.6w}\Big|_{w=5} = \frac{2 + 11 + 5}{4} = 4.5 \\ C_2 &= \frac{2 + 2.2w + 0.2w^2}{1 - 0.2w}\Big|_{w=-5/3} = \frac{2 - 3.6667 + 0.5556}{1.3333} = -0.8333 \end{aligned}

G(z) = -1.6667 + \frac{4.5}{1 - 0.2z^{-1}} - \frac{0.8333}{1 + 0.6z^{-1}}

Check at $z^{-1} = 0$ : $-1.6667 + 4.5 - 0.8333 = 2$ . ✔

Step 3: Difference equations

$w_1(k) = x(k) + 0.2w_1(k-1)$
$w_2(k) = x(k) - 0.6w_2(k-1)$
$y(k) = -1.6667x(k) + 4.5w_1(k) - 0.8333w_2(k)$

Step 4: Block diagram

Overall structure (three branches added):

        +-------------[-1.6667]-------------+
        |                                   v
x(k)-o--+-->[ 4.5/(1-0.2z^-1) ]----------->(S)--> y(k)
        |                                   ^
        +-->[ -0.8333/(1+0.6z^-1) ]---------+

Branch 1:

x(k)-->(S)------------o--[1]-------->(S)->[4.5]-> y1(k)
        ^             |
        |          [z^-1]
        |             | w1(k-1)
        +<[0.2]-------o

Branch 2:

x(k)-->(S)------------o--[1]-------->(S)->[-0.8333]-> y2(k)
        ^             |
        |          [z^-1]
        |             | w2(k-1)
        +<[-0.6]------o

Two delays are used, one per pole.

2072 Asoj · 8 marks

Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²). Realize this filter in the parallel scheme and ladder scheme.

Answer

Given:

G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.12z^{-2}}

(a) Parallel scheme

In parallel programming, $G(z)$ is expanded into partial fractions in $z^{-1}$ , and each term is realized as a separate branch.

Step 1: Factor the denominator

1 + 0.4z^{-1} - 0.12z^{-2} = (1 + 0.6z^{-1})(1 - 0.2z^{-1})

The poles are $z = -0.6$ and $z = 0.2$ .

Step 2: Partial fractions The numerator and denominator have the same degree in $z^{-1}$ , so

G(z) = C_0 + \frac{C_1}{1 - 0.2z^{-1}} + \frac{C_2}{1 + 0.6z^{-1}}

Let $w = z^{-1}$ .

\begin{aligned} C_0 &= \lim_{w\to\infty}\frac{2 + 2.2w + 0.2w^2}{1 + 0.4w - 0.12w^2} = \frac{0.2}{-0.12} = -1.6667 \\ C_1 &= \frac{2 + 2.2w + 0.2w^2}{1 + 0.6w}\Big|_{w=5} = \frac{2 + 11 + 5}{4} = 4.5 \\ C_2 &= \frac{2 + 2.2w + 0.2w^2}{1 - 0.2w}\Big|_{w=-5/3} = \frac{2 - 3.6667 + 0.5556}{1.3333} = -0.8333 \end{aligned}

G(z) = -1.6667 + \frac{4.5}{1 - 0.2z^{-1}} - \frac{0.8333}{1 + 0.6z^{-1}}

Check at $z^{-1} = 0$ : $-1.6667 + 4.5 - 0.8333 = 2$ . ✔

Step 3: Difference equations

$w_1(k) = x(k) + 0.2w_1(k-1)$
$w_2(k) = x(k) - 0.6w_2(k-1)$
$y(k) = -1.6667x(k) + 4.5w_1(k) - 0.8333w_2(k)$

Step 4: Block diagram Overall structure (three branches added):

        +-------------[-1.6667]-------------+
        |                                   v
x(k)-o--+-->[ 4.5/(1-0.2z^-1) ]----------->(S)--> y(k)
        |                                   ^
        +-->[ -0.8333/(1+0.6z^-1) ]---------+

Branch 1:

x(k)-->(S)------------o--[1]-------->(S)->[4.5]-> y1(k)
        ^             |
        |          [z^-1]
        |             | w1(k-1)
        +<[0.2]-------o

Branch 2:

x(k)-->(S)------------o--[1]-------->(S)->[-0.8333]-> y2(k)
        ^             |
        |          [z^-1]
        |             | w2(k-1)
        +<[-0.6]------o

Two delays are used, one per pole.

(b) Ladder scheme

Write the filter in positive powers of $z$ :

G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.12z^{-2}} = \frac{2z^2 + 2.2z + 0.2}{z^2 + 0.4z - 0.12}

In ladder programming, $G(z)$ is expanded as a continued fraction:

G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2}}}}

Each $1/(B_iz)$ is realized by a delay $z^{-1}/B_i$ , and each $A_i$ by a constant.

Step 1: $A_0$ .

G(z) = 2 + \frac{1.4z + 0.44}{z^2 + 0.4z - 0.12}

So $A_0 = 2$ .

Step 2: $B_1$ .

\frac{z^2 + 0.4z - 0.12}{1.4z + 0.44} = 0.7143z + \frac{0.0857z - 0.12}{1.4z + 0.44}

So $B_1 = 1/1.4 = 0.7143$ .

Step 3: $A_1$ .

\frac{1.4z + 0.44}{0.0857z - 0.12} = 16.333 + \frac{2.4}{0.0857z - 0.12}

So $A_1 = 49/3 = 16.333$ .

Step 4: $B_2$ and $A_2$ .

\frac{0.0857z - 0.12}{2.4} = 0.0357z - 0.05

So $B_2 = 1/28 = 0.0357$ and $A_2 = -1/0.05 = -20$ .

G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{16.333 + \cfrac{1}{0.0357z - 0.05}}}

$A_0$	$B_1$	$A_1$	$B_2$	$A_2$
2	0.7143	16.333	0.0357	−20

Ladder diagram:

X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         |
      |                           |
      +<--E2--[1/A2]<-------------+

Signal equations of the ladder ( $o$ = take-off point, (S) = summer):

\begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}Y_2 \\ Y &= A_0X + Y_1 \end{aligned}

With the values: $1/B_1 = 1.4$ , $1/A_1 = 0.06122$ , $1/B_2 = 28$ , $1/A_2 = -0.05$ . Only two delays are needed.

2081 Chaitra · 3 marks

Realize the given digital controller by direct programming: Y(z)/X(z)=(6z²+7z)/(10z³-6z²+7z+5).

Answer

Direct programming realizes the difference equation exactly as written. It uses one chain of delays for past inputs and another for past outputs.

Step 1: Write in powers of $z^{-1}$

Divide the numerator and denominator by $10z^3$ :

\frac{Y(z)}{X(z)} = \frac{0.6z^{-1} + 0.7z^{-2}}{1 - 0.6z^{-1} + 0.7z^{-2} + 0.5z^{-3}}

Step 2: Difference equation

y(k) = 0.6y(k-1) - 0.7y(k-2) - 0.5y(k-3) + 0.6x(k-1) + 0.7x(k-2)

Step 3: Block diagram (direct programming)

x(k)
  o              (S)--------------o--> y(k)
  |               ^               |
[z^-1]            |             [z^-1]
  | x(k-1)        |               | y(k-1)
  o--[0.6]------>(S)<[0.6]--------o
  |               ^               |
[z^-1]            |             [z^-1]
  | x(k-2)        |               | y(k-2)
  o--[0.7]------>(S)<[-0.7]-------o
                  ^               |
                  |             [z^-1]
                  |               | y(k-3)
                  +<[-0.5]--------o

The left delay chain stores $x(k-1)$ and $x(k-2)$ . The right chain stores $y(k-1)$ , $y(k-2)$ and $y(k-3)$ .
Feed-forward gains are 0.6 and 0.7. Feedback gains are $+0.6$ , $-0.7$ and $-0.5$ (the negatives of the denominator coefficients).
There is no direct path from $x(k)$ to $y(k)$ , because the $z^0$ term of the numerator is zero.
Direct programming needs $m + n = 2 + 3 = 5$ delays. Standard programming would need only 3.

2079 Chaitra · 8 marks

Consider the digital filter defined by G(z)=(6z²+7z)/(10z³+6z²+7z+5). Realize this filter in the standard and series scheme.

Answer

Divide by $10z^3$ :

G(z) = \frac{6z^2 + 7z}{10z^3 + 6z^2 + 7z + 5} = \frac{0.6z^{-1} + 0.7z^{-2}}{1 + 0.6z^{-1} + 0.7z^{-2} + 0.5z^{-3}}

(a) Standard programming

Introduce an intermediate variable $W(z) = X(z)/(\text{denominator})$ :

\begin{aligned} w(k) &= x(k) - 0.6w(k-1) - 0.7w(k-2) - 0.5w(k-3) \\ y(k) &= 0.6w(k-1) + 0.7w(k-2) \end{aligned}

x(k)-->(S)------------o
        ^             |
        |          [z^-1]
        |             | w(k-1)
       (S)<[-0.6]-----o--[0.6]------>(S)--> y(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w(k-2)        |
       (S)<[-0.7]-----o--[0.7]--------+
        ^             |
        |          [z^-1]
        |             | w(k-3)
        +<[-0.5]------o

Only $n = 3$ delays are used, the minimum number.

(b) Series programming

Factor the denominator. Its real root, found numerically, is $z = -0.6697$ :

10z^3 + 6z^2 + 7z + 5 = 10(z + 0.6697)(z^2 - 0.0697z + 0.7467)

The quadratic has complex roots $0.0348 \pm j0.8634$ , so it is kept as one second-order block. The numerator is $6z^2 + 7z = 6z(z + 1.1667)$ . So

G(z) = 0.6\cdot\underbrace{\frac{1}{1 + 0.6697z^{-1}}}_{G_1}\cdot\underbrace{\frac{z^{-1} + 1.1667z^{-2}}{1 - 0.0697z^{-1} + 0.7467z^{-2}}}_{G_2}

Check: $0.6(z^{-1} + 1.1667z^{-2}) = 0.6z^{-1} + 0.7z^{-2}$ , and $(1 + 0.6697z^{-1})(1 - 0.0697z^{-1} + 0.7467z^{-2}) = 1 + 0.6z^{-1} + 0.7z^{-2} + 0.5z^{-3}$ . ✔

Difference equations:

$w_1(k) = x(k) - 0.6697w_1(k-1)$ , and $y_1(k) = 0.6w_1(k)$
$w_2(k) = y_1(k) + 0.0697w_2(k-1) - 0.7467w_2(k-2)$
$y(k) = w_2(k-1) + 1.1667w_2(k-2)$

Block $G_1$ (with gain 0.6):

x(k)-->(S)------------o--[1]-------->(S)->[0.6]-> y1(k)
        ^             |
        |          [z^-1]
        |             | w1(k-1)
        +<[-0.6697]---o

Block $G_2$ :

y1(k)-->(S)------------o
         ^             |
         |          [z^-1]
         |             | w2(k-1)
        (S)<[0.0697]---o--[1]-------->(S)--> y(k)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-2)       |
         +<[-0.7467]---o--[1.1667]-----+

The series form also uses 3 delays. Each block is first- or second-order, so it is less sensitive to coefficient rounding.

2076 Bhadra · 4 marks

Realize the given digital controller by direct programming: Y(z)/X(z)=(6z²+7z+16)/(10z³+8z²+9z+15).

Answer

Step 1: Write in powers of $z^{-1}$

Divide the numerator and denominator by $10z^3$ :

\frac{Y(z)}{X(z)} = \frac{0.6z^{-1} + 0.7z^{-2} + 1.6z^{-3}}{1 + 0.8z^{-1} + 0.9z^{-2} + 1.5z^{-3}}

Step 2: Difference equation

y(k) = -0.8y(k-1) - 0.9y(k-2) - 1.5y(k-3) + 0.6x(k-1) + 0.7x(k-2) + 1.6x(k-3)

Step 3: Block diagram (direct programming)

x(k)
  o              (S)--------------o--> y(k)
  |               ^               |
[z^-1]            |             [z^-1]
  | x(k-1)        |               | y(k-1)
  o--[0.6]------>(S)<[-0.8]-------o
  |               ^               |
[z^-1]            |             [z^-1]
  | x(k-2)        |               | y(k-2)
  o--[0.7]------>(S)<[-0.9]-------o
  |               ^               |
[z^-1]            |             [z^-1]
  | x(k-3)        |               | y(k-3)
  o--[1.6]------>(S)<[-1.5]-------o

The input delay chain gives $x(k-1)$ , $x(k-2)$ , $x(k-3)$ , with gains 0.6, 0.7, 1.6.
The output delay chain gives $y(k-1)$ , $y(k-2)$ , $y(k-3)$ , fed back with gains $-0.8$ , $-0.9$ , $-1.5$ .
All products are added in the summer column to give $y(k)$ . There is no direct $x(k)$ term.

Direct programming uses $3 + 3 = 6$ delays. (The denominator has roots $-1.1507$ and $0.1754 \pm j1.1282$ , which are outside the unit circle, so this filter is unstable as given. The question asks only for its realization.)

2075 Baisakh · 8 marks

Realize the digital filter by ladder programming: G(z)=(128z⁻³+224z⁻²+106z⁻¹+11)/(128z⁻³+160z⁻²+34z⁻¹+1).

Answer

Assumption: the textbook form of this filter has positive powers of $z$ :

G(z) = \frac{128z^3 + 224z^2 + 106z + 11}{128z^3 + 160z^2 + 34z + 1}

This gives a stable filter (poles $-0.9895$ , $-0.2254$ , $-0.0350$ ) and integer ladder constants. If the $z^{-k}$ form is used literally, the same continued-fraction steps apply but the constants are not integers.

For a third-order filter, ladder programming expands $G(z)$ as:

G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2 + \cfrac{1}{B_3z + \cfrac{1}{A_3}}}}}}

Step 1: $A_0$

G(z) = 1 + \frac{64z^2 + 72z + 10}{128z^3 + 160z^2 + 34z + 1}

So $A_0 = 1$ .

Step 2: $B_1$

\frac{128z^3 + 160z^2 + 34z + 1}{64z^2 + 72z + 10} = 2z + \frac{16z^2 + 14z + 1}{64z^2 + 72z + 10}

So $B_1 = 2$ .

Step 3: $A_1$

\frac{64z^2 + 72z + 10}{16z^2 + 14z + 1} = 4 + \frac{16z + 6}{16z^2 + 14z + 1}

So $A_1 = 4$ .

Step 4: $B_2$

\frac{16z^2 + 14z + 1}{16z + 6} = z + \frac{8z + 1}{16z + 6}

So $B_2 = 1$ .

Step 5: $A_2$

\frac{16z + 6}{8z + 1} = 2 + \frac{4}{8z + 1}

So $A_2 = 2$ .

Step 6: $B_3$ and $A_3$

\frac{8z + 1}{4} = 2z + \frac{1}{4}

So $B_3 = 2$ and $A_3 = 4$ .

G(z) = 1 + \cfrac{1}{2z + \cfrac{1}{4 + \cfrac{1}{z + \cfrac{1}{2 + \cfrac{1}{2z + \cfrac{1}{4}}}}}}

$A_0$	$B_1$	$A_1$	$B_2$	$A_2$	$B_3$	$A_3$
1	2	4	1	2	2	4

Ladder block diagram

X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         | +
      |                           v
      o<--E2--[1/A2]<----------- (S4)
      |                           ^ -
      v +                         |
     (S5)-->[z^-1/B3]-------------o Y3
      ^ -                         |
      |                           |
      +<--E3--[1/A3]<-------------+

Signal equations of the ladder ( $o$ = take-off point, (S) = summer):

\begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}(Y_2 - Y_3) \\ Y_3 &= \tfrac{z^{-1}}{B_3}(E_2 - E_3), & E_3 &= \tfrac{1}{A_3}Y_3 \\ Y &= A_0X + Y_1 \end{aligned}

With the values: $1/B_1 = 0.5$ , $1/A_1 = 0.25$ , $1/B_2 = 1$ , $1/A_2 = 0.5$ , $1/B_3 = 0.5$ , $1/A_3 = 0.25$ . Three delays are used.

Simulating these equations gives the same impulse response as $G(z)$ .

2075 Bhadra · 6 marks

Assume that a digital filter is given by the following difference equation: y(k)+a₁y(k-1)+a₂y(k-2)=b₁x(k)+b₂x(k)+b₂x(k-1). Draw block diagrams for the filters using (i) standard programming and (ii) ladder programming.

Answer

The difference equation as printed has a typo. It is read as the standard second-order filter

y(k) + a_1y(k-1) + a_2y(k-2) = b_0x(k) + b_1x(k-1) + b_2x(k-2)

G(z) = \frac{Y(z)}{X(z)} = \frac{b_0 + b_1z^{-1} + b_2z^{-2}}{1 + a_1z^{-1} + a_2z^{-2}} = \frac{b_0z^2 + b_1z + b_2}{z^2 + a_1z + a_2}

(i) Standard programming

Let $W(z) = X(z)/(1 + a_1z^{-1} + a_2z^{-2})$ :

\begin{aligned} w(k) &= x(k) - a_1w(k-1) - a_2w(k-2) \\ y(k) &= b_0w(k) + b_1w(k-1) + b_2w(k-2) \end{aligned}

x(k)-->(S)------------o--[b0]------->(S)--> y(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w(k-1)        |
       (S)<[-a1]------o--[b1]------->(S)
        ^             |               ^
        |          [z^-1]             |
        |             | w(k-2)        |
        +<[-a2]-------o--[b2]---------+

Only 2 delays are used, the minimum for second order.

(ii) Ladder programming

Expand as a continued fraction:

G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2}}}}

Define $c_1 = b_1 - a_1b_0$ , $c_2 = b_2 - a_2b_0$ , $d = a_1c_1 - c_2$ and $e = c_2 - \dfrac{a_2c_1^2}{d}$ . Then:

\begin{aligned} G(z) &= b_0 + \frac{c_1z + c_2}{z^2 + a_1z + a_2} \\ A_0 &= b_0, \quad B_1 = \frac{1}{c_1}, \quad A_1 = \frac{c_1^2}{d}, \quad B_2 = \frac{d}{c_1e}, \quad A_2 = \frac{e}{a_2} \end{aligned}

X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         |
      |                           |
      +<--E2--[1/A2]<-------------+

\begin{aligned} y_1(k) &= \tfrac{1}{B_1}[x(k-1) - e_1(k-1)], & e_1(k) &= \tfrac{1}{A_1}[y_1(k) - y_2(k)] \\ y_2(k) &= \tfrac{1}{B_2}[e_1(k-1) - e_2(k-1)], & e_2(k) &= \tfrac{1}{A_2}y_2(k) \\ y(k) &= A_0x(k) + y_1(k) \end{aligned}

Example: For $b_0 = 2$ , $b_1 = 2.2$ , $b_2 = 0.2$ , $a_1 = 0.4$ , $a_2 = -0.12$ : $c_1 = 1.4$ , $c_2 = 0.44$ , $d = 0.12$ , $e = 2.4$ . This gives $A_0 = 2$ , $B_1 = 0.7143$ , $A_1 = 16.33$ , $B_2 = 0.0357$ , $A_2 = -20$ .

2073 Bhadra · 8 marks

Realize the given digital controller by ladder programming whose transfer function is (8z³+7z²+10z+6)/(4z³+6z²+9z+10).

Answer

Ladder programming expands $G(z)$ as a continued fraction:

G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2 + \cfrac{1}{B_3z + \cfrac{1}{A_3}}}}}}

with $G(z) = \dfrac{8z^3 + 7z^2 + 10z + 6}{4z^3 + 6z^2 + 9z + 10}$ .

Step 1: $A_0 = 8/4 = 2$

G(z) = 2 + \frac{-5z^2 - 8z - 14}{4z^3 + 6z^2 + 9z + 10}

Step 2: $B_1$

\frac{4z^3 + 6z^2 + 9z + 10}{-5z^2 - 8z - 14} = -0.8z + \frac{-0.4z^2 - 2.2z + 10}{-5z^2 - 8z - 14}

So $B_1 = -0.8$ .

Step 3: $A_1$

\frac{-5z^2 - 8z - 14}{-0.4z^2 - 2.2z + 10} = 12.5 + \frac{19.5z - 139}{-0.4z^2 - 2.2z + 10}

So $A_1 = 12.5$ .

Step 4: $B_2$

\frac{-0.4z^2 - 2.2z + 10}{19.5z - 139} = -0.02051z + \frac{-5.0513z + 10}{19.5z - 139}

So $B_2 = -0.02051$ .

Step 5: $A_2$

\frac{19.5z - 139}{-5.0513z + 10} = -3.8604 + \frac{-100.396}{-5.0513z + 10}

So $A_2 = -3.8604$ .

Step 6: $B_3$ and $A_3$

\frac{-5.0513z + 10}{-100.396} = 0.05031z - 0.09961

So $B_3 = 0.05031$ and $A_3 = -1/0.09961 = -10.04$ .

$A_0$	$B_1$	$A_1$	$B_2$	$A_2$	$B_3$	$A_3$
2	−0.8	12.5	−0.02051	−3.8604	0.05031	−10.04

Ladder block diagram

X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         | +
      |                           v
      o<--E2--[1/A2]<----------- (S4)
      |                           ^ -
      v +                         |
     (S5)-->[z^-1/B3]-------------o Y3
      ^ -                         |
      |                           |
      +<--E3--[1/A3]<-------------+

Signal equations of the ladder ( $o$ = take-off point, (S) = summer):

\begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}(Y_2 - Y_3) \\ Y_3 &= \tfrac{z^{-1}}{B_3}(E_2 - E_3), & E_3 &= \tfrac{1}{A_3}Y_3 \\ Y &= A_0X + Y_1 \end{aligned}

Numerical gains: $1/B_1 = -1.25$ , $1/A_1 = 0.08$ , $1/B_2 = -48.75$ , $1/A_2 = -0.2590$ , $1/B_3 = 19.875$ , $1/A_3 = -0.0996$ .

The expansion was checked by substituting back. It reproduces $G(z)$ exactly, and the ladder's impulse response matches the direct difference equation. Three delays are used.

2071 Bhadra · 5 marks

Construct the block diagram for the following pulse-transfer function system (a digital filter) by direct programming: G(z)=(2-0.6z⁻¹)/(1+0.5z⁻¹).

Answer

Step 1: Difference equation

\frac{Y(z)}{X(z)} = \frac{2 - 0.6z^{-1}}{1 + 0.5z^{-1}} \;\Rightarrow\; Y(z)(1 + 0.5z^{-1}) = X(z)(2 - 0.6z^{-1})

y(k) = -0.5y(k-1) + 2x(k) - 0.6x(k-1)

Step 2: Direct programming block diagram

Each delayed input and output term is formed with its own delay element and added in a summer:

x(k)
  o--[2]-------->(S)--------------o--> y(k)
  |               ^               |
[z^-1]            |             [z^-1]
  | x(k-1)        |               | y(k-1)
  o--[-0.6]----->(S)<[-0.5]-------o

Feed-forward: $x(k)$ with gain 2, and $x(k-1)$ with gain $-0.6$ .
Feedback: $y(k-1)$ with gain $-0.5$ .
Direct programming uses $1 + 1 = 2$ delays.

For comparison, standard programming needs only one delay: $w(k) = x(k) - 0.5w(k-1)$ and $y(k) = 2w(k) - 0.6w(k-1)$ .

The pole at $z = -0.5$ is inside the unit circle, so the filter is stable.

2068 Magh · 8 marks

Realize system having pulse transfer function G(z)=(z+3)(z²-2z+10)/((z-5)(z²-z+8)) by series programming.

Answer

In series programming, the pulse transfer function is split into a cascade of first- and second-order sections. Each section is realized by standard programming.

Step 1: Split into sections (in powers of $z^{-1}$ )

G(z) = \underbrace{\frac{1 + 3z^{-1}}{1 - 5z^{-1}}}_{G_1(z)}\cdot\underbrace{\frac{1 - 2z^{-1} + 10z^{-2}}{1 - z^{-1} + 8z^{-2}}}_{G_2(z)}

The quadratics have complex roots, so they stay as one second-order section with real coefficients.

Step 2: Difference equations

Section 1 (input $x$ , output $y_1$ ):

$w_1(k) = x(k) + 5w_1(k-1)$
$y_1(k) = w_1(k) + 3w_1(k-1)$

Section 2 (input $y_1$ , output $y$ ):

$w_2(k) = y_1(k) + w_2(k-1) - 8w_2(k-2)$
$y(k) = w_2(k) - 2w_2(k-1) + 10w_2(k-2)$

Step 3: Block diagram

Section 1:

x(k)-->(S)------------o--[1]-------->(S)--> y1(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w1(k-1)       |
        +<[5]---------o--[3]----------+

Section 2:

y1(k)-->(S)------------o--[1]-------->(S)--> y(k)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-1)       |
        (S)<[1]--------o--[-2]------->(S)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-2)       |
         +<[-8]--------o--[10]---------+

x(k) --> [ G1(z) ] --y1(k)--> [ G2(z) ] --> y(k)

A total of 3 delays are used. The order of the sections can be swapped.

Note: The poles are $z = 5$ and $z = 0.5 \pm j2.78$ (with $|z| = 2.83$ ), all outside the unit circle. So this system is unstable, although it can still be realized.

2068 Jestha · 8 marks

Realize the following transfer function by series programming: X(z)=(z+1)(z²+3z+7)/((z-2)(z²+z+9)).

Answer

In series programming, $X(z)$ (the given pulse transfer function) is split into cascaded low-order sections. Each section is realized by standard programming.

Step 1: Sections in powers of $z^{-1}$

X(z) = \underbrace{\frac{1 + z^{-1}}{1 - 2z^{-1}}}_{G_1(z)}\cdot\underbrace{\frac{1 + 3z^{-1} + 7z^{-2}}{1 + z^{-1} + 9z^{-2}}}_{G_2(z)}

Both quadratics have complex roots, so each is kept as one second-order factor.

Step 2: Difference equations

Let $u(k)$ be the input and $y(k)$ the output.

Section 1:

$w_1(k) = u(k) + 2w_1(k-1)$
$y_1(k) = w_1(k) + w_1(k-1)$

Section 2:

$w_2(k) = y_1(k) - w_2(k-1) - 9w_2(k-2)$
$y(k) = w_2(k) + 3w_2(k-1) + 7w_2(k-2)$

Step 3: Block diagram

Section 1:

u(k)-->(S)------------o--[1]-------->(S)--> y1(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w1(k-1)       |
        +<[2]---------o--[1]----------+

Section 2:

y1(k)-->(S)------------o--[1]-------->(S)--> y(k)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-1)       |
        (S)<[-1]-------o--[3]-------->(S)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-2)       |
         +<[-9]--------o--[7]----------+

u(k) --> [ G1(z) ] --y1(k)--> [ G2(z) ] --> y(k)

Three delay elements are used. Each section can be scaled and checked separately.

Note: The poles $z = 2$ and $z = -0.5 \pm j2.96$ lie outside the unit circle, so the system is unstable. The series structure is still a valid realization.

2067 Mangsir · 10 marks

Realize the following transfer function by ladder programming: X(z)=(z³+4z²+10z+7)/(z³+3z²+11z+18).

Answer

Ladder programming expands $X(z) = \dfrac{z^3 + 4z^2 + 10z + 7}{z^3 + 3z^2 + 11z + 18}$ as a continued fraction:

G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2 + \cfrac{1}{B_3z + \cfrac{1}{A_3}}}}}}

Step 1: $A_0 = 1$

X(z) = 1 + \frac{z^2 - z - 11}{z^3 + 3z^2 + 11z + 18}

Step 2: $B_1$

\frac{z^3 + 3z^2 + 11z + 18}{z^2 - z - 11} = z + \frac{4z^2 + 22z + 18}{z^2 - z - 11}

So $B_1 = 1$ .

Step 3: $A_1$

\frac{z^2 - z - 11}{4z^2 + 22z + 18} = 0.25 + \frac{-6.5z - 15.5}{4z^2 + 22z + 18}

So $A_1 = 0.25$ .

Step 4: $B_2$

\frac{4z^2 + 22z + 18}{-6.5z - 15.5} = -0.6154z + \frac{12.4615z + 18}{-6.5z - 15.5}

So $B_2 = -0.6154$ .

Step 5: $A_2$

\frac{-6.5z - 15.5}{12.4615z + 18} = -0.5216 + \frac{-6.1111}{12.4615z + 18}

So $A_2 = -0.5216$ .

Step 6: $B_3$ and $A_3$

\frac{12.4615z + 18}{-6.1111} = -2.0392z - 2.9455

So $B_3 = -2.0392$ and $A_3 = -1/2.9455 = -0.3395$ .

$A_0$	$B_1$	$A_1$	$B_2$	$A_2$	$B_3$	$A_3$
1	1	0.25	−0.6154	−0.5216	−2.0392	−0.3395

Exact values: $B_2 = -8/13$ , $A_2 = -169/324$ , $B_3 = -1458/715$ , $A_3 = -55/162$ .

Ladder block diagram

X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         | +
      |                           v
      o<--E2--[1/A2]<----------- (S4)
      |                           ^ -
      v +                         |
     (S5)-->[z^-1/B3]-------------o Y3
      ^ -                         |
      |                           |
      +<--E3--[1/A3]<-------------+

Signal equations of the ladder ( $o$ = take-off point, (S) = summer):

\begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}(Y_2 - Y_3) \\ Y_3 &= \tfrac{z^{-1}}{B_3}(E_2 - E_3), & E_3 &= \tfrac{1}{A_3}Y_3 \\ Y &= A_0X + Y_1 \end{aligned}

Gains in the diagram: $1/B_1 = 1$ , $1/A_1 = 4$ , $1/B_2 = -1.625$ , $1/A_2 = -1.917$ , $1/B_3 = -0.4904$ , $1/A_3 = -2.9455$ .

Substituting back reproduces $X(z)$ exactly. Three delays are used.

2079 Chaitra · 12 marks

Consider the digital control system shown in figure below. In the z plane, design a digital controller such that the dominant closed-loop poles have a damping ratio of 0.5 and a settling time of 2 seconds. The sampling period is assumed to be 0.2 sec. Also, obtain the static velocity error constant of the system. [Figure: unity-feedback loop: r(t) → summing junction → e(t) → sampler → e(kT) → digital controller → ZOH → 1/s(s+2) → C(t)]

Answer

Step 1: Desired closed-loop poles

With $\zeta = 0.5$ and $t_s = 4/(\zeta\omega_n) = 2$ s:

\zeta\omega_n = 2, \quad \omega_n = 4, \quad \omega_d = 3.464\ \text{rad/s}

With $T = 0.2$ s:

\begin{aligned} |z| &= e^{-0.4} = 0.6703, \qquad \angle z = \omega_dT = 39.69^\circ \\ z_{1,2} &= 0.5158 \pm j0.4281 \end{aligned}

Step 2: ZOH + plant $1/[s(s+2)]$

G(z) = \frac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} = \frac{0.01758(z + 0.8753)}{(z-1)(z-0.6703)}

Step 3: Angle deficiency

At $z_1$ : the zero gives $+17.11^\circ$ , the pole at 1 gives $-138.52^\circ$ , and the pole at 0.6703 gives $-109.85^\circ$ . The total is $-231.26^\circ$ . The controller must supply $+51.26^\circ$ (lead).

Step 4: Lead controller $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$

Choose $\alpha = 0.6703$ to cancel the plant pole. This contributes $+109.85^\circ$ . The controller pole must then contribute $109.85^\circ - 51.26^\circ = 58.59^\circ$ :

\beta = 0.5158 - \frac{0.4281}{\tan 58.59^\circ} = 0.2543

Step 5: Gain

K = \frac{|z_1-1|\,|z_1-0.2543|}{0.01758\,|z_1+0.8753|} = \frac{0.6464 \times 0.5017}{0.01758 \times 1.4555} = 12.67

G_D(z) = 12.67\,\frac{z - 0.6703}{z - 0.2543}

Step 6: Check

G_DG = \frac{0.2228(z + 0.8753)}{(z-1)(z-0.2543)}

The characteristic equation $z^2 - 1.0316z + 0.4493 = 0$ gives $z = 0.5158 \pm j0.4281$ . ✔

Step 7: Static velocity error constant

\begin{aligned} K_v &= \lim_{z\to1}\frac{(1-z^{-1})G_D(z)G(z)}{T} = \frac{1}{0.2}\cdot\frac{0.2228(1 + 0.8753)}{1 - 0.2543} \\ &= 5 \times \frac{0.4178}{0.7457} = 2.80\ \text{s}^{-1} \end{aligned}

Answer: $G_D(z) = 12.67\dfrac{z-0.6703}{z-0.2543}$ and $K_v = 2.80\ \text{s}^{-1}$ . The steady-state error for a unit ramp is $1/K_v = 0.357$ .

2082 Chaitra (new course) · 9 marks

Consider a digital control system shown in figure below. The plant is a second order system described as G(s)=1/s(s+1). The sampling period of the system is taken to be T = 1 s. Design a digital controller such that the dominant closed loop poles of the system will have damping ratio ζ=0.5 and number of samples per cycle of damped sinusoidal oscillation to be 8. Following the design, obtain the unit step response of the designed system and static velocity error constant Kᵥ. [Figure: unity-feedback loop: R(s) → summing junction → sampler (T = 1 s) → digital controller → ZOH → G(s) → Y(s)]

Answer

Step 1: Plant pulse transfer function ( $T = 1$ s)

G(z) = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)} = \frac{0.3679(z + 0.7183)}{(z-1)(z-0.3679)}

Step 2: Desired poles

With 8 samples per cycle, $\omega_dT = 2\pi/8 = 45^\circ$ , and

\begin{aligned} |z| &= \exp\left(-\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{1}{8}\right) = e^{-0.4534} = 0.6354 \\ z_{1,2} &= 0.6354\angle \pm 45^\circ = 0.4493 \pm j0.4493 \end{aligned}

Step 3: Angle condition

Factor	Angle at $z_1$
zero at $-0.7183$	$+21.05^\circ$
pole at $1$	$-140.79^\circ$
pole at $0.3679$	$-79.73^\circ$
Total	$-199.47^\circ$

The controller must add $+19.47^\circ$ , so a lead compensator is needed.

Step 4: Lead controller $G_D(z) = K\dfrac{z-0.3679}{z-\beta}$

The zero cancels the plant pole at 0.3679 and adds $+79.73^\circ$ . The controller pole must contribute $79.73^\circ - 19.47^\circ = 60.26^\circ$ :

\beta = 0.4493 - \frac{0.4493}{\tan 60.26^\circ} = 0.1926

Step 5: Gain

K = \frac{|z_1-1|\,|z_1-0.1926|}{0.3679\,|z_1+0.7183|} = \frac{0.7107 \times 0.5175}{0.3679 \times 1.2511} = 0.7991

G_D(z) = 0.7991\,\frac{z - 0.3679}{z - 0.1926}

Step 6: Closed-loop pulse transfer function

\begin{aligned} G_DG &= \frac{0.2940(z + 0.7183)}{(z-1)(z-0.1926)} \\ \frac{C(z)}{R(z)} &= \frac{0.2940z + 0.2112}{z^2 - 0.8986z + 0.4038} \end{aligned}

The poles are $0.4493 \pm j0.4493$ , as desired. ✔

Step 7: Unit step response

With $R(z) = z/(z-1)$ , the difference equation is $c(k) = 0.8986c(k-1) - 0.4038c(k-2) + 0.2940r(k-1) + 0.2112r(k-2)$ , where $r(k) = 1$ :

$k$	0	1	2	3	4	5	6	7	8	9	10
$c(k)$	0	0.294	0.769	1.078	1.163	1.115	1.038	0.987	0.973	0.981	0.994

The maximum overshoot is about 16.3% at $k = 4$ , which matches $\zeta = 0.5$ . One damped cycle takes about 8 samples, and $c(\infty) = 1$ (type-1 system).

Step 8: Static velocity error constant

K_v = \lim_{z\to1}\frac{(1-z^{-1})G_DG}{T} = \frac{0.2940 \times 1.7183}{1 \times (1 - 0.1926)} = 0.626\ \text{s}^{-1}

Answer: $G_D(z) = 0.7991\dfrac{z-0.3679}{z-0.1926}$ . The step response overshoots by 16.3% and settles to 1, and $K_v = 0.626\ \text{s}^{-1}$ .

2073 Magh · 12 marks

Using the root locus method in z-plane determine a digital controller for a system shown below so that the dominant closed loop poles of the compensated system will have damping ratio of 0.5 and number of samples per cycle of damped sinusoidal oscillation to be 8. Assume that the sampling period T as 0.2 sec. [Figure: unity-feedback loop: R(z) → summing junction → sampler → digital controller → zero order hold → 1/s(s+1) → C(z)]

Answer

Step 1: Desired dominant poles

With 8 samples per cycle, $\omega_dT = 2\pi/8 = 45^\circ$ , and

\begin{aligned} |z| &= \exp\left(-\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{\omega_d}{\omega_s}\right) = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac18\right) = 0.6354 \\ z_{1,2} &= 0.4493 \pm j0.4493 \end{aligned}

With $T = 0.2$ s, this means $\omega_d = \omega_s/8 = 3.927$ rad/s, $\omega_n = 4.534$ rad/s, and $t_s \approx 4/(\zeta\omega_n) = 1.76$ s.

Step 2: Plant with ZOH ( $T = 0.2$ s)

\begin{aligned} G(z) &= \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})} = \frac{0.01873z + 0.01752}{(z-1)(z-0.8187)} \\ &= \frac{0.01873(z + 0.9355)}{(z-1)(z-0.8187)} \end{aligned}

Step 3: Angle deficiency at $z_1$

Factor	Angle
zero at $-0.9355$	$+17.98^\circ$
pole at $1$	$-140.79^\circ$
pole at $0.8187$	$-129.43^\circ$
Total	$-252.24^\circ$

The deficiency is $72.24^\circ$ , so a lead compensator is needed.

Step 4: Controller $G_D(z) = K\dfrac{z-0.8187}{z-\beta}$

The zero cancels the pole at 0.8187 and adds $+129.43^\circ$ . The controller pole must then contribute $129.43^\circ - 72.24^\circ = 57.19^\circ$ :

\beta = 0.4493 - \frac{0.4493}{\tan 57.19^\circ} = 0.1596

Step 5: Gain

K = \frac{|z_1-1|\,|z_1-0.1596|}{0.01873\,|z_1+0.9355|} = \frac{0.7107 \times 0.5346}{0.01873 \times 1.4559} = 13.93

G_D(z) = 13.93\,\frac{z - 0.8187}{z - 0.1596}

Step 6: Check

G_DG = \frac{0.2610(z + 0.9355)}{(z-1)(z-0.1596)}

The characteristic equation is $z^2 - 0.8986z + 0.4038 = 0$ , with roots $0.4493 \pm j0.4493$ . ✔

K_v = \frac{1}{T}\cdot\frac{0.2610 \times 1.9355}{1 - 0.1596} = 3.01\ \text{s}^{-1}

Answer: $G_D(z) = 13.93\dfrac{z-0.8187}{z-0.1596}$ , a lead compensator, with $K_v \approx 3.0\ \text{s}^{-1}$ .

2078 Chaitra · 12 marks

Design a suitable digital controller that includes an integral control action. The design specifications are that the damping ratio of the closed loop poles be 0.5 and there will be at least eight samples per cycle of the damped sinusoidal oscillations. The sampling time is assumed to be 0.2 sec. [Figure: unity-feedback loop: R(z) → summing junction → sampler (T) → digital controller → (1-e⁻ᵀˢ)/s → 10/((s+1)(s+5)) → C(z)]

Answer

Step 1: Plant with ZOH ( $T = 0.2$ s)

\begin{aligned} \frac{10}{s(s+1)(s+5)} &= \frac{2}{s} - \frac{2.5}{s+1} + \frac{0.5}{s+5} \\ G(z) &= (1-z^{-1})\left[\frac{2z}{z-1} - \frac{2.5z}{z-e^{-0.2}} + \frac{0.5z}{z-e^{-1}}\right] \\ &= \frac{0.1371z + 0.0921}{(z-0.8187)(z-0.3679)} = \frac{0.1371(z + 0.6714)}{(z-0.8187)(z-0.3679)} \end{aligned}

Step 2: Desired poles

With $\zeta = 0.5$ and 8 samples per cycle:

|z| = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac18\right) = 0.6354, \quad \angle z = 45^\circ, \quad z_1 = 0.4493 + j0.4493

Step 3: Try PI alone

Integral action requires a controller pole at $z = 1$ . With a PI controller $K(z-0.8187)/(z-1)$ cancelling the slow plant pole, the angle at $z_1$ is:

\underbrace{+21.85^\circ}_{\text{zero } -0.6714} \underbrace{-140.79^\circ}_{\text{pole } 1} \underbrace{-79.73^\circ}_{\text{pole } 0.3679} = -198.67^\circ

This is not $-180^\circ$ . There is $18.67^\circ$ too much lag, so PI alone cannot place the poles. A lead part is added, which makes the controller PID-like.

Step 4: Controller with integral + lead action

G_D(z) = K\frac{(z - 0.8187)(z - 0.3679)}{(z-1)(z-\beta)}

Both plant poles are cancelled. The angle condition at $z_1$ is:

21.85^\circ - 140.79^\circ - \angle(z_1 - \beta) = -180^\circ \;\Rightarrow\; \angle(z_1-\beta) = 61.06^\circ

\beta = 0.4493 - \frac{0.4493}{\tan 61.06^\circ} = 0.2009

Step 5: Gain

K = \frac{|z_1-1|\,|z_1-0.2009|}{0.1371\,|z_1+0.6714|} = \frac{0.7107 \times 0.5134}{0.1371 \times 1.2074} = 2.204

G_D(z) = 2.204\,\frac{(z-0.8187)(z-0.3679)}{(z-1)(z-0.2009)}

Step 6: Check

G_DG = \frac{0.3022(z + 0.6714)}{(z-1)(z-0.2009)}

The characteristic equation is $z^2 - 0.8986z + 0.4038 = 0$ , with roots $z = 0.4493 \pm j0.4493$ . ✔ That is $\zeta = 0.5$ with exactly 8 samples per cycle.

K_v = \frac{1}{T}\cdot\frac{0.3022 \times 1.6714}{1 - 0.2009} = 3.16\ \text{s}^{-1}

The pole at $z = 1$ gives zero steady-state error for a step input.

Answer: $G_D(z) = 2.204\dfrac{(z-0.8187)(z-0.3679)}{(z-1)(z-0.2009)}$ , with $K_v = 3.16\ \text{s}^{-1}$ .

2077 Chaitra · 12 marks

Consider a digital control system shown below. By choosing a reasonable sampling period T, design a digital PI controller such that the dominant closed loop poles have a damping ratio ζ of 0.5 and the number of samples per cycle of damped sinusoidal oscillation is 10. Also find the static velocity error constant. [Figure: unity-feedback loop: R(z) → summing junction → sampler (T) → digital PI controller → (1-e⁻ᵀˢ)/s → e⁻⁵ˢ/(s+0.4) → C(z)]

Answer

Step 1: Choice of sampling period

The plant has time constant $1/0.4 = 2.5$ s and dead time 5 s. Choose $T = 2.5$ s:

The dead time is then exactly $2T$ , giving a clean $z^{-2}$ .
$e^{-0.4T} = e^{-1}$ , which gives simple numbers.
With 10 samples per cycle, the oscillation period is 25 s. This is reasonable for a process with 5 s dead time.

A smaller $T$ such as 1 s is not suitable. The dead time becomes $z^{-5}$ , the phase lag at the desired point exceeds what a single PI zero can make up, and the angle condition cannot be met.

\begin{aligned} G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+0.4)}\right] = z^{-2}\cdot\frac{(1-e^{-1})}{0.4}\cdot\frac{z^{-1}}{1-e^{-1}z^{-1}} \\ &= \frac{1.5803}{z^2(z - 0.3679)} \end{aligned}

Step 2: Desired poles (ζ = 0.5, 10 samples per cycle)

|z| = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac{1}{10}\right) = 0.6958, \qquad \angle z = 36^\circ, \qquad z_1 = 0.5629 + j0.4090

Step 3: PI controller $G_D(z) = K\dfrac{z-\alpha}{z-1}$

Angles at $z_1$ :

double pole at origin: $-72^\circ$
pole at 0.3679: $-64.51^\circ$
pole at 1: $-136.91^\circ$

The sum is $-273.41^\circ$ , so the PI zero must add $+93.41^\circ$ :

\alpha = 0.5629 - \frac{0.4090}{\tan 93.41^\circ} = 0.5873

Step 4: Gain

K = \frac{|z_1|^2\,|z_1-0.3679|\,|z_1-1|}{1.5803\,|z_1-0.5873|} = \frac{0.4841 \times 0.4531 \times 0.5986}{1.5803 \times 0.4097} = 0.2028

G_D(z) = 0.2028\,\frac{z - 0.5873}{z - 1}

so $K_P = K\alpha = 0.1191$ and $K_I = K(1-\alpha) = 0.0837$ .

Step 5: Static velocity error constant

\begin{aligned} K_v &= \lim_{z\to1}\frac{(1-z^{-1})G_DG}{T} = \frac{1}{2.5}\cdot K(1-\alpha)\cdot\frac{1.5803}{1-0.3679} \\ &= \frac{0.0837 \times 2.5}{2.5} = 0.0837\ \text{s}^{-1} \end{aligned}

The steady-state error for a unit ramp is $1/K_v = 11.9$ .

Check: the closed-loop roots are $0.5629 \pm j0.4090$ (as designed), $0.7562$ and $-0.5141$ . The real pole near the PI zero (0.5873) is only partly cancelled and adds a slow tail.

Answer (with $T = 2.5$ s): $G_D(z) = 0.2028\dfrac{z-0.5873}{z-1}$ and $K_v = 0.0837\ \text{s}^{-1}$ .

2072 Asoj · 12 marks

Consider the digital control system as shown in figure below, where the plant is of the first order and has a dead time of 5 second. By choosing reasonable sampling time T, design a digital PI controller such that the dominant closed loop poles have a damping ratio ζ of 0.5 and the number of samples per cycle of damped oscillation is 10. After the controller is designed, determine the response of the system to a unit step input. [Figure: unity-feedback loop: r(t)/R(z) → summing junction → sampler (δT = T) → digital PI controller → ZOH → e⁻⁵ˢ/(s+0.4) → c(t)/C(z)]

Answer

A digital PI controller $G_D(z)=K_p+\dfrac{K_i}{1-z^{-1}}=K\dfrac{z-a}{z-1}$ is designed by the root-locus (angle and magnitude) method in the z-plane. The plant is $G_p(s)=\dfrac{e^{-5s}}{s+0.4}$ (time constant 2.5 s, dead time 5 s).

Choice of sampling period

The dead time should be a whole number of sampling periods, so $T = 5/N$ .
Ten samples per cycle fixes the angle of the desired pole at $\omega_d T = 2\pi/10 = 36^\circ$ , whatever T is. Each sample of delay ( $z^{-1}$ ) then adds $36^\circ$ of phase lag at that pole.
With $T = 1$ s ( $N = 5$ ) the delay alone gives $5\times36^\circ = 180^\circ$ . The poles at $z=1$ and $z=e^{-0.4}$ add more than $180^\circ$ again, and one PI zero cannot make up the difference, so no PI controller exists.
Choose $T = 2.5$ s: this equals the plant time constant, the dead time is $2T$ , and the angle condition can be met.

Pulse transfer function of the plant (with ZOH)

With $e^{-0.4T}=e^{-1}=0.3679$ :

\begin{aligned} G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+0.4)}\right] = z^{-2}\,\frac{(1-e^{-1})/0.4}{z-e^{-1}}\\ &= \frac{1.5803}{z^{2}(z-0.3679)} \end{aligned}

Desired closed-loop poles

\begin{aligned} |z| &= \exp\left(-\frac{2\pi\zeta}{10\sqrt{1-\zeta^2}}\right) = \exp\left(-\frac{0.6283\times0.5}{0.8660}\right) = 0.6958\\ z_1 &= 0.6958\angle 36^\circ = 0.5629 + j0.4090 \end{aligned}

(This gives $\omega_d = 0.2513$ rad/s and $\omega_n = 0.2902$ rad/s.)

Angle condition

Open loop: $G_D(z)G(z) = \dfrac{1.5803K(z-a)}{z^2(z-0.3679)(z-1)}$ . Pole angles at $z_1$ :

Pole	Angle at $z_1$
$z=0$ (double)	$2\times36^\circ = 72.0^\circ$
$z=0.3679$	$64.5^\circ$
$z=1$	$136.9^\circ$
Total	$273.4^\circ$

The zero must give $273.4^\circ - 180^\circ = 93.4^\circ$ :

a = 0.5629 - \frac{0.4090}{\tan 93.4^\circ} = 0.5873

Magnitude condition

K = \left|\frac{z_1^2(z_1-0.3679)(z_1-1)}{1.5803(z_1-0.5873)}\right| = 0.2028

G_D(z) = \frac{0.2028(z-0.5873)}{z-1} = 0.1191 + \frac{0.0837}{1-z^{-1}}

So $K_p = Ka = 0.1191$ and $K_i = K(1-a) = 0.0837$ .

Unit-step response

\frac{C(z)}{R(z)} = \frac{0.3204(z-0.5873)}{z^4-1.3679z^3+0.3679z^2+0.3204z-0.1882}

Closed-loop poles: $0.5629\pm j0.4090$ (designed), $0.7562$ and $-0.5141$ . Solving the difference equation with $r(k)=1$ :

k	0–2	3	4	5	6	7	8	10	12	15	20	29
c(kT)	0	0.320	0.571	0.795	0.907	0.958	0.962	0.947	0.957	0.987	0.997	1.000

Output is zero for 3 samples (2 for the dead time, 1 for ZOH plus computation).
The extra real pole at 0.7562 slows the final approach, so the response rises to about 0.96, dips slightly and creeps up to 1 with almost no overshoot.
Because of the integral action, steady-state error to a step is zero: $c(\infty)=1$ . $K_v = K(1-a)(1.5803)/[T(1-0.3679)] = 0.0837\ \text{s}^{-1}$ .

Answer: $T = 2.5$ s, $G_D(z) = 0.2028(z-0.5873)/(z-1)$ ( $K_p = 0.1191$ , $K_i = 0.0837$ ); the step response settles at 1 with zero steady-state error.

2075 Baisakh · 16 marks

A control system block diagram shown in figure below is of satellite communication system. Design a lead compensator D(z) to stabilize a satellite position. The sampling rate of digital controller is 10 samples per second. Consider design criteria: (i) damping ratio must be greater than 0.6, (ii) damped natural frequency should not exceed 1 Hz i.e. ωd<0.1ωₙ, and the time constant for the closed loop control system should be less than 0.5 second with 5% criterion for settling time. [Figure: summing junction → D(z) → ZOH → 0.02/s (volts to rps) → 1/s (rps to rev) → output; feedback through gain 360 (degrees) to the negative input; sampler T = 0.1 s; sketch of a satellite dish]

Answer

The satellite is a double integrator. With proportional control alone the loop is only marginally stable, so a lead compensator $D(z)=K\dfrac{z-a}{z-b}$ ( $b<a$ ) is designed by the z-plane root-locus method.

Plant and loop gain

The loop is ZOH, then $0.02/s$ , then $1/s$ , then the feedback gain 360. The loop transfer function is

G_p(s) = \frac{0.02\times360}{s^2} = \frac{7.2}{s^2}, \qquad T = 0.1\ \text{s}

For a ZOH followed by $1/s^2$ :

\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{7.2}{s^3}\right] = 7.2\,\frac{T^2(z+1)}{2(z-1)^2}\\ &= \frac{0.036(z+1)}{(z-1)^2} \end{aligned}

The two poles at $z=1$ give two branches that leave the unit circle, so the uncompensated system cannot be stabilised by gain alone.

Translating the specifications

$\zeta > 0.6$ : choose $\zeta = 0.707$ .
Time constant $1/(\zeta\omega_n) < 0.5$ s (5% settling time $3/\zeta\omega_n < 1.5$ s): choose $\sigma = \zeta\omega_n = 3\ \text{s}^{-1}$ , so the time constant is 0.333 s and $t_s \approx 1$ s.
$\omega_d \le 0.1\,\omega_s = 0.1(2\pi/0.1) = 2\pi$ rad/s (1 Hz): with $\zeta=0.707$ , $\omega_d = \sigma = 3$ rad/s (0.48 Hz). This is satisfied.

So $\omega_n = 4.243$ rad/s and

\begin{aligned} |z_1| &= e^{-\sigma T} = e^{-0.3} = 0.7408, \quad \angle z_1 = \omega_d T = 0.3\ \text{rad} = 17.19^\circ\\ z_1 &= 0.7077 + j0.2189 \end{aligned}

Angle deficiency

Factor	Angle at $z_1$
zero at $z=-1$	$+7.31^\circ$
double pole at $z=1$	$-2\times143.16^\circ = -286.33^\circ$
Plant total	$-279.02^\circ$

The compensator must add $-180^\circ - (-279.02^\circ) = +99.02^\circ$ of phase lead.

Placing the zero and pole

Put the zero at $a = 0.8$ (near $z=1$ , which reduces the pull of the double pole):

angle of $(z_1-0.8)$ is $112.85^\circ$
the pole must give $112.85^\circ - 99.02^\circ = 13.83^\circ$ :

b = 0.7077 - \frac{0.2189}{\tan 13.83^\circ} = -0.1816

Gain

K = \left|\frac{(z_1-1)^2(z_1+0.1816)}{0.036(z_1+1)(z_1-0.8)}\right| = 8.294

D(z) = \frac{8.294(z-0.8)}{z+0.1816}

Check

Characteristic equation $(z-1)^2(z+0.1816) + 0.2986(z+1)(z-0.8) = 0$ has roots $0.7077 \pm j0.2189$ (designed) and $0.1044$ (fast, negligible).
Simulated step response (T = 0.1 s): settles within 5% in about 1.1 s, which meets $t_s < 1.5$ s. The peak is about 1.36 at $t = 0.5$ s. This overshoot is larger than $\zeta=0.707$ alone predicts because of the closed-loop zeros at 0.8 and −1, which are usual for a type-2 loop.
The acceleration constant is $K_a = 10.1\ \text{s}^{-2}$ , so steady-state error is zero for step and ramp commands.

 r -->(+)-->[D(z)]-->[ZOH]-->[0.02/s]-->[1/s]--+--> c
       ^-                                       |
       +-----------------[360]<-----------------+

Answer: $D(z) = 8.294\,\dfrac{z-0.8}{z+0.1816}$ , giving dominant poles $0.7077\pm j0.2189$ ( $\zeta=0.707$ , $\omega_n=4.24$ rad/s, time constant 0.33 s).

2075 Bhadra · 4 marks

Discuss the situations where you prefer phase lag/lead controllers and PID controllers.

Answer

Lag/lead compensators and PID controllers both reshape the loop, but each fits a different situation.

Prefer a phase-lead compensator when

the system is stable but has a poor transient response (low phase margin, large overshoot, slow rise);
you need a higher bandwidth and faster response, and some extra noise sensitivity is acceptable;
the plant has an integrator or a double integrator (e.g. satellite or position servo) that needs phase lead to stabilise it.

Prefer a phase-lag compensator when

the transient response is already satisfactory but the steady-state error is too large (low $K_p$ , $K_v$ );
low-frequency gain must be raised without changing the crossover frequency much;
high-frequency noise must be attenuated, and a slower response is acceptable.

A lag–lead compensator is used when both the transient and the steady-state response must be improved.

Prefer a PID controller when

the plant model is not accurately known and the controller must be tuned on site (Ziegler–Nichols rules, for example);
zero steady-state error to a step is needed (integral action) together with damping (derivative action);
the plant is a process plant (temperature, pressure, flow, level) with slow dynamics or dead time, where PID is the industry standard;
the controller must be simple, with three parameters an operator can understand.

Situation	Preferred controller
Good steady state, poor transient	Lead
Good transient, poor steady state	Lag
Both poor, model known	Lag–lead
Model unknown, tuning in field	PID
Process plants with dead time	PI / PID
Noisy measurement	Lag or PI (avoid D)

2072 Asoj · 6 marks

Write down the general procedures for frequency response design in discrete time control system.

Answer

Frequency-response design of a discrete-time system is done in the w-plane. The bilinear transformation maps the unit circle of the z-plane onto the imaginary axis of the w-plane, so the usual Bode-diagram methods can be used.

Procedure

Find the pulse transfer function of the plant with the zero-order hold: $G(z) = (1-z^{-1})\,\mathcal{Z}[G_p(s)/s]$ .
Transform to the w-plane with the bilinear transformation: $z = \frac{1 + (T/2)w}{1 - (T/2)w}, \qquad G(w) = G(z)\Big|_{z=\frac{1+Tw/2}{1-Tw/2}}$ Write $G(w)$ in time-constant form. The fictitious frequency $\nu$ ( $w=j\nu$ ) is related to the actual frequency by $\nu = \frac{2}{T}\tan\frac{\omega T}{2}$ .
Choose the controller gain $K$ to meet the static error constant ( $K_p$ , $K_v$ or $K_a$ ), using $\lim_{w\to0}$ in the w-plane, which gives the same constants as in the s-plane.
Draw the Bode diagram of $KG(j\nu)$ and read the uncompensated gain crossover frequency, phase margin and gain margin.
Select the compensator $G_D(w) = \dfrac{1+\tau w}{1+\alpha\tau w}$ $G_{D} (w) = \frac{1 + τ w}{1 + α τ w}$ :
- lead ( $\alpha<1$ ): needed phase $\phi_m = \text{PM}_{\text{required}} - \text{PM}_{\text{existing}} + 5^\circ\text{ to }12^\circ$ ; $\sin\phi_m = \frac{1-\alpha}{1+\alpha}$ ; place $\phi_m$ at the new crossover where $|KG(j\nu)| = \sqrt{\alpha}$ ; $\tau = 1/(\nu_c\sqrt{\alpha})$ ;
- lag ( $\alpha>1$ in this form): put the new crossover where the phase gives the required PM, and place the corner frequencies one decade below it.
Check the compensated Bode diagram for phase margin, gain margin and bandwidth. Repeat step 5 if needed.
Transform back to the z-plane with $w = \frac{2}{T}\,\frac{z-1}{z+1}$ to get $G_D(z)$ , which is then realised as a difference equation in the computer.
Verify with the closed-loop step response (simulation), and check that the sampling period is small compared with the closed-loop bandwidth (8–10 samples per cycle).

 G(s) --ZOH--> G(z) --bilinear--> G(w) --Bode--> GD(w)
                                                   |
       GD(z)  <---------- w = (2/T)(z-1)/(z+1) ----+

Points to remember

The w-plane magnitude is not bounded at $\nu \to \infty$ as in the s-plane: $G(w)$ usually has a zero at $w = 2/T$ (non-minimum phase), which adds lag at high frequency.
Since $\nu \approx \omega$ for $\omega T \ll 1$ , the design reads like a continuous-time design when sampling is fast.

2072 Magh · 16 marks

Design a digital controller for the system shown below. Use the Bode diagram approach in s plane. The design specifications are that the phase margin be 55°, the gain margin be at least 10 dB and the static velocity error constant be 5 sec⁻¹. The sampling period is specified as 0.1 sec. [Figure: unity-feedback loop: R(z) → summing junction → sampler → digital controller → ZOH → plant 1/s(s+2) → C(z)]

Answer

The design is done in the w-plane (bilinear transformation), where the Bode diagram can be used as in the s-plane. A lead compensator $G_D(w) = K\dfrac{1+\tau w}{1+\alpha\tau w}$ is used.

Step 1: Pulse transfer function of the plant

With $T = 0.1$ s and $e^{-0.2} = 0.8187$ :

\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+2)}\right]\\ &= \frac{0.004683z + 0.004381}{(z-1)(z-0.8187)} = \frac{0.004683(z+0.9355)}{(z-1)(z-0.8187)} \end{aligned}

Step 2: Transform to the w-plane

Substitute $z = \dfrac{1+0.05w}{1-0.05w}$ :

G(w) = \frac{0.5\left(1-\frac{w}{20}\right)\left(1+\frac{w}{600.4}\right)}{w\,(1+0.5017w)}

Step 3: Gain for $K_v = 5$

K_v = \lim_{w\to0} w\,G_D(w)G(w) = K\times0.5 = 5 \;\Rightarrow\; K = 10

Step 4: Uncompensated Bode data ( $10\,G(j\nu)$ )

$\nu$ (rad/s)	1	2	4	6	10	20
Magnitude (dB)	13.0	5.0	−4.9	−11.2	−19.2	−29.1
Phase (deg)	−119.4	−140.6	−164.4	−177.8	−194.3	−217.4

Gain crossover is at $\nu = 2.88$ rad/s and the phase margin is only 26.8°. About 28° or more of lead is needed. The w-plane zero at $w=20$ adds extra lag at high frequency, so a larger lead is needed than in a continuous design.

Step 5: Lead compensator

Choose the compensator zero to cancel the plant pole at $w = -1.993$ , so $\tau = 0.5017$ . Then reduce $\alpha$ until PM = 55°:

$\alpha$	New crossover $\nu_c$	PM	GM
0.25	4.47	48.6°	12.3 dB
0.20	4.65	52.3°	12.3 dB
0.17	4.76	55.0°	12.3 dB

G_D(w) = 10\,\frac{1 + 0.5017w}{1 + 0.0853w}

Compensated loop: $G_D(w)G(w) = \dfrac{5(1-w/20)(1+w/600.4)}{w(1+0.0853w)}$ . Phase margin = 55°, phase crossover at $\nu = 15.7$ rad/s, gain margin = 12.3 dB (≥ 10 dB).

Step 6: Back to the z-plane

Substitute $w = \dfrac{2}{T}\dfrac{z-1}{z+1} = 20\dfrac{z-1}{z+1}$ :

\begin{aligned} G_D(z) &= 10\,\frac{1+10.033\frac{z-1}{z+1}}{1+1.706\frac{z-1}{z+1}} = 40.78\,\frac{z-0.8187}{z-0.2608} \end{aligned}

Check of $K_v$

K_v = \frac{1}{T}\lim_{z\to1}(z-1)G_D(z)G(z) = \frac{1}{0.1}\times40.78\times\frac{0.1813}{0.7392}\times\frac{0.009063}{0.1813} = 5.0\ \text{s}^{-1}

Answer: $G_D(z) = 40.78\,\dfrac{z-0.8187}{z-0.2608}$ , giving PM ≈ 55°, GM ≈ 12.3 dB and $K_v = 5\ \text{s}^{-1}$ .

2071 Bhadra · 6 marks

Draw root locus plot for the given system below. [Figure: unity-feedback loop: R(s)=1/s → summing junction → sampler (T = 0.5 sec) → ZOH → 1/s(s+1) → C(s)]

Answer

The root locus is drawn in the z-plane for the open-loop pulse transfer function with gain K in the forward path.

Open-loop pulse transfer function

With $T = 0.5$ s and $e^{-0.5} = 0.6065$ :

\begin{aligned} G(z) &= K(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = K\,\frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})}\\ &= \frac{K(0.1065z + 0.0902)}{(z-1)(z-0.6065)} = \frac{0.1065K(z+0.8467)}{(z-1)(z-0.6065)} \end{aligned}

Open-loop poles: $z = 1$ , $z = 0.6065$
Open-loop zero: $z = -0.8467$ (the other branch ends at $-\infty$ )

Construction

Real-axis locus: between 0.6065 and 1, and to the left of −0.8467.
Breakaway / break-in points from $dK/dz = 0$ , where $K = -\dfrac{(z-1)(z-0.6065)}{0.1065(z+0.8467)}$ : $z^2 + 1.6935z - 1.9667 = 0 \;\Rightarrow\; z = 0.7915\ \ (K = 0.221),\quad z = -2.4850\ \ (K = 61.7)$
Complex part: for one real zero and two real poles the complex part is a circle centred at the zero: centre $= -0.8467$ , radius $= \sqrt{(1+0.8467)(0.6065+0.8467)} = 1.638$ .
Crossing of the unit circle: characteristic equation $z^2 + (0.1065K - 1.6065)z + (0.6065 + 0.0902K) = 0$ For complex roots $|z|^2 = 0.6065 + 0.0902K = 1$ , so $K_c = 4.36$ . The roots are then $z = 0.571 \pm j0.821$ (angle 55.2°, $\omega = 1.93$ rad/s).

Sketch

              Im
              |      . unit circle
          .   |   x  <- K=4.36, z=0.57+j0.82
      .       |       .
    o---------+----x--x---> Re
  -2.48  -0.85|  0.61 | 1
  break-in  zero      breakaway 0.79
      .       |       .
          .   |   x  <- K=4.36
              |
  circle: centre -0.85, radius 1.64

Result

For $0 < K < 0.221$ : two real closed-loop poles between 0.6065 and 1 (overdamped).
For $0.221 < K < 61.7$ : complex poles on the circle. They lie inside the unit circle only while $K < 4.36$ .
The system is stable for $0 < K < 4.36$ ; at $K = 4.36$ it oscillates at about 1.93 rad/s.
The input $R(s) = 1/s$ does not change the locus; it only decides the step response.

2071 Bhadra · 5 marks

Explain how PID controller is realized in discrete time domain. Discuss the role of P, I and D parts.

Answer

A digital PID controller is obtained by replacing the integral and derivative of the analog PID law with sums and differences of the sampled error $e(kT)$ .

Analog PID law

m(t) = K\left[e(t) + \frac{1}{T_i}\int_0^t e(t)\,dt + T_d\frac{de(t)}{dt}\right]

Discretization

Integral by the trapezoidal rule, derivative by the backward difference:

m(kT) = K\left\{e(kT) + \frac{T}{T_i}\sum_{h=1}^{k}\frac{e((h-1)T)+e(hT)}{2} + \frac{T_d}{T}\big[e(kT)-e((k-1)T)\big]\right\}

Taking the z-transform:

\begin{aligned} \frac{M(z)}{E(z)} = G_D(z) &= K_P + \frac{K_I}{1-z^{-1}} + K_D(1-z^{-1})\\ K_P = K - \frac{KT}{2T_i},\quad K_I &= \frac{KT}{T_i},\quad K_D = \frac{KT_d}{T} \end{aligned}

Realization (velocity / recursive form)

Multiplying by $(1-z^{-1})$ gives an algorithm that needs only the last two errors and the last output:

m(k) = m(k-1) + (K_P+K_I+K_D)e(k) - (K_P+2K_D)e(k-1) + K_De(k-2)

          +--> [ KP ] -----------------+
          |                            v
 e(k) ----+--> [ KI/(1-z^-1) ] ---->( + )---> m(k)
          |                            ^
          +--> [ KD(1-z^-1) ] ---------+

The computer runs this equation once every sampling period, and the D/A converter with a zero-order hold applies $m(k)$ to the plant.

Role of each part

P (proportional): gives an output proportional to the present error. A larger $K_P$ makes the response faster and reduces steady-state error, but too much gain causes overshoot and instability.
I (integral): adds a pole at $z=1$ and accumulates past error. It removes steady-state error to a step (raises the system type by one) but adds phase lag, so it can make the response more oscillatory. Anti-windup is needed when the actuator saturates.
D (derivative): responds to the rate of change of error, so it anticipates future error. It adds damping, reduces overshoot and improves stability, but it amplifies high-frequency measurement noise. It is often applied to the measured output instead of the error, with a filter.

2068 Magh · 16 marks

Design a digital controller for the system shown in figure below so that dominant closed loop poles of the system will have damping ratio of 0.5. It is required that the peak time of the response tₚ≤ 1.2 sec. Assume sampling time T = 0.2 sec. [Figure: unity-feedback loop: R(z) → summing junction → sampler → digital controller → m*(kT) → ZOH → 1/s(s+1) → C(z)]

Answer

A lead-type digital controller $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$ is designed by the z-plane root-locus method. Its zero cancels the slow plant pole.

Plant pulse transfer function ( $T = 0.2$ s)

With $e^{-0.2} = 0.8187$ :

\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})}\\ &= \frac{0.01873(z+0.9355)}{(z-1)(z-0.8187)} \end{aligned}

Desired dominant poles

\begin{aligned} t_p &= \frac{\pi}{\omega_d} = 1.2 \Rightarrow \omega_d = 2.618\ \text{rad/s}\\ \omega_n &= \frac{\omega_d}{\sqrt{1-0.5^2}} = 3.023\ \text{rad/s},\quad \zeta\omega_n = 1.511\\ |z| &= e^{-\zeta\omega_nT} = e^{-0.3023} = 0.7391, \quad \angle z = \omega_dT = 0.5236\ \text{rad} = 30^\circ\\ z_1 &= 0.6401 + j0.3696 \end{aligned}

That is 12 samples per cycle of damped oscillation, which is adequate.

Angle deficiency

Factor	Angle at $z_1$
zero at $-0.9355$	$+13.20^\circ$
pole at $1$	$-134.24^\circ$
pole at $0.8187$	$-115.80^\circ$
Total	$-236.84^\circ$

The controller must add $+56.84^\circ$ .

Controller zero and pole

Zero at $\alpha = 0.8187$ cancels the plant pole and removes its $-115.80^\circ$ .
Then the controller pole must give $13.20^\circ - 134.24^\circ - \theta_\beta = -180^\circ$ , so $\theta_\beta = 58.96^\circ$ :

\beta = 0.6401 - \frac{0.3696}{\tan 58.96^\circ} = 0.4177

Gain

K = \left|\frac{(z_1-1)(z_1-0.4177)}{0.01873(z_1+0.9355)}\right| = 7.340

G_D(z) = 7.340\,\frac{z-0.8187}{z-0.4177}

Check

Characteristic equation: $z^2 - 1.2802z + 0.5463 = 0$ , roots $0.6401 \pm j0.3696$ , which is the design point.
Step response: $c(kT) = 0,\ 0.138,\ 0.442,\ 0.757,\ 0.994,\ 1.125,\ 1.163,\ 1.141,\ \ldots$ . The peak is 1.163 at $t = 6T = 1.2$ s, so $t_p \le 1.2$ s is met.
Velocity error constant:

K_v = \frac{1}{T}\lim_{z\to1}(z-1)G_D(z)G(z) = \frac{7.340\times0.03625}{0.2\times(1-0.4177)} = 2.28\ \text{s}^{-1}

Answer: $G_D(z) = 7.340\,\dfrac{z-0.8187}{z-0.4177}$ , giving closed-loop poles $0.6401\pm j0.3696$ ( $\zeta=0.5$ , $t_p=1.2$ s).

2068 Jestha · 16 marks

Consider the system shown below. Design a digital controller such that the dominant closed loop poles of the system will have damping ratio ζ of 0.55. It is required that the peak time of the response be 1.2 sec. Assume that the sampling period T be 0.2. Also obtain Kᵥ for the designed system. [Figure: unity-feedback loop: r(t)/R(z) → summing junction → sampler → digital controller → (1-e⁻ᵀˢ)/s → 1/s(s+1) → c(t)/C(z)]

Answer

A lead-type digital controller $G_D(z)=K\dfrac{z-\alpha}{z-\beta}$ is designed by the root-locus method in the z-plane. Its zero cancels the plant pole at $e^{-T}$ .

Plant pulse transfer function ( $T = 0.2$ s)

G(z) = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{0.01873(z+0.9355)}{(z-1)(z-0.8187)}

Desired dominant poles

\begin{aligned} \omega_d &= \frac{\pi}{t_p} = \frac{\pi}{1.2} = 2.618\ \text{rad/s}\\ \omega_n &= \frac{2.618}{\sqrt{1-0.55^2}} = 3.135\ \text{rad/s},\quad \zeta\omega_n = 1.724\\ |z| &= e^{-1.724\times0.2} = 0.7083,\quad \angle z = 2.618\times0.2 = 0.5236\ \text{rad} = 30^\circ\\ z_1 &= 0.6134 + j0.3542 \end{aligned}

Angle condition

Factor	Angle at $z_1$
zero at $-0.9355$	$+12.88^\circ$
pole at $1$	$-137.50^\circ$
pole at $0.8187$	$-120.10^\circ$
Total	$-244.72^\circ$

The angle deficiency is $64.72^\circ$ , so lead compensation is needed.

Place the controller zero at $\alpha = 0.8187$ to cancel the plant pole.
The controller pole angle must be $\theta_\beta = 12.88^\circ - 137.50^\circ + 180^\circ = 55.38^\circ$ :

\beta = 0.6134 - \frac{0.3542}{\tan 55.38^\circ} = 0.3689

Gain from the magnitude condition

K = \left|\frac{(z_1-1)(z_1-0.3689)}{0.01873(z_1+0.9355)}\right| = 7.582

G_D(z) = 7.582\,\frac{z-0.8187}{z-0.3689}

Check

Closed-loop characteristic equation: $z^2 - 1.2269z + 0.5018 = 0$ , roots $0.6134 \pm j0.3542$ .
Step response peak $c = 1.126$ at $k = 6$ , i.e. $t_p = 1.2$ s.

Velocity error constant

\begin{aligned} K_v &= \frac{1}{T}\lim_{z\to1}(z-1)G_D(z)G(z) = \frac{1}{0.2}\cdot\frac{7.582\times0.01873(1.9355)}{1-0.3689}\\ &= 2.18\ \text{s}^{-1} \end{aligned}

So the steady-state error to a unit ramp is $1/K_v = 0.459$ .

Answer: $G_D(z) = 7.582\,\dfrac{z-0.8187}{z-0.3689}$ ; $K_v = 2.18\ \text{s}^{-1}$ .

2067 Mangsir · 16 marks

Consider the system shown below. Design a digital PI controller such that the dominant closed loop poles have a damping ratio of 0.55 and number of samples per cycle of damped oscillation is 10. Also find Kᵥ for the designed system. [Figure: unity-feedback loop: R(z) → summing junction → sampler → GD(z) → (1-e⁻ᵀˢ)/s → e⁻²ˢ/(s+2) → C(z)]

Answer

A digital PI controller $G_D(z) = K_p + \dfrac{K_i}{1-z^{-1}} = K\dfrac{z-a}{z-1}$ is designed with the root-locus angle and magnitude conditions. The sampling period is not given; $T = 1$ s is assumed. It makes the 2 s dead time exactly two samples and allows a feasible PI design.

Plant pulse transfer function

With $e^{-2T} = e^{-2} = 0.1353$ :

\begin{aligned} G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+2)}\right] = z^{-2}\,\frac{(1-e^{-2})/2}{z-e^{-2}}\\ &= \frac{0.4323}{z^2(z-0.1353)} \end{aligned}

Desired dominant poles

10 samples per cycle gives $\omega_dT = 2\pi/10 = 36^\circ$ :

\begin{aligned} |z| &= \exp\left(-\frac{\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{2\pi}{10}\right) = \exp\left(-\frac{0.55\times0.6283}{0.8352}\right) = 0.6611\\ z_1 &= 0.6611\angle36^\circ = 0.5349 + j0.3886 \end{aligned}

Angle condition

Pole	Angle at $z_1$
double pole at 0	$72.00^\circ$
pole at 0.1353	$44.21^\circ$
PI pole at 1	$140.12^\circ$
Total	$256.33^\circ$

The PI zero must contribute $256.33^\circ - 180^\circ = 76.33^\circ$ :

a = 0.5349 - \frac{0.3886}{\tan 76.33^\circ} = 0.4403

Magnitude condition

K = \left|\frac{z_1^2(z_1-0.1353)(z_1-1)}{0.4323(z_1-0.4403)}\right| = 0.8540

G_D(z) = 0.8540\,\frac{z-0.4403}{z-1} = 0.3760 + \frac{0.4780}{1-z^{-1}}

So $K_p = Ka = 0.3760$ and $K_i = K(1-a) = 0.4780$ .

Check

Characteristic equation: $z^4 - 1.1353z^3 + 0.1353z^2 + 0.3692z - 0.1626 = 0$ , with roots $0.5349 \pm j0.3886$ (designed), $0.6435$ and $-0.5780$ . Step response: $c(k) = 0, 0, 0, 0.369, 0.626, 0.867, 0.970, 1.020, 1.015, 1.003, \ldots \to 1$ . The peak is about 2% and there is zero steady-state error.

Velocity error constant

\begin{aligned} K_v &= \frac{1}{T}\lim_{z\to1}(1-z^{-1})G_D(z)G(z) = \frac{K(1-a)\times0.4323}{T(1-0.1353)}\\ &= \frac{0.4780\times0.4323}{0.8647} = 0.239\ \text{s}^{-1} \end{aligned}

The steady-state error to a unit ramp is $1/K_v = 4.18$ .

Answer: $G_D(z) = 0.8540\,\dfrac{z-0.4403}{z-1}$ ( $K_p = 0.376$ , $K_i = 0.478$ , with T = 1 s); $K_v = 0.239\ \text{s}^{-1}$ .

2068 Magh · 4 marks

Write a short note on phase-lead/phase-lag compensator.

Answer

A compensator is an extra block placed in the loop to change the root locus or the frequency response so that the specifications are met. Its digital form is

G_D(z) = K\,\frac{z - z_c}{z - p_c}

Phase-lead compensator

The zero is nearer to $z = 1$ than the pole ( $p_c < z_c$ ), so it gives positive phase, at most $\phi_m$ where $\sin\phi_m = \frac{1-\alpha}{1+\alpha}$ (w-plane form $\frac{1+\tau w}{1+\alpha\tau w}$ , $\alpha<1$ ).
It moves the root locus to the left (towards the origin of the z-plane), so it increases phase margin, damping and bandwidth.
Result: faster response and less overshoot, but more high-frequency noise.

Phase-lag compensator

The pole is nearer to $z = 1$ than the zero ( $p_c > z_c$ ), both close to 1. It gives negative phase but high low-frequency gain.
It raises the static error constants ( $K_p$ , $K_v$ ) by about $\frac{1-p_c}{1-z_c}$ without much change to the transient response.
Result: lower steady-state error and less noise, but a slower response.

Feature	Lead	Lag
Phase added	Positive	Negative
Main effect	Transient response	Steady-state error
Bandwidth	Increases	Decreases
Noise	Amplified	Attenuated
Pole–zero order	Zero closer to z=1	Pole closer to z=1

A lag–lead compensator combines the two when both transient and steady-state performance must be improved.

2068 Magh · 4 marks

Write a short note on advantage of digital filter.

Answer

A digital filter is a filter carried out as a difference equation on sampled data in a processor, e.g. $y(k) = -a_1y(k-1) + b_0x(k) + b_1x(k-1)$ . In control systems the digital controller itself is a digital filter.

Advantages

Flexibility: the characteristics are changed by editing coefficients or software, with no rewiring.
No drift: performance does not change with temperature, ageing or component tolerance, unlike R, L, C and op-amp circuits.
Accuracy and repeatability: set by word length; every unit built behaves identically.
Complex and adaptive filters: linear phase (FIR), very sharp cut-offs, notch filters and adaptive or time-varying filters are easy to build.
Very low frequencies can be handled without the huge capacitors an analog filter would need.
Multiplexing: one processor can filter many channels.
Storage and integration: data can be stored, and filtering, control, logging and communication run in the same computer.
No impedance-matching or loading problems between stages.

Limitations

Sampling and quantization errors, aliasing (an anti-aliasing analog filter is still needed), a limited bandwidth set by the sampling rate, and a computation delay.

Past questions and answers

1.Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²) (equivalently GD(z)=(2z²+2.2z+0.2)/(z²+0.4z-0.12)). Realize this filter in the ladder scheme.

Step 1: Constant term A0A_0A0​

Step 2: B1B_1B1​

Step 3: A1A_1A1​

Step 4: B2B_2B2​ and A2A_2A2​

Ladder block diagram

2.Design a digital controller such that the compensated system should have damping ratio ζ=0.5 and settling time tₛ=2 s using root locus approach. [Figure: unity-feedback loop: R(s) → summing junction → E(s) → sampler (T = 0.2) → digital controller → ZOH → 1/s(s+2) → C(s)]

Step 1: Desired closed-loop poles

Step 2: Plant pulse transfer function

Step 3: Angle deficiency at z1=0.5158+j0.4281z_1 = 0.5158 + j0.4281z1​=0.5158+j0.4281

Step 4: Lead controller

Step 5: Gain from the magnitude condition

Step 6: Check

Step 1: Plant pulse transfer function (T=1T = 1T=1 s)

Step 2: Desired dominant poles

Step 3: PI controller and angle condition

Step 4: Gain KKK

Step 5: KvK_vKv​ and steady-state error for a ramp

4.Realize the given digital controller by series programming: G(z)=4(z-1)(z²+1.2z+1)/((z+0.1)(z²-0.3z+0.8)).

Step 1: Factor into sections (divide each factor by the highest power of zzz)

Step 2: Difference equations

Step 3: Block diagram (two standard-programmed sections in cascade)

Why series programming

5.Consider the digital filter defined by parallel programming: Y(z)/X(z)=4(z-1)(z²+1.2z+1)/((z+0.1)(z²-0.3z+0.8)).

Step 1: Write in powers of z−1z^{-1}z−1

Step 2: Partial fractions (let w=z−1w = z^{-1}w=z−1)

Step 3: Difference equations of the branches

Step 4: Block diagram

Step 1: Desired poles and choice of TTT

Step 2: Plant with ZOH

Step 3: Angle condition at z1z_1z1​

Step 4: Lead controller GD(z)=Kz−αz−βG_D(z) = K\dfrac{z-\alpha}{z-\beta}GD​(z)=Kz−βz−α​

Step 5: Gain

Step 6: Check

Step 1: Plant with ZOH (T=0.1T = 0.1T=0.1 s)

Step 2: Desired closed-loop poles

Step 3: Digital PD controller

Step 4: Angle condition at z1z_1z1​

Step 5: Magnitude condition

Step 6: PD gains

Step 7: Check

8.Realize the given digital filter by ladder programming: G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.1z⁻²).

Step 1: A0A_0A0​

Step 2: B1B_1B1​

Step 3: A1A_1A1​

Step 4: B2B_2B2​ and A2A_2A2​

Ladder block diagram

9.Higher order pulse transfer functions are generally realized by decomposing it in to several lower order pulse transfer function why? Consider a digital filter shown below. Realize this filter in ladder scheme. G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²)

Why higher-order pulse transfer functions are decomposed

Ladder realization

10.Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²). Realize this filter in the parallel scheme.

Step 1: Factor the denominator

Step 2: Partial fractions

Step 3: Difference equations

Step 4: Block diagram

11.Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²). Realize this filter in the parallel scheme and ladder scheme.

(a) Parallel scheme

(b) Ladder scheme

12.Realize the given digital controller by direct programming: Y(z)/X(z)=(6z²+7z)/(10z³-6z²+7z+5).

Step 1: Write in powers of z−1z^{-1}z−1

Step 2: Difference equation

Step 3: Block diagram (direct programming)

13.Consider the digital filter defined by G(z)=(6z²+7z)/(10z³+6z²+7z+5). Realize this filter in the standard and series scheme.

(a) Standard programming

(b) Series programming

14.Realize the given digital controller by direct programming: Y(z)/X(z)=(6z²+7z+16)/(10z³+8z²+9z+15).

Step 1: Write in powers of z−1z^{-1}z−1

Step 2: Difference equation

Step 3: Block diagram (direct programming)

15.Realize the digital filter by ladder programming: G(z)=(128z⁻³+224z⁻²+106z⁻¹+11)/(128z⁻³+160z⁻²+34z⁻¹+1).

Step 1: A0A_0A0​

Step 2: B1B_1B1​

Step 3: A1A_1A1​

Step 4: B2B_2B2​

Step 5: A2A_2A2​

Step 6: B3B_3B3​ and A3A_3A3​

Ladder block diagram

16.Assume that a digital filter is given by the following difference equation: y(k)+a₁y(k-1)+a₂y(k-2)=b₁x(k)+b₂x(k)+b₂x(k-1). Draw block diagrams for the filters using (i) standard programming and (ii) ladder programming.

(i) Standard programming

Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²) (equivalently GD(z)=(2z²+2.2z+0.2)/(z²+0.4z-0.12)). Realize this filter in the ladder scheme.

Step 1: Constant term $A_0$

Step 2: $B_1$

Step 3: $A_1$

Step 4: $B_2$ and $A_2$

Design a digital controller such that the compensated system should have damping ratio ζ=0.5 and settling time tₛ=2 s using root locus approach. [Figure: unity-feedback loop: R(s) → summing junction → E(s) → sampler (T = 0.2) → digital controller → ZOH → 1/s(s+2) → C(s)]

Step 3: Angle deficiency at $z_1 = 0.5158 + j0.4281$

Step 1: Plant pulse transfer function ( $T = 1$ s)

Step 4: Gain $K$

Step 5: $K_v$ and steady-state error for a ramp

Realize the given digital controller by series programming: G(z)=4(z-1)(z²+1.2z+1)/((z+0.1)(z²-0.3z+0.8)).

Step 1: Factor into sections (divide each factor by the highest power of $z$ )

Consider the digital filter defined by parallel programming: Y(z)/X(z)=4(z-1)(z²+1.2z+1)/((z+0.1)(z²-0.3z+0.8)).

Step 1: Write in powers of $z^{-1}$

Step 2: Partial fractions (let $w = z^{-1}$ )

Step 1: Desired poles and choice of $T$

Step 3: Angle condition at $z_1$

Step 4: Lead controller $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$

Step 1: Plant with ZOH ( $T = 0.1$ s)

Step 4: Angle condition at $z_1$

Realize the given digital filter by ladder programming: G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.1z⁻²).

Step 1: $A_0$

Step 2: $B_1$

Step 3: $A_1$

Step 4: $B_2$ and $A_2$

Higher order pulse transfer functions are generally realized by decomposing it in to several lower order pulse transfer function why? Consider a digital filter shown below. Realize this filter in ladder scheme. G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²)

Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²). Realize this filter in the parallel scheme.

Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²). Realize this filter in the parallel scheme and ladder scheme.

Realize the given digital controller by direct programming: Y(z)/X(z)=(6z²+7z)/(10z³-6z²+7z+5).

Step 1: Write in powers of $z^{-1}$

Consider the digital filter defined by G(z)=(6z²+7z)/(10z³+6z²+7z+5). Realize this filter in the standard and series scheme.

Realize the given digital controller by direct programming: Y(z)/X(z)=(6z²+7z+16)/(10z³+8z²+9z+15).

Step 1: Write in powers of $z^{-1}$

Realize the digital filter by ladder programming: G(z)=(128z⁻³+224z⁻²+106z⁻¹+11)/(128z⁻³+160z⁻²+34z⁻¹+1).

Step 1: $A_0$

Step 2: $B_1$

Step 3: $A_1$

Step 4: $B_2$

Step 5: $A_2$

Step 6: $B_3$ and $A_3$

Assume that a digital filter is given by the following difference equation: y(k)+a₁y(k-1)+a₂y(k-2)=b₁x(k)+b₂x(k)+b₂x(k-1). Draw block diagrams for the filters using (i) standard programming and (ii) ladder programming.

Realize the given digital controller by ladder programming whose transfer function is (8z³+7z²+10z+6)/(4z³+6z²+9z+10).

Step 1: $A_0 = 8/4 = 2$

Step 2: $B_1$

Step 3: $A_1$

Step 4: $B_2$

Step 5: $A_2$

Step 6: $B_3$ and $A_3$

Construct the block diagram for the following pulse-transfer function system (a digital filter) by direct programming: G(z)=(2-0.6z⁻¹)/(1+0.5z⁻¹).

Realize system having pulse transfer function G(z)=(z+3)(z²-2z+10)/((z-5)(z²-z+8)) by series programming.

Step 1: Split into sections (in powers of $z^{-1}$ )

Realize the following transfer function by series programming: X(z)=(z+1)(z²+3z+7)/((z-2)(z²+z+9)).

Step 1: Sections in powers of $z^{-1}$

Realize the following transfer function by ladder programming: X(z)=(z³+4z²+10z+7)/(z³+3z²+11z+18).

Step 1: $A_0 = 1$

Step 2: $B_1$

Step 3: $A_1$

Step 4: $B_2$

Step 5: $A_2$

Step 6: $B_3$ and $A_3$

Step 2: ZOH + plant $1/[s(s+2)]$

Step 4: Lead controller $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$

Step 1: Plant pulse transfer function ( $T = 1$ s)

Step 4: Lead controller $G_D(z) = K\dfrac{z-0.3679}{z-\beta}$

Step 2: Plant with ZOH ( $T = 0.2$ s)

Step 3: Angle deficiency at $z_1$

Step 4: Controller $G_D(z) = K\dfrac{z-0.8187}{z-\beta}$

Step 1: Plant with ZOH ( $T = 0.2$ s)

Step 3: PI controller $G_D(z) = K\dfrac{z-\alpha}{z-1}$