GATE EC · EE · IN

z-transform and discrete-time systems

The z-transform turns a sequence into a function of a complex variable $z$ , so that convolution becomes multiplication and a difference equation becomes an algebraic equation. These notes cover recovering a sequence from $X (z)$ and its region of convergence (ROC), reading causality and stability off the pole positions, solving difference equations with initial conditions, getting the frequency response on the unit circle and, for IN, finding the pulse transfer function of a sampled-data loop.

12 previous-year questions · 12 solved · 6 practice questions

1. Definition, conventions and the region of convergence

The z-transform does for sequences what the Laplace transform does for continuous-time signals. Convolution becomes multiplication, a linear constant-coefficient difference equation becomes an algebraic equation, and the poles of a system function decide causality and stability. Conventions in these notes: the transform is bilateral unless it is marked unilateral ( $X^{+}$ ). Discrete-time frequency is the normalised digital frequency $Ω$ in rad/sample, and the DTFT is $X (e^{j Ω}) = \sum_{n} x [n] e^{- j Ω n}$ .

BILATERAL Z-TRANSFORM

X (z) = Z {x [n]} = n = - \infty \sum \infty x [n] z^{- n}

Defined only for $z$ in the ROC, the set of $z$ where the sum converges absolutely. Always quote $X (z)$ together with its ROC.

UNILATERAL Z-TRANSFORM

X^{+} (z) = n = 0 \sum \infty x [n] z^{- n}

Ignores $x [n]$ for $n < 0$ ; it equals the bilateral transform of $x [n] u [n]$ . When it exists, its ROC is the exterior of a circle, $∣ z ∣ > r$ (including $z = \infty$ ), so it is usually not written. For a causal sequence the two transforms are identical.

LINK TO THE DTFT

X (e^{j Ω}) = X (z)_{z = e^{j Ω}}

Valid only when the ROC contains the unit circle $∣ z ∣ = 1$ . $X (e^{j Ω})$ is periodic in $Ω$ with period $2 π$ .

Writing $z = r e^{j Ω}$ turns the sum into $\sum_{n} (x [n] r^{- n}) e^{- j Ω n}$ . So the z-transform is the DTFT of $x [n]$ weighted by $r^{- n}$ , and whether it converges absolutely depends only on $r = ∣ z ∣$ . That is why every ROC is a ring centred on the origin. The ROC is part of the answer: the same algebraic $X (z)$ with a different ROC is a different sequence.

Properties of the ROC

For a rational $X (z)$ the poles are what bound the ROC.

The ROC is a ring $r_{1} < ∣ z ∣ < r_{2}$ centred on the origin; $r_{1}$ may be $0$ and $r_{2}$ may be $\infty$ .
The ROC contains no poles. For a rational $X (z)$ its boundaries pass through poles (or it extends to $0$ or $\infty$ ).
Finite-length $x [n]$ : the ROC is the whole z-plane, except $z = 0$ if $x [n] \neq = 0$ for some $n > 0$ , and except $z = \infty$ if $x [n] \neq = 0$ for some $n < 0$ .
Right-sided $x [n]$ (zero for $n < N_{1}$ ): the ROC is outside the outermost pole, $∣ z ∣ > r_{m a x}$ . It includes $z = \infty$ exactly when the sequence is causal ( $N_{1} \geq 0$ ).
Left-sided $x [n]$ (zero for $n > N_{2}$ ): the ROC is inside the innermost non-zero pole, $∣ z ∣ < r_{m i n}$ . It includes $z = 0$ exactly when $N_{2} \leq 0$ .
Two-sided $x [n]$ : the ROC is an annulus between two consecutive pole radii, or it is empty (no z-transform exists) when the rings of the right-sided and left-sided parts do not overlap.
The ROC is one connected region. If the poles lie on $k$ distinct non-zero radii, a rational $X (z)$ has $k + 1$ possible ROCs, and each gives a different $x [n]$ .

Type of sequence and its ROC

Sequence	Example	ROC
Finite length, causal	$δ [n] + 2 δ [n - 3]$	all $z$ except $z = 0$
Finite length, two-sided	$δ [n + 1] + δ [n - 1]$	$0 < ∣ z ∣ < \infty$
Right-sided, causal	$(0.5)^{n} u [n]$	$∣ z ∣ > 0.5$ , including $z = \infty$
Left-sided, anticausal	$- (0.5)^{n} u [- n - 1]$	$∣ z ∣ < 0.5$
Two-sided	$(0.5)^{n} u [n] - 2^{n} u [- n - 1]$	$0.5 < ∣ z ∣ < 2$
Two-sided, no transform	$2^{n} u [n] - (0.5)^{n} u [- n - 1]$	$∣ z ∣ > 2$ and $∣ z ∣ < 0.5$ do not overlap: none

READ THE ROC BEFORE YOU INVERT

$\frac{z}{z - a}$ is the transform of $a^{n} u [n]$ when $∣ z ∣ > ∣ a ∣$ and of $- a^{n} u [- n - 1]$ when $∣ z ∣ < ∣ a ∣$ . When a question gives only $X (z)$ , the ROC is fixed by whatever else it states: causal (the region outside the outermost pole), stable (the region containing $∣ z ∣ = 1$ ), or absolutely summable (same as stable).

2. Standard pairs, properties and the value theorems

RIGHT-SIDED EXPONENTIAL

a^{n} u [n] \leftrightarrow \frac{1}{1 - a z ^{- 1}} = \frac{z}{z - a}, ∣ z ∣ > ∣ a ∣

With $a = 1$ : $u [n] \leftrightarrow \frac{z}{z - 1}$ , $∣ z ∣ > 1$ . With $a = 0$ (limit): $δ [n] \leftrightarrow 1$ for all $z$ .

LEFT-SIDED EXPONENTIAL

- a^{n} u [- n - 1] \leftrightarrow \frac{z}{z - a}, ∣ z ∣ < ∣ a ∣

Same algebraic transform as $a^{n} u [n]$ ; only the ROC differs. Note the minus sign and the $u [- n - 1]$ (it starts at $n = - 1$ ).

RAMP-WEIGHTED EXPONENTIAL

n a^{n} u [n] \leftrightarrow \frac{a z ^{- 1}}{( 1 - a z ^{- 1} ) ^{2}} = \frac{a z}{( z - a ) ^{2}}, ∣ z ∣ > ∣ a ∣

Follows from differentiation in $z$ . Useful variants: $n u [n] \leftrightarrow \frac{z}{( z - 1 ) ^{2}}$ and $(n + 1) a^{n} u [n] \leftrightarrow \frac{z ^{2}}{( z - a ) ^{2}}$ , both right-sided.

DAMPED COSINE AND SINE

r^{n} cos (Ω_{0} n) u [n] r^{n} sin (Ω_{0} n) u [n] \leftrightarrow \frac{z ( z - r cos Ω _{0} )}{z ^{2} - 2 r z cos Ω _{0} + r ^{2}} \leftrightarrow \frac{r z sin Ω _{0}}{z ^{2} - 2 r z cos Ω _{0} + r ^{2}}

Here $r > 0$ and $Ω_{0}$ are real. ROC $∣ z ∣ > r$ ; poles at $r e^{\pm j Ω_{0}}$ . Put $r = 1$ for the undamped $cos (Ω_{0} n) u [n]$ and $sin (Ω_{0} n) u [n]$ , whose poles lie on the unit circle (ROC $∣ z ∣ > 1$ ).

Standard pairs

$x [n]$	$X (z)$	ROC
$δ [n]$	$1$	all $z$
$δ [n - k]$ , $k > 0$	$z^{- k}$	all $z$ except $0$
$δ [n + k]$ , $k > 0$	$z^{k}$	all finite $z$ (excludes $\infty$ )
$u [n]$	$\frac{z}{z - 1}$	$∣ z ∣ > 1$
$- u [- n - 1]$	$\frac{z}{z - 1}$	$∣ z ∣ < 1$
$a^{n} u [n]$	$\frac{z}{z - a}$	$∣ z ∣ > ∣ a ∣$
$- a^{n} u [- n - 1]$	$\frac{z}{z - a}$	$∣ z ∣ < ∣ a ∣$
$a^{- n} u [- n]$	$\frac{1}{1 - a z}$	$∣ z ∣ < 1/∣ a ∣$
$n a^{n} u [n]$	$\frac{a z}{( z - a ) ^{2}}$	$∣ z ∣ > ∣ a ∣$
$- n a^{n} u [- n - 1]$	$\frac{a z}{( z - a ) ^{2}}$	$∣ z ∣ < ∣ a ∣$
$n a^{n - 1} u [n]$	$\frac{z}{( z - a ) ^{2}}$	$∣ z ∣ > ∣ a ∣$
$a^{n} (u [n] - u [n - N])$	$\frac{1 - a ^{N} z ^{- N}}{1 - a z ^{- 1}}$	$∣ z ∣ > 0$
$cos (Ω_{0} n) u [n]$	$\frac{z ( z - cos Ω _{0} )}{z ^{2} - 2 z cos Ω _{0} + 1}$	$∣ z ∣ > 1$
$sin (Ω_{0} n) u [n]$	$\frac{z sin Ω _{0}}{z ^{2} - 2 z cos Ω _{0} + 1}$	$∣ z ∣ > 1$
$a^{∣ n ∣}$ , $∣ a ∣ < 1$	$\frac{( 1 - a ^{2} ) z}{( z - a ) ( 1 - a z )}$	$∣ a ∣ < ∣ z ∣ < 1/∣ a ∣$
causal periodic $x [n]$ , period $N$ ; $X_{1} (z)$ is the transform of one period $x [n] (u [n] - u [n - N])$	$\frac{X _{1} ( z )}{1 - z ^{- N}}$	$∣ z ∣ > 1$

TIME SHIFT (BILATERAL)

x [n - k] \leftrightarrow z^{- k} X (z)

ROC unchanged except possibly for $z = 0$ or $z = \infty$ . A delay of $k$ samples multiplies by $z^{- k}$ . For the unilateral version with initial conditions see Section 5.

TIME REVERSAL

x [- n] \leftrightarrow X (1/ z)

The ROC inverts: $r_{1} < ∣ z ∣ < r_{2}$ becomes $1/ r_{2} < ∣ z ∣ < 1/ r_{1}$ . A pole at $p$ moves to $1/ p$ .

SCALING IN Z (MULTIPLICATION BY AN EXPONENTIAL)

a^{n} x [n] \leftrightarrow X (\frac{z}{a})

ROC scales to $∣ a ∣ r_{1} < ∣ z ∣ < ∣ a ∣ r_{2}$ ; every pole and zero moves from $p$ to $a p$ . With $a = e^{j Ω_{0}}$ (modulation) the pole-zero plot rotates by $Ω_{0}$ .

DIFFERENTIATION IN Z

n x [n] \leftrightarrow - z \frac{d X ( z )}{d z}

Same ROC. Apply it twice for $n^{2} x [n]$ .

CONVOLUTION

x_{1} [n] * x_{2} [n] \leftrightarrow X_{1} (z) X_{2} (z)

ROC contains $R_{1} \cap R_{2}$ (it can be larger if a pole cancels against a zero).

ACCUMULATION

k = - \infty \sum n x [k] \leftrightarrow \frac{X ( z )}{1 - z ^{- 1}} = \frac{z X ( z )}{z - 1}

ROC contains $R \cap {∣ z ∣ > 1}$ . The inverse operation, the first difference $x [n] - x [n - 1]$ , gives $(1 - z^{- 1}) X (z)$ .

CONJUGATION

x^{*} [n] \leftrightarrow X^{*} (z^{*})

Same ROC. For a real sequence $X (z^{*}) = X^{*} (z)$ , so complex poles and zeros come in conjugate pairs.

Properties at a glance

Property	Sequence	Transform	ROC
Linearity	$a x_{1} [n] + b x_{2} [n]$	$a X_{1} (z) + b X_{2} (z)$	contains $R_{1} \cap R_{2}$
Time shift	$x [n - k]$	$z^{- k} X (z)$	$R$ , except possibly $0$ or $\infty$
Time reversal	$x [- n]$	$X (1/ z)$	$1/ r_{2} < ∣ z ∣ < 1/ r_{1}$
Scaling in $z$	$a^{n} x [n]$	$X (z / a)$	$∣ a ∣ r_{1} < ∣ z ∣ < ∣ a ∣ r_{2}$
Differentiation in $z$	$n x [n]$	$- z d X / d z$	$R$
Convolution	$x_{1} [n] * x_{2} [n]$	$X_{1} (z) X_{2} (z)$	contains $R_{1} \cap R_{2}$
Accumulation	$\sum_{k = - \infty}^{n} x [k]$	$\frac{z X ( z )}{z - 1}$	contains $R \cap {∣ z ∣ > 1}$
First difference	$x [n] - x [n - 1]$	$(1 - z^{- 1}) X (z)$	contains $R \cap {∣ z ∣ > 0}$
Conjugation	$x^{*} [n]$	$X^{} (z^{})$	$R$
Time expansion	$x [n / k]$ for $n$ a multiple of $k$ , else $0$	$X (z^{k})$	$r_{1}^{1/ k} < ∣ z ∣ < r_{2}^{1/ k}$

INITIAL VALUE THEOREM

x [0] = z \to \infty lim X (z)

For a causal $x [n]$ ( $x [n] = 0$ for $n < 0$ ), or for any $x [n]$ with the unilateral transform. The next sample is $x [1] = lim_{z \to \infty} z (X (z) - x [0])$ . A rational $X (z)$ whose numerator degree in $z$ exceeds the denominator degree cannot be the transform of a causal sequence, and the theorem does not apply. For an anticausal $x [n]$ ( $x [n] = 0$ for $n > 0$ ) the mirror result is $x [0] = lim_{z \to 0} X (z)$ .

FINAL VALUE THEOREM

n \to \infty lim x [n] = z \to 1 lim (z - 1) X (z) = z \to 1 lim (1 - z^{- 1}) X (z)

For a causal $x [n]$ (or the unilateral transform of any $x [n]$ ), and only when the limit exists: every pole of $(z - 1) X (z)$ must lie strictly inside the unit circle. $X (z)$ itself may have one simple pole at $z = 1$ .

THE FINAL VALUE THEOREM TRAP

Applied blindly to $X (z) = \frac{z}{z - 2}$ , the theorem gives $lim_{z \to 1} (z - 1) \frac{z}{z - 2} = 0$ , yet $x [n] = 2^{n} u [n]$ grows without bound. Likewise $cos (Ω_{0} n) u [n]$ has poles at $e^{\pm j Ω_{0}}$ on the unit circle and no final value. Check the poles of $(z - 1) X (z)$ first.

SUMS WITHOUT INVERTING

If the ROC contains the unit circle: $\sum_{n} x [n] = X (1)$ , $\sum_{n} (- 1)^{n} x [n] = X (- 1)$ and $\sum_{n} n x [n] = - \frac{d X}{d z}_{z = 1}$ . These come straight from the definition and save an inversion.

3. Inverse z-transform

There are three methods. Partial fractions is the standard route for a rational $X (z)$ : split it into table pairs and let the ROC decide, term by term, whether each pole gives a right-sided or a left-sided sequence. Long division gives the samples one at a time, which is enough when a question asks for only $x [0]$ , $x [1]$ , $x [2]$ . Residues evaluate the inversion integral directly.

Partial fractions, step by step

Form $X (z) / z$ and make sure it is a proper rational function of $z$ (divide first if it is not). If $X (z)$ has no zero at the origin, $X (z) / z$ has a pole at $z = 0$ ; its term $\frac{A _{0}}{z}$ becomes $A_{0} δ [n]$ after multiplying back by $z$ .
Factor the denominator and expand: $\frac{X ( z )}{z} = \sum_{k} \frac{A _{k}}{z - p _{k}}$ .
Multiply back by $z$ : $X (z) = \sum_{k} A_{k} \frac{z}{z - p _{k}}$ . Every term is now a table pair.
Use the ROC $r_{1} < ∣ z ∣ < r_{2}$ term by term. A pole with $∣ p_{k} ∣ \leq r_{1}$ (inside the ring's inner edge) gives the right-sided $A_{k} p_{k}^{n} u [n]$ . A pole with $∣ p_{k} ∣ \geq r_{2}$ (outside the outer edge) gives the left-sided $- A_{k} p_{k}^{n} u [- n - 1]$ .
For a double pole at $p$ , include both $\frac{A}{z - p}$ and $\frac{B}{( z - p ) ^{2}}$ in the expansion of $X (z) / z$ . After multiplying by $z$ , use $\frac{z}{( z - p ) ^{2}} \leftrightarrow n p^{n - 1} u [n]$ if the pole is inside the ROC's inner edge, or $- n p^{n - 1} u [- n - 1]$ if it is outside the outer edge.

PARTIAL-FRACTION COEFFICIENT (SIMPLE POLE)

A_{k} = (z - p_{k}) \frac{X ( z )}{z}_{z = p_{k}}

Equivalent form in $z^{- 1}$ : $X (z) = \sum_{k} \frac{A _{k}}{1 - p _{k} z ^{- 1}}$ with $A_{k} = (1 - p_{k} z^{- 1}) X (z)_{z = p_{k}}$ . That form needs the numerator degree in $z^{- 1}$ to be less than the denominator degree; otherwise divide out the polynomial part first (it gives impulses).

INVERSION INTEGRAL AND RESIDUES

x [n] = \frac{1}{2 π j} \oint_{C} X (z) z^{n - 1} d z = p_{i} inside C \sum Res [X (z) z^{n - 1}, p_{i}]

$C$ is a counter-clockwise circle centred on the origin and lying in the ROC. For a causal $x [n]$ it encloses every pole. For small $n$ , check whether $z^{n - 1}$ adds a pole at $z = 0$ (for example at $n = 0$ when $X (z)$ has no zero at the origin) and include its residue.

RESIDUE AT A POLE OF ORDER M

Res_{z = p} = \frac{1}{( m - 1 )!} z \to p lim \frac{d ^{m - 1}}{d z ^{m - 1}} [(z - p)^{m} X (z) z^{n - 1}]

For a simple pole ( $m = 1$ ) this is just $(z - p) X (z) z^{n - 1}$ evaluated at $z = p$ .

Long division. For a right-sided (causal) sequence, write numerator and denominator in descending powers of $z$ (equivalently ascending powers of $z^{- 1}$ ) and divide. The coefficient of $z^{- n}$ in the quotient is $x [n]$ . For a left-sided sequence, write both in ascending powers of $z$ , so that the quotient runs in positive powers of $z$ . Division rarely gives a closed form, but it is the quickest check on a partial-fraction answer.

WORKED EXAMPLE · EXAMPLE 1: ONE X(Z), THREE SEQUENCES

$X (z) = \frac{z}{( z - 0.5 ) ( z - 2 )}$ . Find $x [n]$ for every possible ROC, and say which one is causal and which one is stable.

Solution

The poles are at $0.5$ and $2$ , so there are three possible ROCs: $∣ z ∣ > 2$ , $0.5 < ∣ z ∣ < 2$ and $∣ z ∣ < 0.5$ .
$\frac{X ( z )}{z} = \frac{1}{( z - 0.5 ) ( z - 2 )} = \frac{A}{z - 0.5} + \frac{B}{z - 2}$ , with $A = \frac{1}{0.5 - 2} = - \frac{2}{3}$ and $B = \frac{1}{2 - 0.5} = \frac{2}{3}$ .
So $X (z) = - \frac{2}{3} \frac{z}{z - 0.5} + \frac{2}{3} \frac{z}{z - 2}$ .
$∣ z ∣ > 2$ : both terms are right-sided, so $x [n] = \frac{2}{3} (2^{n} - 0. 5^{n}) u [n]$ . This one is causal. It is not stable, because the ROC excludes $∣ z ∣ = 1$ .
$0.5 < ∣ z ∣ < 2$ : the pole at $0.5$ is inside the ring (right-sided) and the pole at $2$ is outside it (left-sided), so $x [n] = - \frac{2}{3} (0.5)^{n} u [n] - \frac{2}{3} 2^{n} u [- n - 1]$ . This one is stable (the ring contains $∣ z ∣ = 1$ ) but two-sided, so not causal.
$∣ z ∣ < 0.5$ : both terms are left-sided, so $x [n] = \frac{2}{3} (0. 5^{n} - 2^{n}) u [- n - 1]$ . This one is anticausal and not stable.
Check the causal case by long division: $\frac{z}{z ^{2} - 2.5 z + 1} = z^{- 1} + 2.5 z^{- 2} + 5.25 z^{- 3} + \dots$ The formula gives $x [1] = \frac{2}{3} (2 - 0.5) = 1$ , $x [2] = \frac{2}{3} (4 - 0.25) = 2.5$ and $x [3] = \frac{2}{3} (8 - 0.125) = 5.25$ .

Answer: $∣ z ∣ > 2$ : $\frac{2}{3} (2^{n} - 0. 5^{n}) u [n]$ , causal, unstable. $0.5 < ∣ z ∣ < 2$ : $- \frac{2}{3} (0.5)^{n} u [n] - \frac{2}{3} 2^{n} u [- n - 1]$ , stable, not causal. $∣ z ∣ < 0.5$ : $\frac{2}{3} (0. 5^{n} - 2^{n}) u [- n - 1]$ , anticausal, unstable.

EXPAND X(Z)/Z, NOT X(Z)

Expanding $X (z) / z$ and multiplying back by $z$ makes every term $\frac{z}{z - p}$ , which is directly in the table. Expanding $X (z)$ itself gives terms $\frac{A}{z - p} = A z^{- 1} \frac{z}{z - p}$ , whose inverse is the delayed $A p^{n - 1} u [n - 1]$ . Forgetting that delay is a common slip.

4. Discrete-time LTI systems: system function, causality, stability, frequency response

For an LTI system $y [n] = x [n] * h [n]$ , so $Y (z) = H (z) X (z)$ . The system function $H (z) = Z {h [n]}$ is the ratio $Y (z) / X (z)$ with the system initially at rest. Its poles, zeros and ROC fix causality, stability and the frequency response.

SYSTEM FUNCTION OF A DIFFERENCE EQUATION

k = 0 \sum N a_{k} y [n - k] = k = 0 \sum M b_{k} x [n - k] \Rightarrow H (z) = \frac{\sum _{k = 0}^{M} b _{k} z ^{- k}}{\sum _{k = 0}^{N} a _{k} z ^{- k}}

Take the transform term by term with $x [n - k] \leftrightarrow z^{- k} X (z)$ and zero initial conditions. Sign convention: every output term is on the left. If the equation is written $y [n] = 0.5 y [n - 1] + x [n]$ , the feedback coefficient changes sign when moved across: $H (z) = \frac{1}{1 - 0.5 z ^{- 1}}$ .

Causality and stability from H(z)

Property	Condition on $h [n]$	Condition on $H (z)$ and its ROC
Causal	$h [n] = 0$ for $n < 0$	ROC is the exterior of a circle and includes $z = \infty$ . For a rational $H (z)$ : ROC outside the outermost pole, and numerator degree in $z$ not greater than the denominator degree.
BIBO stable	$\sum_{n} ∣ h [n] ∣ < \infty$	ROC contains the unit circle $∣ z ∣ = 1$ .
Causal and stable	both of the above	For a rational $H (z)$ : the causal ROC $∣ z ∣ > r_{m a x}$ with $r_{m a x} < 1$ , i.e. every pole strictly inside the unit circle, and numerator degree in $z$ not greater than the denominator degree.

FREQUENCY RESPONSE

H (e^{j Ω}) = H (z)_{z = e^{j Ω}}

Exists when the ROC contains the unit circle (a stable system). $Ω = 0$ is the point $z = 1$ (DC) and $Ω = π$ is $z = - 1$ (the highest digital frequency). $H (e^{j Ω})$ has period $2 π$ .

STEADY-STATE RESPONSE TO A SINUSOID

x [n] = A cos (Ω_{0} n + θ) \Rightarrow y [n] = A H (e^{j Ω_{0}}) cos (Ω_{0} n + θ + ∠ H (e^{j Ω_{0}}))

For a stable LTI system with real $h [n]$ . It is exact for an input present for all $n$ ; for an input switched on at $n = 0$ it is the response once the transients have died out. A constant input $A$ gives the output $A H (1)$ .

Reading the pole-zero plot. Write $H (z) = K \frac{\prod _{i} ( z - z _{i} )}{\prod _{i} ( z - p _{i} )}$ . Then $∣ H (e^{j Ω}) ∣ = ∣ K ∣ \frac{\prod _{i} ∣ e ^{j Ω} - z _{i} ∣}{\prod _{i} ∣ e ^{j Ω} - p _{i} ∣}$ , the product of distances from the point $e^{j Ω}$ on the unit circle to the zeros, divided by the product of distances to the poles. A zero on the unit circle at angle $Ω_{1}$ gives an exact null at $Ω = Ω_{1}$ . A pole close to the unit circle gives a sharp peak near its angle. Poles and zeros at the origin change only the phase (they are pure delays or advances).

Pole position and the mode it adds to a causal h[n] (real coefficients)

Pole of $H (z)$	Term in $h [n]$ , $n \geq 0$	Behaviour
at $z = 0$	impulses $δ [n - k]$	lasts a finite number of samples
real $p$ , $0 < p < 1$	$p^{n}$	decays without oscillating
real $p$ , $- 1 < p < 0$	$p^{n}$	decays, alternating in sign every sample
complex pair $r e^{\pm j θ}$ , $r < 1$	$r^{n} cos (θ n + ϕ)$	decaying oscillation at $θ$ rad/sample
simple pole on $∣ z ∣ = 1$	$1$ at $z = 1$ ; $(- 1)^{n}$ at $z = - 1$ ; $cos (θ n + ϕ)$ for a pair $e^{\pm j θ}$	neither decays nor grows; not BIBO stable
repeated pole on $∣ z ∣ = 1$ , order $m$	terms up to $n^{m - 1}$ times the simple-pole mode	grows without bound
$∣ p ∣ > 1$	$p^{n}$ , or $r^{n} cos (θ n + ϕ)$ for a pair	grows without bound

FIR and IIR systems

	FIR	IIR
Difference equation	non-recursive: $y [n] = \sum_{k = 0}^{M} b_{k} x [n - k]$ (causal FIR)	recursive: past outputs are fed back ( $a_{k} \neq = 0$ for some $k \geq 1$ )
$H (z)$	a polynomial in $z^{- 1}$ ; for a causal FIR every pole is at $z = 0$	a ratio of polynomials with at least one pole away from the origin after common factors cancel
$h [n]$	finite length	infinite length
Stability	always BIBO stable	causal IIR: stable if and only if every pole is inside the unit circle
Exact linear phase	possible, with symmetric or antisymmetric $h [n]$	not achievable by a causal, stable IIR filter

RECURSIVE IS NOT THE SAME AS IIR

A recursive equation can still have a finite-length $h [n]$ when its poles are cancelled by zeros. $y [n] = y [n - 1] + x [n] - x [n - N]$ has $H (z) = \frac{1 - z ^{- N}}{1 - z ^{- 1}} = 1 + z^{- 1} + \dots + z^{- (N - 1)}$ : the pole at $z = 1$ cancels, and $h [n] = u [n] - u [n - N]$ is an $N$ -point running sum, an FIR system.

FIRST-ORDER ALLPASS SECTION

H_{a p} (z) = \frac{z ^{- 1} - a ^{*}}{1 - a z ^{- 1}}, H_{a p} (e^{j Ω}) = 1

Pole at $a$ , zero at $1/ a^{*}$ (its mirror image in the unit circle). Causal and stable for $∣ a ∣ < 1$ . Any pole-zero pair placed as mirror images in the unit circle has a flat magnitude response.

A causal, stable system is minimum phase when all its zeros are also inside the unit circle. The zeros of $H (z)$ become the poles of the inverse system $1/ H (z)$ , so a causal and stable inverse needs every zero of $H$ inside the unit circle. It also needs the numerator and denominator of $H (z)$ to have equal degree in $z$ (in the $z^{- 1}$ form, $b_{0} \neq = 0$ ). The pure delay $H (z) = z^{- 1}$ has no finite zeros, but its inverse $z$ is an advance, which is not causal. When both conditions hold, the inverse is causal and stable.

RIGHT-SIDED IS NOT THE SAME AS CAUSAL

$H (z) = \frac{z ^{2}}{z - 0.5}$ with $∣ z ∣ > 0.5$ is right-sided, but $h [n] = (0.5)^{n + 1} u [n + 1]$ is non-zero at $n = - 1$ . A rational $H (z)$ with more finite zeros than finite poles cannot be causal.

STABLE DOES NOT NEED CAUSAL

The ROC $0.5 < ∣ z ∣ < 2$ in Example 1 contains the unit circle, so that system is stable although it is two-sided. Poles at the origin (from pure delays) never affect stability.

A POLE ON THE UNIT CIRCLE IS NOT BIBO STABLE

The accumulator $y [n] = y [n - 1] + x [n]$ has $H (z) = \frac{1}{1 - z ^{- 1}}$ , a simple pole at $z = 1$ , and $h [n] = u [n]$ , which is not absolutely summable. The bounded input $x [n] = u [n]$ gives $y [n] = (n + 1) u [n]$ , which grows without bound. A causal system with any pole on the unit circle is not BIBO stable; some texts call the simple-pole case marginally stable.

POLE-ZERO SHORTCUTS

A zero at $z = - 1$ kills $Ω = π$ (for example $1 + z^{- 1}$ , a lowpass shape). A zero at $z = 1$ blocks DC (for example $1 - z^{- 1}$ , a highpass shape). A system with real coefficients has its complex poles and zeros in conjugate pairs.

5. Unilateral z-transform and difference equations with initial conditions

The bilateral transform assumes the system is at rest. When a difference equation comes with initial conditions $y [- 1], y [- 2], \dots$ , use the unilateral transform: its shift property brings those values into the algebra automatically.

UNILATERAL DELAY PROPERTY

x [n - k] \leftrightarrow z^{- k} X^{+} (z) + m = 1 \sum k x [- m] z^{- (k - m)}

The cases used most: $x [n - 1] \leftrightarrow z^{- 1} X^{+} (z) + x [- 1]$ and $x [n - 2] \leftrightarrow z^{- 2} X^{+} (z) + z^{- 1} x [- 1] + x [- 2]$ .

UNILATERAL ADVANCE PROPERTY

x [n + k] \leftrightarrow z^{k} X^{+} (z) - m = 0 \sum k - 1 x [m] z^{k - m}

For $k = 1$ : $x [n + 1] \leftrightarrow z X^{+} (z) - z x [0]$ . For $k = 2$ : $z^{2} X^{+} (z) - z^{2} x [0] - z x [1]$ .

Solving a difference equation with initial conditions

Take the unilateral transform of both sides, using the delay property with the given $y [- 1], y [- 2], \dots$ For an input applied at $n = 0$ , $x [- 1] = x [- 2] = \dots = 0$ .
Solve the resulting algebraic equation for $Y^{+} (z)$ .
Separate the terms: those produced by the initial conditions form the zero-input response; the term $H (z) X^{+} (z)$ is the zero-state response.
Invert by partial fractions. Every term is right-sided, and the result holds for $n \geq 0$ .
Check $y [0]$ (and $y [1]$ ) by substituting directly into the difference equation, and check the final value with the final value theorem if the poles allow it.

WORKED EXAMPLE · EXAMPLE 2: FIRST-ORDER SYSTEM WITH AN INITIAL CONDITION

A system obeys $y [n] - 0.5 y [n - 1] = x [n]$ . With $x [n] = u [n]$ and $y [- 1] = 4$ , find $y [n]$ for $n \geq 0$ , and separate the zero-input and zero-state responses.

Solution

Unilateral transform: $Y^{+} (z) - 0.5 [z^{- 1} Y^{+} (z) + y [- 1]] = \frac{1}{1 - z ^{- 1}}$ .
So $(1 - 0.5 z^{- 1}) Y^{+} (z) = 2 + \frac{1}{1 - z ^{- 1}}$ , and $Y^{+} (z) = \frac{2}{1 - 0.5 z ^{- 1}} + \frac{1}{( 1 - 0.5 z ^{- 1} ) ( 1 - z ^{- 1} )}$ .
Zero-input part: $\frac{2}{1 - 0.5 z ^{- 1}} \leftrightarrow 2 (0.5)^{n} u [n]$ .
Zero-state part: $\frac{1}{( 1 - 0.5 z ^{- 1} ) ( 1 - z ^{- 1} )} = \frac{A}{1 - 0.5 z ^{- 1}} + \frac{B}{1 - z ^{- 1}}$ , with $A = \frac{1}{1 - z ^{- 1}}_{z = 0.5} = - 1$ and $B = \frac{1}{1 - 0.5 z ^{- 1}}_{z = 1} = 2$ . So the zero-state response is $[2 - (0.5)^{n}] u [n]$ .
Total: $y [n] = 2 (0.5)^{n} + 2 - (0.5)^{n} = 2 + (0.5)^{n}$ for $n \geq 0$ .
Check: from the equation, $y [0] = 0.5 (4) + 1 = 3$ and $y [1] = 0.5 (3) + 1 = 2.5$ ; the formula gives $3$ and $2.5$ . Final value: $lim_{z \to 1} (z - 1) Y^{+} (z) = \frac{1}{1 - 0.5} = 2$ , valid because the only other pole, $0.5$ , is inside the unit circle.

Answer: $y [n] = 2 + (0.5)^{n}$ for $n \geq 0$ : zero-input response $2 (0.5)^{n}$ , zero-state response $2 - (0.5)^{n}$ .

ZERO-STATE IS NOT THE SAME AS FORCED

The zero-state response $2 - (0.5)^{n}$ in Example 2 itself contains a natural-mode term $(0.5)^{n}$ , so the zero-input response is not the whole transient. Read carefully whether a question asks for the zero-input, zero-state, natural or forced response.

6. From s to z: sampling, the mapping and the pulse transfer function

Sampling $x (t)$ every $T$ seconds gives $x [n] = x (n T)$ . The Laplace transform of the impulse-sampled signal $x^{*} (t) = \sum_{n} x (n T) δ (t - n T)$ is $X^{*} (s) = \sum_{n} x (n T) e^{- s n T}$ , which is exactly $X (z)$ with $z = e^{s T}$ . This is the bridge between continuous-time and discrete-time pole locations. For a signal that jumps at $t = 0$ , the sample $x (0)$ is taken as $x (0^{+})$ ; the standard tables (for example $e^{- a t} u (t) \to 1$ at $n = 0$ ) use this convention.

THE S-TO-Z MAPPING

z = e^{s T}, s = σ + j ω \Rightarrow ∣ z ∣ = e^{σ T}, ∠ z = ω T

$T$ is the sampling period. The digital frequency is $Ω = ω T = 2 π f / f_{s}$ rad/sample. A continuous-time pole at $s = p$ appears at $z = e^{pT}$ in the transform of the sampled signal.

SAMPLED EXPONENTIAL

e^{- a t} u (t) \to e^{- an T} u [n] \leftrightarrow \frac{z}{z - e ^{- a T}}, ∣ z ∣ > e^{- a T}

For real $a$ . The pole $s = - a$ maps to $z = e^{- a T}$ , inside the unit circle when $a > 0$ .

How regions of the s-plane map under z = e^{sT}

s-plane	z-plane
imaginary axis, $σ = 0$	unit circle, $∣ z ∣ = 1$
left half-plane, $σ < 0$	inside the unit circle
right half-plane, $σ > 0$	outside the unit circle
$s = 0$	$z = 1$
negative real axis, $s = - a$ ( $a > 0$ )	the segment $0 < z < 1$ of the real axis
vertical line $σ = σ_{0}$	circle of radius $e^{σ_{0} T}$
horizontal line $ω = ω_{0}$	radial line at angle $ω_{0} T$
$ω = \pm π / T$ (half the sampling frequency)	negative real axis
primary strip $- π / T < ω \leq π / T$	the whole z-plane except $z = 0$
line of constant damping ratio $ζ$ , $0 < ζ < 1$	a logarithmic spiral into $z = 0$

THE MAPPING IS MANY-TO-ONE

$s$ and $s + j 2 π k / T$ land on the same $z$ for every integer $k$ : that is aliasing, and only the primary strip maps one-to-one. Poles map by $z = e^{pT}$ ; zeros, in general, do not.

BILINEAR TRANSFORMATION

s = \frac{2}{T} \frac{1 - z ^{- 1}}{1 + z ^{- 1}}, ω = \frac{2}{T} tan \frac{Ω}{2}

A different mapping, used to design a digital filter from an analog one. It also sends the $j ω$ axis onto the unit circle and the left half-plane inside it, one-to-one, so there is no aliasing, but the frequency axis is warped as shown. It is not $z = e^{s T}$ .

Pulse transfer function (IN). In a sampled-data system a sampler (period $T$ ) and a zero-order hold (ZOH) drive a continuous plant $G_{p} (s)$ , and the output is read at the sampling instants. The pulse transfer function is $G (z) = C (z) / R (z)$ with zero initial conditions, where $C (z)$ is the z-transform of $c (n T)$ . Notation: $Z {G (s)}$ means the z-transform of the samples $g (n T)$ of $g (t) = L^{- 1} {G (s)}$ .

ZERO-ORDER HOLD WITH PLANT

G_{h} (s) = \frac{1 - e ^{- s T}}{s}, G (z) = Z {\frac{1 - e ^{- s T}}{s} G_{p} (s)} = (1 - z^{- 1}) Z {\frac{G _{p} ( s )}{s}}

Exact at the sampling instants: $G (z)$ is $(1 - z^{- 1})$ times the transform of the sampled step response of the plant.

ZOH WITH A FIRST-ORDER PLANT

G_{p} (s) = \frac{K}{s + a} \Rightarrow G (z) = \frac{K}{a} \cdot \frac{1 - e ^{- a T}}{z - e ^{- a T}}

For $a \neq = 0$ . From $\frac{K}{s ( s + a )} = \frac{K}{a} (\frac{1}{s} - \frac{1}{s + a})$ . Without the hold, $Z {G_{p} (s)} = \frac{K z}{z - e ^{- a T}}$ , a different answer.

CASCADED BLOCKS IN A SAMPLED-DATA SYSTEM

sampler between the blocks: G_{1} (z) G_{2} (z), no sampler between: G_{1} G_{2} (z) = Z {G_{1} (s) G_{2} (s)}

When a sampler separates $G_{1} (s)$ and $G_{2} (s)$ , their pulse transfer functions multiply. When none does, multiply in $s$ first and then transform; in general $G_{1} G_{2} (z) \neq = G_{1} (z) G_{2} (z)$ . Example with $G_{1} (s) = G_{2} (s) = \frac{1}{s + 1}$ and $T = 1$ : with a sampler between them, $\frac{z ^{2}}{( z - e ^{- 1} ) ^{2}}$ ; without, $Z {\frac{1}{( s + 1 ) ^{2}}} = \frac{e ^{- 1} z}{( z - e ^{- 1} ) ^{2}}$ .

CLOSED LOOP WITH THE SAMPLER ON THE ERROR

\frac{C ( z )}{R ( z )} = \frac{G ( z )}{1 + G H ( z )}, G H (z) = Z {G (s) H (s)}

$G (s)$ is the forward path after the sampler (hold included) and $H (s)$ the continuous feedback path. Because $G (s)$ and $H (s)$ are not separated by a sampler, they are multiplied before transforming. The loop is stable when every root of $1 + G H (z) = 0$ lies inside the unit circle.

STEADY-STATE ERROR OF A UNITY-FEEDBACK SAMPLED LOOP

e_{ss}_{r = u (t)} = \frac{1}{1 + lim _{z \to 1} G ( z )}, e_{ss}_{r = t u (t)} = \frac{T}{lim _{z \to 1} ( z - 1 ) G ( z )}

Unity feedback, the sampler on the error, $G (z)$ the pulse transfer function of the forward path (hold included), and a stable closed loop (otherwise the final value theorem does not apply). From $E (z) = \frac{R ( z )}{1 + G ( z )}$ and the final value theorem, with $R (z) = \frac{z}{z - 1}$ for a step and $R (z) = \frac{T z}{( z - 1 ) ^{2}}$ for the sampled ramp. The constants are often written $K_{p} = lim_{z \to 1} G (z)$ and $K_{v} = \frac{1}{T} lim_{z \to 1} (z - 1) G (z)$ , so that $e_{ss} = \frac{1}{1 + K _{p}}$ for a step and $\frac{1}{K _{v}}$ for a ramp. Each pole of $G (z)$ at $z = 1$ plays the part of an integrator.

STABILITY TEST FOR A SECOND-ORDER CHARACTERISTIC POLYNOMIAL

P (z) = z^{2} + a_{1} z + a_{0} : ∣ a_{0} ∣ < 1, P (1) = 1 + a_{1} + a_{0} > 0, P (- 1) = 1 - a_{1} + a_{0} > 0

For real coefficients, all three conditions together are necessary and sufficient for both roots to lie strictly inside the unit circle (the Jury conditions for order 2). For $a_{2} z^{2} + a_{1} z + a_{0}$ with $a_{2} > 0$ the first condition becomes $∣ a_{0} ∣ < a_{2}$ . At any order $n$ , with a positive leading coefficient, $P (1) > 0$ and $(- 1)^{n} P (- 1) > 0$ are necessary. For a full test at higher orders, use the Jury table, or substitute $z = \frac{1 + w}{1 - w}$ and apply the Routh test to the polynomial in $w$ .

PREVIOUS-YEAR QUESTIONS

12 questions from the EC, EE and IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2026 · GATE ECE 2025 · GATE ECE 2024 · GATE ECE 2020 · GATE ECE 2018 · GATE EE 2026 · GATE EE 2024 · GATE EE 2023 · GATE EE 2020 · GATE IN 2024 · GATE IN 2018

GATE 2026 EC Q.16

GATE 2026 · ECQ.16 · 1 mark · MCQ

The response of a discrete time system $y [n]$ obeys the following relation:

y [n] = \frac{5}{6} y [n - 1] - \frac{1}{6} y [n - 2] + x [n] .

The input to the system is $x [n] = δ [n] - \frac{1}{3} δ [n - 1]$ .

Which of the following options is TRUE for $y [n]$ ?

(A)
Stable and causal response
(B)
Stable and non-causal response
(C)
Unstable and causal response
(D)
Unstable and non-causal response

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GATE 2026 EE Q.11

GATE 2026 · EEQ.11 · 1 mark · MCQ

Consider the infinite-length, discrete-time sequence $x [n] = 0. 9^{∣ n ∣}$ , where $n$ is an integer. The region of convergence of its Z-transform $X (z)$ is given by:

(Note: $z$ is a complex variable)

(A)
$∣ z ∣ > 0.9$
(B)
$∣ z ∣ < 0.9$
(C)
$0.9 < ∣ z ∣ < 1/0.9$
(D)
${z such that ∣ z ∣ < 0.9} \cup {z such that ∣ z ∣ > 1/0.9}$

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GATE 2025 EC Q.27

GATE 2025 · ECQ.27 · 1 mark · MSQ

Let $x [n]$ be a discrete-time signal whose $z$ -transform is $X (z)$ .

Which of the following statements is/are TRUE?

(A)
The discrete-time Fourier transform (DTFT) of $x [n]$ always exists
(B)
The region of convergence (RoC) of $X (z)$ contains neither poles nor zeros
(C)
The discrete-time Fourier transform (DTFT) exists if the region of convergence (RoC) contains the unit circle
(D)
If $x [n] = α δ [n]$ , where $δ [n]$ is the unit impulse and $α$ is a scalar, then the region of convergence (RoC) is the entire $z$ -plane

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GATE 2024 EC Q.24

GATE 2024 · ECQ.24 · 1 mark · MSQ

For a causal discrete-time LTI system with transfer function
$H (z) = \frac{2 z ^{2} + 3}{( z + \frac{1}{3} ) ( z - \frac{1}{3} )}$
which of the following statements is/are true?

(A)
The system is stable.
(B)
The system is a minimum phase system.
(C)
The initial value of the impulse response is 2.
(D)
The final value of the impulse response is 0.

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GATE 2024 EE Q.39

GATE 2024 · EEQ.39 · 2 marks · MCQ

If the $Z$ -transform of a finite-duration discrete-time signal $x [n]$ is $X (z)$ , then the $Z$ -transform of the signal $y [n] = x [2 n]$ is

(A)
$Y (z) = X (z^{2})$
(B)
$Y (z) = \frac{1}{2} [X (z^{- 1/2}) + X (- z^{- 1/2})]$
(C)
$Y (z) = \frac{1}{2} [X (z^{1/2}) + X (- z^{1/2})]$
(D)
$Y (z) = \frac{1}{2} [X (z^{2}) + X (- z^{2})]$

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GATE 2024 IN Q.38

GATE 2024 · INQ.38 · 2 marks · MCQ

Consider a discrete-time sequence

x [n] = {(0.2)^{n}, 0, 0 \leq n \leq 7 otherwise

The region of convergence of $X (z)$ , the $z$ -transform of $x [n]$ , consists of

(A)
all values of $z$ except $z = 0.2$
(B)
all values of $z$
(C)
all values of $z$ except $z = 0$
(D)
all values of $z$ except $z = \infty$

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GATE 2023 EE Q.19

GATE 2023 · EEQ.19 · 1 mark · MCQ

The $Z$ -transform of a discrete signal $x [n]$ is

X (z) = \frac{4 z}{( z - \frac{1}{5} ) ( z - \frac{2}{3} ) ( z - 3 )} with R O C = R .

Which one of the following statements is true?

(A)
Discrete-time Fourier transform of x[n] converges if $R$ is $∣ z ∣ > 3$
(B)
Discrete-time Fourier transform of x[n] converges if $R$ is $\frac{2}{3} < ∣ z ∣ < 3$
(C)
Discrete-time Fourier transform of x[n] converges if $R$ is such that x[n] is a left-sided sequence
(D)
Discrete-time Fourier transform of x[n] converges if $R$ is such that x[n] is a right-sided sequence

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GATE 2020 EC Q.14

GATE 2020 · ECQ.14 · 1 mark · MCQ

Which one of the following pole-zero plots corresponds to the transfer function of an LTI system characterized by the input-output difference equation given below?

y [n] = k = 0 \sum 3 (- 1)^{k} x [n - k]

GATE 2020 EC question 14: Each option is a z-plane pole-zero plot: horizontal axis Re, vertical axis Im, with the unit circle drawn and the points… — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
Figure — zeros (o) at $z = 1$ , $z = j$ and $z = - j$ on the unit circle; a cross at the origin labelled "3rd order pole".
(B)
Figure — zeros (o) at $z = - 1$ , $z = j$ and $z = - j$ on the unit circle; a cross at the origin labelled "3rd order pole".
(C)
Figure — zeros (o) at $z = - 1$ and at two conjugate points on the unit circle in the right half-plane, at about $\pm 6 0^{\circ}$ (i.e. $e^{\pm j π /3}$ ); a cross at the origin labelled "3rd order pole".
(D)
Figure — zeros (o) at $z = 1$ , $z = - 1$ , $z = j$ and $z = - j$ on the unit circle; a cross at the origin labelled "4th order pole".

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GATE 2020 EC Q.14: answer and solution

GATE 2020 EC Q.53

GATE 2020 · ECQ.53 · 2 marks · NAT

The transfer function of a stable discrete-time LTI system is $H (z) = \frac{K ( z - α )}{z + 0.5}$ , where $K$ and $α$ are real numbers. The value of $α$ (rounded off to one decimal place) with $∣ α ∣ > 1$ , for which the magnitude response of the system is constant over all frequencies, is ________.

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GATE 2020 EE Q.4

GATE 2020 · EEQ.4 · 1 mark · MCQ

Consider a signal $x [n] = (\frac{1}{2})^{n} 1 [n]$ , where $1 [n] = 0$ if $n < 0$ , and $1 [n] = 1$ if $n \geq 0$ . The z-transform of $x [n - k]$ , $k > 0$ is $\frac{z ^{- k}}{1 - \frac{1}{2} z ^{- 1}}$ with region of convergence being

(A)
$∣ z ∣ < 2$
(B)
$∣ z ∣ > 2$
(C)
$∣ z ∣ < 1/2$
(D)
$∣ z ∣ > 1/2$

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GATE 2018 EC Q.13

GATE 2018 · ECQ.13 · 1 mark · MCQ

A discrete-time all-pass system has two of its poles at $0.25∠ 0^{\circ}$ and $2∠3 0^{\circ}$ . Which one of the following statements about the system is TRUE?

(A)
It has two more poles at $0.5∠3 0^{\circ}$ and $4∠ 0^{\circ}$ .
(B)
It is stable only when the impulse response is two-sided.
(C)
It has constant phase response over all frequencies.
(D)
It has constant phase response over the entire $z$ -plane.

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GATE 2018 IN Q.34

GATE 2018 · INQ.34 · 2 marks · MCQ

Let $y [n] = x [n] * h [n]$ , where $*$ denotes convolution and $x [n]$ and $h [n]$ are two discrete time sequences. Given that the z-transform of $y [n]$ is $Y (z) = 2 + 3 z^{- 1} + z^{- 2}$ , the z-transform of $p [n] = x [n] * h [n - 2]$ is

(A)
$2 + 3 z + z^{- 2}$
(B)
$3 z + z^{- 2}$
(C)
$2 z^{2} + 3 z + 1$
(D)
$2 z^{- 2} + 3 z^{- 3} + z^{- 4}$

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PRACTICE

6 questions to test yourself

PRACTICE 11 mark · MCQ

Under the mapping $z = e^{s T}$ with sampling period $T = 0.5$ s, the region $Re (s) < - 2$ of the s-plane maps onto

(A)
$0 < ∣ z ∣ < e^{- 1}$
(B)
$∣ z ∣ > e^{- 1}$
(C)
$0 < ∣ z ∣ < e^{- 2}$
(D)
$e^{- 1} < ∣ z ∣ < 1$

Answer and solution

Answer: (A)

With $s = σ + j ω$ , $∣ z ∣ = e^{σ T}$ depends only on $σ$ , and $∠ z = ω T$ depends only on $ω$ .
$σ < - 2$ gives $∣ z ∣ < e^{- 2 \times 0.5} = e^{- 1}$ . As $σ \to - \infty$ , $∣ z ∣ \to 0$ without reaching it.
Every angle $ω T$ occurs, so the image is the whole punctured disc $0 < ∣ z ∣ < e^{- 1} \approx 0.368$ .

PRACTICE 21 mark · NAT

A causal sequence $x [n]$ has the z-transform $X (z) = \frac{z ^{2} + 1}{( z - 1 ) ( z - 0.5 )}$ . The value of $lim_{n \to \infty} x [n]$ is ______.

Answer and solution

Answer: $4$ (accepted range $\pm 0.01$ )

$(z - 1) X (z) = \frac{z ^{2} + 1}{z - 0.5}$ has its only pole at $z = 0.5$ , inside the unit circle, so the final value theorem applies.
$lim_{n \to \infty} x [n] = lim_{z \to 1} \frac{z ^{2} + 1}{z - 0.5} = \frac{2}{0.5} = 4$ .
Check by partial fractions: $X (z) = 2 + \frac{4 z}{z - 1} - \frac{5 z}{z - 0.5}$ , so $x [n] = 2 δ [n] + [4 - 5 (0.5)^{n}] u [n]$ , which tends to $4$ .

PRACTICE 32 marks · MSQ

A causal LTI system has the system function $H (z) = \frac{1 - 4 z ^{- 2}}{1 - 0.25 z ^{- 2}}$ . Which of the following statements is/are correct?

(A)
The system is BIBO stable.
(B)
The system is minimum phase (all its poles and zeros lie inside the unit circle).
(C)
$H (e^{j Ω}) = 4$ for every $Ω$ .
(D)
There is an inverse system $1/ H (z)$ that is both causal and stable.

Answer and solution

Answer: (A), (C)

Poles: $z^{2} = 0.25$ , so $z = \pm 0.5$ . Zeros: $z^{2} = 4$ , so $z = \pm 2$ .
The system is causal, so the ROC is $∣ z ∣ > 0.5$ , which contains the unit circle. It is stable: (A) is correct.
The zeros at $\pm 2$ lie outside the unit circle, so the system is not minimum phase: (B) is wrong.
On the unit circle let $w = e^{- j 2Ω}$ , so $∣ w ∣ = 1$ and $w^{- 1} = w^{*}$ . Then $∣1 - 4 w ∣ = ∣ w ∣ ∣ w^{*} - 4∣ = ∣4 - w ∣$ and $∣1 - 0.25 w ∣ = 0.25 ∣4 - w ∣$ , so $∣ H (e^{j Ω}) ∣ = 4$ for every $Ω$ . Spot check: $H (1) = \frac{- 3}{0.75} = - 4$ . (C) is correct.
The inverse has poles at $\pm 2$ . Its causal version (ROC $∣ z ∣ > 2$ ) excludes the unit circle and is unstable; its stable version (ROC $∣ z ∣ < 2$ ) is anticausal. (D) is wrong.

PRACTICE 42 marks · NAT

$X (z) = \frac{3.5 z}{( z + 0.5 ) ( z - 3 )}$ is the z-transform of an absolutely summable sequence $x [n]$ . The value of $x [- 1]$ is ______ (rounded to three decimal places).

Answer and solution

Answer: $- 0.333$ (accepted range $\pm 0.005$ )

Absolutely summable means the ROC contains $∣ z ∣ = 1$ , so the ROC is $0.5 < ∣ z ∣ < 3$ .
$\frac{X ( z )}{z} = \frac{3.5}{( z + 0.5 ) ( z - 3 )} = \frac{- 1}{z + 0.5} + \frac{1}{z - 3}$ (coefficients $\frac{3.5}{- 0.5 - 3} = - 1$ and $\frac{3.5}{3 + 0.5} = 1$ ).
$X (z) = - \frac{z}{z + 0.5} + \frac{z}{z - 3}$ . The pole at $- 0.5$ is inside the ring, giving $- (- 0.5)^{n} u [n]$ . The pole at $3$ is outside it, giving $- 3^{n} u [- n - 1]$ .
At $n = - 1$ only the left-sided term contributes: $x [- 1] = - 3^{- 1} = - 0.333$ .

PRACTICE 52 marks · MCQ

In a sampled-data system, a zero-order hold with sampling period $T = 1$ s drives the plant $G_{p} (s) = \frac{1}{s ( s + 1 )}$ . Taking $e^{- 1} = 0.368$ , the pulse transfer function $G (z)$ of the hold and plant together is

(A)
$\frac{0.368 z + 0.264}{( z - 1 ) ( z - 0.368 )}$
(B)
$\frac{0.632 z}{( z - 1 ) ( z - 0.368 )}$
(C)
$\frac{0.632}{z - 0.368}$
(D)
$\frac{0.264 z + 0.368}{( z - 1 ) ( z - 0.368 )}$

Answer and solution

Answer: (A)

$G (z) = (1 - z^{- 1}) Z {\frac{1}{s ^{2} ( s + 1 )}}$ .
$\frac{1}{s ^{2} ( s + 1 )} = \frac{1}{s ^{2}} - \frac{1}{s} + \frac{1}{s + 1}$ . Its time function sampled at $t = n T$ is $n T - 1 + e^{- n T}$ , with transform $\frac{T z}{( z - 1 ) ^{2}} - \frac{z}{z - 1} + \frac{z}{z - e ^{- T}}$ .
Multiplying by $\frac{z - 1}{z}$ and putting $T = 1$ : $G (z) = \frac{1}{z - 1} - 1 + \frac{z - 1}{z - 0.368}$ .
Over the common denominator $(z - 1) (z - 0.368)$ the numerator is $(z - 0.368) - (z - 1) (z - 0.368) + (z - 1)^{2} = 0.368 z + 0.264$ .
Option B is $Z {G_{p} (s)}$ with the hold left out; option C is the hold with $\frac{1}{s + 1}$ only, dropping the integrator; option D swaps the numerator coefficients.

PRACTICE 62 marks · NAT

A unity-feedback sampled-data loop has the sampler on the error signal, sampling period $T = 1$ s, and forward-path pulse transfer function (hold and plant together) $G (z) = \frac{0.368 z + 0.264}{( z - 1 ) ( z - 0.368 )}$ . For the input $r (t) = t u (t)$ , the steady-state error at the sampling instants is ______ (rounded to two decimal places).

Answer and solution

Answer: $1$ (accepted range $\pm 0.02$ )

The final value theorem needs a stable closed loop, so check that first. Characteristic polynomial: $(z - 1) (z - 0.368) + 0.368 z + 0.264 = z^{2} - z + 0.632$ .
Second-order Jury conditions: $∣0.632∣ < 1$ ; $P (1) = 1 - 1 + 0.632 = 0.632 > 0$ ; $P (- 1) = 1 + 1 + 0.632 = 2.632 > 0$ . Both roots are inside the unit circle, so the loop is stable.
$lim_{z \to 1} (z - 1) G (z) = \frac{0.368 + 0.264}{1 - 0.368} = \frac{0.632}{0.632} = 1$ .
$e_{ss} = \frac{T}{lim _{z \to 1} ( z - 1 ) G ( z )} = \frac{1}{1} = 1$ , i.e. $K_{v} = 1$ . As a check, the continuous plant $\frac{1}{s ( s + 1 )}$ that this $G (z)$ comes from has the same velocity constant, $lim_{s \to 0} s \cdot \frac{1}{s ( s + 1 )} = 1$ .

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EC

z-transform

GATE EE

Z transform

GATE IN

z-transforms … pulse transfer function

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