Digital Control System — Chapter 2: The Z-Transform

Asked 5 times
2082 Chaitra (new course) · 4 marks
2082 Chaitra · 6 marks
2081 Chaitra · 5 marks
2076 Bhadra · 4 marks
2073 Magh · 8 marks

Solve the following difference equation by use of Z-transform method: x(k+2)+3x(k+1)+2x(k)=0, given that x(0)=0, x(1)=1.

Answer

Take the z-transform of both sides using the shifting theorems:

$\mathcal{Z}[x(k+1)] = zX(z) - zx(0)$ , $\quad\mathcal{Z}[x(k+2)] = z^2X(z) - z^2x(0) - zx(1)$

Step 1: Transform

[z^2X(z) - z^2x(0) - zx(1)] + 3[zX(z) - zx(0)] + 2X(z) = 0

Substituting $x(0)=0$ , $x(1)=1$ :

\begin{aligned} z^2X(z) - z + 3zX(z) + 2X(z) &= 0 \\ (z^2+3z+2)X(z) &= z \\ X(z) &= \frac{z}{(z+1)(z+2)} \end{aligned}

Step 2: Partial fractions of $X(z)/z$

\frac{X(z)}{z} = \frac{1}{(z+1)(z+2)} = \frac{A}{z+1} + \frac{B}{z+2}

$A = \left.\frac{1}{z+2}\right|_{z=-1} = 1$ , $\quad B = \left.\frac{1}{z+1}\right|_{z=-2} = -1$

X(z) = \frac{z}{z+1} - \frac{z}{z+2}

Step 3: Inverse z-transform

Using $\mathcal{Z}^{-1}\left[\frac{z}{z-a}\right] = a^k$ :

x(k) = (-1)^k - (-2)^k, \quad k = 0, 1, 2, \dots

Check

k	0	1	2	3	4	5
$(-1)^k-(-2)^k$	0	1	-3	7	-15	31
Recursion $x(k+2)=-3x(k+1)-2x(k)$	0	1	-3	7	-15	31

Both agree, and the initial conditions are satisfied. The pole at $z=-2$ lies outside the unit circle, so the sequence grows in magnitude with alternating sign.

Answer: $x(k) = (-1)^k - (-2)^k$ , i.e. $x(k) = 0, 1, -3, 7, -15, 31, \dots$

Asked 4 times
2082 Chaitra · 5 marks
2082 Kartik · 5 marks
2078 Chaitra · 6 marks
2071 Bhadra · 4 marks

Obtain the inverse z-transform of X(z)=(2z³+z)/((z-2)²(z-1)) by using partial fraction expansion method.

Answer

Since $X(z)$ has a zero at $z=0$ , expand $X(z)/z$ in partial fractions.

\frac{X(z)}{z} = \frac{2z^2+1}{(z-2)^2(z-1)} = \frac{A}{z-1} + \frac{B}{(z-2)^2} + \frac{C}{z-2}

Coefficients

\begin{aligned} A &= \left.\frac{2z^2+1}{(z-2)^2}\right|_{z=1} = \frac{2+1}{1} = 3 \\ B &= \left.\frac{2z^2+1}{z-1}\right|_{z=2} = \frac{8+1}{1} = 9 \\ C &= \left.\frac{d}{dz}\left[\frac{2z^2+1}{z-1}\right]\right|_{z=2} = \left.\frac{4z(z-1) - (2z^2+1)}{(z-1)^2}\right|_{z=2} \\ &= \frac{8 - 9}{1} = -1 \end{aligned}

So

X(z) = \frac{3z}{z-1} + \frac{9z}{(z-2)^2} - \frac{z}{z-2}

Inverse transforms used

$\mathcal{Z}^{-1}\left[\frac{z}{z-1}\right] = 1$
$\mathcal{Z}^{-1}\left[\frac{z}{z-a}\right] = a^k$
$\mathcal{Z}^{-1}\left[\frac{z}{(z-a)^2}\right] = k\,a^{k-1}$

Result

x(k) = 3 + 9k\,2^{k-1} - 2^k, \quad k = 0, 1, 2, \dots

Check (first few values)

k	0	1	2	3	4
$x(k)$	2	10	35	103	275

$x(0) = \lim_{z\to\infty}X(z) = 2$ (initial value theorem), which agrees. Long division of $X(z) = \frac{2z^3+z}{z^3-5z^2+8z-4}$ also gives $2 + 10z^{-1} + 35z^{-2} + 103z^{-3} + \cdots$

Answer: $x(k) = 3 + 9k(2^{k-1}) - 2^k$ , which can also be written $x(k) = 3 + (4.5k - 1)\,2^k$ .

Asked 3 times
2081 Chaitra · 5 marks
2077 Chaitra · 4 marks
2074 Bhadra · 4 marks

Obtain the inverse z-transform of X(z)=z(1-e⁻ᵃᵀ)/((z-1)(z-e⁻ᵃᵀ)) using inversion integral method.

Answer

The inversion integral gives

x(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum \text{residues of } X(z)z^{k-1} \text{ at its poles}

Form $X(z)z^{k-1}$

X(z)z^{k-1} = \frac{(1-e^{-aT})\,z^k}{(z-1)(z-e^{-aT})}

For $k = 0, 1, 2, \dots$ there is no pole at $z=0$ ; the poles are $z = 1$ and $z = e^{-aT}$ (both simple, assuming $a > 0$ ).

Residues

At $z = 1$ :

K_1 = \left.(z-1)\frac{(1-e^{-aT})z^k}{(z-1)(z-e^{-aT})}\right|_{z=1} = \frac{1-e^{-aT}}{1-e^{-aT}} = 1

At $z = e^{-aT}$ :

K_2 = \left.\frac{(1-e^{-aT})z^k}{z-1}\right|_{z=e^{-aT}} = \frac{(1-e^{-aT})e^{-akT}}{e^{-aT}-1} = -e^{-akT}

Result

x(k) = K_1 + K_2 = 1 - e^{-akT}, \quad k = 0, 1, 2, \dots

Check: $x(0) = 0$ and $\lim_{z\to\infty}X(z) = 0$ (initial value theorem) agree; $x(\infty) = 1$ and $\lim_{z\to1}(z-1)X(z) = 1$ (final value theorem) agree. This is the sampled form of $x(t) = 1 - e^{-at}$ .

Answer: $x(kT) = 1 - e^{-akT}$ .

Asked 3 times
2080 Chaitra · 4 marks
2072 Magh · 4 marks
2070 Magh · 4 marks

Obtain the inverse Z-transform of X(z)=z(z+2)/(z-1)² by inversion integral method.

Answer

By the inversion integral method,

x(k) = \sum \text{residues of } X(z)z^{k-1} \text{ at the poles of } X(z)z^{k-1}

X(z)z^{k-1} = \frac{z(z+2)z^{k-1}}{(z-1)^2} = \frac{z^k(z+2)}{(z-1)^2}

For $k \ge 0$ the only pole is a double pole at $z = 1$ (no pole at the origin).

Residue at a double pole

\begin{aligned} x(k) &= \frac{1}{(2-1)!}\lim_{z\to1}\frac{d}{dz}\left[(z-1)^2\frac{z^k(z+2)}{(z-1)^2}\right] \\ &= \lim_{z\to1}\frac{d}{dz}\left[z^{k+1} + 2z^k\right] \\ &= \lim_{z\to1}\left[(k+1)z^k + 2kz^{k-1}\right] \\ &= (k+1) + 2k = 3k + 1 \end{aligned}

Check

k	0	1	2	3
$3k+1$	1	4	7	10

Long division: $X(z) = \frac{z^2+2z}{z^2-2z+1} = 1 + 4z^{-1} + 7z^{-2} + 10z^{-3} + \cdots$ , which agrees.

Answer: $x(k) = 3k + 1$ , $k = 0, 1, 2, \dots$

Asked 3 times
2077 Chaitra · 4 marks
2075 Bhadra · 4 marks
2075 Baisakh · 4 marks

Obtain the z-transform of x(t)={sin ωt, t≥ 0; 0, t<0}

Answer

Sampling with period $T$ : $x(kT) = \sin\omega kT$ for $k \ge 0$ . Use Euler's formula:

\sin\omega kT = \frac{e^{j\omega kT} - e^{-j\omega kT}}{2j}

Derivation

Since $\mathcal{Z}[a^k] = \sum_{k=0}^{\infty} a^k z^{-k} = \frac{1}{1-az^{-1}}$ for $|z| > |a|$ :

\begin{aligned} X(z) &= \frac{1}{2j}\left[\frac{1}{1-e^{j\omega T}z^{-1}} - \frac{1}{1-e^{-j\omega T}z^{-1}}\right] \\ &= \frac{1}{2j}\cdot\frac{(e^{j\omega T} - e^{-j\omega T})z^{-1}}{1 - (e^{j\omega T}+e^{-j\omega T})z^{-1} + z^{-2}} \\ &= \frac{z^{-1}\sin\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} \end{aligned}

using $e^{j\omega T} - e^{-j\omega T} = 2j\sin\omega T$ and $e^{j\omega T} + e^{-j\omega T} = 2\cos\omega T$ .

Multiplying numerator and denominator by $z^2$ :

X(z) = \frac{z\sin\omega T}{z^2 - 2z\cos\omega T + 1}, \quad |z| > 1

Answer: $\mathcal{Z}[\sin\omega t] = \dfrac{z\sin\omega T}{z^2 - 2z\cos\omega T + 1}$ . The poles $z = e^{\pm j\omega T}$ lie on the unit circle, as expected for an undamped sinusoid.

Asked 2 times
2082 Kartik · 3 marks
2081 Chaitra · 3 marks

Find the Z-transform of x(t)=t²e⁻ᵃᵗ (for k≥ 0).

Answer

Sampled form: $x(kT) = (kT)^2 e^{-akT} = T^2 k^2 (e^{-aT})^k$ . Let $r = e^{-aT}$ .

Step 1: $\mathcal{Z}[k^2]$

From $\mathcal{Z}[1] = \frac{z}{z-1}$ and $\mathcal{Z}[k\,x(k)] = -z\frac{d}{dz}X(z)$ :

\mathcal{Z}[k] = \frac{z}{(z-1)^2}, \qquad \mathcal{Z}[k^2] = -z\frac{d}{dz}\frac{z}{(z-1)^2} = \frac{z(z+1)}{(z-1)^3}

Step 2: multiply by $r^k$ (scaling: replace $z$ by $z/r$ )

\mathcal{Z}[k^2 r^k] = \frac{(z/r)(z/r+1)}{(z/r-1)^3} = \frac{rz(z+r)}{(z-r)^3}

Result

X(z) = \frac{T^2 e^{-aT} z\,(z + e^{-aT})}{(z - e^{-aT})^3}

Answer: $\mathcal{Z}[t^2e^{-at}] = \dfrac{T^2e^{-aT}z(z+e^{-aT})}{(z-e^{-aT})^3}$ (equivalently $\dfrac{T^2e^{-aT}z^{-1}(1+e^{-aT}z^{-1})}{(1-e^{-aT}z^{-1})^3}$ ).

Asked 2 times
2082 Chaitra (new course) · 4 marks
2075 Bhadra · 8 marks

Find the inverse z-transform of X(z)=z²/((z-1)²(z-e⁻ᵃᵀ)) by using inversion integral method.

Answer

Let $p = e^{-aT}$ (with $a > 0$ , so $p \ne 1$ ). By the inversion integral method,

x(k) = \sum \text{residues of } X(z)z^{k-1} = \sum \text{Res}\left[\frac{z^{k+1}}{(z-1)^2(z-p)}\right]

For $k \ge 0$ there is no pole at the origin. Poles: a double pole at $z=1$ and a simple pole at $z=p$ .

Residue at $z = p$ (simple pole)

K_1 = \left.\frac{z^{k+1}}{(z-1)^2}\right|_{z=p} = \frac{p^{k+1}}{(1-p)^2} = \frac{e^{-a(k+1)T}}{(1-e^{-aT})^2}

Residue at $z = 1$ (double pole)

\begin{aligned} K_2 &= \lim_{z\to1}\frac{d}{dz}\left[\frac{z^{k+1}}{z-p}\right] \\ &= \lim_{z\to1}\frac{(k+1)z^k(z-p) - z^{k+1}}{(z-p)^2} \\ &= \frac{(k+1)(1-p) - 1}{(1-p)^2} = \frac{k+1}{1-p} - \frac{1}{(1-p)^2} \end{aligned}

Result

\begin{aligned} x(k) &= K_1 + K_2 = \frac{k+1}{1-e^{-aT}} - \frac{1 - e^{-a(k+1)T}}{(1-e^{-aT})^2} \\ &= \frac{k - (k+1)e^{-aT} + e^{-a(k+1)T}}{(1-e^{-aT})^2} \end{aligned}

Check

$k=0$ : numerator $= 0 - p + p = 0$ , so $x(0) = 0$ . Initial value theorem: $\lim_{z\to\infty}X(z) = 0$ . ✔
$k=1$ : numerator $= 1 - 2p + p^2 = (1-p)^2$ , so $x(1) = 1$ . Long division gives $X(z) = z^{-1} + (2+p)z^{-2} + \cdots$ ✔
$k=2$ : $x(2) = \frac{2-3p+p^3}{(1-p)^2} = p + 2$ ✔

Answer: $x(k) = \dfrac{k - (k+1)e^{-aT} + e^{-a(k+1)T}}{(1-e^{-aT})^2}$ , $k = 0, 1, 2, \dots$

Asked 2 times
2080 Chaitra · 8 marks
2073 Bhadra · 8 marks

Solve the given difference equation and hence determine the output x(k), if u(t) is a unit step input function and x(t) is an output function: x(k+2)-x(k+1)+0.25x(k)=u(k+2), where x(0)=1 and x(1)=2.

Answer

Given $x(k+2) - x(k+1) + 0.25x(k) = u(k+2)$ , $x(0)=1$ , $x(1)=2$ , $u(k) = 1$ for $k \ge 0$ .

Step 1: z-transform of both sides

Shifting theorems: $\mathcal{Z}[x(k+2)] = z^2X(z) - z^2x(0) - zx(1)$ , $\mathcal{Z}[x(k+1)] = zX(z) - zx(0)$ .

Right side: $u(k+2) = 1$ for all $k \ge 0$ , so $\mathcal{Z}[u(k+2)] = z^2U(z) - z^2u(0) - zu(1) = \frac{z^3}{z-1} - z^2 - z = \frac{z}{z-1}$ .

Left side:

\begin{aligned} &[z^2X - z^2(1) - z(2)] - [zX - z(1)] + 0.25X \\ &= (z^2 - z + 0.25)X(z) - z^2 - z \end{aligned}

Step 2: solve for $X(z)$

\begin{aligned} (z-0.5)^2X(z) &= z^2 + z + \frac{z}{z-1} = \frac{(z^2+z)(z-1) + z}{z-1} = \frac{z^3}{z-1} \\ X(z) &= \frac{z^3}{(z-1)(z-0.5)^2} \end{aligned}

Step 3: inverse by residues of $X(z)z^{k-1} = \frac{z^{k+2}}{(z-1)(z-0.5)^2}$

At $z = 1$ :

K_1 = \left.\frac{z^{k+2}}{(z-0.5)^2}\right|_{z=1} = \frac{1}{0.25} = 4

At $z = 0.5$ (double pole):

\begin{aligned} K_2 &= \left.\frac{d}{dz}\left[\frac{z^{k+2}}{z-1}\right]\right|_{z=0.5} = \left.\frac{z^{k+1}[(k+2)(z-1) - z]}{(z-1)^2}\right|_{z=0.5} \\ &= \frac{0.5^{k+1}[-0.5(k+2) - 0.5]}{0.25} = -2(0.5)^{k+1}(k+3) \\ &= -(k+3)(0.5)^k \end{aligned}

Result

x(k) = 4 - (k+3)(0.5)^k, \quad k = 0, 1, 2, \dots

Check with the recursion $x(k+2) = x(k+1) - 0.25x(k) + 1$

k	0	1	2	3	4	5	6
Formula	1	2	2.75	3.25	3.5625	3.75	3.8594
Recursion	1	2	2.75	3.25	3.5625	3.75	3.8594

Final value: $\lim_{z\to1}(z-1)X(z) = 1/0.25 = 4$ , which matches $x(\infty) = 4$ .

Answer: $x(k) = 4 - (k+3)(0.5)^k$ ; the output rises from 1 and settles at 4.

Asked 2 times
2075 Baisakh · 8 marks
2070 Magh · 8 marks

Solve the following difference equation by use of Z-transform method: x(k+2)+1.379x(k+1)+0.3679x(k)=0.3679u(k), where u(k) is step input and x(0)=0, x(1)=0.3679.

Answer

Given (as printed) $x(k+2) + 1.379x(k+1) + 0.3679x(k) = 0.3679u(k)$ , $u(k) = 1$ for $k \ge 0$ , $x(0) = 0$ , $x(1) = 0.3679$ .

Step 1: z-transform

[z^2X - z^2x(0) - zx(1)] + 1.379[zX - zx(0)] + 0.3679X = 0.3679\frac{z}{z-1}

\begin{aligned} (z^2 + 1.379z + 0.3679)X(z) &= 0.3679z + \frac{0.3679z}{z-1} = \frac{0.3679z^2}{z-1} \\ X(z) &= \frac{0.3679z^2}{(z-1)(z^2+1.379z+0.3679)} \end{aligned}

Step 2: roots of the characteristic equation

z = \frac{-1.379 \pm \sqrt{1.379^2 - 4(0.3679)}}{2} = \frac{-1.379 \pm \sqrt{0.4300}}{2} = \frac{-1.379 \pm 0.6558}{2}

$p_1 = -0.3616$ , $\quad p_2 = -1.0174$

Step 3: partial fractions of $X(z)/z = \frac{0.3679z}{(z-1)(z-p_1)(z-p_2)}$

\begin{aligned} A &= \frac{0.3679(1)}{(1-p_1)(1-p_2)} = \frac{0.3679}{2.7469} = 0.1339 \\ B &= \frac{0.3679p_1}{(p_1-1)(p_1-p_2)} = \frac{0.3679(-0.3616)}{(-1.3616)(0.6558)} = 0.1490 \\ C &= \frac{0.3679p_2}{(p_2-1)(p_2-p_1)} = \frac{0.3679(-1.0174)}{(-2.0174)(-0.6558)} = -0.2829 \end{aligned}

X(z) = 0.1339\frac{z}{z-1} + 0.1490\frac{z}{z+0.3616} - 0.2829\frac{z}{z+1.0174}

Result

x(k) = 0.1339 + 0.1490(-0.3616)^k - 0.2829(-1.0174)^k

Check

k	0	1	2	3	4	5
$x(k)$	0	0.3679	-0.1394	0.4248	-0.1666	0.4414

These agree with the recursion $x(k+2) = -1.379x(k+1) - 0.3679x(k) + 0.3679$ . Since $|p_2| > 1$ , the response oscillates with slowly growing amplitude, so the final value theorem cannot be applied.

Answer: $x(k) = 0.1339 + 0.1490(-0.3616)^k - 0.2829(-1.0174)^k$ .

Note: this equation is usually printed in textbooks with $-1.3679$ (poles at $1$ and $0.3679$ ). With $x(k+2) - 1.3679x(k+1) + 0.3679x(k) = 0.3679u(k)$ and the same data, $X(z) = \frac{0.3679z^2}{(z-1)^2(z-0.3679)}$ and $x(k) = 0.5820k - 0.3388 + 0.3388(0.3679)^k$ , i.e. $0, 0.3679, 0.8712, 1.4242, \dots$ (a ramp-like rise).

2082 Chaitra (new course) · 4 marks

Obtain the Z-transform of x(k)=k²aᵏ⁻¹.

Answer

Use the known pair $\mathcal{Z}[a^k] = \frac{z}{z-a}$ and the multiplication by $k$ property $\mathcal{Z}[k\,x(k)] = -z\frac{d}{dz}X(z)$ .

Step 1: $\mathcal{Z}[k a^{k-1}]$

\mathcal{Z}[k a^k] = -z\frac{d}{dz}\left(\frac{z}{z-a}\right) = -z\cdot\frac{(z-a) - z}{(z-a)^2} = \frac{az}{(z-a)^2}

so $\mathcal{Z}[k a^{k-1}] = \frac{1}{a}\cdot\frac{az}{(z-a)^2} = \frac{z}{(z-a)^2}$ .

Step 2: $\mathcal{Z}[k^2 a^{k-1}]$

\begin{aligned} \mathcal{Z}[k\cdot k a^{k-1}] &= -z\frac{d}{dz}\left[\frac{z}{(z-a)^2}\right] \\ &= -z\cdot\frac{(z-a)^2 - 2z(z-a)}{(z-a)^4} \\ &= -z\cdot\frac{(z-a) - 2z}{(z-a)^3} = \frac{z(z+a)}{(z-a)^3} \end{aligned}

Check

Series of $\frac{z(z+a)}{(z-a)^3} = z^{-1} + 4a z^{-2} + 9a^2 z^{-3} + \cdots$ , i.e. coefficients $k^2a^{k-1}$ ✔.

Answer: $\mathcal{Z}[k^2a^{k-1}] = \dfrac{z(z+a)}{(z-a)^3} = \dfrac{z^{-1}(1+az^{-1})}{(1-az^{-1})^3}$ , $|z| > |a|$ .

2082 Chaitra · 5 marks

Determine the Z-transform for (i) x(t)=t²e⁻ᵃᵗ (ii) x(t)=t· x(t-nT).

Answer

(i) $x(t) = t^2e^{-at}$

Sampled: $x(kT) = T^2k^2(e^{-aT})^k$ . Let $r = e^{-aT}$ .

From $\mathcal{Z}[1] = \frac{z}{z-1}$ and $\mathcal{Z}[k\,x(k)] = -z\frac{dX}{dz}$ :

\mathcal{Z}[k] = \frac{z}{(z-1)^2}, \qquad \mathcal{Z}[k^2] = \frac{z(z+1)}{(z-1)^3}

Multiplying by $r^k$ replaces $z$ by $z/r$ :

\mathcal{Z}[k^2r^k] = \frac{(z/r)(z/r+1)}{(z/r-1)^3} = \frac{rz(z+r)}{(z-r)^3}

X(z) = \frac{T^2e^{-aT}z(z+e^{-aT})}{(z-e^{-aT})^3}

(ii) $y(t) = t\cdot x(t-nT)$

Assume $x(t) = 0$ for $t < 0$ and $\mathcal{Z}[x(t)] = X(z)$ .

Step 1 (real translation): $\mathcal{Z}[x(t-nT)] = z^{-n}X(z)$ .

Step 2 (multiplication by $t$ ): for sampled signals $t = kT$ , and $\mathcal{Z}[kT\,f(kT)] = -Tz\frac{d}{dz}F(z)$ .

\begin{aligned} Y(z) &= -Tz\frac{d}{dz}\left[z^{-n}X(z)\right] \\ &= -Tz\left[-nz^{-n-1}X(z) + z^{-n}\frac{dX(z)}{dz}\right] \\ &= z^{-n}\left[nT\,X(z) - Tz\frac{dX(z)}{dz}\right] \end{aligned}

Example: if $x(t) = 1(t)$ , $X(z) = \frac{z}{z-1}$ and $\frac{dX}{dz} = \frac{-1}{(z-1)^2}$ , so $Y(z) = z^{-n}\left[\frac{nTz}{z-1} + \frac{Tz}{(z-1)^2}\right]$ , the transform of $kT$ for $k \ge n$ .

Answers: (i) $\dfrac{T^2e^{-aT}z(z+e^{-aT})}{(z-e^{-aT})^3}$ ; (ii) $z^{-n}\left[nTX(z) - Tz\dfrac{dX(z)}{dz}\right]$ .

2082 Kartik · 3 marks

Find the Z-transform of x(k)=k·aᵏ⁻¹.

Answer

Use $\mathcal{Z}[a^k] = \frac{z}{z-a}$ and the property $\mathcal{Z}[k\,x(k)] = -z\frac{d}{dz}X(z)$ .

\begin{aligned} \mathcal{Z}[k a^k] &= -z\frac{d}{dz}\left(\frac{z}{z-a}\right) = -z\cdot\frac{(z-a) - z}{(z-a)^2} = \frac{az}{(z-a)^2} \end{aligned}

Dividing by $a$ :

\mathcal{Z}[k a^{k-1}] = \frac{z}{(z-a)^2}, \quad |z| > |a|

Alternative (differentiate with respect to $a$ ): $\sum_{k=0}^{\infty} a^k z^{-k} = \frac{z}{z-a}$ ; differentiating both sides with respect to $a$ gives $\sum k a^{k-1}z^{-k} = \frac{z}{(z-a)^2}$ .

Check: $\frac{z}{(z-a)^2} = z^{-1} + 2az^{-2} + 3a^2z^{-3} + \cdots$ , i.e. coefficients $k a^{k-1}$ ✔.

Answer: $\mathcal{Z}[k a^{k-1}] = \dfrac{z}{(z-a)^2} = \dfrac{z^{-1}}{(1-az^{-1})^2}$ .

2072 Asoj · 4 marks

Obtain the inverse z-transform of X(z)=z²/((z-2)²(z-e⁻ᵃᵀ)).

Answer

The inverse z-transform is found by the residue (inversion integral) method, since $X(z)$ has a double pole at $z = 2$ and a simple pole at $z = e^{-aT}$ . Let $b = e^{-aT}$ for short.

x(k) = \sum \text{residues of } X(z)\,z^{k-1} = \sum \text{residues of } \frac{z^{k+1}}{(z-2)^2 (z-b)}

For $k \ge 0$ the factor $z^{k+1}$ adds no pole at $z = 0$ , so only two poles count.

Residue at the simple pole $z = b$

R_1 = \left[\frac{z^{k+1}}{(z-2)^2}\right]_{z=b} = \frac{b^{k+1}}{(b-2)^2} = \frac{e^{-a(k+1)T}}{(2-e^{-aT})^2}

Residue at the double pole $z = 2$

\begin{aligned} R_2 &= \frac{d}{dz}\left[\frac{z^{k+1}}{z-b}\right]_{z=2} = \left[\frac{(k+1)z^{k}(z-b) - z^{k+1}}{(z-b)^2}\right]_{z=2} \\ &= \frac{2^{k}\left[(k+1)(2-b) - 2\right]}{(2-b)^2} = \frac{2^{k}\left[(2-b)k - b\right]}{(2-b)^2} \end{aligned}

Result

x(k) = \frac{1}{(2-e^{-aT})^2}\Big[(2-e^{-aT})\,k\,2^{k} - e^{-aT}\,2^{k} + e^{-a(k+1)T}\Big], \quad k = 0,1,2,\dots

Check

$k = 0$ : $\dfrac{-b + b}{(2-b)^2} = 0$ . Since the numerator degree (2) is one less than the denominator degree (3), $X(z)$ starts with $z^{-1}$ , so $x(0) = 0$ is correct.
$k = 1$ : $\dfrac{2(2-b) - 2b + b^2}{(2-b)^2} = \dfrac{(2-b)^2}{(2-b)^2} = 1$ , which matches the first long-division term $z^{-1}$ .
Numerical check with $b = 0.5$ : the formula gives $0, 1, 4.5, 14.25, 39.125$ for $k = 0$ to $4$ , the same as long division of $X(z)$ .

Answer: $x(k) = \dfrac{(2-e^{-aT})k\,2^k - e^{-aT}2^k + e^{-a(k+1)T}}{(2-e^{-aT})^2}$ for $k \ge 0$ (and $x(k) = 0$ for $k < 0$ ).

2082 Kartik · 5 marks

Consider the function y(k), which is a sum of functions x(h), where h=0,1,2,…,k, such that y(k) = Σ x(h) (sum from h = 0 to k), k=0,1,2,…, where y(k)=0 for k<0. Obtain Y(z).

Answer

$y(k)$ is the running sum (accumulation) of $x(k)$ . Its z-transform is $Y(z) = \dfrac{X(z)}{1-z^{-1}} = \dfrac{z}{z-1}X(z)$ .

Step 1: Write a difference equation

\begin{aligned} y(k) &= x(0) + x(1) + \dots + x(k-1) + x(k) \\ y(k-1) &= x(0) + x(1) + \dots + x(k-1) \end{aligned}

Subtracting:

y(k) - y(k-1) = x(k), \qquad k = 0, 1, 2, \dots

This also holds at $k = 0$ because $y(-1) = 0$ (given $y(k) = 0$ for $k < 0$ ), so $y(0) = x(0)$ .

Step 2: Take the z-transform

Use the right-shift (delay) theorem $\mathcal{Z}[y(k-1)] = z^{-1}Y(z)$ , which needs $y(k) = 0$ for $k < 0$ (given):

Y(z) - z^{-1}Y(z) = X(z)

Y(z) = \frac{1}{1-z^{-1}}X(z) = \frac{z}{z-1}X(z)

Alternative view (convolution)

$y(k) = \sum_{h=0}^{k} x(h)\cdot 1(k-h)$ is the convolution of $x(k)$ with the unit step $1(k)$ . Convolution in time is multiplication in $z$ :

Y(z) = X(z)\cdot \mathcal{Z}[1(k)] = X(z)\cdot\frac{1}{1-z^{-1}}

Example

Let $x(k) = 1$ for all $k \ge 0$ (unit step), so $X(z) = \dfrac{z}{z-1}$ . Then $y(k) = k+1$ and

Y(z) = \frac{z}{z-1}\cdot\frac{z}{z-1} = \frac{z^2}{(z-1)^2}

Check: $\mathcal{Z}[k+1] = \dfrac{z}{(z-1)^2} + \dfrac{z}{z-1} = \dfrac{z + z(z-1)}{(z-1)^2} = \dfrac{z^2}{(z-1)^2}$ , which agrees.

Remarks

$\dfrac{1}{1-z^{-1}}$ is the pulse transfer function of a digital accumulator (summer). It has a pole at $z = 1$ .
This result is the discrete version of integration ( $1/s$ ) and is used in the integral term of a digital PID controller.

Answer: $Y(z) = \dfrac{X(z)}{1 - z^{-1}} = \dfrac{z\,X(z)}{z-1}$ .

2073 Magh · 6 marks

Obtain the inverse z-transform of the following system using partial fraction expansion method: x(z)=z(1-eᵃᵀ)/((z-1)(z-eᵃᵀ)).

Answer

The function is solved by partial fractions of $X(z)/z$ , because every entry in the z-transform table has a factor $z$ in its numerator. The question is solved as printed, with $e^{aT}$ . (The usual textbook form has $e^{-aT}$ . That version is given at the end.)

Let $c = e^{aT}$ :

X(z) = \frac{(1-c)\,z}{(z-1)(z-c)}

Step 1: Divide by $z$ and expand

\frac{X(z)}{z} = \frac{1-c}{(z-1)(z-c)} = \frac{A}{z-1} + \frac{B}{z-c}

\begin{aligned} A &= \left[\frac{1-c}{z-c}\right]_{z=1} = \frac{1-c}{1-c} = 1 \\ B &= \left[\frac{1-c}{z-1}\right]_{z=c} = \frac{1-c}{c-1} = -1 \end{aligned}

Step 2: Multiply back by $z$

X(z) = \frac{z}{z-1} - \frac{z}{z-e^{aT}} = \frac{1}{1-z^{-1}} - \frac{1}{1-e^{aT}z^{-1}}

Step 3: Use standard pairs

$X(z)$	$x(k)$
$\dfrac{z}{z-1}$	$1$
$\dfrac{z}{z-e^{aT}}$	$e^{akT}$

x(k) = 1 - e^{akT}, \qquad k = 0, 1, 2, \dots

Check by long division

$X(z) = \dfrac{(1-c)z}{z^2 - (1+c)z + c} = (1-c)z^{-1} + (1-c)(1+c)z^{-2} + \dots$

$x(0) = 1 - 1 = 0$ (matches)
$x(1) = 1 - c$ (matches)
$x(2) = 1 - c^2 = (1-c)(1+c)$ (matches)

Note on the usual form

For $a > 0$ the printed form grows without bound: the pole $z = e^{aT}$ lies outside the unit circle. The standard textbook function is $X(z) = \dfrac{(1-e^{-aT})z}{(z-1)(z-e^{-aT})}$ , which is $\mathcal{Z}\left[\dfrac{a}{s(s+a)}\right]$ . The same steps give $A = 1$ , $B = -1$ and

x(k) = 1 - e^{-akT}

This is a sampled exponential rise toward 1.

Answer (as printed): $x(k) = 1 - e^{akT}$ for $k \ge 0$ . With $e^{-aT}$ : $x(k) = 1 - e^{-akT}$ .

2081 Chaitra · 3 marks

Obtain the inverse z-transform of X(z)=(1+z⁻¹-z⁻²)/(1-z⁻¹).

Answer

Split $X(z)$ into a polynomial part plus a proper part, then read each term from the z-transform table.

Step 1: Divide the numerator by the denominator (in powers of $z^{-1}$ )

Write $1 + z^{-1} - z^{-2} = (1 - z^{-1})(\alpha + \beta z^{-1}) + \gamma$ :

(1 - z^{-1})(\alpha + \beta z^{-1}) + \gamma = (\alpha + \gamma) + (\beta - \alpha)z^{-1} - \beta z^{-2}

Comparing coefficients:

$z^{-2}$ : $-\beta = -1 \Rightarrow \beta = 1$
$z^{-1}$ : $\beta - \alpha = 1 \Rightarrow \alpha = 0$
constant: $\alpha + \gamma = 1 \Rightarrow \gamma = 1$

X(z) = z^{-1} + \frac{1}{1 - z^{-1}}

Step 2: Inverse transform

$z^{-1} \;\leftrightarrow\; \delta(k-1)$ (a unit pulse at $k = 1$ )
$\dfrac{1}{1-z^{-1}} \;\leftrightarrow\; 1(k)$ (a unit step)

x(k) = 1(k) + \delta(k-1)

So:

$k$	0	1	2	3	4	...
$x(k)$	1	2	1	1	1	...

Check by direct division

$(1 + z^{-1} - z^{-2}) \div (1 - z^{-1}) = 1 + 2z^{-1} + z^{-2} + z^{-3} + \dots$ , which gives the same sequence.

Answer: $x(0) = 1$ , $x(1) = 2$ , $x(k) = 1$ for $k \ge 2$ , i.e. $x(k) = 1(k) + \delta(k-1)$ .

2080 Chaitra · 4 marks

The waveform below shows the input signal to a digital controller. Obtain the Z-transform of the input x(t). [Figure: x(t) rises linearly from 0 at t = 0 to 3 at t = 3 (passing 1 at t = 1 and 2 at t = 2), then stays constant at 3; dotted sample lines at t = 1, 2, …, 7]

Answer

From the figure, $x(t)$ is a unit-slope ramp that stops rising at $t = 3$ and then holds at 3. Take the sampling period $T = 1$ s, from the sample lines at $t = 1, 2, \dots$ .

Step 1: Write $x(t)$ as a ramp minus a delayed ramp

x(t) = t\cdot 1(t) - (t-3)\cdot 1(t-3)

For $t \le 3$ only the first term acts, giving $x = t$ . For $t > 3$ the result is $t - (t-3) = 3$ .

Step 2: Sampled values ( $T = 1$ )

$k$	0	1	2	3	4	5	...
$x(k)$	0	1	2	3	3	3	...

Step 3: Take the z-transform

Use $\mathcal{Z}[k] = \dfrac{z^{-1}}{(1-z^{-1})^2}$ (with $T = 1$ ) and the shift theorem $\mathcal{Z}[x(k-n)] = z^{-n}X(z)$ :

\begin{aligned} X(z) &= \frac{z^{-1}}{(1-z^{-1})^2} - z^{-3}\frac{z^{-1}}{(1-z^{-1})^2} = \frac{z^{-1}(1 - z^{-3})}{(1-z^{-1})^2} \end{aligned}

Since $1 - z^{-3} = (1 - z^{-1})(1 + z^{-1} + z^{-2})$ :

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

Step 4: Check by series

X(z) = z^{-1} + 2z^{-2} + 3z^{-3}\left(1 + z^{-1} + z^{-2} + \dots\right) = z^{-1} + 2z^{-2} + 3z^{-3} + 3z^{-4} + \dots

The coefficients $0, 1, 2, 3, 3, \dots$ match the samples.

Final value check: $\lim_{z\to1}(1-z^{-1})X(z) = 1\cdot(1+1+1) = 3$ , which is correct.

Answer: $X(z) = \dfrac{z^{-1}(1 - z^{-3})}{(1 - z^{-1})^2} = \dfrac{z^2 + z + 1}{z^2(z-1)}$ (with $T = 1$ s).

2079 Chaitra · 4 marks

Obtain the Z-transform of {x(k)=9k(2ᵏ⁻¹)-2ᵏ+3}, for k≥ 0.

Answer

Use linearity and three standard pairs.

Standard pairs used

$x(k)$	$X(z)$
$1$	$\dfrac{z}{z-1}$
$a^k$	$\dfrac{z}{z-a}$
$k\,a^{k-1}$	$\dfrac{z}{(z-a)^2}$

The last pair comes from $\mathcal{Z}[k\,x(k)] = -z\dfrac{d}{dz}X(z)$ :

-z\frac{d}{dz}\left(\frac{z}{z-a}\right) = -z\cdot\frac{-a}{(z-a)^2} = \frac{az}{(z-a)^2} = \mathcal{Z}[k a^k]

Dividing by $a$ gives $\mathcal{Z}[k a^{k-1}] = \dfrac{z}{(z-a)^2}$ .

Transform each term ( $a = 2$ )

\begin{aligned} \mathcal{Z}[9k\,2^{k-1}] &= \frac{9z}{(z-2)^2} \\ \mathcal{Z}[-2^k] &= -\frac{z}{z-2} \\ \mathcal{Z}[3] &= \frac{3z}{z-1} \end{aligned}

X(z) = \frac{9z}{(z-2)^2} - \frac{z}{z-2} + \frac{3z}{z-1}

Combine over a common denominator $(z-2)^2(z-1)$

\begin{aligned} \text{Numerator} &= 9z(z-1) - z(z-2)(z-1) + 3z(z-2)^2 \\ &= z\left[9z - 9 - (z^2 - 3z + 2) + 3(z^2 - 4z + 4)\right] \\ &= z\left[2z^2 + 0\cdot z + 1\right] \end{aligned}

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

Check

From the formula: $x(0) = 0 - 1 + 3 = 2$ , $x(1) = 9 - 2 + 3 = 10$ , $x(2) = 36 - 4 + 3 = 35$ , $x(3) = 108 - 8 + 3 = 103$ .

Long division of $X(z)$ gives $2 + 10z^{-1} + 35z^{-2} + 103z^{-3} + \dots$ , which matches.

Answer: $X(z) = \dfrac{9z}{(z-2)^2} - \dfrac{z}{z-2} + \dfrac{3z}{z-1} = \dfrac{z(2z^2+1)}{(z-2)^2(z-1)}$ , ROC $|z| > 2$ .

2079 Chaitra · 4 marks

Obtain the inverse z-transform of X(z)=z⁻³/((1-z⁻¹)(1-0.2z⁻¹)).

Answer

Remove the pure delay $z^{-3}$ first. Invert the remaining part by partial fractions, then shift the result by 3 samples.

Step 1: Write $X(z) = z^{-3}F(z)$

F(z) = \frac{1}{(1-z^{-1})(1-0.2z^{-1})} = \frac{z^2}{(z-1)(z-0.2)}

Step 2: Partial fractions of $F(z)/z$

\frac{F(z)}{z} = \frac{z}{(z-1)(z-0.2)} = \frac{A}{z-1} + \frac{B}{z-0.2}

A = \frac{1}{1-0.2} = 1.25, \qquad B = \frac{0.2}{0.2-1} = -0.25

F(z) = \frac{1.25z}{z-1} - \frac{0.25z}{z-0.2} \;\Rightarrow\; f(k) = 1.25 - 0.25(0.2)^k = 1.25\left[1 - (0.2)^{k+1}\right]

Step 3: Apply the shift theorem

$z^{-3}F(z) \leftrightarrow f(k-3)\,1(k-3)$ :

x(k) = \begin{cases} 0, & k = 0, 1, 2 \\[4pt] 1.25\left[1 - (0.2)^{k-2}\right], & k \ge 3 \end{cases}

Equivalently $x(k) = \left[1.25 - 0.25(0.2)^{k-3}\right]1(k-3)$ .

Values

$k$	0	1	2	3	4	5	6	$\infty$
$x(k)$	0	0	0	1	1.2	1.24	1.248	1.25

Check

Long division: $X(z) = z^{-3} + 1.2z^{-4} + 1.24z^{-5} + 1.248z^{-6} + \dots$ , which matches.
Final value theorem: $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1}{1-0.2} = 1.25$ , which matches.

Answer: $x(k) = 1.25\left[1 - (0.2)^{k-2}\right]$ for $k \ge 3$ , and $x(k) = 0$ for $k < 3$ .

2079 Chaitra · 8 marks

Solve the difference equation x(k+2)-x(k+1)-x(k)=0, where x(0)=0 and x(1)=1, and obtain the general form of solution. After obtaining the general form of solution determine the series up to 8th term.

Answer

This is the Fibonacci difference equation. Solve it by z-transform, then list the terms.

Step 1: Take the z-transform

Use $\mathcal{Z}[x(k+2)] = z^2X(z) - z^2x(0) - zx(1)$ and $\mathcal{Z}[x(k+1)] = zX(z) - zx(0)$ , with $x(0) = 0$ and $x(1) = 1$ :

\left[z^2X(z) - z\right] - zX(z) - X(z) = 0

X(z) = \frac{z}{z^2 - z - 1}

Step 2: Factorise the denominator

z^2 - z - 1 = 0 \;\Rightarrow\; z_{1,2} = \frac{1 \pm \sqrt{5}}{2}, \quad z_1 = 1.6180,\; z_2 = -0.6180

Step 3: Partial fractions of $X(z)/z$

\frac{X(z)}{z} = \frac{1}{(z-z_1)(z-z_2)} = \frac{1}{z_1 - z_2}\left[\frac{1}{z-z_1} - \frac{1}{z-z_2}\right], \quad z_1 - z_2 = \sqrt{5}

X(z) = \frac{1}{\sqrt5}\left[\frac{z}{z - z_1} - \frac{z}{z - z_2}\right]

Step 4: General (closed-form) solution

x(k) = \frac{1}{\sqrt5}\left[\left(\frac{1+\sqrt5}{2}\right)^k - \left(\frac{1-\sqrt5}{2}\right)^k\right], \quad k = 0, 1, 2, \dots

Here $1/\sqrt5 = 0.4472$ . Check: $x(0) = 0$ , and $x(1) = \frac{1}{\sqrt5}\cdot\sqrt5 = 1$ .

Step 5: Series up to the 8th term

From the closed form, or from the recursion $x(k+2) = x(k+1) + x(k)$ :

Term	1st	2nd	3rd	4th	5th	6th	7th	8th
$k$	0	1	2	3	4	5	6	7
$x(k)$	0	1	1	2	3	5	8	13

Example check with the formula for $k = 7$ : $\frac{1}{\sqrt5}\left[(1.618)^7 - (-0.618)^7\right] = \frac{1}{2.2361}\left[29.034 + 0.034\right] = 13.0$ .

(If the series is counted from $k = 1$ , the 8th term is $x(8) = 21$ .)

Remarks

One pole ( $z = 1.618$ ) is outside the unit circle, so the sequence grows without bound.
For large $k$ the second term dies out, so $x(k) \approx 0.4472(1.618)^k$ . The ratio of successive terms tends to the golden ratio $1.6180$ .

Answer: $x(k) = \dfrac{1}{\sqrt5}\left[\left(\dfrac{1+\sqrt5}{2}\right)^k - \left(\dfrac{1-\sqrt5}{2}\right)^k\right]$ . Series: $0, 1, 1, 2, 3, 5, 8, 13$ .

2073 Magh · 8 marks

Find X(k) when x(k+2)=x(k+1)+x(k), x(0)=0 and x(1)=1, and show that the limiting value of x(k+1)/x(k)=1.6180 when k tends to infinity.

Answer

This is the Fibonacci equation $x(k+2) - x(k+1) - x(k) = 0$ . Solve it by z-transform, then take the limit of the ratio.

Step 1: Take the z-transform

Use $\mathcal{Z}[x(k+2)] = z^2X(z) - z^2x(0) - zx(1)$ and $\mathcal{Z}[x(k+1)] = zX(z) - zx(0)$ , with $x(0) = 0$ and $x(1) = 1$ :

\left[z^2X(z) - z\right] - zX(z) - X(z) = 0

X(z) = \frac{z}{z^2 - z - 1}

Step 2: Factorise the denominator

z^2 - z - 1 = 0 \;\Rightarrow\; z_{1,2} = \frac{1 \pm \sqrt{5}}{2}, \quad z_1 = 1.6180,\; z_2 = -0.6180

Step 3: Partial fractions of $X(z)/z$

\frac{X(z)}{z} = \frac{1}{(z-z_1)(z-z_2)} = \frac{1}{z_1 - z_2}\left[\frac{1}{z-z_1} - \frac{1}{z-z_2}\right], \quad z_1 - z_2 = \sqrt{5}

X(z) = \frac{1}{\sqrt5}\left[\frac{z}{z - z_1} - \frac{z}{z - z_2}\right]

Step 4: General (closed-form) solution

x(k) = \frac{1}{\sqrt5}\left[\left(\frac{1+\sqrt5}{2}\right)^k - \left(\frac{1-\sqrt5}{2}\right)^k\right], \quad k = 0, 1, 2, \dots

Here $1/\sqrt5 = 0.4472$ . Check: $x(0) = 0$ , and $x(1) = \frac{1}{\sqrt5}\cdot\sqrt5 = 1$ .

First terms: $0, 1, 1, 2, 3, 5, 8, 13, 21, 34, \dots$

Step 5: Limiting value of $x(k+1)/x(k)$

Let $z_1 = 1.6180$ and $z_2 = -0.6180$ :

\frac{x(k+1)}{x(k)} = \frac{z_1^{k+1} - z_2^{k+1}}{z_1^{k} - z_2^{k}} = \frac{z_1 - z_2\left(\dfrac{z_2}{z_1}\right)^{k}}{1 - \left(\dfrac{z_2}{z_1}\right)^{k}}

Since $\left|\dfrac{z_2}{z_1}\right| = \dfrac{0.618}{1.618} = 0.382 < 1$ , we get $\left(\dfrac{z_2}{z_1}\right)^k \to 0$ as $k \to \infty$ :

\lim_{k\to\infty}\frac{x(k+1)}{x(k)} = z_1 = \frac{1+\sqrt5}{2} = 1.6180

This is the golden ratio.

Numerical check

$k$	5	6	7	8	9
$x(k+1)/x(k)$	8/5 = 1.600	13/8 = 1.625	21/13 = 1.6154	34/21 = 1.6190	55/34 = 1.6176

The ratio oscillates about 1.6180 and settles there.

Answer: $x(k) = \dfrac{1}{\sqrt5}\left[(1.6180)^k - (-0.6180)^k\right]$ , and $\lim_{k\to\infty} x(k+1)/x(k) = \dfrac{1+\sqrt5}{2} = 1.6180$ .

2072 Asoj · 8 marks

Consider the difference equation x(k+2)=x(k+1)+x(k) where x(0)=0, x(1)=1 and x(2)=2. Obtain the general solution x(k) in a closed form. Find the limiting value of x(k+2)/x(k) when sequence variable approaches infinity.

Answer

This is the Fibonacci equation. The data are not fully consistent: with $x(0) = 0$ and $x(1) = 1$ , the equation itself gives $x(2) = x(1) + x(0) = 1$ , not 2. So $x(0) = 0$ and $x(1) = 1$ are taken as the initial conditions. A note at the end covers the case $x(2) = 2$ .

Step 1: Take the z-transform

Use $\mathcal{Z}[x(k+2)] = z^2X(z) - z^2x(0) - zx(1)$ and $\mathcal{Z}[x(k+1)] = zX(z) - zx(0)$ , with $x(0) = 0$ and $x(1) = 1$ :

\left[z^2X(z) - z\right] - zX(z) - X(z) = 0

X(z) = \frac{z}{z^2 - z - 1}

Step 2: Factorise the denominator

z^2 - z - 1 = 0 \;\Rightarrow\; z_{1,2} = \frac{1 \pm \sqrt{5}}{2}, \quad z_1 = 1.6180,\; z_2 = -0.6180

Step 3: Partial fractions of $X(z)/z$

\frac{X(z)}{z} = \frac{1}{(z-z_1)(z-z_2)} = \frac{1}{z_1 - z_2}\left[\frac{1}{z-z_1} - \frac{1}{z-z_2}\right], \quad z_1 - z_2 = \sqrt{5}

X(z) = \frac{1}{\sqrt5}\left[\frac{z}{z - z_1} - \frac{z}{z - z_2}\right]

Step 4: General (closed-form) solution

x(k) = \frac{1}{\sqrt5}\left[\left(\frac{1+\sqrt5}{2}\right)^k - \left(\frac{1-\sqrt5}{2}\right)^k\right], \quad k = 0, 1, 2, \dots

Here $1/\sqrt5 = 0.4472$ . Check: $x(0) = 0$ , and $x(1) = \frac{1}{\sqrt5}\cdot\sqrt5 = 1$ .

First terms: $0, 1, 1, 2, 3, 5, 8, 13, 21, \dots$

Step 5: Limiting value of $x(k+2)/x(k)$

Let $z_1 = 1.6180$ and $z_2 = -0.6180$ :

\frac{x(k+2)}{x(k)} = \frac{z_1^{k+2} - z_2^{k+2}}{z_1^{k} - z_2^{k}} = \frac{z_1^2 - z_2^2\left(\dfrac{z_2}{z_1}\right)^{k}}{1 - \left(\dfrac{z_2}{z_1}\right)^{k}}

Since $|z_2/z_1| = 0.382 < 1$ , the bracketed powers go to 0:

\lim_{k\to\infty}\frac{x(k+2)}{x(k)} = z_1^2 = \left(\frac{1+\sqrt5}{2}\right)^2 = \frac{3+\sqrt5}{2} = 2.6180

Note that $z_1^2 = z_1 + 1$ (from $z^2 - z - 1 = 0$ ), so the answer is $1.6180 + 1 = 2.6180$ .

Numerical check: $x(10)/x(8) = 55/21 = 2.619$ .

Note on $x(2) = 2$

If $x(1) = 1$ and $x(2) = 2$ are taken instead, the equation forces $x(0) = 1$ . The sequence $1, 1, 2, 3, 5, \dots$ is the same series shifted by one step:

x(k) = \frac{1}{\sqrt5}\left[(1.618)^{k+1} - (-0.618)^{k+1}\right]

The limiting ratio is still $2.6180$ , because the ratio depends only on the dominant root.

Answer: $x(k) = \dfrac{1}{\sqrt5}\left[\left(\dfrac{1+\sqrt5}{2}\right)^k - \left(\dfrac{1-\sqrt5}{2}\right)^k\right]$ , and $\lim_{k\to\infty} x(k+2)/x(k) = 2.6180$ .

2078 Chaitra · 4 marks

Obtain the z-transform of x(t) for which time response is given by (assume sampling period T = 1 s). [Figure: continuous time signal x(t) rising linearly from 0 at t = 0 to 1 at t = 4 (0.5 marked on the axis), then constant at 1; time axis marked 1 to 13]

Answer

Step 1: Express $x(t)$ as ramp minus delayed ramp

The ramp has slope $\dfrac{1}{4}$ and reaches 1 at $t = 4$ s, after which it holds at 1:

x(t) = \frac{1}{4}t\cdot1(t) - \frac{1}{4}(t-4)\cdot1(t-4)

Step 2: Sampled values ( $T = 1$ s)

$k$	0	1	2	3	4	5	6	...
$x(k)$	0	0.25	0.5	0.75	1	1	1	...

Step 3: Z-transform

With $\mathcal{Z}[t] = \dfrac{Tz^{-1}}{(1-z^{-1})^2}$ ( $T = 1$ ) and delay of 4 samples $\to z^{-4}$ :

\begin{aligned} X(z) &= \frac{1}{4}\cdot\frac{z^{-1}}{(1-z^{-1})^2} - \frac{1}{4}z^{-4}\cdot\frac{z^{-1}}{(1-z^{-1})^2} \\ &= \frac{1}{4}\cdot\frac{z^{-1}(1 - z^{-4})}{(1-z^{-1})^2} \end{aligned}

Using $1 - z^{-4} = (1-z^{-1})(1+z^{-1}+z^{-2}+z^{-3})$ :

X(z) = \frac{z^{-1}(1+z^{-1}+z^{-2}+z^{-3})}{4(1-z^{-1})} = \frac{z^3+z^2+z+1}{4z^3(z-1)}

Step 4: Check

Series: $X(z) = 0.25z^{-1} + 0.5z^{-2} + 0.75z^{-3} + z^{-4} + z^{-5} + \dots$ , which matches the table.

Final value: $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1\cdot 4}{4} = 1$ , which is correct.

Answer: $X(z) = \dfrac{z^{-1}(1-z^{-4})}{4(1-z^{-1})^2} = \dfrac{z^3+z^2+z+1}{4z^3(z-1)}$ .

2073 Magh · 4 marks

Obtain the z-transform of x(t) for which time response is given by: [Figure: continuous time signal x(t) rising linearly from 0 at the origin to 1.0 at the tick labelled 6, then constant at 1.0; 0.5 marked on the x(t) axis; time-axis ticks labelled 1 (just left of the origin), 2 (at the origin), then 3 to 13]

Answer

Reading the figure: the tick labels are offset. "2" sits at the origin, so the ramp starts at $t = 0$ and reaches 1.0 four divisions later (the tick labelled 6), then stays at 1.0. Each division is taken as one sampling period, $T = 1$ s. So $x(t)$ rises linearly from 0 to 1 in 4 s and then holds at 1. This is the standard textbook signal $x(t) = \tfrac14 t - \tfrac14(t-4)1(t-4)$ .

Step 1: Express $x(t)$ as ramp minus delayed ramp

The ramp has slope $\dfrac{1}{4}$ and reaches 1 at $t = 4$ s, after which it holds at 1:

x(t) = \frac{1}{4}t\cdot1(t) - \frac{1}{4}(t-4)\cdot1(t-4)

Step 2: Sampled values ( $T = 1$ s)

$k$	0	1	2	3	4	5	6	...
$x(k)$	0	0.25	0.5	0.75	1	1	1	...

Step 3: Z-transform

With $\mathcal{Z}[t] = \dfrac{Tz^{-1}}{(1-z^{-1})^2}$ ( $T = 1$ ) and delay of 4 samples $\to z^{-4}$ :

\begin{aligned} X(z) &= \frac{1}{4}\cdot\frac{z^{-1}}{(1-z^{-1})^2} - \frac{1}{4}z^{-4}\cdot\frac{z^{-1}}{(1-z^{-1})^2} \\ &= \frac{1}{4}\cdot\frac{z^{-1}(1 - z^{-4})}{(1-z^{-1})^2} \end{aligned}

Using $1 - z^{-4} = (1-z^{-1})(1+z^{-1}+z^{-2}+z^{-3})$ :

X(z) = \frac{z^{-1}(1+z^{-1}+z^{-2}+z^{-3})}{4(1-z^{-1})} = \frac{z^3+z^2+z+1}{4z^3(z-1)}

Step 4: Check

Series: $X(z) = 0.25z^{-1} + 0.5z^{-2} + 0.75z^{-3} + z^{-4} + z^{-5} + \dots$ , which matches the table.

Final value: $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1\cdot 4}{4} = 1$ , which is correct.

If the signal were instead read as reaching 1 after $n$ sampling periods, the same method gives $X(z) = \dfrac{z^{-1}(1-z^{-n})}{n(1-z^{-1})^2}$ .

Answer: $X(z) = \dfrac{z^{-1}(1-z^{-4})}{4(1-z^{-1})^2} = \dfrac{z^3+z^2+z+1}{4z^3(z-1)}$ (with $T = 1$ s).

2076 Bhadra · 4 marks

Find the z-transform of the following: x(t)={0, t<0; 0, t≤ 4; 1, t>4}

Answer

$x(t)$ is a unit step that switches on just after $t = 4$ . No sampling period is given, so take $T = 1$ s (the usual choice for such problems).

Step 1: Sampled sequence

$x(kT) = 1$ only when $kT > 4$ . At $t = 4$ exactly the value is 0 (since $x = 0$ for $t \le 4$ ).

$k$	0	1	2	3	4	5	6	...
$x(k)$	0	0	0	0	0	1	1	...

So $x(k) = 1(k-5)$ , a unit step delayed by 5 samples.

Step 2: Z-transform from the definition

\begin{aligned} X(z) &= \sum_{k=0}^{\infty}x(k)z^{-k} = z^{-5} + z^{-6} + z^{-7} + \dots \\ &= z^{-5}\left(1 + z^{-1} + z^{-2} + \dots\right) = \frac{z^{-5}}{1 - z^{-1}}, \quad |z| > 1 \end{aligned}

The same result follows from the shift theorem: $\mathcal{Z}[1(k-5)] = z^{-5}\mathcal{Z}[1(k)]$ .

X(z) = \frac{z^{-5}}{1-z^{-1}} = \frac{1}{z^4(z-1)}

Remarks

If the step were defined as 1 for $t \ge 4$ , the sample at $k = 4$ would be 1, and $X(z) = \dfrac{z^{-4}}{1-z^{-1}}$ . The strict inequality $t > 4$ is what moves the first nonzero sample to $k = 5$ .
For a general $T$ with $4 = nT$ ( $n$ an integer), the first nonzero sample is $k = n+1$ , so $X(z) = \dfrac{z^{-(n+1)}}{1-z^{-1}}$ .

Answer ( $T = 1$ s): $X(z) = \dfrac{z^{-5}}{1 - z^{-1}} = \dfrac{1}{z^4(z-1)}$ .

2075 Baisakh · 4 marks

Find the z-transform of x(t)=1/4t-1/4(t-4)1(t-4).

Answer

$x(t)$ is a ramp of slope $\tfrac14$ minus the same ramp delayed by 4 s. It rises from 0 to 1 over 4 s and then stays at 1.

Step 1: Standard pairs

$\mathcal{Z}[t\cdot1(t)] = \dfrac{Tz^{-1}}{(1-z^{-1})^2}$
Real translation (shift) theorem: if $4 = nT$ , then $\mathcal{Z}[x(t - nT)1(t-nT)] = z^{-n}X(z)$

Step 2: General sampling period $T$ (4 an integer multiple of $T$ )

X(z) = \frac{1}{4}\cdot\frac{Tz^{-1}}{(1-z^{-1})^2}\left(1 - z^{-4/T}\right)

Step 3: With $T = 1$ s (usual case, $n = 4$ )

\begin{aligned} X(z) &= \frac{1}{4}\cdot\frac{z^{-1}(1-z^{-4})}{(1-z^{-1})^2} \\ &= \frac{z^{-1}(1+z^{-1}+z^{-2}+z^{-3})}{4(1-z^{-1})} \\ &= \frac{z^3+z^2+z+1}{4z^3(z-1)} \end{aligned}

Here $1 - z^{-4} = (1-z^{-1})(1+z^{-1}+z^{-2}+z^{-3})$ was used.

Step 4: Check with the samples

$k$	0	1	2	3	4	5	...
$x(k)$	0	0.25	0.5	0.75	1	1	...

Long division gives $X(z) = 0.25z^{-1} + 0.5z^{-2} + 0.75z^{-3} + z^{-4} + z^{-5} + \dots$ , which matches. The final value from the theorem, $\lim_{z\to1}(1-z^{-1})X(z) = 1$ , is also correct.

Answer ( $T = 1$ s): $X(z) = \dfrac{z^{-1}(1-z^{-4})}{4(1-z^{-1})^2} = \dfrac{z^3+z^2+z+1}{4z^3(z-1)}$ .

2068 Magh · 8 marks

Obtain Z transform of curve x(t) shown in figure below. [Figure: x(t) rises linearly from 0 at t = 0 through 0.5 at t = 1 to 1 at t = 2, then stays constant at 1; time axis marked 0 to 4]

Answer

From the figure, $x(t)$ rises linearly with slope $0.5$ from 0 at $t = 0$ to 1 at $t = 2$ , and then stays at 1. Take the sampling period $T = 1$ s, as marked on the axis. The general- $T$ form is also given.

Step 1: Express the curve with standard functions

x(t) = 0.5\,t\cdot1(t) - 0.5\,(t-2)\cdot1(t-2)

For $0 \le t \le 2$ : $x = 0.5t$
For $t > 2$ : $x = 0.5t - 0.5t + 1 = 1$

 x(t)
  1 |        ________________
    |       /
0.5 |     /
    |   /
  0 +--/-----+-----+-----+----> t
    0     1     2     3     4

Step 2: Sampled values ( $T = 1$ )

$k$	0	1	2	3	4	...
$x(k)$	0	0.5	1	1	1	...

Step 3: Z-transform

Use $\mathcal{Z}[t] = \dfrac{Tz^{-1}}{(1-z^{-1})^2}$ and the real translation theorem $\mathcal{Z}[x(t-nT)1(t-nT)] = z^{-n}X(z)$ , with $n = 2/T$ :

X(z) = 0.5\cdot\frac{Tz^{-1}}{(1-z^{-1})^2}\left(1 - z^{-2/T}\right)

With $T = 1$ , $n = 2$ :

\begin{aligned} X(z) &= \frac{0.5\,z^{-1}(1 - z^{-2})}{(1-z^{-1})^2} = \frac{0.5\,z^{-1}(1 + z^{-1})}{1 - z^{-1}} \\ &= \frac{0.5(z + 1)}{z(z-1)} \end{aligned}

Step 4: Verify by the definition

Directly from $X(z) = \sum x(k)z^{-k}$ :

\begin{aligned} X(z) &= 0.5z^{-1} + z^{-2} + z^{-3} + z^{-4} + \dots \\ &= 0.5z^{-1} + \frac{z^{-2}}{1-z^{-1}} = \frac{0.5z^{-1}(1 - z^{-1}) + z^{-2}}{1 - z^{-1}} = \frac{0.5z^{-1} + 0.5z^{-2}}{1-z^{-1}} \end{aligned}

This is the same as Step 3.

Step 5: Checks with the limit theorems

Initial value: $x(0) = \lim_{z\to\infty}X(z) = 0$ , which is correct.
Final value: $\lim_{z\to1}(1-z^{-1})X(z) = 0.5(1+1) = 1$ , which is correct.

Note for other $T$

If $T = 0.5$ s, then $n = 4$ and $X(z) = \dfrac{0.25z^{-1}(1 - z^{-4})}{(1-z^{-1})^2}$ , with samples $0, 0.25, 0.5, 0.75, 1, 1, \dots$

Answer ( $T = 1$ s): $X(z) = \dfrac{0.5z^{-1}(1 - z^{-2})}{(1-z^{-1})^2} = \dfrac{0.5(z+1)}{z(z-1)}$ .

2078 Chaitra · 6 marks

Consider the difference equation: x(k+2)-1.3679x(k+1)+0.3679x(k)=0.3679u(k+1)+0.2642u(k), where x(k) is the output and x(k)=0 for k≤ 0, where u(k) is the input and is given by u(k)=0, k<0; u(0)=1; u(1)=0.2142; u(2)=0.2142; u(k)=0, k=3,4,5,6,… Determine the output x(k).

Answer

The output is found by solving the equation recursively, then confirmed with a closed form from the z-transform.

Step 1: Initial values

$x(k) = 0$ for $k \le 0$ , so $x(0) = 0$ . Put $k = -1$ in the equation, with $x(-1) = 0$ , $u(-1) = 0$ , $u(0) = 1$ :

x(1) - 1.3679x(0) + 0.3679x(-1) = 0.3679u(0) + 0.2642u(-1) \;\Rightarrow\; x(1) = 0.3679

Step 2: Pulse transfer function

With zero initial conditions (the $x(1)$ and $u(0)$ terms cancel), taking z-transforms gives:

G(z) = \frac{X(z)}{U(z)} = \frac{0.3679z + 0.2642}{z^2 - 1.3679z + 0.3679} = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)}

The input is a finite sequence:

U(z) = 1 + 0.2142z^{-1} + 0.2142z^{-2} = \frac{z^2 + 0.2142z + 0.2142}{z^2}

Step 3: Recursive solution

x(k+2) = 1.3679x(k+1) - 0.3679x(k) + 0.3679u(k+1) + 0.2642u(k)

$k$	Calculation of $x(k+2)$	$x(k+2)$
0	$1.3679(0.3679) - 0 + 0.3679(0.2142) + 0.2642(1)$	0.8463
1	$1.3679(0.8463) - 0.3679(0.3679) + 0.3679(0.2142) + 0.2642(0.2142)$	1.1576
2	$1.3679(1.1576) - 0.3679(0.8463) + 0 + 0.2642(0.2142)$	1.3288
3	$1.3679(1.3288) - 0.3679(1.1576)$	1.3918
4	$1.3679(1.3918) - 0.3679(1.3288)$	1.4149
5	$1.3679(1.4149) - 0.3679(1.3918)$	1.4234

Step 4: Closed form for $k \ge 3$

$X(z) = G(z)U(z) = \dfrac{(0.3679z+0.2642)(z^2+0.2142z+0.2142)}{z^2(z-1)(z-0.3679)}$ . For $k \ge 3$ only the poles at $z = 1$ and $z = 0.3679$ contribute to the residues of $X(z)z^{k-1}$ :

\begin{aligned} \text{Res}_{z=1} &= \frac{(0.6321)(1.4284)}{1\,(0.6321)} = 1.4284 \\ \text{Res}_{z=0.3679} &= \frac{(0.3995)(0.4284)}{(0.3679)^2(-0.6321)}(0.3679)^{k-1} = -5.4375(0.3679)^k \end{aligned}

x(k) = 1.4284 - 5.4375(0.3679)^k, \quad k \ge 3

Check: $k = 3$ gives $1.4284 - 5.4375(0.04979) = 1.1576$ , which matches.

Result

$k$	0	1	2	3	4	5	6	$\infty$
$x(k)$	0	0.3679	0.8463	1.1576	1.3288	1.3918	1.4149	1.4284

The final value from the theorem is $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{0.6321(1 + 0.2142 + 0.2142)}{0.6321} = 1.4284$ .

Answer: $x(0) = 0$ , $x(1) = 0.3679$ , $x(2) = 0.8463$ , and $x(k) = 1.4284 - 5.4375(0.3679)^k$ for $k \ge 3$ . The output settles at 1.4284.

(With $u(2) = -0.2142$ , the textbook deadbeat case, the output is exactly 1 for all $k \ge 3$ .)

2071 Magh · 4 marks

Consider the difference equation x(k+2)-1.3679x(k+1)+0.3679x(k)=0.3679u(k+1)+0.2642u(k), where x(k) is the output and x(k)=0 for k≤ 0, where u(k) is the input and is given by u(k)=0, k<0; u(0)=1; u(1)=0.2142; u(2)=-0.2142; u(k)=0, k=3,4,5,6,… Determine the output x(k).

Answer

The output is found by recursion, and the z-transform then shows why it becomes constant.

Step 1: Initial values

$x(k) = 0$ for $k \le 0$ , so $x(0) = 0$ . Put $k = -1$ in the equation, with $x(-1) = 0$ , $u(-1) = 0$ , $u(0) = 1$ :

x(1) - 1.3679x(0) + 0.3679x(-1) = 0.3679u(0) + 0.2642u(-1) \;\Rightarrow\; x(1) = 0.3679

Step 2: Pulse transfer function

With zero initial conditions (the $x(1)$ and $u(0)$ terms cancel), taking z-transforms gives:

G(z) = \frac{X(z)}{U(z)} = \frac{0.3679z + 0.2642}{z^2 - 1.3679z + 0.3679} = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)}

The input is a finite sequence:

U(z) = 1 + 0.2142z^{-1} - 0.2142z^{-2} = \frac{z^2 + 0.2142z - 0.2142}{z^2}

Step 3: Recursive solution

x(k+2) = 1.3679x(k+1) - 0.3679x(k) + 0.3679u(k+1) + 0.2642u(k)

$k$	Calculation	$x(k+2)$
0	$1.3679(0.3679) + 0.3679(0.2142) + 0.2642(1)$	0.8463
1	$1.3679(0.8463) - 0.3679(0.3679) - 0.3679(0.2142) + 0.2642(0.2142)$	1.0000
2	$1.3679(1) - 0.3679(0.8463) - 0.2642(0.2142)$	1.0000
3	$1.3679(1) - 0.3679(1)$	1.0000

Step 4: Why the output stays at 1 (z-transform)

X(z) = \frac{(0.3679z + 0.2642)(z^2 + 0.2142z - 0.2142)}{z^2(z-1)(z-0.3679)}

The input numerator has a root at $z = 0.3679$ :

(0.3679)^2 + 0.2142(0.3679) - 0.2142 = 0.13535 + 0.07880 - 0.2142 \approx 0

So $z^2 + 0.2142z - 0.2142 \approx (z - 0.3679)(z + 0.5821)$ . The plant pole at $0.3679$ is cancelled:

X(z) = \frac{(0.3679z + 0.2642)(z + 0.5821)}{z^2(z-1)}

Only the pole at $z = 1$ remains, apart from poles at the origin, which give a finite transient. Expanding:

X(z) = 0.3679z^{-1} + 0.8463z^{-2} + z^{-3} + z^{-4} + \dots

Result

$k$	0	1	2	3	4	5	...
$x(k)$	0	0.3679	0.8463	1	1	1	...

This is a deadbeat response: the output reaches its final value 1 in three sampling periods and stays there with no ripple. The chosen input sequence cancels the slow plant mode.

Answer: $x(0) = 0$ , $x(1) = 0.3679$ , $x(2) = 0.8463$ , and $x(k) = 1$ for $k \ge 3$ .

2077 Chaitra · 8 marks

Consider the difference equation x(k+3)-2.2x(k+2)+1.57x(k+1)-0.36x(k)=u(k), where u(k)=1 for k≥ 0 and x(0)=x(1)=x(2)=0. Obtain the general solution x(k) in a closed form. Find the limiting value of x(k+2)/x(k) when sequence variable approaches infinity.

Answer

Solve by the z-transform method, with $u(k)$ a unit step.

Step 1: Z-transform (zero initial conditions)

With $x(0) = x(1) = x(2) = 0$ , all initial-condition terms vanish. The input transform is $U(z) = \dfrac{z}{z-1}$ :

(z^3 - 2.2z^2 + 1.57z - 0.36)X(z) = \frac{z}{z-1}

Step 2: Factorise the characteristic polynomial

Trial roots: $P(0.5) = 0.125 - 0.55 + 0.785 - 0.36 = 0$ , so $z = 0.5$ is a root. Dividing out gives $z^2 - 1.7z + 0.72 = (z-0.8)(z-0.9)$ .

z^3 - 2.2z^2 + 1.57z - 0.36 = (z-0.5)(z-0.8)(z-0.9)

X(z) = \frac{z}{(z-1)(z-0.5)(z-0.8)(z-0.9)}

Step 3: Partial fractions of $X(z)/z$

\frac{X(z)}{z} = \frac{A}{z-1} + \frac{B}{z-0.5} + \frac{C}{z-0.8} + \frac{D}{z-0.9}

\begin{aligned} A &= \frac{1}{(0.5)(0.2)(0.1)} = 100 \\ B &= \frac{1}{(-0.5)(-0.3)(-0.4)} = -16.667 \\ C &= \frac{1}{(-0.2)(0.3)(-0.1)} = 166.667 \\ D &= \frac{1}{(-0.1)(0.4)(0.1)} = -250 \end{aligned}

Step 4: Closed-form solution

x(k) = 100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k, \quad k = 0, 1, 2, \dots

The exact fractions are $-\tfrac{50}{3}$ and $\tfrac{500}{3}$ .

Check

$k$	0	1	2	3	4	5	6
Formula	0	0	0	1	3.2	6.47	10.57
Recursion	0	0	0	1	3.2	6.47	10.57

The recursion used is $x(k+3) = 2.2x(k+2) - 1.57x(k+1) + 0.36x(k) + 1$ . For example, $x(4) = 2.2(1) + 1 = 3.2$ .

The final value theorem gives $\lim_{z\to1}(z-1)X(z) = \dfrac{1}{0.5\times0.2\times0.1} = 100$ , which equals $A$ .

Step 5: Limiting value of $x(k+2)/x(k)$

All the transient terms have $|z| < 1$ , so $(0.5)^k, (0.8)^k, (0.9)^k \to 0$ :

\lim_{k\to\infty}\frac{x(k+2)}{x(k)} = \lim_{k\to\infty}\frac{100 - 16.667(0.5)^{k+2} + 166.667(0.8)^{k+2} - 250(0.9)^{k+2}}{100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k} = \frac{100}{100} = 1

The system is stable (poles 0.5, 0.8, 0.9 lie inside the unit circle), so $x(k)$ settles at 100 and successive values become equal.

Answer: $x(k) = 100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k$ , and $\lim_{k\to\infty} x(k+2)/x(k) = 1$ .

2074 Bhadra · 8 marks

Solve the following difference equation using z transform method. Also determine the value of x(k+2)/x(k+1) as k approaches infinity: x(k+3)-2.2x(k+2)+1.57x(k+1)-0.36x(k)=u(k), where u(k)=1 for all k≥ 0, and x(0)=x(1)=x(2)=0.

Answer

Solve by the z-transform method, with $u(k)$ a unit step.

Step 1: Z-transform (zero initial conditions)

With $x(0) = x(1) = x(2) = 0$ , all initial-condition terms vanish. The input transform is $U(z) = \dfrac{z}{z-1}$ :

(z^3 - 2.2z^2 + 1.57z - 0.36)X(z) = \frac{z}{z-1}

Step 2: Factorise the characteristic polynomial

Trial roots: $P(0.5) = 0.125 - 0.55 + 0.785 - 0.36 = 0$ , so $z = 0.5$ is a root. Dividing out gives $z^2 - 1.7z + 0.72 = (z-0.8)(z-0.9)$ .

z^3 - 2.2z^2 + 1.57z - 0.36 = (z-0.5)(z-0.8)(z-0.9)

X(z) = \frac{z}{(z-1)(z-0.5)(z-0.8)(z-0.9)}

Step 3: Partial fractions of $X(z)/z$

\frac{X(z)}{z} = \frac{A}{z-1} + \frac{B}{z-0.5} + \frac{C}{z-0.8} + \frac{D}{z-0.9}

\begin{aligned} A &= \frac{1}{(0.5)(0.2)(0.1)} = 100 \\ B &= \frac{1}{(-0.5)(-0.3)(-0.4)} = -16.667 \\ C &= \frac{1}{(-0.2)(0.3)(-0.1)} = 166.667 \\ D &= \frac{1}{(-0.1)(0.4)(0.1)} = -250 \end{aligned}

Step 4: Closed-form solution

x(k) = 100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k, \quad k = 0, 1, 2, \dots

The exact fractions are $-\tfrac{50}{3}$ and $\tfrac{500}{3}$ .

Check

$k$	0	1	2	3	4	5	6
Formula	0	0	0	1	3.2	6.47	10.57
Recursion	0	0	0	1	3.2	6.47	10.57

The recursion used is $x(k+3) = 2.2x(k+2) - 1.57x(k+1) + 0.36x(k) + 1$ . For example, $x(4) = 2.2(1) + 1 = 3.2$ .

The final value theorem gives $\lim_{z\to1}(z-1)X(z) = \dfrac{1}{0.5\times0.2\times0.1} = 100$ , which equals $A$ .

Step 5: Limiting value of $x(k+2)/x(k+1)$

The poles $0.5$ , $0.8$ and $0.9$ are inside the unit circle, so every transient term dies out and $x(k) \to 100$ :

\lim_{k\to\infty}\frac{x(k+2)}{x(k+1)} = \frac{100 + 0}{100 + 0} = 1

The ratio approaches 1 slowly, because the slowest mode $(0.9)^k$ dominates the transient.

Answer: $x(k) = 100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k$ , and $\lim_{k\to\infty} x(k+2)/x(k+1) = 1$ .

2071 Magh · 4 marks

Obtain the z-transform of x(t)={cos ωt, t≥ 0; 0, t<0}

Answer

The z-transform of the sampled cosine is

X(z) = \frac{1 - z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} = \frac{z(z - \cos\omega T)}{z^2 - 2z\cos\omega T + 1}

The derivation follows.

Step 1: Sample the signal

With sampling period $T$ , $x(kT) = \cos\omega kT$ for $k \ge 0$ . Write it with Euler's formula:

\cos\omega kT = \frac{1}{2}\left(e^{j\omega kT} + e^{-j\omega kT}\right)

Step 2: Use the exponential pair

\mathcal{Z}\left[e^{-akT}\right] = \sum_{k=0}^{\infty}e^{-akT}z^{-k} = \frac{1}{1 - e^{-aT}z^{-1}}, \quad |z| > |e^{-aT}|

With $a = \mp j\omega$ :

X(z) = \frac{1}{2}\left[\frac{1}{1 - e^{j\omega T}z^{-1}} + \frac{1}{1 - e^{-j\omega T}z^{-1}}\right]

Step 3: Combine the fractions

\begin{aligned} X(z) &= \frac{1}{2}\cdot\frac{(1 - e^{-j\omega T}z^{-1}) + (1 - e^{j\omega T}z^{-1})}{1 - (e^{j\omega T} + e^{-j\omega T})z^{-1} + z^{-2}} \\ &= \frac{1}{2}\cdot\frac{2 - 2z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} \\ &= \frac{1 - z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} \end{aligned}

Here $e^{j\omega T} + e^{-j\omega T} = 2\cos\omega T$ was used.

Multiplying numerator and denominator by $z^2$ :

X(z) = \frac{z(z - \cos\omega T)}{z^2 - 2z\cos\omega T + 1}, \quad |z| > 1

Remarks

The poles are at $z = e^{\pm j\omega T}$ , on the unit circle. This is expected for an undamped oscillation.
Check: $x(0) = \lim_{z\to\infty}X(z) = 1 = \cos 0$ .
In the same way, $\mathcal{Z}[\sin\omega kT] = \dfrac{z\sin\omega T}{z^2 - 2z\cos\omega T + 1}$ .

Answer: $X(z) = \dfrac{1 - z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} = \dfrac{z(z-\cos\omega T)}{z^2 - 2z\cos\omega T + 1}$ .

2071 Magh · 4 marks

Obtain the z-transform of X(s)=1/s(s+1).

Answer

Expand $X(s)$ in partial fractions, take the z-transform of each sampled time function, and combine.

Step 1: Partial fractions

X(s) = \frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1}

Step 2: Time function and its samples

x(t) = 1 - e^{-t}, \quad t \ge 0 \;\Rightarrow\; x(kT) = 1 - e^{-kT}

Step 3: Z-transform of each term

$X(s)$	$x(kT)$	$X(z)$
$\dfrac{1}{s}$	$1$	$\dfrac{1}{1-z^{-1}}$
$\dfrac{1}{s+1}$	$e^{-kT}$	$\dfrac{1}{1-e^{-T}z^{-1}}$

\begin{aligned} X(z) &= \frac{1}{1 - z^{-1}} - \frac{1}{1 - e^{-T}z^{-1}} \\ &= \frac{(1 - e^{-T}z^{-1}) - (1 - z^{-1})}{(1-z^{-1})(1-e^{-T}z^{-1})} \\ &= \frac{(1 - e^{-T})z^{-1}}{(1-z^{-1})(1-e^{-T}z^{-1})} \end{aligned}

In positive powers of $z$ :

X(z) = \frac{(1 - e^{-T})z}{(z-1)(z - e^{-T})}

Numerical form for $T = 1$ s

$e^{-1} = 0.3679$ , so

X(z) = \frac{0.6321z}{(z-1)(z-0.3679)} = \frac{0.6321z}{z^2 - 1.3679z + 0.3679}

Check

Initial value: $\lim_{z\to\infty}X(z) = 0 = x(0)$ .
Final value: $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1-e^{-T}}{1-e^{-T}} = 1 = x(\infty)$ .

Answer: $X(z) = \dfrac{(1-e^{-T})z^{-1}}{(1-z^{-1})(1-e^{-T}z^{-1})} = \dfrac{(1-e^{-T})z}{(z-1)(z-e^{-T})}$ .

2076 Bhadra · 4 marks

By using inversion integral method, obtain inverse Z-transform of X(z)=z⁻¹(1-z⁻²)/(1+z⁻¹)².

Answer

In the inversion integral method, $x(k)$ is the sum of the residues of $X(z)z^{k-1}$ at all its poles inside the contour (which encloses every pole):

x(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum \text{residues of } X(z)z^{k-1}

Step 1: Simplify $X(z)$

$1 - z^{-2} = (1 - z^{-1})(1 + z^{-1})$ , so

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

X(z)z^{k-1} = \frac{(z-1)z^{k-2}}{z+1}

The pole at $z = 0$ exists only for $k = 0$ and $k = 1$ , so those cases are handled separately.

Case $k = 0$ : $\dfrac{z-1}{z^2(z+1)}$

Double pole at $z = 0$ : $\dfrac{d}{dz}\left[\dfrac{z-1}{z+1}\right]_{z=0} = \left[\dfrac{2}{(z+1)^2}\right]_{z=0} = 2$
Pole at $z = -1$ : $\left[\dfrac{z-1}{z^2}\right]_{z=-1} = -2$

x(0) = 2 - 2 = 0

Case $k = 1$ : $\dfrac{z-1}{z(z+1)}$

Pole at $z = 0$ : $\left[\dfrac{z-1}{z+1}\right]_{z=0} = -1$
Pole at $z = -1$ : $\left[\dfrac{z-1}{z}\right]_{z=-1} = 2$

x(1) = -1 + 2 = 1

Case $k \ge 2$ : only the pole at $z = -1$

x(k) = \left[(z-1)z^{k-2}\right]_{z=-1} = (-2)(-1)^{k-2} = -2(-1)^k

Result

x(k) = \begin{cases} 0, & k = 0 \\ 1, & k = 1 \\ -2(-1)^k, & k \ge 2 \end{cases}

$k$	0	1	2	3	4	5
$x(k)$	0	1	−2	2	−2	2

Check by long division

$\dfrac{z^{-1}(1-z^{-1})}{1+z^{-1}} = z^{-1}(1 - 2z^{-1} + 2z^{-2} - 2z^{-3} + \dots) = z^{-1} - 2z^{-2} + 2z^{-3} - \dots$ , which matches.

The pole at $z = -1$ lies on the unit circle, so the sequence keeps oscillating between $\pm2$ .

Answer: $x(0) = 0$ , $x(1) = 1$ , $x(k) = -2(-1)^k$ for $k \ge 2$ .

2075 Bhadra · 4 marks

Find the region of convergence for x(k)=-aᵏu(-k-1), where u(k) is the unit step function.

Answer

$x(k) = -a^k u(-k-1)$ is a left-sided (anti-causal) sequence. It is nonzero only for $k \le -1$ , where it equals $-a^k$ . Its z-transform is $\dfrac{z}{z-a}$ with ROC $|z| < |a|$ .

Step 1: Apply the definition

X(z) = \sum_{k=-\infty}^{\infty}x(k)z^{-k} = -\sum_{k=-\infty}^{-1}a^k z^{-k}

Put $m = -k$ , so $m$ runs from 1 to $\infty$ :

X(z) = -\sum_{m=1}^{\infty}a^{-m}z^{m} = -\sum_{m=1}^{\infty}\left(\frac{z}{a}\right)^m

Step 2: Sum the geometric series

The series converges only when $\left|\dfrac{z}{a}\right| < 1$ , i.e. $|z| < |a|$ . Then

X(z) = -\frac{z/a}{1 - z/a} = -\frac{z}{a - z} = \frac{z}{z-a} = \frac{1}{1 - az^{-1}}

Step 3: Region of convergence

\text{ROC: } |z| < |a|

This is the inside of a circle of radius $|a|$ , centred at the origin. The pole at $z = a$ lies on the outer boundary.

           Im z
            |
        .---+---.
      /     |     \
     |  ROC: |z|<|a|
  ---+------0------x---- Re z
     |      |      a  (pole on
      \     |     /    boundary)
        '---+---'

Comparison with the right-sided sequence

Sequence	$X(z)$	ROC
$a^k u(k)$ (causal)	$\dfrac{z}{z-a}$	$\lvert z\rvert > \lvert a\rvert$
$-a^k u(-k-1)$ (anti-causal)	$\dfrac{z}{z-a}$	$\lvert z\rvert < \lvert a\rvert$

Both sequences have the same algebraic $X(z)$ . Only the ROC tells them apart, so an ROC must always be stated with a z-transform. For a left-sided sequence the ROC is always the inside of a circle bounded by the innermost pole.

Answer: $X(z) = \dfrac{z}{z-a}$ , ROC $|z| < |a|$ .

2071 Magh · 4 marks

Find ROC for x(k)=(1/2)ᵏ u(k)+(-1/3)ᵏ u(k), where u(k)=2 for t≥ 0 and u(k)=0 for t<0.

Answer

Both terms are right-sided (causal) exponentials. Each converges outside a circle, so the ROC of the sum is the overlap: $|z| > \tfrac12$ .

$u(k)$ is taken as the unit step. The printed " $u(k) = 2$ " would only multiply $X(z)$ by 2 and does not change the ROC.

Step 1: First term

\mathcal{Z}\left[\left(\tfrac12\right)^k u(k)\right] = \sum_{k=0}^{\infty}\left(\tfrac12 z^{-1}\right)^k = \frac{1}{1 - \frac12 z^{-1}} = \frac{z}{z - \frac12}

This converges when $\left|\tfrac12 z^{-1}\right| < 1$ , i.e. $|z| > \tfrac12$ .

Step 2: Second term

\mathcal{Z}\left[\left(-\tfrac13\right)^k u(k)\right] = \sum_{k=0}^{\infty}\left(-\tfrac13 z^{-1}\right)^k = \frac{z}{z + \frac13}

This converges when $|z| > \tfrac13$ .

Step 3: Combine

\begin{aligned} X(z) &= \frac{z}{z - \frac12} + \frac{z}{z + \frac13} = \frac{z\left(z + \frac13\right) + z\left(z - \frac12\right)}{\left(z - \frac12\right)\left(z + \frac13\right)} \\ &= \frac{2z\left(z - \frac{1}{12}\right)}{\left(z - \frac12\right)\left(z + \frac13\right)} \end{aligned}

Poles: $z = \tfrac12$ , $z = -\tfrac13$
Zeros: $z = 0$ , $z = \tfrac{1}{12}$

Step 4: ROC of the sum

The ROC is the intersection of the two ROCs:

\{|z| > \tfrac12\} \cap \{|z| > \tfrac13\} = \{|z| > \tfrac12\}

            Im z
             |   ROC: outside
        .----+----.   circle r=1/2
       /     |     \
  ----x------o--x---+---- Re z
     -1/3    0  1/2  1
       \     |     /
        '----+----'

Remarks

For a causal sequence the ROC lies outside the circle through the outermost pole ( $|z| = \tfrac12$ here).
The ROC contains the unit circle, so the sequence is absolutely summable (stable).

Answer: $X(z) = \dfrac{z}{z-\frac12} + \dfrac{z}{z+\frac13}$ , ROC $|z| > \tfrac12$ .

2075 Bhadra · 1+3 marks

State and prove final value theorem of z-transform.

Answer

Statement

Let $x(k) = 0$ for $k < 0$ , with z-transform $X(z)$ . Suppose all poles of $X(z)$ lie inside the unit circle, except possibly a simple pole at $z = 1$ . Then

\lim_{k\to\infty}x(k) = \lim_{z\to1}\left[(1 - z^{-1})X(z)\right] = \lim_{z\to1}\left[\frac{z-1}{z}X(z)\right]

This condition means $(1 - z^{-1})X(z)$ has no poles on or outside the unit circle, so that $x(k)$ actually settles to a final value.

Proof

Consider the sequences $x(k)$ and $x(k-1)$ . By definition and by the shift theorem:

\mathcal{Z}[x(k)] = \sum_{k=0}^{\infty}x(k)z^{-k} = X(z), \qquad \mathcal{Z}[x(k-1)] = \sum_{k=0}^{\infty}x(k-1)z^{-k} = z^{-1}X(z)

Subtract the two:

\sum_{k=0}^{\infty}x(k)z^{-k} - \sum_{k=0}^{\infty}x(k-1)z^{-k} = X(z) - z^{-1}X(z) = (1 - z^{-1})X(z)

Let $z \to 1$ . Then $z^{-k} \to 1$ :

\lim_{z\to1}(1 - z^{-1})X(z) = \sum_{k=0}^{\infty}\left[x(k) - x(k-1)\right]

The right side is a telescoping sum. Write out the partial sum up to $k = n$ :

\begin{aligned} \sum_{k=0}^{n}\left[x(k) - x(k-1)\right] &= [x(0) - x(-1)] + [x(1) - x(0)] + \dots + [x(n) - x(n-1)] \\ &= x(n) - x(-1) = x(n) \end{aligned}

since $x(-1) = 0$ . Taking $n \to \infty$ :

\lim_{z\to1}(1 - z^{-1})X(z) = \lim_{n\to\infty}x(n)

This proves the theorem.

Example

For $X(z) = \dfrac{0.6321z}{(z-1)(z-0.3679)}$ , the only pole outside $|z| < 1$ is the simple pole at $z = 1$ . So:

x(\infty) = \lim_{z\to1}\frac{z-1}{z}\cdot\frac{0.6321z}{(z-1)(z-0.3679)} = \frac{0.6321}{0.6321} = 1

Use: finding steady-state errors of digital control systems without inverting $X(z)$ .

2068 Magh · 8 marks

Solve the following difference equation: x(K+2)-1.379x(K+1)+0.3679x(K)=0.3679u(K+1)+0.2642u(K), where u(K) is input which has initial values x(0)=0, x(1)=0.3679, u(K)=1 for K≥ 0.

Answer

Assumption: the coefficient "1.379" is read as 1.3679. This is the standard textbook equation, where $1.3679 = 1 + e^{-1}$ and $0.3679 = e^{-1}$ , and it agrees with the given $x(1) = 0.3679$ . A note at the end gives the values for 1.379 as printed.

Step 1: Z-transform with initial conditions

Use $\mathcal{Z}[x(k+2)] = z^2X - z^2x(0) - zx(1)$ , $\mathcal{Z}[x(k+1)] = zX - zx(0)$ , and $\mathcal{Z}[u(k+1)] = zU - zu(0)$ . Substitute $x(0) = 0$ , $x(1) = 0.3679$ , $u(0) = 1$ :

z^2X - 0.3679z - 1.3679zX + 0.3679X = 0.3679zU - 0.3679z + 0.2642U

The $-0.3679z$ terms on both sides cancel:

X(z)\,(z^2 - 1.3679z + 0.3679) = (0.3679z + 0.2642)\,U(z)

Step 2: Substitute the step input $U(z) = \dfrac{z}{z-1}$

$z^2 - 1.3679z + 0.3679 = (z-1)(z-0.3679)$ , so

X(z) = \frac{z(0.3679z + 0.2642)}{(z-1)^2(z-0.3679)}

Step 3: Partial fractions of $X(z)/z$

\frac{X(z)}{z} = \frac{0.3679z + 0.2642}{(z-1)^2(z-0.3679)} = \frac{A}{(z-1)^2} + \frac{B}{z-1} + \frac{C}{z-0.3679}

\begin{aligned} A &= \left[\frac{0.3679z + 0.2642}{z - 0.3679}\right]_{z=1} = \frac{0.6321}{0.6321} = 1 \\ C &= \left[\frac{0.3679z + 0.2642}{(z-1)^2}\right]_{z=0.3679} = \frac{0.1354 + 0.2642}{(0.6321)^2} = \frac{0.3996}{0.3996} = 1 \\ B &= \frac{d}{dz}\left[\frac{0.3679z + 0.2642}{z - 0.3679}\right]_{z=1} = \left[\frac{0.3679(z-0.3679) - (0.3679z+0.2642)}{(z-0.3679)^2}\right]_{z=1} = \frac{-0.3996}{0.3996} = -1 \end{aligned}

X(z) = \frac{z}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z - 0.3679}

Step 4: Inverse transform

Using $\dfrac{z}{(z-1)^2} \to k$ , $\dfrac{z}{z-1} \to 1$ , $\dfrac{z}{z-e^{-1}} \to e^{-k}$ :

x(k) = k - 1 + (0.3679)^k = k - 1 + e^{-k}, \quad k = 0, 1, 2, \dots

Step 5: Check

$k$	0	1	2	3	4	5
Formula	0	0.3679	1.1353	2.0498	3.0183	4.0067
Recursion	0	0.3679	1.1354	2.0498	3.0183	4.0067

The recursion is $x(k+2) = 1.3679x(k+1) - 0.3679x(k) + 0.6321$ . The two agree to rounding. The double pole at $z = 1$ makes the output grow like a ramp ( $x(k) \approx k - 1$ for large $k$ ). This is the step response of an open-loop system containing an integrator.

Note: with 1.379 as printed

The recursion $x(k+2) = 1.379x(k+1) - 0.3679x(k) + 0.6321$ gives $0, 0.3679, 1.1394, 2.0680, 3.0647, 4.0975, \dots$ . The roots of $z^2 - 1.379z + 0.3679$ are then $z = 1.0174$ and $0.3616$ , so the output grows slightly faster than a ramp.

Answer (1.3679): $x(k) = k - 1 + e^{-k} = k - 1 + (0.3679)^k$ for $k \ge 0$ .

2074 Bhadra · 4 marks

Obtain X(z) by the use of convolution integral in the left half of s-plane of the transfer function X(s)=1/s²(s+1).

Answer

Method: convolution integral in the left half-plane

When the denominator of $X(s)$ has at least two more poles than the numerator has zeros, the z-transform of the sampled $x(t)$ is the sum of residues at the poles of $X(s)$ :

X(z) = \sum_{\text{poles of } X(s)} \text{Res}\left[X(s)\,\frac{z}{z - e^{sT}}\right]

For a pole of order $m$ at $s = s_i$ :

\text{Res} = \frac{1}{(m-1)!}\lim_{s\to s_i}\frac{d^{m-1}}{ds^{m-1}}\left[(s-s_i)^m X(s)\frac{z}{z-e^{sT}}\right]

Apply to $X(s) = \dfrac{1}{s^2(s+1)}$

Poles: a double pole at $s = 0$ and a simple pole at $s = -1$ .

Simple pole $s = -1$ :

R_1 = \left[\frac{1}{s^2}\cdot\frac{z}{z - e^{sT}}\right]_{s=-1} = \frac{z}{z - e^{-T}}

Double pole $s = 0$ :

\begin{aligned} R_2 &= \frac{d}{ds}\left[\frac{1}{s+1}\cdot\frac{z}{z - e^{sT}}\right]_{s=0} \\ &= \left[-\frac{1}{(s+1)^2}\cdot\frac{z}{z-e^{sT}} + \frac{1}{s+1}\cdot\frac{zTe^{sT}}{(z-e^{sT})^2}\right]_{s=0} \\ &= -\frac{z}{z-1} + \frac{Tz}{(z-1)^2} \end{aligned}

Sum of residues

X(z) = \frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z - e^{-T}}

Over the common denominator:

X(z) = \frac{z\left[(T - 1 + e^{-T})z + (1 - e^{-T} - Te^{-T})\right]}{(z-1)^2(z - e^{-T})}

Check

$x(t) = t - 1 + e^{-t}$ (from partial fractions $\frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1}$ ). Its z-transform is $\frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}$ , which is the same.

For $T = 1$ s: $X(z) = \dfrac{z(0.3679z + 0.2642)}{(z-1)^2(z-0.3679)}$ .

Answer: $X(z) = \dfrac{Tz}{(z-1)^2} - \dfrac{z}{z-1} + \dfrac{z}{z-e^{-T}} = \dfrac{z[(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})]}{(z-1)^2(z-e^{-T})}$ .

2068 Magh · 8 marks

What do you mean by convolution? Obtain Z-transform of X(s)=1/s²(s+1) by use of the convolution integral in the left half s-plane.

Answer

Convolution

Convolution combines two signals to give the response of a linear time-invariant system. If $g$ is the impulse response and $x$ is the input, the output is the convolution $y = g * x$ .

Continuous time: $y(t) = \displaystyle\int_0^t g(t-\tau)\,x(\tau)\,d\tau$
Discrete time (convolution sum): $y(k) = \displaystyle\sum_{h=0}^{k} g(k-h)\,x(h)$

The key property is that convolution in time becomes multiplication in the transform domain: $Y(s) = G(s)X(s)$ and $Y(z) = G(z)X(z)$ . This is why pulse transfer functions are useful.

Convolution integral for the z-transform

The impulse-sampled signal is $x^*(t) = x(t)\sum_k\delta(t-kT)$ , which is a product in time. Its Laplace transform is therefore a complex convolution of $X(s)$ with the transform of the impulse train, $\dfrac{1}{1 - e^{-Ts}}$ :

X^*(s) = \frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty}X(p)\,\frac{1}{1 - e^{-T(s-p)}}\,dp

The contour is closed in the left half of the p-plane, which encloses the poles of $X(p)$ . This is valid when $X(s)$ has at least two more poles than zeros. Setting $e^{Ts} = z$ gives:

X(z) = \sum_{\text{poles of } X(s)}\text{Res}\left[X(s)\frac{z}{z-e^{sT}}\right]

Apply to $X(s) = \dfrac{1}{s^2(s+1)}$

Poles: a double pole at $s = 0$ and a simple pole at $s = -1$ .

Simple pole $s = -1$ :

R_1 = \left[\frac{1}{s^2}\cdot\frac{z}{z - e^{sT}}\right]_{s=-1} = \frac{z}{z - e^{-T}}

Double pole $s = 0$ :

\begin{aligned} R_2 &= \frac{d}{ds}\left[\frac{1}{s+1}\cdot\frac{z}{z - e^{sT}}\right]_{s=0} \\ &= \left[-\frac{1}{(s+1)^2}\cdot\frac{z}{z-e^{sT}} + \frac{1}{s+1}\cdot\frac{zTe^{sT}}{(z-e^{sT})^2}\right]_{s=0} \\ &= -\frac{z}{z-1} + \frac{Tz}{(z-1)^2} \end{aligned}

Sum of residues

X(z) = \frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z - e^{-T}}

Over the common denominator:

X(z) = \frac{z\left[(T - 1 + e^{-T})z + (1 - e^{-T} - Te^{-T})\right]}{(z-1)^2(z - e^{-T})}

Check

$x(t) = t - 1 + e^{-t}$ (from partial fractions $\frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1}$ ). Its z-transform is $\frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}$ , which is the same.

For $T = 1$ s: $X(z) = \dfrac{z(0.3679z + 0.2642)}{(z-1)^2(z-0.3679)}$ .

Answer: $X(z) = \dfrac{z[(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})]}{(z-1)^2(z-e^{-T})}$ .

2067 Mangsir · 8 marks

Given X(s)=1/((s+1)²(s+3)(s+2)) and T=1. Obtain X(z) by using convolution integral in left half plane.

Answer

Use the convolution integral closed in the left half-plane. The z-transform is the sum of residues of $X(s)\dfrac{z}{z-e^{sT}}$ at the poles of $X(s)$ . This is valid because $X(s)$ has 4 poles and no zeros.

X(z) = \sum_{\text{poles of } X(s)}\text{Res}\left[\frac{1}{(s+1)^2(s+2)(s+3)}\cdot\frac{z}{z - e^{sT}}\right]

Poles: a double pole at $s = -1$ , and simple poles at $s = -2$ and $s = -3$ .

Residue at $s = -2$

R_2 = \left[\frac{1}{(s+1)^2(s+3)}\cdot\frac{z}{z-e^{sT}}\right]_{s=-2} = \frac{1}{(1)(1)}\cdot\frac{z}{z - e^{-2T}} = \frac{z}{z - e^{-2T}}

Residue at $s = -3$

R_3 = \left[\frac{1}{(s+1)^2(s+2)}\cdot\frac{z}{z-e^{sT}}\right]_{s=-3} = \frac{1}{(4)(-1)}\cdot\frac{z}{z - e^{-3T}} = -\frac{1}{4}\cdot\frac{z}{z - e^{-3T}}

Residue at the double pole $s = -1$

Let $F(s) = \dfrac{1}{(s+2)(s+3)}$ , so $F(-1) = \dfrac12$ and

F'(s) = -\frac{2s+5}{(s+2)^2(s+3)^2} \;\Rightarrow\; F'(-1) = -\frac{3}{4}

\begin{aligned} R_1 &= \frac{d}{ds}\left[F(s)\frac{z}{z-e^{sT}}\right]_{s=-1} = F'(-1)\frac{z}{z-e^{-T}} + F(-1)\frac{Tze^{-T}}{(z-e^{-T})^2} \\ &= -\frac{3}{4}\cdot\frac{z}{z-e^{-T}} + \frac{T e^{-T}}{2}\cdot\frac{z}{(z-e^{-T})^2} \end{aligned}

General result

X(z) = \frac{Te^{-T}}{2}\cdot\frac{z}{(z-e^{-T})^2} - \frac{3}{4}\cdot\frac{z}{z-e^{-T}} + \frac{z}{z-e^{-2T}} - \frac{1}{4}\cdot\frac{z}{z-e^{-3T}}

Substitute $T = 1$

$e^{-1} = 0.36788$ , $e^{-2} = 0.13534$ , $e^{-3} = 0.04979$ , and $\frac{Te^{-T}}{2} = 0.18394$ :

X(z) = \frac{0.18394z}{(z-0.36788)^2} - \frac{0.75z}{z-0.36788} + \frac{z}{z-0.13534} - \frac{0.25z}{z-0.04979}

Combined into a single ratio:

X(z) = \frac{0.030919z^3 + 0.023057z^2 + 0.000939z}{z^4 - 0.92088z^3 + 0.27828z^2 - 0.030011z + 0.000912}

The denominator is $(z-0.36788)^2(z-0.13534)(z-0.04979)$ .

Check

The inverse Laplace transform is $x(t) = \tfrac14\left[(2t-3)e^{-t} + 4e^{-2t} - e^{-3t}\right]$ . Its samples are:

$k$	0	1	2	3
$x(k)$	0	0.03092	0.05153	0.03979

Long division of $X(z)$ gives $0.03092z^{-1} + 0.05153z^{-2} + 0.03979z^{-3} + \dots$ , which matches.

Answer: $X(z) = \dfrac{0.18394z}{(z-0.3679)^2} - \dfrac{0.75z}{z-0.3679} + \dfrac{z}{z-0.1353} - \dfrac{0.25z}{z-0.0498}$ .

2073 Bhadra · 4 marks

Obtain the inverse transform of the function X(z)=-3.894z/(z²+0.6065).

Answer

The denominator has complex poles on the imaginary axis. So match $X(z)$ to the damped-sine pair:

\mathcal{Z}\left[r^k\sin k\Omega\right] = \frac{rz\sin\Omega}{z^2 - 2rz\cos\Omega + r^2}

Step 1: Find the poles

z^2 + 0.6065 = 0 \;\Rightarrow\; z = \pm j\sqrt{0.6065} = \pm j\,0.7788

So $r = 0.7788$ and $\Omega = \dfrac{\pi}{2}$ . Note that $0.7788 = e^{-0.25}$ .

Step 2: Match coefficients

With $\Omega = \pi/2$ : $\cos\Omega = 0$ and $\sin\Omega = 1$ . The pair becomes

\mathcal{Z}\left[r^k\sin\frac{k\pi}{2}\right] = \frac{rz}{z^2 + r^2} = \frac{0.7788z}{z^2 + 0.6065}

Rewrite $X(z)$ :

X(z) = -\frac{3.894}{0.7788}\cdot\frac{0.7788z}{z^2 + 0.6065} = -5.0\cdot\frac{0.7788z}{z^2+0.6065}

since $3.894/0.7788 = 5.000$ .

Step 3: Inverse transform

x(k) = -5\,(0.7788)^k\sin\frac{k\pi}{2} = -5\,e^{-0.25k}\sin\frac{k\pi}{2}, \quad k \ge 0

Step 4: Check by long division

$X(z) = \dfrac{-3.894z^{-1}}{1 + 0.6065z^{-2}} = -3.894z^{-1} + 2.3617z^{-3} - 1.4324z^{-5} + 0.8687z^{-7} - \dots$

$k$	0	1	2	3	4	5	6	7
$x(k)$	0	−3.894	0	2.362	0	−1.432	0	0.869

The formula gives the same values; for example, $k = 3$ : $-5(0.4724)(-1) = 2.362$ .

The poles have magnitude $0.7788 < 1$ , so the oscillation decays.

Answer: $x(k) = -5(0.7788)^k\sin\dfrac{k\pi}{2}$ , i.e. $0, -3.894, 0, 2.362, 0, -1.432, \dots$

2072 Asoj · 6 marks

Find x(k) for k=0,1,2,3,4 when X(z) is given by X(z)=(10z+5)/((z-1)(z-2)).

Answer

The first few values are found by direct (long) division. A closed form from partial fractions confirms them.

Method 1: Long division

X(z) = \frac{10z + 5}{(z-1)(z-2)} = \frac{10z + 5}{z^2 - 3z + 2} = \frac{10z^{-1} + 5z^{-2}}{1 - 3z^{-1} + 2z^{-2}}

Divide $10z^{-1} + 5z^{-2}$ by $1 - 3z^{-1} + 2z^{-2}$ :

Step	Quotient term	Remainder after subtraction
1	$10z^{-1}$	$35z^{-2} - 20z^{-3}$
2	$35z^{-2}$	$85z^{-3} - 70z^{-4}$
3	$85z^{-3}$	$185z^{-4} - 170z^{-5}$
4	$185z^{-4}$	$385z^{-5} - 370z^{-6}$

X(z) = 0 + 10z^{-1} + 35z^{-2} + 85z^{-3} + 185z^{-4} + \dots

Equivalently, the recursion $x(k) = 3x(k-1) - 2x(k-2)$ holds for $k \ge 3$ : for example $x(3) = 3(35) - 2(10) = 85$ and $x(4) = 3(85) - 2(35) = 185$ .

Method 2: Closed form (check)

\frac{X(z)}{z} = \frac{10z+5}{z(z-1)(z-2)} = \frac{A}{z} + \frac{B}{z-1} + \frac{C}{z-2}

A = \frac{5}{(-1)(-2)} = 2.5, \quad B = \frac{15}{(1)(-1)} = -15, \quad C = \frac{25}{(2)(1)} = 12.5

X(z) = 2.5 - \frac{15z}{z-1} + \frac{12.5z}{z-2}

x(k) = 2.5\,\delta(k) - 15 + 12.5(2)^k

$x(0) = 2.5 - 15 + 12.5 = 0$
$x(1) = -15 + 25 = 10$
$x(2) = -15 + 50 = 35$
$x(3) = -15 + 100 = 85$
$x(4) = -15 + 200 = 185$

Result

$k$	0	1	2	3	4
$x(k)$	0	10	35	85	185

The pole at $z = 2$ lies outside the unit circle, so $x(k)$ grows without bound.

Answer: $x(0) = 0$ , $x(1) = 10$ , $x(2) = 35$ , $x(3) = 85$ , $x(4) = 185$ . In general $x(k) = -15 + 12.5(2)^k$ for $k \ge 1$ .

2072 Magh · 4 marks

State Complex Translation theorem. Obtain the z-transform of e⁻ᵃᵗsin ωt by using Complex Translation theorem.

Answer

Complex translation theorem

If $\mathcal{Z}[x(t)] = X(z)$ , then

\mathcal{Z}\left[e^{-at}x(t)\right] = X\!\left(ze^{aT}\right)

So multiplying a time function by $e^{-at}$ is the same as replacing $z$ by $ze^{aT}$ in its z-transform.

Proof:

\mathcal{Z}\left[e^{-at}x(t)\right] = \sum_{k=0}^{\infty}x(kT)e^{-akT}z^{-k} = \sum_{k=0}^{\infty}x(kT)\left(ze^{aT}\right)^{-k} = X\!\left(ze^{aT}\right)

Z-transform of $\sin\omega t$

Write $\sin\omega kT = \dfrac{e^{j\omega kT} - e^{-j\omega kT}}{2j}$ :

\begin{aligned} \mathcal{Z}[\sin\omega t] &= \frac{1}{2j}\left[\frac{1}{1 - e^{j\omega T}z^{-1}} - \frac{1}{1 - e^{-j\omega T}z^{-1}}\right] \\ &= \frac{1}{2j}\cdot\frac{(e^{j\omega T} - e^{-j\omega T})z^{-1}}{1 - 2z^{-1}\cos\omega T + z^{-2}} = \frac{z^{-1}\sin\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} \end{aligned}

Apply the theorem

Replace $z$ by $ze^{aT}$ , so that $z^{-1}$ becomes $e^{-aT}z^{-1}$ and $z^{-2}$ becomes $e^{-2aT}z^{-2}$ :

\mathcal{Z}\left[e^{-at}\sin\omega t\right] = \frac{e^{-aT}z^{-1}\sin\omega T}{1 - 2e^{-aT}z^{-1}\cos\omega T + e^{-2aT}z^{-2}}

In positive powers of $z$ :

\mathcal{Z}\left[e^{-at}\sin\omega t\right] = \frac{ze^{-aT}\sin\omega T}{z^2 - 2ze^{-aT}\cos\omega T + e^{-2aT}}

The poles are at $z = e^{-aT}e^{\pm j\omega T}$ , inside the unit circle for $a > 0$ .

Answer: $\dfrac{ze^{-aT}\sin\omega T}{z^2 - 2ze^{-aT}\cos\omega T + e^{-2aT}}$ .

2070 Magh · 4 marks

Find the z-transform of x(t)=e⁻ᵃᵗsin ωt.

Answer

The sampled signal is $x(kT) = e^{-akT}\sin\omega kT$ . It is transformed below directly with Euler's formula and then checked with the complex translation theorem.

Method 1: Euler's formula

e^{-akT}\sin\omega kT = \frac{1}{2j}\left[e^{-(a - j\omega)kT} - e^{-(a + j\omega)kT}\right]

Using $\mathcal{Z}\left[e^{-bkT}\right] = \dfrac{1}{1 - e^{-bT}z^{-1}}$ :

\begin{aligned} X(z) &= \frac{1}{2j}\left[\frac{1}{1 - e^{-aT}e^{j\omega T}z^{-1}} - \frac{1}{1 - e^{-aT}e^{-j\omega T}z^{-1}}\right] \\ &= \frac{1}{2j}\cdot\frac{e^{-aT}\left(e^{j\omega T} - e^{-j\omega T}\right)z^{-1}}{1 - e^{-aT}\left(e^{j\omega T} + e^{-j\omega T}\right)z^{-1} + e^{-2aT}z^{-2}} \\ &= \frac{e^{-aT}z^{-1}\sin\omega T}{1 - 2e^{-aT}z^{-1}\cos\omega T + e^{-2aT}z^{-2}} \end{aligned}

Multiply by $z^2/z^2$ :

X(z) = \frac{ze^{-aT}\sin\omega T}{z^2 - 2ze^{-aT}\cos\omega T + e^{-2aT}}

Method 2: Complex translation theorem (check)

$\mathcal{Z}[\sin\omega t] = \dfrac{z\sin\omega T}{z^2 - 2z\cos\omega T + 1}$ . Since $\mathcal{Z}[e^{-at}x(t)] = X(ze^{aT})$ :

X(z) = \frac{ze^{aT}\sin\omega T}{z^2e^{2aT} - 2ze^{aT}\cos\omega T + 1}

Multiplying numerator and denominator by $e^{-2aT}$ gives the same result as Method 1.

Remarks

The poles are at $z = e^{-aT}e^{\pm j\omega T}$ : radius $e^{-aT}$ , angle $\pm\omega T$ . For $a > 0$ they lie inside the unit circle, giving a decaying oscillation.
Example with $a = 1$ , $\omega = 2$ rad/s, $T = 0.5$ s: $e^{-0.5} = 0.6065$ , $\sin 1 = 0.8415$ , $\cos 1 = 0.5403$ , so $X(z) = \dfrac{0.5104z}{z^2 - 0.6554z + 0.3679}$ .

Answer: $X(z) = \dfrac{ze^{-aT}\sin\omega T}{z^2 - 2ze^{-aT}\cos\omega T + e^{-2aT}}$ .

2072 Magh · 4 marks

If x(k)=1/2ᵏ, for -4≤ k≤ 4, then find Z[x(k)].

Answer

$x(k) = (1/2)^k = 2^{-k}$ is defined only for $-4 \le k \le 4$ , so it is a finite-length, two-sided sequence. Use the two-sided definition $X(z) = \sum_k x(k)z^{-k}$ .

Step 1: List the samples

$k$	−4	−3	−2	−1	0	1	2	3	4
$x(k)$	16	8	4	2	1	0.5	0.25	0.125	0.0625

Step 2: Write the sum

\begin{aligned} X(z) &= \sum_{k=-4}^{4}2^{-k}z^{-k} \\ &= 16z^4 + 8z^3 + 4z^2 + 2z + 1 + 0.5z^{-1} + 0.25z^{-2} + 0.125z^{-3} + 0.0625z^{-4} \end{aligned}

Step 3: Closed form (finite geometric series)

The ratio between successive terms is $r = (2z)^{-1}$ , and the first term is $(2z)^4$ :

X(z) = \sum_{k=-4}^{4}(2z)^{-k} = \frac{(2z)^4\left[1 - (2z)^{-9}\right]}{1 - (2z)^{-1}} = \frac{(2z)^4 - (2z)^{-5}}{1 - \frac{1}{2z}}

Multiplying by $\dfrac{2z}{2z}$ :

X(z) = \frac{(2z)^9 - 1}{(2z)^4\,(2z - 1)} = \frac{512z^9 - 1}{16z^4(2z-1)}

The apparent pole at $z = \tfrac12$ is cancelled by a zero of the numerator, so $X(z)$ is really a finite polynomial in $z$ and $z^{-1}$ .

Step 4: Region of convergence

A finite sum always converges except where a term becomes infinite:

the positive powers $z^4, \dots, z$ blow up at $z = \infty$
the negative powers $z^{-1}, \dots, z^{-4}$ blow up at $z = 0$

\text{ROC: } 0 < |z| < \infty

Check: at $z = 1$ the sum is $16+8+4+2+1+0.5+0.25+0.125+0.0625 = 31.9375$ , and the closed form gives $\dfrac{511}{16} = 31.9375$ .

Answer: $X(z) = 16z^4 + 8z^3 + 4z^2 + 2z + 1 + 0.5z^{-1} + 0.25z^{-2} + 0.125z^{-3} + 0.0625z^{-4} = \dfrac{512z^9 - 1}{16z^4(2z-1)}$ , ROC $0 < |z| < \infty$ .

2071 Bhadra · 6 marks

Solve the following difference equation: 2x(k)-2x(k-1)+x(k-2)=u(k); where x(k)=0 for k<0 and u(k) is unit step function.

Answer

Solve by z-transform with $x(k) = 0$ for $k < 0$ and $U(z) = \dfrac{1}{1-z^{-1}}$ .

Step 1: Z-transform

Since $x(-1) = x(-2) = 0$ , the delay theorem gives $\mathcal{Z}[x(k-n)] = z^{-n}X(z)$ :

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

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

Step 2: Poles

2z^2 - 2z + 1 = 0 \;\Rightarrow\; z = \frac{2 \pm \sqrt{4 - 8}}{4} = 0.5 \pm j0.5 = \frac{1}{\sqrt2}e^{\pm j\pi/4}

So $r = \dfrac{1}{\sqrt2} = 0.7071$ and $\Omega = \dfrac{\pi}{4}$ . Both complex poles are inside the unit circle.

Step 3: Form of the solution

The pole at $z = 1$ gives a constant, and the complex pair gives a damped sinusoid:

x(k) = A_0 + r^k\left(A\cos\frac{k\pi}{4} + B\sin\frac{k\pi}{4}\right)

Steady state: $A_0 = \lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1}{2 - 2 + 1} = 1$
Initial values from the equation: $k = 0$ gives $2x(0) = 1$ , so $x(0) = 0.5$ . $k = 1$ gives $2x(1) - 2(0.5) = 1$ , so $x(1) = 1$ .

\begin{aligned} k = 0:&\quad 1 + A = 0.5 \;\Rightarrow\; A = -0.5 \\ k = 1:&\quad 1 + 0.7071\left(-0.5\times0.7071 + B\times0.7071\right) = 1 \;\Rightarrow\; B = 0.5 \end{aligned}

Step 4: Closed-form solution

x(k) = 1 + \left(\frac{1}{\sqrt2}\right)^k\left[-0.5\cos\frac{k\pi}{4} + 0.5\sin\frac{k\pi}{4}\right], \quad k \ge 0

Since $-0.5\cos\theta + 0.5\sin\theta = \frac{1}{\sqrt2}\sin\left(\theta - \frac{\pi}{4}\right)$ , this simplifies to

x(k) = 1 + \left(\frac{1}{\sqrt2}\right)^{k+1}\sin\frac{(k-1)\pi}{4}

Step 5: Check with the recursion $x(k) = \frac12\left[1 + 2x(k-1) - x(k-2)\right]$

$k$	0	1	2	3	4	5	6	7	8	9
$x(k)$	0.5	1	1.25	1.25	1.125	1	0.9375	0.9375	0.9688	1

The formula gives the same values. For example, $k = 2$ : $1 + (0.7071)^3\sin(\pi/4) = 1 + 0.3536\times0.7071 = 1.25$ .

The output overshoots to 1.25 and settles at 1 with a decaying oscillation.

Answer: $x(k) = 1 + \left(\dfrac{1}{\sqrt2}\right)^{k+1}\sin\dfrac{(k-1)\pi}{4}$ , with $x(\infty) = 1$ .

2071 Bhadra · 6 marks

What are the methods of inverse z-transform? Explain each of them using suitable example.

Answer

The inverse z-transform recovers the sequence $x(k)$ (the sample values $x(kT)$ ) from $X(z)$ . It gives only the values at the sampling instants, not the continuous signal between them. There are four common methods.

1. Direct division method

Write $X(z)$ as a ratio of polynomials in $z^{-1}$ and divide the numerator by the denominator. The coefficient of $z^{-k}$ is $x(k)$ . This is simple and good for the first few values, but it gives no closed form.

Example: $X(z) = \dfrac{10z + 5}{(z-1)(z-0.2)} = \dfrac{10z^{-1} + 5z^{-2}}{1 - 1.2z^{-1} + 0.2z^{-2}}$

X(z) = 10z^{-1} + 17z^{-2} + 18.4z^{-3} + 18.68z^{-4} + \dots

So $x(0) = 0$ , $x(1) = 10$ , $x(2) = 17$ , $x(3) = 18.4$ .

2. Computational method

Treat $X(z)$ as a transfer function excited by a unit pulse. Either write its difference equation and solve it recursively (or with MATLAB filter), or simulate it.

Example (same $X(z)$ ): $x(k) = 1.2x(k-1) - 0.2x(k-2) + 10\delta(k-1) + 5\delta(k-2)$ . This gives $x(1) = 10$ , $x(2) = 1.2(10) + 5 = 17$ , $x(3) = 1.2(17) - 0.2(10) = 18.4$ .

3. Partial-fraction expansion method

Expand $X(z)/z$ into partial fractions, multiply back by $z$ , and use the standard table ( $\frac{z}{z-a} \leftrightarrow a^k$ ). This gives a closed form.

Example:

\frac{X(z)}{z} = \frac{10z + 5}{z(z-1)(z-0.2)} = \frac{25}{z} + \frac{18.75}{z-1} - \frac{43.75}{z-0.2}

x(k) = 25\delta(k) + 18.75 - 43.75(0.2)^k

Check: $x(0) = 25 + 18.75 - 43.75 = 0$ , and $x(1) = 18.75 - 8.75 = 10$ .

4. Inversion integral method

x(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum\text{residues of } X(z)z^{k-1}

The contour $C$ encloses all the poles. This works for repeated poles too, and needs no table.

Example: $X(z) = \dfrac{z}{(z-1)(z-0.5)}$

x(k) = \left[\frac{z^k}{z-0.5}\right]_{z=1} + \left[\frac{z^k}{z-1}\right]_{z=0.5} = 2 - 2(0.5)^k

Comparison

Method	Gives	Best for
Direct division	A few numerical values	Quick checks
Computational	Values by recursion or computer	Long sequences, simulation
Partial fractions	Closed form	Simple, distinct poles
Inversion integral	Closed form	Repeated or awkward poles

2068 Jestha · 8 marks

Find the inverse Z-transform of X(z)=z⁻¹/((1-z⁻¹)(1+1.6z⁻¹+0.64z⁻²)) by inversion integral method.

Answer

In the inversion integral method, $x(k)$ is the sum of the residues of $X(z)z^{k-1}$ at its poles:

x(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum_i\text{Res}\left[X(z)z^{k-1}\right]_{z=z_i}

Step 1: Rewrite $X(z)$ in powers of $z$

$1 + 1.6z^{-1} + 0.64z^{-2} = (1 + 0.8z^{-1})^2$ . Multiply numerator and denominator by $z^3$ :

X(z) = \frac{z^{-1}}{(1 - z^{-1})(1 + 0.8z^{-1})^2} = \frac{z^2}{(z-1)(z+0.8)^2}

X(z)z^{k-1} = \frac{z^{k+1}}{(z-1)(z+0.8)^2}

For $k \ge 0$ there is no pole at the origin. The poles are a simple pole at $z = 1$ and a double pole at $z = -0.8$ .

Step 2: Residue at $z = 1$

R_1 = \left[\frac{z^{k+1}}{(z+0.8)^2}\right]_{z=1} = \frac{1}{(1.8)^2} = \frac{1}{3.24} = 0.3086

Step 3: Residue at the double pole $z = -0.8$

\begin{aligned} R_2 &= \frac{d}{dz}\left[\frac{z^{k+1}}{z-1}\right]_{z=-0.8} = \left[\frac{(k+1)z^{k}(z-1) - z^{k+1}}{(z-1)^2}\right]_{z=-0.8} \\ &= \frac{(-0.8)^k\left[(k+1)(-1.8) - (-0.8)\right]}{(-1.8)^2} = \frac{(-0.8)^k\left(-1.8k - 1\right)}{3.24} \end{aligned}

Step 4: Result

x(k) = R_1 + R_2 = \frac{1}{3.24}\left[1 - (1 + 1.8k)(-0.8)^k\right], \quad k = 0, 1, 2, \dots

Equivalently $x(k) = 0.3086 - (0.3086 + 0.5556k)(-0.8)^k$ .

Step 5: Check

$k$	0	1	2	3	4	5	6	$\infty$
$x(k)$	0	1	−0.6	1.32	−0.728	1.32	−0.646	0.3086

Direct division of $\dfrac{z^{-1}}{1 + 0.6z^{-1} - 0.32z^{-2} - 0.64z^{-3}}$ gives $z^{-1} - 0.6z^{-2} + 1.32z^{-3} - 0.728z^{-4} + \dots$ , which matches. Here the denominator $(1 - z^{-1})(1 + 1.6z^{-1} + 0.64z^{-2})$ has been expanded.

The final value theorem gives $\lim_{z\to1}(1 - z^{-1})X(z) = \dfrac{1}{(1.8)^2} = 0.3086$ , which also matches.

Answer: $x(k) = \dfrac{1}{3.24}\left[1 - (1 + 1.8k)(-0.8)^k\right]$ for $k \ge 0$ .

2067 Mangsir · 8 marks

Starting from the z transform of unit step function, determine the z transform of k²e⁻ᵃᵏ.

Answer

Start from the unit step and build the result in three steps: multiply by $e^{-ak}$ (scaling in the z-domain), then multiply by $k$ twice (differentiation in the z-domain).

Properties used

Unit step: $\mathcal{Z}[1(k)] = \displaystyle\sum_{k=0}^{\infty}z^{-k} = \frac{z}{z-1}$ , $|z| > 1$
Scaling (multiplication by $a^k$ ): $\mathcal{Z}[b^k x(k)] = X\left(\dfrac{z}{b}\right)$
Multiplication by $k$ : $\mathcal{Z}[k\,x(k)] = -z\dfrac{d}{dz}X(z)$

Proof of property 3: $-z\dfrac{d}{dz}\sum x(k)z^{-k} = -z\sum(-k)x(k)z^{-k-1} = \sum k\,x(k)z^{-k}$ .

Step 1: $e^{-ak}$

Use property 2 with $b = e^{-a}$ , writing $E = e^{-a}$ for short:

\mathcal{Z}[e^{-ak}] = \frac{z/E}{z/E - 1} = \frac{z}{z - E} = \frac{z}{z - e^{-a}}

Step 2: $k\,e^{-ak}$

\mathcal{Z}[k\,e^{-ak}] = -z\frac{d}{dz}\left(\frac{z}{z-E}\right) = -z\cdot\frac{(z-E) - z}{(z-E)^2} = \frac{Ez}{(z-E)^2}

Step 3: $k^2e^{-ak}$

\begin{aligned} \mathcal{Z}[k^2e^{-ak}] &= -z\frac{d}{dz}\left[\frac{Ez}{(z-E)^2}\right] \\ &= -z\cdot\frac{E(z-E)^2 - Ez\cdot2(z-E)}{(z-E)^4} \\ &= -z\cdot\frac{E(z-E) - 2Ez}{(z-E)^3} = -z\cdot\frac{-E(z+E)}{(z-E)^3} \\ &= \frac{Ez(z+E)}{(z-E)^3} \end{aligned}

Result

\mathcal{Z}[k^2e^{-ak}] = \frac{e^{-a}z\,(z + e^{-a})}{(z - e^{-a})^3} = \frac{e^{-a}z^{-1}(1 + e^{-a}z^{-1})}{(1 - e^{-a}z^{-1})^3}, \quad |z| > e^{-a}

Check

Take $e^{-a} = 0.5$ . Then $X(z) = \dfrac{0.5z(z+0.5)}{(z-0.5)^3}$ , and its series is $0.5z^{-1} + z^{-2} + 1.125z^{-3} + z^{-4} + \dots$ . Direct values of $k^2(0.5)^k$ for $k = 0$ to $4$ are $0, 0.5, 1, 1.125, 1$ , which match.

As a special case, $a = 0$ gives $\mathcal{Z}[k^2] = \dfrac{z(z+1)}{(z-1)^3}$ , the known result.

Answer: $\mathcal{Z}[k^2e^{-ak}] = \dfrac{e^{-a}z(z + e^{-a})}{(z - e^{-a})^3}$ .

Past questions and answers

1.Solve the following difference equation by use of Z-transform method: x(k+2)+3x(k+1)+2x(k)=0, given that x(0)=0, x(1)=1.

Step 1: Transform

Step 2: Partial fractions of X(z)/zX(z)/zX(z)/z

Step 3: Inverse z-transform

Check

2.Obtain the inverse z-transform of X(z)=(2z³+z)/((z-2)²(z-1)) by using partial fraction expansion method.

Coefficients

Inverse transforms used

Result

Check (first few values)

3.Obtain the inverse z-transform of X(z)=z(1-e⁻ᵃᵀ)/((z-1)(z-e⁻ᵃᵀ)) using inversion integral method.

Form X(z)zk−1X(z)z^{k-1}X(z)zk−1

Residues

Result

4.Obtain the inverse Z-transform of X(z)=z(z+2)/(z-1)² by inversion integral method.

Residue at a double pole

Check

5.Obtain the z-transform of x(t)={sin ωt, t≥ 0; 0, t<0}

Derivation

6.Find the Z-transform of x(t)=t²e⁻ᵃᵗ (for k≥ 0).

Step 1: Z[k2]\mathcal{Z}[k^2]Z[k2]

Step 2: multiply by rkr^krk (scaling: replace zzz by z/rz/rz/r)

Result

7.Find the inverse z-transform of X(z)=z²/((z-1)²(z-e⁻ᵃᵀ)) by using inversion integral method.

Residue at z=pz = pz=p (simple pole)

Residue at z=1z = 1z=1 (double pole)

Result

Check

8.Solve the given difference equation and hence determine the output x(k), if u(t) is a unit step input function and x(t) is an output function: x(k+2)-x(k+1)+0.25x(k)=u(k+2), where x(0)=1 and x(1)=2.

Step 1: z-transform of both sides

Step 2: solve for X(z)X(z)X(z)

Step 3: inverse by residues of X(z)zk−1=zk+2(z−1)(z−0.5)2X(z)z^{k-1} = \frac{z^{k+2}}{(z-1)(z-0.5)^2}X(z)zk−1=(z−1)(z−0.5)2zk+2​

Result

Check with the recursion x(k+2)=x(k+1)−0.25x(k)+1x(k+2) = x(k+1) - 0.25x(k) + 1x(k+2)=x(k+1)−0.25x(k)+1

9.Solve the following difference equation by use of Z-transform method: x(k+2)+1.379x(k+1)+0.3679x(k)=0.3679u(k), where u(k) is step input and x(0)=0, x(1)=0.3679.

Step 1: z-transform

Step 2: roots of the characteristic equation

Step 3: partial fractions of X(z)/z=0.3679z(z−1)(z−p1)(z−p2)X(z)/z = \frac{0.3679z}{(z-1)(z-p_1)(z-p_2)}X(z)/z=(z−1)(z−p1​)(z−p2​)0.3679z​

Result

Check

10.Obtain the Z-transform of x(k)=k²aᵏ⁻¹.

Step 1: Z[kak−1]\mathcal{Z}[k a^{k-1}]Z[kak−1]

Step 2: Z[k2ak−1]\mathcal{Z}[k^2 a^{k-1}]Z[k2ak−1]

Check

11.Determine the Z-transform for (i) x(t)=t²e⁻ᵃᵗ (ii) x(t)=t· x(t-nT).

(i) x(t)=t2e−atx(t) = t^2e^{-at}x(t)=t2e−at

(ii) y(t)=t⋅x(t−nT)y(t) = t\cdot x(t-nT)y(t)=t⋅x(t−nT)

12.Find the Z-transform of x(k)=k·aᵏ⁻¹.

13.Obtain the inverse z-transform of X(z)=z²/((z-2)²(z-e⁻ᵃᵀ)).

Residue at the simple pole z=bz = bz=b

Residue at the double pole z=2z = 2z=2

Result

Check

14.Consider the function y(k), which is a sum of functions x(h), where h=0,1,2,…,k, such that y(k) = Σ x(h) (sum from h = 0 to k), k=0,1,2,…, where y(k)=0 for k<0. Obtain Y(z).

Step 1: Write a difference equation

Step 2: Take the z-transform

Alternative view (convolution)

Example

Remarks

15.Obtain the inverse z-transform of the following system using partial fraction expansion method: x(z)=z(1-eᵃᵀ)/((z-1)(z-eᵃᵀ)).

Step 1: Divide by zzz and expand

Step 2: Multiply back by zzz

Step 3: Use standard pairs

Check by long division

Note on the usual form

16.Obtain the inverse z-transform of X(z)=(1+z⁻¹-z⁻²)/(1-z⁻¹).

Step 1: Divide the numerator by the denominator (in powers of z−1z^{-1}z−1)

Step 2: Inverse transform

Check by direct division

17.The waveform below shows the input signal to a digital controller. Obtain the Z-transform of the input x(t). [Figure: x(t) rises linearly from 0 at t = 0 to 3 at t = 3 (passing 1 at t = 1 and 2 at t = 2), then stays constant at 3; dotted sample lines at t = 1, 2, …, 7]

Step 1: Write x(t)x(t)x(t) as a ramp minus a delayed ramp

Step 2: Sampled values (T=1T = 1T=1)

Step 3: Take the z-transform

Step 4: Check by series

18.Obtain the Z-transform of {x(k)=9k(2ᵏ⁻¹)-2ᵏ+3}, for k≥ 0.

Standard pairs used

Transform each term (a=2a = 2a=2)

Combine over a common denominator (z−2)2(z−1)(z-2)^2(z-1)(z−2)2(z−1)

Check

Solve the following difference equation by use of Z-transform method: x(k+2)+3x(k+1)+2x(k)=0, given that x(0)=0, x(1)=1.

Step 2: Partial fractions of $X(z)/z$

Obtain the inverse z-transform of X(z)=(2z³+z)/((z-2)²(z-1)) by using partial fraction expansion method.

Obtain the inverse z-transform of X(z)=z(1-e⁻ᵃᵀ)/((z-1)(z-e⁻ᵃᵀ)) using inversion integral method.

Form $X(z)z^{k-1}$

Obtain the inverse Z-transform of X(z)=z(z+2)/(z-1)² by inversion integral method.

Obtain the z-transform of x(t)={sin ωt, t≥ 0; 0, t<0}

Find the Z-transform of x(t)=t²e⁻ᵃᵗ (for k≥ 0).

Step 1: $\mathcal{Z}[k^2]$

Step 2: multiply by $r^k$ (scaling: replace $z$ by $z/r$ )

Find the inverse z-transform of X(z)=z²/((z-1)²(z-e⁻ᵃᵀ)) by using inversion integral method.

Residue at $z = p$ (simple pole)

Residue at $z = 1$ (double pole)

Solve the given difference equation and hence determine the output x(k), if u(t) is a unit step input function and x(t) is an output function: x(k+2)-x(k+1)+0.25x(k)=u(k+2), where x(0)=1 and x(1)=2.

Step 2: solve for $X(z)$

Step 3: inverse by residues of $X(z)z^{k-1} = \frac{z^{k+2}}{(z-1)(z-0.5)^2}$

Check with the recursion $x(k+2) = x(k+1) - 0.25x(k) + 1$

Solve the following difference equation by use of Z-transform method: x(k+2)+1.379x(k+1)+0.3679x(k)=0.3679u(k), where u(k) is step input and x(0)=0, x(1)=0.3679.

Step 3: partial fractions of $X(z)/z = \frac{0.3679z}{(z-1)(z-p_1)(z-p_2)}$

Obtain the Z-transform of x(k)=k²aᵏ⁻¹.

Step 1: $\mathcal{Z}[k a^{k-1}]$

Step 2: $\mathcal{Z}[k^2 a^{k-1}]$

Determine the Z-transform for (i) x(t)=t²e⁻ᵃᵗ (ii) x(t)=t· x(t-nT).

(i) $x(t) = t^2e^{-at}$

(ii) $y(t) = t\cdot x(t-nT)$

Find the Z-transform of x(k)=k·aᵏ⁻¹.

Obtain the inverse z-transform of X(z)=z²/((z-2)²(z-e⁻ᵃᵀ)).

Residue at the simple pole $z = b$

Residue at the double pole $z = 2$

Consider the function y(k), which is a sum of functions x(h), where h=0,1,2,…,k, such that y(k) = Σ x(h) (sum from h = 0 to k), k=0,1,2,…, where y(k)=0 for k<0. Obtain Y(z).

Obtain the inverse z-transform of the following system using partial fraction expansion method: x(z)=z(1-eᵃᵀ)/((z-1)(z-eᵃᵀ)).

Step 1: Divide by $z$ and expand

Step 2: Multiply back by $z$

Obtain the inverse z-transform of X(z)=(1+z⁻¹-z⁻²)/(1-z⁻¹).

Step 1: Divide the numerator by the denominator (in powers of $z^{-1}$ )

The waveform below shows the input signal to a digital controller. Obtain the Z-transform of the input x(t). [Figure: x(t) rises linearly from 0 at t = 0 to 3 at t = 3 (passing 1 at t = 1 and 2 at t = 2), then stays constant at 3; dotted sample lines at t = 1, 2, …, 7]

Step 1: Write $x(t)$ as a ramp minus a delayed ramp

Step 2: Sampled values ( $T = 1$ )

Obtain the Z-transform of {x(k)=9k(2ᵏ⁻¹)-2ᵏ+3}, for k≥ 0.

Transform each term ( $a = 2$ )

Combine over a common denominator $(z-2)^2(z-1)$

Obtain the inverse z-transform of X(z)=z⁻³/((1-z⁻¹)(1-0.2z⁻¹)).

Step 1: Write $X(z) = z^{-3}F(z)$

Step 2: Partial fractions of $F(z)/z$

Solve the difference equation x(k+2)-x(k+1)-x(k)=0, where x(0)=0 and x(1)=1, and obtain the general form of solution. After obtaining the general form of solution determine the series up to 8th term.

Step 3: Partial fractions of $X(z)/z$

Find X(k) when x(k+2)=x(k+1)+x(k), x(0)=0 and x(1)=1, and show that the limiting value of x(k+1)/x(k)=1.6180 when k tends to infinity.

Step 3: Partial fractions of $X(z)/z$

Step 5: Limiting value of $x(k+1)/x(k)$

Consider the difference equation x(k+2)=x(k+1)+x(k) where x(0)=0, x(1)=1 and x(2)=2. Obtain the general solution x(k) in a closed form. Find the limiting value of x(k+2)/x(k) when sequence variable approaches infinity.

Step 3: Partial fractions of $X(z)/z$

Step 5: Limiting value of $x(k+2)/x(k)$

Note on $x(2) = 2$

Obtain the z-transform of x(t) for which time response is given by (assume sampling period T = 1 s). [Figure: continuous time signal x(t) rising linearly from 0 at t = 0 to 1 at t = 4 (0.5 marked on the axis), then constant at 1; time axis marked 1 to 13]

Step 1: Express $x(t)$ as ramp minus delayed ramp

Step 2: Sampled values ( $T = 1$ s)

Step 1: Express $x(t)$ as ramp minus delayed ramp

Step 2: Sampled values ( $T = 1$ s)

Find the z-transform of the following: x(t)={0, t<0; 0, t≤ 4; 1, t>4}

Find the z-transform of x(t)=1/4t-1/4(t-4)1(t-4).

Step 2: General sampling period $T$ (4 an integer multiple of $T$ )

Step 3: With $T = 1$ s (usual case, $n = 4$ )

Obtain Z transform of curve x(t) shown in figure below. [Figure: x(t) rises linearly from 0 at t = 0 through 0.5 at t = 1 to 1 at t = 2, then stays constant at 1; time axis marked 0 to 4]

Step 2: Sampled values ( $T = 1$ )

Note for other $T$

Consider the difference equation: x(k+2)-1.3679x(k+1)+0.3679x(k)=0.3679u(k+1)+0.2642u(k), where x(k) is the output and x(k)=0 for k≤ 0, where u(k) is the input and is given by u(k)=0, k<0; u(0)=1; u(1)=0.2142; u(2)=0.2142; u(k)=0, k=3,4,5,6,… Determine the output x(k).

Step 4: Closed form for $k \ge 3$

Consider the difference equation x(k+2)-1.3679x(k+1)+0.3679x(k)=0.3679u(k+1)+0.2642u(k), where x(k) is the output and x(k)=0 for k≤ 0, where u(k) is the input and is given by u(k)=0, k<0; u(0)=1; u(1)=0.2142; u(2)=-0.2142; u(k)=0, k=3,4,5,6,… Determine the output x(k).

Consider the difference equation x(k+3)-2.2x(k+2)+1.57x(k+1)-0.36x(k)=u(k), where u(k)=1 for k≥ 0 and x(0)=x(1)=x(2)=0. Obtain the general solution x(k) in a closed form. Find the limiting value of x(k+2)/x(k) when sequence variable approaches infinity.