Unit 5 · Signals & Systems

Z-Transform —
Discrete-Time System Analysis

B.Tech ECE · Signals & Systems· ~30 min read· 10 GATE Questions· GATE Weightage: 1–2 marks
Contents
  1. 1. Discrete-Time Signals
  2. 2. DTFT — Discrete-Time Fourier Transform
  3. 3. Z-Transform — Definition & ROC
  4. 4. Standard Z-Transform Pairs
  5. 5. Properties of Z-Transform
  6. 6. Inverse Z-Transform
  7. 7. DT-LTI Systems & Transfer Function H(z)
  8. 8. Stability in the z-Plane
  9. 9. GATE Questions (10)
Real-World Context — Where is Z-Transform Used?

Digital audio: Your smartphone's equalizer, echo cancellation, and noise reduction are all DT-LTI filters designed using the Z-transform. The output of a digital filter y[n] = b₀x[n] + b₁x[n-1] + b₂x[n-2] − a₁y[n-1] − a₂y[n-2] is completely characterized by H(z) = (b₀ + b₁z⁻¹ + b₂z⁻²)/(1 + a₁z⁻¹ + a₂z⁻²).

Image processing: Blur and sharpen filters applied to pixel arrays are DT-LTI systems analyzed with 2D Z-transforms.

Control systems (digital): A PI controller implemented in a microcontroller uses the bilinear transform z=(1+sT/2)/(1-sT/2) to convert a continuous H(s) to a discrete H(z), which is then run every sample interval T.

The z-Plane — ROC Regions, Unit Circle & Stability
Re(z) Im(z) Unit Circle |z| = 1 |z| < 1 Stable poles |z| > 1 Unstable poles +1 -1 +j -j pole (stable) |z|=0.6<1 conj. pair |z|≈0.8 pole (unstable) |z|=1.3>1 zero at z=0 ROC: |z| > r₀ (causal signal) s-plane → z-plane mapping z = e^(sT), T = sample interval jω-axis → unit circle |z|=1 LHP (stable) → inside unit circle

The unit circle |z|=1 is the DT analog of the jω-axis. Stability condition: all poles inside unit circle (|pᵢ|<1). The ROC for a causal signal is the region |z|>r₀ (outside a circle). For stability, ROC must include the unit circle.

1. Discrete-Time Signals

A discrete-time (DT) signal x[n] is defined only at integer values of n ∈ ℤ. It is obtained by sampling a CT signal at interval T: x[n] = x_c(nT).

Elementary DT Signals

Standard DT Signals (Must Know for GATE)
Unit Sample (Impulse): δ[n] = 1 for n=0, δ[n] = 0 for n≠0 Unit Step: u[n] = 1 for n≥0, u[n] = 0 for n<0 Unit Ramp: r[n] = n·u[n] Real Exponential: x[n] = aⁿu[n] (|a|<1: decays, |a|>1: grows) Complex Exponential: x[n] = e^(jω₀n) = cos(ω₀n) + j·sin(ω₀n) Sinusoidal: x[n] = A·cos(ω₀n + φ) Relation: u[n] = Σₖ₌₀^∞ δ[n-k] and δ[n] = u[n] - u[n-1]
Elementary DT Signals — δ[n], u[n], and aⁿu[n]
Unit Impulse δ[n] -3 -2 -1 0 1 2 3 = 1 only at n=0 Unit Step u[n] -3 -2 -1 0 1 2 3 n = 1 for n ≥ 0 Decaying: (0.6)ⁿu[n] |a|=0.6<1 → decays

DT signals are defined only at integer n. Key: u[n] = Σδ[n-k] (superposition of shifts). For aⁿu[n]: if |a|<1 → decays (stable); if |a|>1 → grows (unstable); if a=1 → u[n] (unit step, marginally stable).

2. DTFT — Discrete-Time Fourier Transform

DTFT Definition
X(e^jω) = Σₙ₌₋∞^∞ x[n] e^(-jωn) [Analysis equation] x[n] = (1/2π) ∫₋π^π X(e^jω) e^(jωn) dω [Synthesis equation] Properties: · X(e^jω) is PERIODIC with period 2π: X(e^j(ω+2π)) = X(e^jω) · Exists if x[n] is absolutely summable: Σ|x[n]| < ∞ · For real x[n]: |X(e^jω)| is even, ∠X(e^jω) is odd

3. Z-Transform — Definition & ROC

Bilateral Z-Transform
X(z) = Σₙ₌₋∞^∞ x[n] z^(-n) where z ∈ ℂ Write z = re^(jω): X(re^jω) = Σₙ x[n] (re^jω)^(-n) = Σₙ [x[n]r^(-n)] e^(-jωn) → Z-Transform = DTFT of x[n]r^(-n) (exponentially weighted signal) Setting z = e^jω (r=1, on unit circle): X(e^jω) = DTFT of x[n] → FT is Z-transform evaluated ON the unit circle |z|=1

ROC in the z-Plane

Signal TypeROCExample
Right-sided (causal): x[n]=0 for n<N|z| > r₀ (outside a circle)aⁿu[n] → ROC: |z|>|a|
Left-sided: x[n]=0 for n>N|z| < r₀ (inside a circle)−aⁿu[−n−1] → ROC: |z|<|a|
Two-sidedr₁ < |z| < r₂ (annular ring)|z| between two poles
Finite-duration (causal)|z|>0, entire z-plane except z=0FIR filters
All time (δ[n])Entire z-planeX(z)=1

4. Standard Z-Transform Pairs

x[n]X(z)ROC
δ[n]1All z
u[n]1/(1 − z⁻¹) = z/(z−1)|z| > 1
n·u[n]z⁻¹/(1−z⁻¹)² = z/(z−1)²|z| > 1
aⁿu[n]1/(1−az⁻¹) = z/(z−a)|z| > |a|
n·aⁿu[n]az⁻¹/(1−az⁻¹)² = az/(z−a)²|z| > |a|
−aⁿu[−n−1]1/(1−az⁻¹) = z/(z−a)|z| < |a|
cos(ω₀n)u[n](z²−z·cos ω₀)/(z²−2z·cos ω₀+1)|z| > 1
sin(ω₀n)u[n](z·sin ω₀)/(z²−2z·cos ω₀+1)|z| > 1
aⁿcos(ω₀n)u[n](z²−az·cos ω₀)/(z²−2az·cos ω₀+a²)|z| > |a|
δ[n−k]z⁻ᵏAll z (except 0 if k>0)
GATE Trap — Same X(z), Different ROC = Different x[n]

X(z) = z/(z−a) with ROC |z|>|a| → x[n] = aⁿu[n] (causal, right-sided)
X(z) = z/(z−a) with ROC |z|<|a| → x[n] = −aⁿu[−n−1] (anti-causal, left-sided)
The ROC determines which x[n] corresponds to X(z). Always state ROC!

5. Properties of Z-Transform

PropertySequence x[n]Z-Transform X(z)
Linearityax[n] + by[n]aX(z) + bY(z)
Time Shiftingx[n−k]z⁻ᵏ X(z) [delay k → multiply by z⁻ᵏ]
z-Domain Scalingaⁿx[n]X(z/a) [ROC scaled by |a|]
Time Reversalx[−n]X(z⁻¹) [replace z by 1/z]
Differentiation (z-domain)n·x[n]−z·dX(z)/dz
Convolutionx[n] * h[n]X(z) · H(z) [key theorem]
Initial Valuex[0]lim_{z→∞} X(z) [causal x[n]]
Final Valuex[∞]lim_{z→1} (z−1)X(z) [poles inside unit circle]
Parseval'sΣ|x[n]|²(1/2πj)∮ X(z)X*(1/z*)z⁻¹ dz

6. Inverse Z-Transform

Three methods: (1) Partial fraction expansion + table lookup, (2) Power series (long division), (3) Contour integration (Cauchy's residue theorem).

Method 1 — Partial Fraction Expansion (Most Used in GATE)

PFE for Inverse Z-Transform
Step 1: Express X(z)/z (not X(z)) in partial fractions, then multiply by z. This gives cleaner fractions matching the table form z/(z−a). X(z)/z = A/(z−p₁) + B/(z−p₂) + ... X(z) = Az/(z−p₁) + Bz/(z−p₂) + ... Residue: A = [(z−p₁)·X(z)/z]_{z=p₁} Then: az/(z-a) ↔ aⁿu[n] (causal, ROC: |z|>|a|)

Method 2 — Long Division (Power Series)

Power Series Method
For causal x[n]: divide numerator by denominator to get series in z⁻¹: X(z) = x[0] + x[1]z⁻¹ + x[2]z⁻² + ... Example: X(z) = 1/(1 − 0.5z⁻¹) Long division: = 1 + 0.5z⁻¹ + 0.25z⁻² + 0.125z⁻³ + ... → x[n] = (0.5)ⁿu[n] ✓

7. DT-LTI Systems & Transfer Function H(z)

DT-LTI System — Transfer Function
Convolution sum: y[n] = x[n] * h[n] = Σₖ x[k] h[n−k] Z-domain: Y(z) = X(z) · H(z) Transfer function: H(z) = Y(z)/X(z) = Z{h[n]} For a difference equation: Σₖ₌₀^N aₖ y[n−k] = Σₖ₌₀^M bₖ x[n−k] H(z) = (b₀ + b₁z⁻¹ + ... + bₘz⁻ᴹ) / (a₀ + a₁z⁻¹ + ... + aₙz⁻ᴺ) = B(z)/A(z) [rational function of z] Frequency response: H(e^jω) = H(z)|_{z=e^jω} [z on unit circle]
Real-Life Lab — Digital FIR Filter Design

3-tap moving average filter: y[n] = (x[n] + x[n-1] + x[n-2])/3
H(z) = (1 + z⁻¹ + z⁻²)/3
All poles at z=0 (inside unit circle) → always stable! (This is why FIR filters are always BIBO stable.)
Zeros: 1+z⁻¹+z⁻² = 0 → z = e^(±j2π/3), at 120° on unit circle → attenuates those frequencies.
This filter smooths noisy sensor data — used in accelerometers, ECG preprocessing, and stock price averaging.

8. Stability in the z-Plane

BIBO Stability — z-Domain
A causal DT-LTI system is BIBO stable if and only if: All poles of H(z) are strictly INSIDE the unit circle: |pᵢ| < 1 Equivalently: ROC of H(z) includes the unit circle |z|=1. Stability categories: ┌────────────────────────────────────────────────────────────────┐ │ Stable: ALL poles: |pᵢ| < 1 (inside unit circle) │ │ Marginally: Poles on unit circle, non-repeated |pᵢ| = 1 │ │ Unstable: ANY pole: |pᵢ| > 1 (outside unit circle) │ └────────────────────────────────────────────────────────────────┘ Impulse response: h[n] absolutely summable ↔ BIBO stable: Σ|h[n]| < ∞
s-Plane vs z-Plane — Stability Comparison

s-plane: stable ↔ Re(pᵢ) < 0 (poles in left half-plane)
z-plane: stable ↔ |pᵢ| < 1 (poles inside unit circle)
Mapping: z = e^(sT). LHP (σ<0) → inside unit circle (|z|=|e^σT|=e^(σT)<1 since σ<0).

GATE PYQs

GATE Questions — Z-Transform & DT Signals (10)

GATE 2023 2 Marks MCQ
The Z-transform of x[n] = (0.5)ⁿu[n] is:
(A) z/(z − 0.5), |z| > 0.5
(B) z/(z + 0.5), |z| > 0.5
(C) 0.5z/(z − 0.5), |z| > 0.5
(D) 1/(z − 0.5), |z| > 0.5

Standard pair: aⁿu[n] ↔ z/(z−a) = 1/(1−az⁻¹), ROC: |z| > |a|

With a = 0.5: X(z) = z/(z−0.5), ROC: |z| > 0.5

Answer: (A) z/(z−0.5), |z| > 0.5

GATE 2022 2 Marks MCQ
A causal DT-LTI system has H(z) = 1/(1 − 0.8z⁻¹). The system is:
(A) BIBO stable
(B) BIBO unstable
(C) Marginally stable
(D) Cannot determine without ROC

H(z) = 1/(1−0.8z⁻¹) = z/(z−0.8). Pole at z = 0.8.

|0.8| = 0.8 < 1 → pole is inside unit circle → system is BIBO stable.

h[n] = (0.8)ⁿu[n] → Σ(0.8)ⁿ = 1/(1−0.8) = 5 < ∞ ✓

Answer: (A) BIBO stable

GATE 2021 1 Mark MCQ
The time-shift property of Z-transform states that x[n−k] ↔:
(A) z^k · X(z)
(B) z^(-k) · X(z)
(C) X(z−k)
(D) k · X(z)

Time shift: x[n−k] ↔ z⁻ᵏ·X(z) [delay by k samples = multiply by z⁻ᵏ]

Analog: in s-domain, delay by T → e^(−sT)·X(s). In z-domain, delay by k → z⁻ᵏ·X(z).

z⁻¹ is often called the "unit delay" operator — fundamental building block of digital filters.

Answer: (B) z⁻ᵏ·X(z)

GATE 2020 2 Marks MCQ
The inverse Z-transform of X(z) = 1/(1 − 2z⁻¹) with ROC |z| < 2 is:
(A) 2ⁿu[n]
(B) −2ⁿu[−n−1]
(C) 2⁻ⁿu[n]
(D) (1/2)ⁿu[n]

X(z) = 1/(1−2z⁻¹) = z/(z−2) has a pole at z=2.

ROC |z|<2 → left-sided (anti-causal) signal.

Using the pair: −aⁿu[−n−1] ↔ z/(z−a) with ROC |z|<|a|

With a=2: x[n] = −2ⁿu[−n−1]

Contrast: if ROC were |z|>2 → x[n] = 2ⁿu[n] (unstable, causal).

Answer: (B) −2ⁿu[−n−1]

GATE 2019 2 Marks MCQ
For the system y[n] − 0.5y[n−1] = x[n], the transfer function H(z) is:
(A) z/(z − 0.5)
(B) 1/(1 + 0.5z⁻¹)
(C) (z − 0.5)/z
(D) z/(z + 0.5)

Taking Z-transform of y[n] − 0.5y[n−1] = x[n]:

Y(z) − 0.5z⁻¹Y(z) = X(z) Y(z)(1 − 0.5z⁻¹) = X(z) H(z) = Y(z)/X(z) = 1/(1−0.5z⁻¹) = z/(z−0.5)

Pole at z=0.5 (inside unit circle) → stable causal system.

Answer: (A) z/(z−0.5)

GATE 2018 2 Marks MCQ
The DTFT X(e^jω) of x[n] = (0.5)ⁿu[n] at ω = 0 is:
(A) 0.5
(B) 1
(C) 2
(D) ∞

X(e^jω) = Z{x[n]}|_{z=e^jω} = z/(z−0.5)|_{z=e^jω}

At ω=0: z = e^j0 = 1

X(e^j0) = 1/(1−0.5) = 1/0.5 = 2

Alternatively: X(e^j0) = Σₙ (0.5)ⁿu[n] = Σₙ₌₀^∞ (0.5)ⁿ = 1/(1−0.5) = 2 [geometric series]

Answer: (C) 2

GATE 2017 2 Marks MCQ
The convolution x[n] = u[n] * u[n] equals:
(A) u[n]
(B) δ[n]
(C) (n+1)u[n]
(D) n·u[n]

Z-domain: Z{u[n]} = z/(z−1). Convolution → multiplication:

Y(z) = [z/(z-1)]² = z²/(z-1)²

Inverse Z-transform: z²/(z−1)² → (n+1)u[n]

Verify: y[n] = Σₖ₌₀ⁿ 1·1 = (n+1) for n≥0, 0 for n<0 = (n+1)u[n] ✓

Answer: (C) (n+1)u[n]

GATE 2016 1 Mark MCQ
Which statement about the DTFT is correct?
(A) DTFT is a CT function of frequency ω
(B) DTFT is a CT, periodic function of ω with period 2π
(C) DTFT is a DT function of ω
(D) DTFT is aperiodic

X(e^jω) = Σₙ x[n]e^(-jωn). Since e^(-j(ω+2π)n) = e^(-jωn)·e^(-j2πn) = e^(-jωn) for integer n:

X(e^j(ω+2π)) = X(e^jω) → DTFT is periodic with period 2π.

ω is a continuous variable → DTFT is a continuous, periodic function of ω.

Answer: (B)

GATE 2015 2 Marks MCQ
The Z-transform of n·u[n] is:
(A) z/(z−1)
(B) z/(z−1)²
(C) 1/(z−1)²
(D) z²/(z−1)²

Using the z-differentiation property: n·x[n] ↔ −z·dX(z)/dz

X(z) = Z{u[n]} = z/(z−1)

dX/dz = [(z-1)·1 - z·1]/(z-1)² = -1/(z-1)² n·u[n] ↔ -z·(-1/(z-1)²) = z/(z-1)²

Alternatively, direct table: n·u[n] ↔ z/(z−1)² [standard pair for ramp]

Answer: (B) z/(z−1)²

GATE 2014 2 Marks MCQ
A DT-LTI system with h[n] = δ[n] − δ[n−1] has input x[n] = u[n]. Find y[n].
(A) δ[n]
(B) u[n]
(C) n·u[n]
(D) (n+1)u[n]

H(z) = Z{δ[n]−δ[n-1]} = 1 − z⁻¹ = (z−1)/z

X(z) = Z{u[n]} = z/(z−1)

Y(z) = X(z)·H(z) = [z/(z-1)]·[(z-1)/z] = 1

Y(z) = 1 → y[n] = δ[n]

Intuition: h[n] = difference filter. Applying it to u[n] gives u[n]−u[n-1] = δ[n]. The difference filter is the DT analog of differentiation (which converts a step to an impulse).

Answer: (A) δ[n]