10 Chapter Map - Z-Transform & Discrete-Time Analysis
Chapter 10 Overview & Map of Content (MOC)
Complex mapping , convergence annuli, telescoping sum proofs, residue inversions, and unilateral delay difference solvers.
📚 Study Notes Index
Read in order — each note assumes the previous one.
| # | Note | What it covers |
|---|---|---|
| 10.00 | 10.00 Z-Transform and Discrete-Time Analysis Compact Review | Z-Transform Compact Review, Z-Transform Formula Sheet |
| 10.01 | 10.01 Z-Transform Definitions and Region of Convergence ROC | Z-Transform Definitions, Z-Plane Mapping, Region of Convergence Properties, ROC, Laplace to Z Mapping |
| 10.02 | 10.02 Z-Transform Properties and Standard Pairs | Z-Transform Properties, Properties of the Z-Transform, Z-Transform Lookup Table |
| 10.03 | 10.03 Initial and Final Value Theorems in the Z-Domain | Initial & Final Value Theorems in Z-Domain, Z-Domain IVT and FVT, IVT and FVT of Z-Transform |
| 10.04 | 10.04 Inverse Z-Transform Methods | Inverse Z-Transform, Partial Fraction Expansion Z-Domain, Residue Method Z-Domain, Long Division Z-Domain |
| 10.05 | 10.05 Convolution Correlation and Realization in Z-Domain | Discrete Convolution and Correlation, Z-Domain Realization, System Difference Equations Z-Domain |
🎯 Exam Weight
ECE 2107 Exam Relevance
Master the core derivations, mathematical definitions, and problem-solving techniques. Refer to ECE 2107 - Signals and Systems for syllabus boundaries and past year questions.
🔗 Related Resources
- Course Teaching Plan: ECE 2107 - Signals and Systems
- Previous chapter: 09 Chapter Map - Laplace Transform & s-Domain Circuit Applications
- Next chapter: 11 Chapter Map - DFT, FFT, and Frequency-Domain Discrete Analysis
ECE 2107 Master Roadmap | 10.01 Z-Transform Definitions & ROC | 10.02 Z-Transform Properties
Chapter 10: Z-Transform & Discrete Analysis - Compact Review
10.01 Z-Transform Core Definitions
*(Target: Theory Descriptive / 5-Mark Formula)* [10.01]
- Bilateral (Two-Sided) Z-Transform: Maps a discrete-time sequence defined over all integers onto the complex z-plane [10.01]: [10.01]
- Unilateral (One-Sided) Z-Transform: Maps strictly causal signals () [10.01]. This is uniquely suited for solving difference equations with non-zero initial conditions [10.05]: [10.01]
10.02 s-Plane to z-Plane Exponential Mapping
*(Target: Mathematical Proof / 6-Mark Derivation)* [10.01]
- The Mapping Operator: Sampling a continuous-time signal at uniform intervals maps Laplace domain variables to discrete z-plane variables [10.01]: [10.01]
- Let and [10.01]: [10.01]
Table 10.1: s-Plane to z-Plane Boundary Mappings [10.01]
| s-Plane Domain / Path | z-Plane Equivalent Domain / Path | Physical Significance |
|---|---|---|
| Imaginary Axis () | **Unit Circle ($r = 1 \implies \lvert z | |
| vert = 1$)** | Standard DTFT boundary path [10.01] | |
| Left-Half s-Plane () | **Interior of Unit Circle ($r < 1 \implies \lvert z | |
| vert < 1$)** | Stable discrete pole region [10.01] | |
| Right-Half s-Plane () | **Exterior of Unit Circle ($r > 1 \implies \lvert z | |
| vert > 1$)** | Unstable discrete pole region [10.01] | |
| Origin () | DC Unit-Circle Crossing (z = 1 ngle 0^\circ) | Zero-frequency/DC gain point [10.01] |
s-plane (Laplace) z-plane (Z-Transform)
+j\Omega Im(z)
| ^
LHP | RHP _---|___
(Stable) | (Unstable) / * * ~~~~~~~~~* | * | * (o) * | --> Re(z)
~~~~~~~~~ +---------> \sigma | Stable | |z|=1 (Unit Circle)
~~~~~~~~~ | \ * * /
| `---|---'
| (Interior)
[DIAGRAM: Visual mapping showing s-plane boundary lines and shaded areas projecting onto unit circle disk envelopes inside the z-plane] [10.01]
10.03 Region of Convergence (ROC) Core Symmetries
*(Target: Theory Descriptive / 5-to-13 Mark Properties Mapping)* [10.01]
- The Region of Convergence (ROC) is the vertical annulus (ring) centered at the origin of the complex z-plane where the Laurent power series converges absolutely [10.01]:
ight vert < \infty$$ [10.01]
Table 10.2: ROC Geometries & Signal Classifications [10.01]
| Signal Sequence Classification | Time-Domain Constraints | ROC Geometric Bounds in z-Plane | Illustration Placeholder |
|---|---|---|---|
| Finite Causal | for , length | Entire z-plane except | [GRAPH: ROC covering entire plane except center] [10.01] |
| Finite Anticausal | for , length | Entire z-plane except | [GRAPH: ROC covering plane except infinite boundary] [10.01] |
| Finite Two-Sided | non-zero only for | Entire z-plane except and | [GRAPH: ROC plane punctured at origin and outer rim] [10.01] |
| Infinite Right-Sided (Causal) | for | Exterior of a circle: $\lvert z | |
| vert > \lvert p_{ ext{max}} | |||
| vert$ | [GRAPH: Shaded exterior bounding highest pole outward] [10.01] | ||
| Infinite Left-Sided (Anticausal) | for | Interior of a circle: $\lvert z | |
| vert < \lvert p_{ ext{min}} | |||
| vert$ | [GRAPH: Shaded disk bounding lowest pole inward] [10.01] | ||
| Infinite Two-Sided | defined over | Vertical ring/annulus: $r_1 < \lvert z | |
| vert < r_2$ | [GRAPH: Shaded ring bounded between two pole radii] [10.01] |
WARNING
The Pole ROC Boundary Principle: The ROC cannot contain any poles. It must always be bounded by poles or extend to infinity [10.01]. If a system is stable, the ROC must include the unit circle () [10.01].
10.04 Master Z-Transform Properties Matrix
*(Target: Theory Descriptive / Properties Ingestion)* [10.02]
Table 10.3: Bilateral Z-Transform Properties [10.02]
| Property Name | Time-Domain Sequence | z-Domain Function | ROC Bound |
|---|---|---|---|
| Linearity | At least [10.02] | ||
| Time Shifting | Same as except (if ) or (if ) [10.02] | ||
| Time Reversal | Inverted boundary: (i.e., $ rac{1}{r_2} < \lvert z | ||
| vert < rac{1}{r_1}$) [10.02] | |||
| Scaling in z | Scaled boundary: $\lvert a | ||
| vert R\lvert a | |||
| vert r_1 < \lvert z | |||
| vert < \lvert a | |||
| vert r_2$) [10.02] | |||
| Differentiation | -z rac{dX(z)}{dz} | Same as [10.02] | |
| Convolution | At least [10.02] | ||
| Correlation | At least [10.02] | ||
| Complex Conjugate | Unchanged: [10.02] | ||
| Real Part | $ rac{1}{2}\left[ X(z) + X^(z^) | ||
| ight]$ | At least [10.02] | ||
| Imaginary Part | $ rac{1}{2j}\left[ X(z) - X^(z^) | ||
| ight]$ | At least [10.02] |
- Z-Domain Differentiation Property Proof Outline: ext{Given: } X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} \implies rac{dX(z)}{dz} = \sum_{n=-\infty}^{\infty} -n x[n] z^{-n-1} \implies -z rac{dX(z)}{dz} = \sum_{n=-\infty}^{\infty} n x[n] z^{-n} \quad ext{(derivation)} [10.02]
10.05 Master Z-Transform Pair Lookup Sheet
*(Target: Numerical Solving / Rapid Lookups)* [10.02]
Table 10.4: High-Yield Discrete Transform Pairs [10.02]
| Discrete Sequence | z-Domain Transform | Discrete ROC Boundary |
|---|---|---|
| Unit Impulse | Entire z-plane [10.02] | |
| Shifted Impulse | Entire plane except (if ) [10.02] | |
| Unit Step | rac{1}{1-z^{-1}} = rac{z}{z-1} | $\lvert z |
| vert > 1$ [10.02] | ||
| Negative Unit Step | rac{1}{1-z^{-1}} = rac{z}{z-1} | $\lvert z |
| vert < 1$ [10.02] | ||
| Causal Exponential | rac{1}{1-a z^{-1}} = rac{z}{z-a} | $\lvert z |
| vert > \lvert a | ||
| vert$ [10.02] | ||
| Anticausal Exponential | rac{1}{1-a z^{-1}} = rac{z}{z-a} | $\lvert z |
| vert < \lvert a | ||
| vert$ [10.02] | ||
| Ramp Function | rac{z^{-1}}{(1-z^{-1})^2} = rac{z}{(z-1)^2} | $\lvert z |
| vert > 1$ [10.02] | ||
| Scaled Ramp | rac{a z^{-1}}{(1-a z^{-1})^2} = rac{a z}{(z-a)^2} | $\lvert z |
| vert > \lvert a | ||
| vert$ [10.02] | ||
| Discrete Cosine | rac{1 - z^{-1}\cos(\omega_0)}{1 - 2z^{-1}\cos(\omega_0) + z^{-2}} = rac{z(z - \cos(\omega_0))}{z^2 - 2z\cos(\omega_0) + 1} | $\lvert z |
| vert > 1$ [10.02] | ||
| Discrete Sine | rac{z^{-1}\sin(\omega_0)}{1 - 2z^{-1}\cos(\omega_0) + z^{-2}} = rac{z\sin(\omega_0)}{z^2 - 2z\cos(\omega_0) + 1} | $\lvert z |
| vert > 1$ [10.02] | ||
| Exponential Cosine | rac{z(z - a\cos(\omega_0))}{z^2 - 2az\cos(\omega_0) + a^2} | $\lvert z |
| vert > \lvert a | ||
| vert$ [10.02] | ||
| Exponential Sine | rac{az\sin(\omega_0)}{z^2 - 2az\cos(\omega_0) + a^2} | $\lvert z |
| vert > \lvert a | ||
| vert$ [10.02] |
10.06 Initial & Final Value Theorems
*(Target: Mathematical Proof & Numerical Solving)* [10.03]
- Initial Value Theorem (IVT): Determines the starting value of a causal sequence directly from its Z-transform without requiring inverse operations [10.03]: [10.03]
- Final Value Theorem (FVT): Finds the steady-state DC limit of a stable sequence [10.03]:
ight) X(z) \quad ext{(derivation)}$$ [10.03]
WARNING
The Critical FVT Stability Constraint: The final value theorem yields a correct steady-state limit if and only if all poles of lie strictly inside the unit circle () [10.03]. If the system has multiple poles on the unit circle or poles outside the unit circle (e.g., an oscillatory system like or an unstable system like ), the FVT mathematically fails and yields invalid values [10.03].
10.07 Unilateral Z-Transform Shift properties
*(Target: Numerical Solving / Recursive Difference Equations)* [10.05]
When solving linear constant-coefficient difference equations (LCCDE) with non-zero initial conditions (e.g., ), the standard bilateral shift theorem fails. We must apply the Unilateral Shift properties [10.05]:
- First-Order Delay Shift (): [10.05]
- Second-Order Delay Shift (): [10.05]
- General Delay Shift ():
ight] \quad ext{(derivation)}$$ [10.05]
10.08 Methods for Inverse Z-Transform Solving
*(Target: Theory Descriptive & Numerical Solving)* [10.04]
1. Cauchy’s Residue Integration Method
- Concept: Resolves the inverse transform by evaluating contour integrals over a closed counter-clockwise path in the ROC [10.04]:
ight] \quad ext{(derivation)}$$ [10.04]
- Residue Formula at a Simple Pole ():
ight} = \lim_{z o p_i} (z - p_i) X(z) z^{n-1}$$ [10.04]
- Residue Formula at a Pole of Multiplicity :
ight} = rac{1}{(m-1)!} \lim_{z o p_i} rac{d^{m-1}}{dz^{m-1}} \left[ (z - p_i)^m X(z) z^{n-1} ight]$$ [10.04]
2. Partial Fraction Expansion Method ()
- Concept: Rather than expanding directly (which creates factors in numerators), expand the ratio rac{X(z)}{z}. Once decomposed, multiply back by to recover standard forms [10.04]:
ight] u[n] \quad ext{(derivation)}$$ [10.04]
3. Power Series Expansion Method (Long Division)
- Causal Case (): Divide the numerator by the denominator arranged in descending powers of to obtain a power series in negative powers of () [10.04]: [10.04]
- Anticausal Case (): Divide the numerator by the denominator arranged in ascending powers of to obtain a power series in positive powers of [10.04]: [10.04]
10.09 System Realization Architectures
*(Target: Theory Descriptive & Graphical Diagramming)* [10.05]
Discrete transfer functions H(z) = rac{Y(z)}{X(z)} = rac{b_0 + b_1 z^{-1} + b_2 z^{-2}}{1 + a_1 z^{-1} + a_2 z^{-2}} are mapped to hardware block elements (Adders, Multipliers, and Unit Delay Registers ) using two principal topological structures [10.05]:
- Direct Form I realization: Implements poles and zeros separately [10.05]. It requires delay registers (non-canonic) and is highly immune to coefficient quantization noise [10.05].
- Direct Form II canonic realization: Merges feedback and feedforward delay registers [10.05]. It minimizes total register count to exactly delays (canonic form) but is more sensitive to internal register overflow [10.05].
Direct Form I (Non-Canonic) Direct Form II (Canonic)
x[n] --->(+)----------->(+)---> y[n] x[n] --->(+)------------->(+)---> y[n]
| ^ | | ^
[z-1] [z-1] [z-1] | [z-1]
| | |<-a1 | b1-> |
( )*-a1 ( )*b1 v | v
| | [z-1] v [z-1]
[z-1] [z-1] |<-a2 |
| | v v
( )*-a2 ( )*b2
[DIAGRAM: Non-canonic Direct Form I vs canonic Direct Form II delay element merging pathways] [10.05]
10.10 Common Mistakes That Cost Marks
- The Unilateral Delay Shift Trap: Accidentally writing when initial conditions are non-zero. You must include the initial condition step [10.05]!
- The FVT Pole-Location Oversight: Applying the Final Value Theorem to an oscillatory signal like or an unstable sequence. You must check that the poles of lie strictly inside the unit circle () before using the theorem [10.03].
- Long Division Order Confusion: Sorting polynomials incorrectly during long division. If seeking a causal sequence, write polynomials in descending powers of (or ascending powers of ) [10.04]. If anticausal, reverse the order [10.04].
- Failing to scale the partial fraction expansion: Trying to expand directly rather than rac{X(z)}{z}, which leads to algebraically intensive terms that do not match standard causal pairs [10.04].
10.11 Verbatim PYQ Bank & Answer Plans
[PYQ 2025, Question 5a - 5 Marks]
- Question: Find the z-transform of the sequence and its ROC.
- Answer Plan:
- Express the window sequence as a sum of impulses or a finite geometric series: for .
- Apply the definition: .
- Express in closed ratio form: X(z) = rac{1 - z^{-5}}{1 - z^{-1}} = rac{z^5 - 1}{z^4(z-1)}.
- State the ROC: Since it is a finite-duration causal sequence, the ROC is the entire z-plane except .
[PYQ 2024, Question 5b - 10 Marks]
- Question: Prove that the final values of for X(z) = rac{z^2}{(z-1)(z-0.2)} is 1.25 and its initial value is unity.
- Answer Plan:
- Initial Value Proof: Apply the IVT: x[0] = \lim_{z o \infty} X(z) = \lim_{z o \infty} rac{z^2}{z^2 - 1.2z + 0.2} = 1.
- FVT Prerequisite Check: Locate the poles of (1-z^{-1})X(z) = rac{z-1}{z} rac{z^2}{(z-1)(z-0.2)} = rac{z}{z-0.2}. The only pole is at , which lies strictly inside the unit circle (). Thus, FVT is valid.
- Final Value Proof: Apply the FVT: x[\infty] = \lim_{z o 1} (1-z^{-1}) X(z) = \lim_{z o 1} rac{z}{z-0.2} = rac{1}{1-0.2} = rac{1}{0.8} = 1.25.
[PYQ 2023, Question 5b - 5 Marks]
- Question: If is causal, find the inverse z-transform of X(z) = rac{1}{z(z-0.8)(z+0.4)}.
- Answer Plan:
- Rewrite X(z) = z^{-1} \left[ rac{1}{(z-0.8)(z+0.4)} ight] or expand rac{X(z)}{z} = rac{1}{z^2(z-0.8)(z+0.4)} using multiple pole residues.
- Alternatively, perform partial fraction expansion on the core bracket first: F(z) = rac{z}{(z-0.8)(z+0.4)} = rac{2/3 z}{z-0.8} + rac{1/3 z}{z+0.4} \implies f[n] = \left[ rac{2}{3}(0.8)^n + rac{1}{3}(-0.4)^n ight] u[n].
- Apply the time-shifting property: Since , we delay by 2 samples: x[n] = f[n-2] = \left[ rac{2}{3}(0.8)^{n-2} + rac{1}{3}(-0.4)^{n-2} ight] u[n-2].
10.12 Interactive Self-Check Revision List
- Do you know the exact unit-circle boundary coordinates of the s-plane imaginary axis projection ()?
- Can you state the 3 distinct finite-duration ROC cases (causal, anticausal, double-sided)?
- Do you remember the unilateral shifting terms for without looking?
- Have you verified the pole locations of before solving any FVT?
- Do you know why Direct Form II canonic structures use exactly half the delay elements of Direct Form I?
← 9.08 Transient Response of Series & Parallel RLC Networks | Chapter 10 Map | 10.02 Z-Transform Properties & Standard Pairs →
10.01 Z-Transform: Definitions & Region of Convergence (ROC)
Alright — let’s cross the boundary from continuous-time systems into transform-domain discrete-time analysis! Just as the Laplace Transform generalized the Continuous-Time Fourier Transform (CTFT) to analyze unstable or growing continuous waveforms, the Z-Transform generalizes the Discrete-Time Fourier Transform (DTFT) to analyze discrete-time sequences by introducing a complex radial scaling factor.
By mapping discrete sequences from the time-domain index into the complex -plane, we transform linear difference equations into simple algebraic expressions. Let’s derive the fundamental definitions, prove how the continuous -plane maps to the discrete -plane, establish Region of Convergence (ROC) properties, and master the core “exam-killer” numericals to secure full marks on your exam!
1. Unilateral vs. Bilateral Z-Transform Definitions
Depending on the causal boundaries of your signal, the Z-transform is mathematically defined in two ways:
1.1 Bilateral (Two-Sided) Z-Transform
The bilateral Z-transform of an arbitrary, infinite-duration discrete-time signal is defined as:
where is a continuous complex variable represented in polar coordinates as: Here, is the radial distance (magnitude) from the origin of the complex -plane, and \omega = ngle z represents the angular frequency in radians per sample.
1.2 Unilateral (One-Sided) Z-Transform
The unilateral Z-transform is appropriate for causal signals and systems {where for all } and is heavily used to solve difference equations with non-zero initial conditions:
1.3 Structural Comparison
Let’s look at the key operational differences between these two domains:
| Feature | Bilateral (Two-Sided) Z-Transform | Unilateral (One-Sided) Z-Transform |
|---|---|---|
| Summation Limits | Sums over | Sums strictly over |
| Sequence Handling | Evaluates general causal, anticausal, or bilateral sequences. | Assumes causality {forces for }. |
| ROC Complexity | Must specify explicit ROC boundaries to ensure uniqueness. | The ROC is always the exterior of a circle (no need to specify). |
| Initial Conditions | Cannot handle systems with non-zero initial states. | Excellent for solving difference equations with initial conditions. |
| Laplace Counterpart | Analogous to the Bilateral Laplace transform. | Analogous to the Unilateral Laplace transform. |
2. Mathematical Derivation: Laplace-to-Z Mapping ()
NOTE
This derivation is a highly tested exam classic [PYQ 2024, 2019, 2017, 2015] commanding 6 to 8 marks. It mathematically bridges continuous-time and discrete-time transform spaces.
Let’s begin with a continuous-time signal . When we sample uniformly at intervals of seconds, we can model the physical sampled signal mathematically as a continuous-time multiplication of with an ideal periodic impulse train:
Now, let’s take the Unilateral Laplace transform of this continuous sampled impulse train :
ight] e^{-st} \, dt$$ Because the integral operator is linear, we can slide the integration inside the summation: $$X^*(s) = \sum_{n=0}^{\infty} x(nT_s) \int_{0^{-}}^{\infty} \delta(t - nT_s) e^{-st} \, dt$$ Applying the **Sifting Property** of the unit impulse function {$\int \delta(t - t_0)f(t)dt = f(t_0)$}: $$X^*(s) = \sum_{n=0}^{\infty} x(nT_s) e^{-n s T_s} \quad [148]$$ Let's look at this equation side-by-side with the definition of the unilateral Z-transform of the discrete-time sampled sequence $x[n] = x(nT_s)$: $$X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} \quad [149]$$ By direct matching of the exponential terms: $$z^{-n} = e^{-n s T_s} \implies z = e^{s T_s} \quad [147]$$ ### 2.1 Mapping the Complex Plane Territories Let's analyze how this complex exponential relationship $z = e^{s T_s}$ maps the continuous complex $s$-plane ($s = \sigma + j\Omega$) into the discrete complex $z$-plane: $$z = e^{(\sigma + j\Omega)T_s} = e^{\sigma T_s} e^{j\Omega T_s} \quad [147]$$ Expressing this mapped point in polar coordinates $z = r e^{j\omega}$ reveals the matching relationships: 1. **Magnitude Mapping:** $r = |z| = e^{\sigma T_s}$ 2. **Frequency Mapping:** $\omega = ngle z = \Omega T_s$ This math yields three absolute geometric mappings: #### Case 1: The imaginary $j\Omega$-axis of the $s$-plane maps to the unit circle $|z| = 1$ * On the imaginary axis, the real attenuation factor is strictly zero ($\sigma = 0$). * Therefore, the magnitude of $z$ becomes: $$|z| = e^{0 \cdot T_s} = 1 \quad [147]$$ * As the continuous frequency $\Omega$ sweeps from $-\infty$ to $+\infty$, the mapped point wraps around the unit circle $|z| = 1$ in the $z$-plane periodically with period $2\pi/T_s$. #### Case 2: The Left-Half Plane (LHP) of the $s$-plane maps to the interior of the unit circle $|z| < 1$ * For stable systems, poles must lie in the LHP, meaning the real part is negative ($\sigma < 0$). * Therefore, the radial magnitude in the $z$-plane is bounded by: $$|z| = e^{\sigma T_s} < 1 \quad ( ext{since } \sigma < 0) \quad [147]$$ * This proves that the entire stable LHP of continuous-time systems maps directly **inside the unit circle** in the $z$-plane. #### Case 3: The Right-Half Plane (RHP) of the $s$-plane maps to the exterior of the unit circle $|z| > 1$ * Unstable system poles lie in the RHP, where the real growth factor is positive ($\sigma > 0$). * Therefore, the radial magnitude in the $z$-plane is: $$|z| = e^{\sigma T_s} > 1 \quad ( ext{since } \sigma > 0) \quad [147]$$ * This maps the unstable continuous territory **outside the unit circle**. ``` Continuous s-Plane Discrete z-Plane jΩ Im(z) ▲ ▲ LHP │ RHP _.-'''''''-._ ROC: Exterior (Stable) │ (Unstable) .-' _.._ '-. (Causal Stable) ◄─────────┼─────────► σ .' .' '. '. xxxxxxxxx │ Z(s)=eˢᵀ │ │ o │ │ xxxxxxxxx │ ────────►│◄────┼────*───┼───────┼► Re(z) Poles (x) │ │ │ z=0 │Pole(x)│ ◄── σ < 0 │ ──► σ > 0 '. '. .' .' |z|=1 │ '-._ '..' _.-' (Unit Circle) ▼ '-._____.-' ``` --- ## 3. Core Concept: The Region of Convergence (ROC) The **Region of Convergence (ROC)** of a Z-transform is defined as the set of values of $z$ in the complex $z$-plane for which the infinite summation converges to a finite, bounded value {i.e., $|X(z)| < \infty$}. If the summation diverges to infinity, the Z-transform does not exist for those values of $z$. ### 3.1 Mathematical Convergence Condition For $X(z)$ to converge, the series must be absolutely summable: $$\sum_{n=-\infty}^{\infty} |x[n] z^{-n}| < \infty \implies \sum_{n=-\infty}^{\infty} |x[n]| r^{-n} < \infty \quad [37, 153, 154]$$ This absolute convergence summation represents the Discrete-Time Fourier Transform (DTFT) of the exponentially weighted sequence $x[n]r^{-n}$. ### 3.2 The 8 Golden Properties of the ROC These properties are critical for identifying system boundaries during exam problems: 1. **The ROC is a ring or disk in the $z$-plane centered at the origin:** Since convergence depends strictly on $|z| = r$, the ROC is bounded by concentric circles. 2. **The ROC cannot contain any poles:** By definition, a pole is a root where $X(z) o \infty$. Since the ROC contains only points where $X(z)$ is finite, poles always form the physical boundaries of the ROC. 3. **ROC of a Finite-Duration Causal Sequence:** The ROC is the **entire $z$-plane except $z = 0$**. *{Explanation: If $x[n]$ exists only from $n = 0$ to $N-1$, $X(z) = x[0] + x[1]z^{-1} + \dots + x[N-1]z^{-N+1}$. The negative powers of $z$ blow up only when $z = 0$}.* 4. **ROC of a Finite-Duration Anticausal Sequence:** The ROC is the **entire $z$-plane except $z = \infty$**. *{Explanation: If $x[n]$ exists only for negative indices, $X(z) = x[-N]z^{N} + \dots + x[-1]z^1$. The positive powers of $z$ blow up only when $z o \infty$}.* 5. **ROC of a Finite-Duration Two-Sided Sequence:** The ROC is the **entire $z$-plane except $z = 0$ and $z = \infty$**. 6. **ROC of an Infinite-Duration Right-Sided (Causal) Sequence:** The ROC is the **exterior of a circle** $|z| > r_1$, extending out to infinity. 7. **ROC of an Infinite-Duration Left-Sided (Anticausal) Sequence:** The ROC is the **interior of a circle** $|z| < r_2$. 8. **ROC of an Infinite-Duration Two-Sided Sequence:** The ROC is an **open annular ring** $r_1 < |z| < r_2$, bounded by poles on both the inner and outer circles. --- ## 4. High-Yield Worked "Exam Killers" [PYQ Bank] Let's work through the highest-yielding Z-transform calculation problems step-by-step to guarantee maximum marks on your papers. ### 4.1 The Causal Window Sequence [PYQ 2025] > [!question] **2025 Exam Section B Q. 7b (05 Marks)** > > Find the Z-transform of the sequence $x[n] = [u[n] - u[n-5]]$ and specify its Region of Convergence (ROC). #### Step 1: Write down the sequence values The unit step difference represents a finite-duration causal pulse of length 5: $$x[n] = egin{cases} 1, & 0 \le n \le 4 \ 0, & ext{otherwise} \end{cases}$$ $$x[n] = \{1, 1, 1, 1, 1\}$$ #### Step 2: Apply the Z-transform definition $$X(z) = \sum_{n=0}^{4} x[n] z^{-n} = 1 + z^{-1} + z^{-2} + z^{-3} + z^{-4} \quad [74]$$ This is a finite geometric series. Applying the sum of a finite geometric progression ($S_N = rac{1 - q^N}{1 - q}$): $$X(z) = rac{1 - z^{-5}}{1 - z^{-1}} = rac{z^5 - 1}{z^4 (z - 1)} \quad [74]$$ #### Step 3: Determine the ROC and Pole-Zero behavior * **Zeros:** Roots of $z^5 - 1 = 0 \implies z = e^{j 2\pi k / 5}$ for $k = 0, 1, 2, 3, 4$. * **Poles:** There is a 4th-order pole at the origin $z = 0$, and a simple pole at $z = 1$. * **Pole-Zero Cancellation:** Notice that for $k=0$, the zero is $z = 1$. This zero exactly cancels the simple pole at $z = 1$! * Therefore, the function has no poles outside the origin. * **Final ROC:** **The entire $z$-plane except $z = 0$**. This perfectly matches **Property 3** {finite causal sequence}. --- ### 4.2 Causal Decaying Exponential with Shifting [PYQ 2020] > [!question] **2020 Exam Section B Q. 6b (03 Marks)** > > Find the Z-transform of the discrete-time signal including the Region of Convergence: > $$x[n] = e^{-4n} u[n-1]$$ #### Step 1: Set up the Z-transform summation $$X(z) = \sum_{n=-\infty}^{\infty} e^{-4n} u[n-1] z^{-n} = \sum_{n=1}^{\infty} e^{-4n} z^{-n}$$ We can group the base terms inside the exponent: $$X(z) = \sum_{n=1}^{\infty} \left( e^{-4} z^{-1} ight)^n$$ #### Step 2: Adjust index to start from $m = 0$ Let $m = n - 1 \implies n = m + 1$: $$X(z) = \sum_{m=0}^{\infty} \left( e^{-4} z^{-1} ight)^{m+1} = \left( e^{-4} z^{-1} ight) \sum_{m=0}^{\infty} \left( e^{-4} z^{-1} ight)^m$$ #### Step 3: Evaluate the infinite sum Applying the infinite geometric series formula ($\sum_{m=0}^{\infty} q^m = rac{1}{1-q}$ for $|q| < 1$): $$X(z) = \left( e^{-4} z^{-1} ight) \left[ rac{1}{1 - e^{-4} z^{-1}} ight] = rac{e^{-4} z^{-1}}{1 - e^{-4} z^{-1}}$$ Multiplying the numerator and denominator by $z$: $$X(z) = rac{e^{-4}}{z - e^{-4}}$$ #### Step 4: Find the ROC The geometric series converges if and only if the common ratio has a magnitude less than 1: $$|e^{-4} z^{-1}| < 1 \implies e^{-4} |z|^{-1} < 1 \implies |z| > e^{-4}$$ * **Poles:** $z = e^{-4} pprox 0.0183$. * **Zeros:** No finite zeros (or a zero at infinity). * **ROC:** **$|z| > e^{-4}$** {exterior of a circle of radius $e^{-4}$}. --- ### 4.3 Left-Sided Anticausal Exponential Sequence [PYQ 2022] > [!question] **2022 Exam Section B Q. 5b (05 Marks)** > > Calculate the Z-transform and ROC of: > $$x[n] = \left(- rac{1}{3} ight)^n u[-n]$$ #### Step 1: Formulate the Z-transform summation The anticausal step $u[-n]$ is defined as $1$ for $n \le 0$ and $0$ for $n > 0$. Therefore: $$X(z) = \sum_{n=-\infty}^{0} \left(- rac{1}{3} ight)^n z^{-n}$$ #### Step 2: Use variable substitution to flip the index Let $k = -n \implies$ as $n o -\infty$, $k o +\infty$: $$X(z) = \sum_{k=0}^{\infty} \left(- rac{1}{3} ight)^{-k} z^{k} = \sum_{k=0}^{\infty} (-3)^k z^k = \sum_{k=0}^{\infty} (-3z)^k$$ #### Step 3: Evaluate the sum Applying the infinite geometric series formula: $$X(z) = rac{1}{1 - (-3z)} = rac{1}{1 + 3z}$$ #### Step 4: Determine the ROC The geometric series converges if and only if the absolute value of the ratio is less than 1: $$|-3z| < 1 \implies 3|z| < 1 \implies |z| < rac{1}{3}$$ * **Poles:** $z = - rac{1}{3}$. * **Zeros:** None (or a zero at infinity). * **ROC:** **$|z| < rac{1}{3}$** {interior of a circle of radius $1/3$, which matches Property 7 for anticausal sequences}. --- ### 4.4 Multi-Impulse Array with Time Shift [PYQ 2022, 2018] > [!question] **2022 / 2018 Exam Q. 5a (02 to 05 Marks)** > > Determine the Z-transform of the following signal and specify its ROC: > $$x[n] = \delta[n+1] + 3\delta[n] + 7\delta[n-3] - 3\delta[n-4]$$ #### Step 1: Take Z-transform of individual terms Using the definition of the Z-transform of an impulse $\mathcal{Z}\{\delta[n-k]\} = z^{-k}$: * $\mathcal{Z}\{\delta[n+1]\} = z^1 = z$ * $\mathcal{Z}\{3\delta[n]\} = 3 z^0 = 3$ * $\mathcal{Z}\{7\delta[n-3]\} = 7 z^{-3}$ * $\mathcal{Z}\{-3\delta[n-4]\} = -3 z^{-4}$ #### Step 2: Assemble the final expression By linearity: $$X(z) = z + 3 + 7z^{-3} - 3z^{-4} \quad [168]$$ #### Step 3: Establish the ROC * The term $z$ has a pole at $z = \infty$ {magnitude blows up as $z o \infty$}. * The terms $z^{-3}$ and $z^{-4}$ have poles of order 3 and 4 at the origin $z = 0$ {magnitude blows up as $z o 0$}. * **Final ROC:** **The entire $z$-plane except $z = 0$ and $z = \infty$**. This matches **Property 5** for two-sided finite sequences. --- ## 5. Common Mistakes That Cost Marks > [!WARNING] **Critical Exam Pitfalls** > > 1. **Omitting the ROC Statement:** In discrete-time systems, a Z-domain expression $X(z)$ without an ROC does **not** represent a unique time-domain sequence! For example, $X(z) = rac{z}{z-a}$ can represent the causal sequence $a^n u[n]$ (if $|z| > |a|$) or the anticausal sequence $-a^n u[-n-1]$ (if $|z| < |a|$). **Always state the ROC explicitly, or you will lose 2 to 3 marks immediately!** > 2. **Algebraic Sign Errors in Anticausal Substitutions:** When evaluating anticausal geometric series like $\sum_{n=-\infty}^{0} a^n z^{-n}$, students often forget that replacing $n$ with $-k$ flips the exponent of both $a$ and $z$ {yielding $(a^{-1}z)^k$, not $(az)^k$}. Work through the substitution steps explicitly. > 3. **Confusing the Stability Boundary with Laplace:** Do not mix up the stability regions of the $s$-plane and the $z$-plane! In Laplace, the imaginary axis $\sigma = 0$ separates stable and unstable territories. In the Z-transform, the **unit circle $|z| = 1$** acts as this boundary. A discrete-time LTI system is BIBO stable if and only if the ROC of its transfer function **includes the unit circle $|z| = 1$**. --- ## 6. PYQ Bank — Verbatim Questions & Answer Plans ### Q1: Relationship between Laplace and Z-Transform [PYQ 2024, 2019, 2017, 2015] * **Question:** How is the Z-transform obtained from the Laplace transform? Describe the relationship between Laplace transform and Z-transform. **(08 Marks)** * **Answer Plan:** 1. Model the discrete sequence $x[n]$ as a sampled continuous signal $x^*(t)$ convolved with an impulse train. 2. Take the Laplace transform of $x^*(t)$ and prove that $X^*(s) = \sum x(nT_s) e^{-n s T_s}$. 3. Compare this to the Z-transform definition $\sum x[n] z^{-n}$ and prove the mapping equation $z = e^{s T_s}$. 4. Draw the $s$-plane to $z$-plane mapping diagram, detailing why the imaginary axis maps to $|z|=1$, the LHP maps to $|z|<1$, and the RHP maps to $|z|>1$. ### Q2: Bilateral vs. Unilateral Z-Transforms [PYQ 2020] * **Question:** Briefly discuss the properties of two-sided Z-transform and compare them with one-sided Z-transform. **(06 Marks)** * **Answer Plan:** 1. Write the mathematical equations for both bilateral ($\sum_{-\infty}^{\infty}$) and unilateral ($\sum_{0}^{\infty}$) Z-transforms. 2. Provide a structured comparison table analyzing limits, ROC behavior, causality assumptions, and their continuous Laplace analogues. --- ## 7. Self-Check Before Moving On - [ ] Can you define the mathematical differences between bilateral and unilateral Z-transforms? - [ ] Can you derive the mapping relationship $z = e^{s T_s}$ and sketch the mapped LHP and imaginary axis territories? - [ ] Do you know why a rational Z-transform cannot have poles inside its Region of Convergence? - [ ] Can you calculate the Z-transform and ROC of a causal decaying exponential sequence? - [ ] Can you calculate the Z-transform and ROC of an anticausal left-sided sequence using geometric series convergence rules? --- ## 8. Source Citations * *(k.Deergha Rao) signals and systems.pdf, Springer Nature, 2018* * *04 Z-transform.pdf, Digital Signal Processing lecture slides* * *Signals, Systems, and Networks MOC & Checklist* --- [[10.01_Z-Transform_Definitions_and_Region_of_Convergence_ROC|← 10.01 Z-Transform: Definitions & Region of Convergence (ROC)]] | [[10.00_Chapter_Map_-_Z-Transform_and_Discrete_Analysis|Chapter 10 Map]] | [[10.03_Initial_and_Final_Value_Theorems_in_the_Z-Domain|10.03 Initial & Final Value Theorems in the Z-Domain →]] # 10.02 Z-Transform Properties & Standard Pairs Alright — let's proceed to **Note 10.02: Z-Transform Properties & Standard Pairs**. In discrete-time analysis, evaluating the Z-transform of complex sequences directly from the infinite Laurent sum can be mathematically tedious. By leveraging a robust set of algebraic **Z-transform properties** {symmetries and operators mapping time-domain manipulations to the z-domain}, we can bypass direct summation and solve complex systems by inspection. This note provides the rigorous mathematical proofs of these properties, compiles a comprehensive **Master Z-Transform Lookup Table**, and solves high-yield composite sequence transforms to secure top marks in your ECE 2107 exam. --- ## 1. Linearity Property ### 1.1 Mathematical Statement If $x_1[n] \overset{\mathcal{Z}}{\leftrightarrow} X_1(z)$ with ROC $R_1$, and $x_2[n] \overset{\mathcal{Z}}{\leftrightarrow} X_2(z)$ with ROC $R_2$, then: $$a_1 x_1[n] + a_2 x_2[n] \overset{\mathcal{Z}}{\leftrightarrow} a_1 X_1(z) + a_2 X_2(z)$$ The Region of Convergence (ROC) is at least the intersection: $$ ext{ROC} \supseteq R_1 \cap R_2$$ *{Where $a_1$ and $a_2$ are arbitrary scalar constants [8.21, 101].}* ### 1.2 Mathematical Proof Starting directly from the Bilateral Z-transform summation definition: $$\mathcal{Z}\left\{ a_1 x_1[n] + a_2 x_2[n] ight\} = \sum_{n=-\infty}^{\infty} \left( a_1 x_1[n] + a_2 x_2[n] ight) z^{-n}$$ Distributing the complex exponential variable $z^{-n}$ and splitting the summation {justified by absolute convergence of the series within their common region $R_1 \cap R_2$}: $$\mathcal{Z}\left\{ a_1 x_1[n] + a_2 x_2[n] ight\} = a_1 \sum_{n=-\infty}^{\infty} x_1[n] z^{-n} + a_2 \sum_{n=-\infty}^{\infty} x_2[n] z^{-n}$$ By identifying the individual Z-transform integrals: $$\mathcal{Z}\left\{ a_1 x_1[n] + a_2 x_2[n] ight\} = a_1 X_1(z) + a_2 X_2(z) \quad lacksquare$$ --- ## 2. Time Shifting Property (Time Delay & Advance) ### 2.1 Mathematical Statement If $x[n] \overset{\mathcal{Z}}{\leftrightarrow} X(z)$ with ROC $R$, then delaying the sequence by an integer $k$ yields: $$x[n-k] \overset{\mathcal{Z}}{\leftrightarrow} z^{-k} X(z)$$ The ROC remains identical to $R$, except possibly for the inclusion or exclusion of the boundaries $z = 0$ {if $k > 0$, multiplying by $z^{-k}$ adds poles at the origin} or $z = \infty$ {if $k < 0$, multiplying by $z^{-k}$ adds poles at infinity} [8.27, 109]. ### 2.2 Mathematical Proof Let $y[n] = x[n-k]$. Its Bilateral Z-transform is: $$Y(z) = \sum_{n=-\infty}^{\infty} x[n-k] z^{-n}$$ To evaluate this sum, we apply a change-of-variable substitution. Let: $$m = n - k \implies n = m + k$$ Substituting these indices into the summation limits {since $n$ ranges from $-\infty$ to $\infty$, $m$ also ranges from $-\infty$ to $\infty$}: $$Y(z) = \sum_{m=-\infty}^{\infty} x[m] z^{-(m+k)}$$ Factoring out the term $z^{-k}$ {which is independent of the summation index $m$}: $$Y(z) = z^{-k} \sum_{m=-\infty}^{\infty} x[m] z^{-m}$$ Since the dummy summation index $m$ can be rewritten as $n$: $$Y(z) = z^{-k} X(z) \quad lacksquare$$ > [!WARNING] > > **The Unilateral Delay Trap (Initial Conditions):** > Note that for the **one-sided** unilateral transform, the shifting property is not a simple multiplication by $z^{-k}$ unless the signal is strictly causal (i.e., $x[n] = 0$ for $n < 0$). If initial conditions are non-zero, unilateral time shifting introduces initial-state correction terms: > $$\mathcal{Z}_+\{x[n-1]\} = z^{-1}X_+(z) + x[-1]$$ > Forgetting to add these initial conditions will cost you 3–4 marks on difference equation solving questions [8.129]! --- ## 3. Scaling in the z-Domain (Frequency Scaling) ### 3.1 Mathematical Statement If $x[n] \overset{\mathcal{Z}}{\leftrightarrow} X(z)$ with ROC $r_1 < |z| < r_2$, then scaling the sequence by a complex exponential $a^n$ yields: $$a^n x[n] \overset{\mathcal{Z}}{\leftrightarrow} X(a^{-1}z)$$ The scaled ROC is modified by the magnitude of $a$: $$ ext{ROC: } |a|r_1 < |z| < |a|r_2$$ ### 3.2 Mathematical Proof Starting from the definition of the Z-transform of the scaled sequence: $$\mathcal{Z}\left\{ a^n x[n] ight\} = \sum_{n=-\infty}^{\infty} a^n x[n] z^{-n}$$ We can group the exponential terms together: $$\mathcal{Z}\left\{ a^n x[n] ight\} = \sum_{n=-\infty}^{\infty} x[n] \left( a^{-1} z ight)^{-n}$$ This sum is structurally identical to the Z-transform of $x[n]$ evaluated at the variable $w = a^{-1}z$: $$\mathcal{Z}\left\{ a^n x[n] ight\} = X(a^{-1}z) \quad lacksquare$$ For convergence, the scaled variable must lie within the original ROC bounds: $$r_1 < |a^{-1}z| < r_2 \implies |a|r_1 < |z| < |a|r_2 \quad [8.31, 114]$$ --- ## 4. Time Reversal Property ### 4.1 Mathematical Statement If $x[n] \overset{\mathcal{Z}}{\leftrightarrow} X(z)$ with ROC $r_1 < |z| < r_2$, then folding the signal in time yields: $$x[-n] \overset{\mathcal{Z}}{\leftrightarrow} X(z^{-1})$$ The folded ROC is inverted: $$ ext{ROC: } rac{1}{r_2} < |z| < rac{1}{r_1} \quad [8.24, 106]$$ ### 4.2 Mathematical Proof Let $y[n] = x[-n]$. Its Z-transform is: $$Y(z) = \sum_{n=-\infty}^{\infty} x[-n] z^{-n}$$ Applying the index substitution $m = -n \implies n = -m$: $$Y(z) = \sum_{m=-\infty}^{\infty} x[m] z^{m} = \sum_{m=-\infty}^{\infty} x[m] \left(z^{-1} ight)^{-m}$$ This matches the definition of $X(z)$ evaluated at $z^{-1}$: $$Y(z) = X(z^{-1}) \quad lacksquare$$ --- ## 5. Rigorous Proof: Differentiation in the z-Domain This is a **highly tested recurring KUET exam classic [PYQ 2019, 2015]** worth **5 Marks**. ### 5.1 Mathematical Statement If $x[n] \overset{\mathcal{Z}}{\leftrightarrow} X(z)$ with ROC $R$, then multiplying the time sequence by the ramp index $n$ corresponds to differentiation in the complex $z$-plane: $$n x[n] \overset{\mathcal{Z}}{\leftrightarrow} -z rac{dX(z)}{dz}$$ The ROC remains identical to $R$ [8.33, 116]. ### 5.2 Mathematical Proof ## Step 0: Write down the baseline transform By definition, the Z-transform of our sequence $x[n]$ is: $$X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}$$ ## Step 1: Differentiate with respect to the complex variable $z$ Take the derivative of both sides of the equation with respect to $z$. Under the uniform convergence constraints of the Laurent series inside the ROC, we can interchange the order of differentiation and summation: $$ rac{dX(z)}{dz} = rac{d}{dz}\left( \sum_{n=-\infty}^{\infty} x[n] z^{-n} ight) = \sum_{n=-\infty}^{\infty} x[n] \left( rac{d}{dz} z^{-n} ight)$$ ## Step 2: Apply the power rule to the derivative term Using the calculus power rule $ rac{d}{dz} z^{-n} = -n z^{-n-1}$: $$ rac{dX(z)}{dz} = \sum_{n=-\infty}^{\infty} x[n] \left( -n z^{-n-1} ight)$$ ## Step 3: Isolate the target discrete sum Factor out the constant negative sign and the term $z^{-1}$ from the summation index $n$: $$ rac{dX(z)}{dz} = -z^{-1} \sum_{n=-\infty}^{\infty} \left( n x[n] ight) z^{-n}$$ ## Step 4: Multiply both sides by $-z$ to simplify Multiply both sides of the expression by $-z$ to clear the coefficients on the right-hand side: $$-z rac{dX(z)}{dz} = \sum_{n=-\infty}^{\infty} \left( n x[n] ight) z^{-n}$$ Since the right-hand side is exactly the Z-transform summation of the sequence $n x[n]$: $$\mathcal{Z}\left\{ n x[n] ight\} = -z rac{dX(z)}{dz} \quad lacksquare$$ --- ## 6. Discrete Convolution Property ### 6.1 Mathematical Statement If $x_1[n] \overset{\mathcal{Z}}{\leftrightarrow} X_1(z)$ with ROC $R_1$, and $x_2[n] \overset{\mathcal{Z}}{\leftrightarrow} X_2(z)$ with ROC $R_2$, then time-domain linear convolution corresponds to direct algebraic multiplication in the z-domain: $$x_1[n] * x_2[n] \overset{\mathcal{Z}}{\leftrightarrow} X_1(z) X_2(z)$$ The resulting ROC is at least the intersection of the individual ROCs: $$ ext{ROC} \supseteq R_1 \cap R_2$$ ### 6.2 Mathematical Proof Let $y[n] = x_1[n] * x_2[n]$. Expanding the discrete linear convolution sum: $$y[n] = \sum_{k=-\infty}^{\infty} x_1[k] x_2[n-k]$$ Taking the Z-transform of both sides: $$Y(z) = \sum_{n=-\infty}^{\infty} \left[ \sum_{k=-\infty}^{\infty} x_1[k] x_2[n-k] ight] z^{-n}$$ Interchanging the order of summation {valid within the region of absolute convergence $R_1 \cap R_2$}: $$Y(z) = \sum_{k=-\infty}^{\infty} x_1[k] \left[ \sum_{n=-\infty}^{\infty} x_2[n-k] z^{-n} ight]$$ Applying the **Time Shifting Property** (Section 2) to the inner summation: $$\sum_{n=-\infty}^{\infty} x_2[n-k] z^{-n} = z^{-k} X_2(z)$$ Substitute this result back into the outer summation: $$Y(z) = \sum_{k=-\infty}^{\infty} x_1[k] \left( z^{-k} X_2(z) ight)$$ Since $X_2(z)$ is independent of the summation index $k$, factor it out of the sum: $$Y(z) = X_2(z) \sum_{k=-\infty}^{\infty} x_1[k] z^{-k}$$ The remaining sum is exactly the Z-transform of $x_1[k]$: $$Y(z) = X_1(z) X_2(z) \quad lacksquare$$ --- ## 7. Master Z-Transform Lookup Table This lookup matrix compiles the standard discrete-time signal pairs heavily utilized in ECE 2107 exam scripts: | Time Sequence $x[n]$ | Z-Transform $X(z)$ | Region of Convergence (ROC) | Physical Signal Type | | :--- | :--- | :--- | :--- | | **$\delta[n]$** | $1$ | Entire $z$-plane | Unit Impulse (Sample) [8.60] | | **$\delta[n-k]$** | $z^{-k}$ | All $z$-plane except $z=0$ (if $k>0$) | Time Shifted Impulse [8.63] | | **$u[n]$** | $ rac{1}{1 - z^{-1}} = rac{z}{z - 1}$ | $|z| > 1$ | Causal Step (DC switch) [8.62] | | **$-u[-n-1]$** | $ rac{1}{1 - z^{-1}} = rac{z}{z - 1}$ | $|z| < 1$ | Anticausal Step (Negative time) | | **$n u[n]$** | $ rac{z^{-1}}{(1 - z^{-1})^2} = rac{z}{(z - 1)^2}$ | $|z| > 1$ | Causal Ramp [8.37] | | **$a^n u[n]$** | $ rac{1}{1 - a z^{-1}} = rac{z}{z - a}$ | $|z| > |a|$ | Causal Exponential Decay [8.13] | | **$-a^n u[-n-1]$** | $ rac{1}{1 - a z^{-1}} = rac{z}{z - a}$ | $|z| < |a|$ | Anticausal Exponential Growth | | **$n a^n u[n]$** | $ rac{a z^{-1}}{(1 - a z^{-1})^2} = rac{a z}{(z - a)^2}$ | $|z| > |a|$ | Ramp-weighted Exponential | | **$\cos(\omega_0 n) u[n]$** | $ rac{1 - z^{-1}\cos(\omega_0)}{1 - 2z^{-1}\cos(\omega_0) + z^{-2}}$ | $|z| > 1$ | Damped Cosine Oscillation [8.39] | | **$\sin(\omega_0 n) u[n]$** | $ rac{z^{-1}\sin(\omega_0)}{1 - 2z^{-1}\cos(\omega_0) + z^{-2}}$ | $|z| > 1$ | Damped Sine Oscillation [8.38] | --- ## 8. High-Yield Solved "Exam Killers" Let's work through some actual exam numericals step-by-step to lock down your properties algebra! ### 8.1 Complex Left-Sided Anticausal Sequence with Shift > [!question] **2022 Exam Section B Q. 8c (ii)** > > Determine the Z-transform and Region of Convergence (ROC) of the sequence: > $$x[n] = \left(- rac{1}{3} ight)^n u[-n]$$ #### Step 1: Represent in Terms of Standard Anticausal Pair Our master lookup table lists the standard anticausal exponential pair as: $$-a^n u[-n-1] \overset{\mathcal{Z}}{\leftrightarrow} rac{1}{1 - a z^{-1}} \quad ext{ROC: } |z| < |a|$$ However, our target sequence is $x[n] = \left(- rac{1}{3} ight)^n u[-n]$. Notice that the unit step is $u[-n]$ instead of $u[-n-1]$. Let's expand the summation directly to handle this boundary index shift cleanly: $$X(z) = \sum_{n=-\infty}^{\infty} \left(- rac{1}{3} ight)^n u[-n] z^{-n}$$ Since $u[-n] = 1$ only for $n \le 0$: $$X(z) = \sum_{n=-\infty}^{0} \left(- rac{1}{3} ight)^n z^{-n}$$ #### Step 2: Swap Summation Indices Let $k = -n \implies n = -k$. As $n$ goes from $-\infty$ to $0$, $k$ goes from $0$ to $\infty$: $$X(z) = \sum_{k=0}^{\infty} \left(- rac{1}{3} ight)^{-k} z^{k}$$ $$X(z) = \sum_{k=0}^{\infty} \left(-3 ight)^{k} z^{k} = \sum_{k=0}^{\infty} (-3z)^k$$ #### Step 3: Apply the Geometric Series Formula This is an infinite geometric series with a common ratio $r = -3z$. The series converges to $ rac{1}{1-r}$ if and only if the magnitude of the ratio is strictly less than unity: $$X(z) = rac{1}{1 - (-3z)} = rac{1}{1 + 3z}$$ #### Step 4: Determine the ROC Bound The convergence condition requires: $$|r| < 1 \implies |-3z| < 1 \implies 3|z| < 1 \implies |z| < rac{1}{3}$$ **Final Exam Answer:** $$X(z) = rac{1}{1 + 3z} \quad ext{ROC: } |z| < rac{1}{3}$$ --- ### 8.2 Using Differentiation Property to Solve Ramp Exponential > [!question] **Exam Practice Problem** > > Find the Z-transform and ROC of $x[n] = n \left( rac{1}{2} ight)^n u[n]$ using the differentiation property. #### Step 1: Start with the Baseline Exponential Decay Transform Let $x_1[n] = \left( rac{1}{2} ight)^n u[n]$. From our standard pairs: $$X_1(z) = rac{1}{1 - 0.5 z^{-1}} \quad ext{ROC: } |z| > 0.5$$ #### Step 2: Apply the Z-Domain Differentiation Operator Since $x[n] = n x_1[n]$, we apply: $$X(z) = -z rac{dX_1(z)}{dz}$$ Let's compute the derivative of $X_1(z)$ with respect to $z$: $$ rac{dX_1(z)}{dz} = rac{d}{dz} \left( 1 - 0.5 z^{-1} ight)^{-1}$$ $$ rac{dX_1(z)}{dz} = -1 \left( 1 - 0.5 z^{-1} ight)^{-2} \cdot rac{d}{dz}\left( -0.5 z^{-1} ight)$$ $$ rac{dX_1(z)}{dz} = -1 \left( 1 - 0.5 z^{-1} ight)^{-2} \cdot \left( 0.5 z^{-2} ight) = rac{-0.5 z^{-2}}{\left( 1 - 0.5 z^{-1} ight)^2}$$ #### Step 3: Multiply by $-z$ to Complete the Property Operator $$X(z) = -z \cdot \left[ rac{-0.5 z^{-2}}{\left( 1 - 0.5 z^{-1} ight)^2} ight] = rac{0.5 z^{-1}}{\left( 1 - 0.5 z^{-1} ight)^2}$$ Multiplying the numerator and denominator by $z^2$: $$X(z) = rac{0.5 z}{(z - 0.5)^2}$$ **Final Answer:** $$X(z) = rac{0.5 z}{(z - 0.5)^2} \quad ext{ROC: } |z| > 0.5$$ --- ## 9. Common Mistakes That Cost Marks > [!WARNING] **Critical Exam Pitfalls** > > 1. **Forgetting to Scale the ROC in Complex Modulation:** When finding the transform of $a^n x[n]$, students correctly update the algebraic formula to $X(a^{-1}z)$ but forget to scale the ROC boundary. Remember: **you must multiply your ROC boundaries by $|a|$**! Failing to scale the ROC boundary leads to an immediate deduction of 2 marks. > 2. **Unilateral Shift Initial Condition Omission:** Applying $z^{-k}X(z)$ to a unilateral shifting problem where initial conditions are non-zero. The simple multiplication rule is **only** valid for the bilateral transform or strictly causal systems. For non-causal systems with active historical memory, you must use the unilateral summation limits to capture initial values [8.129]. > 3. **Inverting the Time-Reversal ROC Inequality:** When folding a signal $x[-n]$, the ROC is inverted ($1/R$). Students often invert the numbers but forget to flip the inequality signs. For example, if the causal ROC is $|z| > 2$, the time-reversed ROC becomes $|z| < 1/2$. Keeping it as $|z| > 1/2$ is a major conceptual error. --- ## 10. Verbatim PYQ Bank * **[PYQ 2022 - 5 Marks]:** Define Z-transform and write down some properties. * **[PYQ 2021 - 4 Marks]:** Mention some properties of the Z-transform. * **[PYQ 2019 - 5 Marks]:** Prove the Z-domain differentiation property (multiplication by $n$): $$\mathcal{Z}\{n x[n]\} = -z rac{dX(z)}{dz}$$ --- ## 11. Self-Check Checklist - [ ] Can you rigorously prove the Z-domain differentiation theorem without skipping steps? - [ ] Do you know the difference between the bilateral time-shift property and unilateral initial condition shifts? - [ ] Can you derive the transform of $x[n] = (-1/3)^n u[-n]$ and sketch its LHP boundary circle on the s-plane? --- *Source: [[(k.Deergha Rao) signals and systems.pdf]], [[04 Z-transform.pdf]], [[signal checklist.md]]* --- [[10.02_Z-Transform_Properties_and_Standard_Pairs|← 10.02 Z-Transform Properties & Standard Pairs]] | [[10.00_Chapter_Map_-_Z-Transform_and_Discrete_Analysis|Chapter 10 Map]] | [[10.04_Inverse_Z-Transform_Methods|10.04 Inverse Z-Transform Methods →]] # 10.03 Initial & Final Value Theorems in the Z-Domain Welcome back! Today, we are going to master one of the most reliable scoring sections of Chapter 10 under **Instructor 1**: **Initial & Final Value Theorems (IVT & FVT) in the Z-Domain**. Just like their continuous-time Laplace counterparts, these discrete boundary theorems allow us to calculate the immediate initial sample $x[0]$ and the long-term steady-state final value $x[\infty]$ directly from the $z$-domain algebraic expression $X(z)$ [8.66, 8.132]. This completely bypasses the tedious process of executing partial fractions or long division to invert the function [8.90]! In KUET examinations, this topic routinely commands **5 to 15 marks**. Let's walk through the strict mathematical proofs, map out the s-plane/z-plane stability boundaries, and solve the classic exam-killer problems step-by-step to lock in maximum marks! --- ## 1. The Initial Value Theorem (IVT) The **Initial Value Theorem** determines the first term $x[0]$ of a causal discrete sequence directly from its $z$-domain representation [8.88]. ### 1.1 Theorem Statement If a sequence $x[n]$ is causal {meaning $x[n] = 0$ for all $n < 0$}, and its Z-transform $X(z)$ converges for $|z| > r$ [8.88], then: $$\mathbf{x[0] = \lim_{z o \infty} X(z)}$$ --- ### 1.2 Mathematically Rigorous Proof [5 Marks] To prove this theorem, we start with the fundamental definition of the unilateral Z-transform for a causal sequence [8.128, 8.88]: $$X(z) = \sum_{n=0}^{\infty} x[n] z^{-n}$$ Expanding this power series term-by-term [8.88]: $$X(z) = x[0] z^0 + x[1] z^{-1} + x[2] z^{-2} + x[3] z^{-3} + \dots$$ $$X(z) = x[0] + rac{x[1]}{z} + rac{x[2]}{z^2} + rac{x[3]}{z^3} + \dots$$ Now, let's evaluate the limit of both sides as $z o \infty$ [8.88]: $$\lim_{z o \infty} X(z) = \lim_{z o \infty} \left[ x[0] + rac{x[1]}{z} + rac{x[2]}{z^2} + rac{x[3]}{z^3} + \dots ight]$$ Since the sequence terms $x[n]$ are finite values, the limit of each rational term with $z$ in the denominator approaches zero: $$\lim_{z o \infty} rac{x[n]}{z^n} = 0 \quad ext{for all } n \ge 1$$ Substituting these zero limits back into the expanded series: $$\lim_{z o \infty} X(z) = x[0] + 0 + 0 + 0 + \dots$$ $$\mathbf{x[0] = \lim_{z o \infty} X(z)} \quad \mathbf{[ ext{Q.E.D.}]}$$ --- ## 2. The Final Value Theorem (FVT) The **Final Value Theorem** evaluates the steady-state, long-term behavior of a causal sequence as discrete time approaches infinity ($n o \infty$) [8.132, 151]. ### 2.1 Theorem Statement If a sequence $x[n]$ is causal, and all poles of its scaled transform $(z-1)X(z)$ lie strictly **inside the unit circle** in the $z$-plane [8.132], then the steady-state value $x[\infty]$ is given by: $$\mathbf{x[\infty] = \lim_{n o \infty} x[n] = \lim_{z o 1} \left(1 - z^{-1} ight) X(z) = \lim_{z o 1} rac{z-1}{z} X(z)}$$ --- ### 2.2 Mathematically Rigorous Proof [8-10 Marks] To prove FVT, we analyze the Z-transform of a first-difference sequence $d[n] = x[n] - x[n-1]$ [8.131]: $$\mathcal{Z}\left\{ x[n] - x[n-1] ight\} = \sum_{n=0}^{\infty} \left( x[n] - x[n-1] ight) z^{-n}$$ Using the **Linearity** [8.21] and **Time-Shifting (Delay)** [8.27] properties of unilateral transforms: $$\mathcal{Z}\left\{ x[n] - x[n-1] ight\} = X(z) - z^{-1} X(z) = \left( 1 - z^{-1} ight) X(z)$$ Now, let's equate this algebraic expression back to the expanded summation definition: $$\sum_{n=0}^{\infty} \left( x[n] - x[n-1] ight) z^{-n} = \left( 1 - z^{-1} ight) X(z)$$ Let's evaluate the limit of both sides as $z o 1$: $$\lim_{z o 1} \sum_{n=0}^{\infty} \left( x[n] - x[n-1] ight) z^{-n} = \lim_{z o 1} \left( 1 - z^{-1} ight) X(z)$$ Because $\lim_{z o 1} z^{-n} = 1^{-n} = 1$ for all finite $n$, we can substitute $z = 1$ directly into the summation terms: $$\sum_{n=0}^{\infty} \left( x[n] - x[n-1] ight) = \lim_{z o 1} \left( 1 - z^{-1} ight) X(z)$$ The infinite summation on the left is a **telescoping series**. Let's write it as the limit of its partial sums as $N o \infty$: $$\lim_{N o \infty} \sum_{n=0}^{N} \left( x[n] - x[n-1] ight) = \lim_{z o 1} \left( 1 - z^{-1} ight) X(z)$$ Let's expand this partial sum explicitly: $$\sum_{n=0}^{N} \left( x[n] - x[n-1] ight) = (x[0] - x[-1]) + (x[1] - x[0]) + (x[2] - x[1]) + \dots + (x[N] - x[N-1])$$ Notice how all intermediate terms cancel out perfectly {telescoping behavior}: * $x[0]$ cancels with $-x[0]$ * $x[1]$ cancels with $-x[1]$ * $x[N-1]$ cancels with $-x[N-1]$ This leaves only the boundary elements: $$\sum_{n=0}^{N} \left( x[n] - x[n-1] ight) = x[N] - x[-1]$$ Since the sequence $x[n]$ is causal, the initial boundary condition is strictly zero: $x[-1] = 0$. Substituting this back: $$\lim_{N o \infty} x[N] = \lim_{z o 1} \left( 1 - z^{-1} ight) X(z)$$ By definition, $\lim_{N o \infty} x[N] = x[\infty]$. Therefore: $$\mathbf{x[\infty] = \lim_{z o 1} \left( 1 - z^{-1} ight) X(z)} \quad \mathbf{[ ext{Q.E.D.}]}$$ --- ## 3. The Crucial Pole-Stability Boundary Criterion In continuous-time Laplace analysis, the Final Value Theorem is bounded by the imaginary $j\Omega$-axis. In the discrete $z$-domain, **FVT is strictly bounded by the Unit Circle ($|z| = 1$)**. ``` Complex z-Plane Stability Boundaries Im(z) ▲ │ Unstable Region ┌───┴───┐ (|z| > 1) ┌──┘ │││ └──┐ ┌─┘ │││ └─┐ │ Stable Region │ ──────┼───────┼─────────┼──────► Re(z) │ (|z| < 1) │ Pole at z = 1 └─┐ │││ ┌─┘ Allowed for FVT └──┐ │││ ┌──┘ └───┼───┘ │ ``` > [!CAUTION] > > **Strict Mathematical Prerequisites for FVT:** > The Final Value Theorem is valid **if and only if** the limit $x[\infty]$ exists as a single, finite steady-state value [8.132]. This imposes three absolute rules on the poles of $X(z)$: > 1. **Stable Open Region:** All poles of $X(z)$ must lie strictly **inside** the unit circle ($|p_i| < 1$). > 2. **Stable Step Offset:** A single, non-repeated pole is permitted exactly **on** the unit circle at $z = 1$ {representing a constant step DC level like $u[n]$} [8.131]. > 3. **The Instability Trap:** If any poles of $X(z)$ lie **on** the unit circle other than $z = 1$ {e.g., oscillatory poles at $z = e^{\pm j\omega_0}$ like sines/cosines} or **outside** the unit circle ($|p_i| > 1$ {representing growing exponentials}), **FVT fails completely**! The algebraic limit will yield a meaningless number. --- ### Proof of FVT Failure on Oscillatory Signals Let $x[n] = \cos\left(n rac{\pi}{2} ight) u[n] = \{1, 0, -1, 0, 1, 0, -1, 0, \dots\}$. The sequence oscillates forever and has no steady-state limit. Its Z-transform is: $$X(z) = rac{z^2}{z^2 + 1} \quad ext{with poles at } p_{1,2} = \pm j \quad (|p_i| = 1)$$ If we blindly apply FVT without checking poles: $$x[\infty] = \lim_{z o 1} \left(1 - z^{-1} ight) rac{z^2}{z^2 + 1} = \lim_{z o 1} \left( rac{z-1}{z} ight) rac{z^2}{z^2 + 1} = rac{(0) \cdot 1}{1 + 1} = 0$$ This algebraic result ($0$) is completely false because the physical signal never decays to zero! This is why conducting a **Pole Stability Audit** is your mandatory first step on exam questions to avoid losing marks. --- ## 4. High-Yield Solved "Exam Killers" Let's solve the most heavily tested, recurring KUET past year questions with complete exam-scoring layout. ### 4.1 The Finite-Duration Sequence [5 Marks] > [!question] **KUET 2025 Fig 16(c) / 2022 / 2020 / 2019** > > If the Z-transform of a causal sequence $x[n]$ is given by: > $$X(z) = 2 + 3z^{-1} + 4z^{-2}$$ > Find the initial value $x[0]$ and the final value $x[\infty]$ of the corresponding sequence. #### Step 1: Calculate the Initial Value $x[0]$ Apply the Z-domain Initial Value Theorem [8.132]: $$x[0] = \lim_{z o \infty} X(z)$$ $$x[0] = \lim_{z o \infty} \left[ 2 + 3z^{-1} + 4z^{-2} ight] = \lim_{z o \infty} \left[ 2 + rac{3}{z} + rac{4}{z^2} ight]$$ Since $ rac{3}{\infty} o 0$ and $ rac{4}{\infty^2} o 0$: $$\mathbf{x[0] = 2} \quad \mathbf{[ ext{Verified}]}$$ --- #### Step 2: Conduct the Pole Stability Audit To apply FVT, we must verify pole stability [8.132]. Let's rewrite $X(z)$ in positive powers of $z$: $$X(z) = rac{2z^2 + 3z + 4}{z^2}$$ The denominator roots yield a **double pole at the origin**: $$z^2 = 0 \implies p_{1,2} = 0$$ Since the pole magnitudes $|0| = 0 < 1$ lie strictly inside the unit circle, the system is absolutely stable. The steady-state value converges, making FVT strictly valid! --- #### Step 3: Calculate the Final Value $x[\infty]$ Apply the Z-domain Final Value Theorem [8.132]: $$x[\infty] = \lim_{z o 1} \left(1 - z^{-1} ight) X(z)$$ $$x[\infty] = \lim_{z o 1} \left(1 - rac{1}{z} ight) \left[ 2 + rac{3}{z} + rac{4}{z^2} ight]$$ Substitute $z = 1$: $$x[\infty] = \left(1 - rac{1}{1} ight) \left[ 2 + rac{3}{1} + rac{4}{1} ight] = (0) \cdot (9)$$ $$\mathbf{x[\infty] = 0} \quad \mathbf{[ ext{Verified}]}$$ --- #### Step 4: Double-Domain Inspection Verification Let's verify our boundary limits by directly expanding $X(z)$ to obtain the discrete sequence $x[n]$ [8.106]: $$X(z) = 2z^0 + 3z^{-1} + 4z^{-2} \leftrightarrow x[n] = \{ \mathbf{2}, 3, 4, 0, 0, 0, \dots \}$$ * Initial element at $n = 0$ is $x[0] = 2$ {Matches IVT perfectly!}. * As $n o \infty$, the finite sequence has settled to $0$, so $x[\infty] = 0$ {Matches FVT perfectly!}. This complete cross-verification proves our analytical methods are 100% correct! --- ### 4.2 The Rational Fractional Proof [10 Marks] > [!question] **KUET 2024 / 2019 / 2015 (10-Mark Classic - Heavily Tested)** > > Prove that the final value of $x[n]$ for: > $$X(z) = rac{z^2}{(z-1)(z-0.2)}$$ > is $1.25$ and its initial value is unity. #### Step 1: Prove the Initial Value is Unity Apply the Z-domain Initial Value Theorem [8.132]: $$x[0] = \lim_{z o \infty} X(z) = \lim_{z o \infty} rac{z^2}{(z-1)(z-0.2)}$$ To evaluate this high-order limit, divide both the numerator and the denominator by the highest power of $z$, which is $z^2$: $$X(z) = rac{z^2}{z^2 - 1.2z + 0.2} = rac{1}{1 - 1.2z^{-1} + 0.2z^{-2}}$$ Now, apply the limit as $z o \infty$: $$x[0] = \lim_{z o \infty} rac{1}{1 - rac{1.2}{z} + rac{0.2}{z^2}}$$ $$x[0] = rac{1}{1 - 0 + 0} = 1$$ $$\mathbf{x[0] = 1 \quad [ ext{PROVED}]}$$ --- #### Step 2: Conduct the Pole Stability Audit We must isolate the roots of the denominator of $X(z)$ [8.132]: $$(z-1)(z-0.2) = 0 \implies p_1 = 1, \ p_2 = 0.2$$ * The pole $p_2 = 0.2$ lies strictly inside the unit circle ($|0.2| = 0.2 < 1$). * The pole $p_1 = 1$ is a simple, non-repeated pole located exactly on the unit circle. * Since all poles are within $|z| \le 1$ with no multiple or oscillatory poles on the boundary, the sequence is stable and settles to a constant DC value. FVT is strictly valid! --- #### Step 3: Prove the Final Value is 1.25 Apply the Z-domain Final Value Theorem [8.132]: $$x[\infty] = \lim_{z o 1} \left(1 - z^{-1} ight) X(z)$$ $$x[\infty] = \lim_{z o 1} \left( rac{z-1}{z} ight) \left[ rac{z^2}{(z-1)(z-0.2)} ight]$$ Let's simplify this algebraic expression before taking the limit: * Cancel out the $(z-1)$ term in both the numerator and denominator: $$x[\infty] = \lim_{z o 1} \left( rac{1}{z} ight) \left[ rac{z^2}{z-0.2} ight]$$ * Simplify the $z$ terms: $$x[\infty] = \lim_{z o 1} rac{z}{z - 0.2}$$ Now substitute the boundary limit $z = 1$ directly: $$x[\infty] = rac{1}{1 - 0.2} = rac{1}{0.8}$$ $$x[\infty] = rac{10}{8} = 1.25$$ $$\mathbf{x[\infty] = 1.25 \quad [ ext{PROVED}]}$$ Both boundaries have been rigorously proven, securing a flawless 10/10 mark allocation! --- ## 5. Common Mistakes That Cost Marks > [!warning] **Critical Exam Pitfalls** > > 1. **Omitting the Pole Stability Check:** Applying FVT directly to an unstable transform. For example, if $X(z) = rac{z}{z-2}$, applying FVT blindly yields $\lim_{z o 1} (1-z^{-1}) rac{z}{z-2} = 0$. But the actual sequence is $x[n] = 2^n u[n]$ which grows exponentially to $\infty$. Omitting this stability audit will result in a **3 to 4 mark deduction**! > 2. **Confusing IVT Limits:** Attempting to evaluate the initial value of $x[n]$ by setting $z o 0$ instead of $z o \infty$. Remember, the high-frequency limit in the time-continuous domain ($s o \infty$) maps to the infinite radial exterior of the $z$-plane ($z o \infty$) due to $z = e^{s T_s}$ [8.131]. > 3. **Dropping the $(1-z^{-1})$ Multiplier in FVT:** Evaluating $\lim_{z o 1} X(z)$ instead of $\lim_{z o 1} (1-z^{-1}) X(z)$. If $X(z)$ has a pole at $z = 1$, evaluating $X(1)$ will divide by zero and explode, leaving you stuck in the middle of your exam calculus. --- ## 6. Verbatim PYQ Reference Bank * **KUET 2025 Q6(b):** If $X(z) = 2 + 3z^{-1} + 4z^{-2}$, find the initial and final values of the corresponding sequence, $x[n]$. **(15 Marks)** * **KUET 2024 Q7(b):** Prove that the final value of $x[n]$ for $X(z) = rac{z^2}{(z-1)(z-0.2)}$ is 1.25 and its initial value is unity. **(10 Marks)** * **KUET 2022 Q6(b):** If $X(z) = 2 + 3z^{-1} + 4z^{-2}$, find the initial and final values of the corresponding sequence $x[n]$. **(05 Marks)** * **KUET 2020 Q6(b):** Find the initial and final values of the sequence $x[n]$ if its Z-transform is $X(z) = 2 + 3z^{-1} + 4z^{-2}$. **(05 Marks)** * **KUET 2019 Q7(b):** Prove that the final value of $x[n]$ for $X(z) = rac{z^2}{(z-1)(z-0.2)}$ is 1.25 and its initial value is unity. **(07 Marks)** * **KUET 2015 Q7(b):** Prove that the final value of $x[n]$ for $X(z) = rac{z^2}{(z-1)(z-0.2)}$ is 1.25 and its initial value is unity. **(10 Marks)** --- ## 7. Interactive Self-Check Checklist - [ ] Can you state and write down the term-by-term proof of the Z-domain Initial Value Theorem? - [ ] Can you prove the discrete telescoping series cancellation that establishes the Final Value Theorem? - [ ] Do you know how to perform a Pole Stability Audit to check if FVT is mathematically valid? - [ ] Why does FVT fail for oscillatory sequences (e.g., unit circle poles other than $z = 1$)? - [ ] Can you solve the classic 10-mark $1.25$ proof without checking partial fractions? --- *Source: signals-and-systems textbook (K. Deergha Rao) Ch 8 [8.131] / Rabiul sir class notes (Lec-16) / KUET QB 24-15.* --- [[10.03_Initial_and_Final_Value_Theorems_in_the_Z-Domain|← 10.03 Initial & Final Value Theorems in the Z-Domain]] | [[10.00_Chapter_Map_-_Z-Transform_and_Discrete_Analysis|Chapter 10 Map]] | [[10.05_Convolution,_Correlation_and_Realization_in_the_Z-Domain|10.05 Convolution, Correlation & Realization in the Z-Domain →]] # 10.04 Inverse Z-Transform Methods Alright — let's tackle one of the most high-yield computational topics under Instructor 1: **Inverse Z-Transform Methods**! This topic alone frequently commands **10 to 15 marks** on the semester final exam. The Inverse Z-transform allows us to map a discrete-time signal's frequency-domain algebraic representation $X(z)$ back to its original time-domain sequence $x[n]$. While the forward transform is an infinite summation, the inverse transform is fundamentally defined as a **complex contour integral**. Because computing a contour integral directly is mathematically tedious, we employ four distinct engineering methods to compute the inverse transform quickly: 1. **Partial Fraction Expansion** {decomposing rational functions into standard first-order and second-order terms}. 2. **Power Series Expansion via Long Division** {explicitly dividing polynomials to yield a sequence of sample coefficients}. 3. **Cauchy's Residue Theorem** {using complex analysis residues on poles enclosed by the ROC boundary}. 4. **Time-Shifting Property Cascade** {rewriting fractions with negative powers of $z$ and applying shift-delay operators}. Let's prove the core residue integration formula, dissect each of these four methods step-by-step, solve every classic past year question (PYQ), and secure full marks on exam day! --- ## 1. Mathematical Derivation of the Inverse Z-Transform Contour Integral Before diving into the algorithms, we must understand where the inverse Z-transform comes from mathematically. It is not an arbitrary formula; it is a direct consequence of **Cauchy's Integral Theorem**. > [!theorem] **Cauchy's Residue Integration Proof [PYQ 2019 - 5 Marks]** > > Prove that the inverse Z-transform of a complex function $X(z)$ is given by: > $$x[n] = \frac{1}{2\pi j} \oint_{C} X(z) z^{n-1} \, dz = \sum \left[ \text{Residues of } X(z)z^{n-1} \text{ at internal poles} \right]$$ > where $C$ is a counter-clockwise closed circular contour lying within the Region of Convergence (ROC) of $X(z)$ and enclosing the origin. ### Step-by-Step Proof: 1. **Recall the forward Z-transform definition:** $$X(z) = \sum_{m=-\infty}^{\infty} x[m] z^{-m} \quad [44, 303]$$ where $z$ is a complex variable enclosing the origin within its ROC. 2. **Multiply both sides of the forward transform equation by $z^{n-1}$:** $$X(z) z^{n-1} = \left( \sum_{m=-\infty}^{\infty} x[m] z^{-m} \right) z^{n-1}$$ $$X(z) z^{n-1} = \sum_{m=-\infty}^{\infty} x[m] z^{n-m-1} \quad [44]$$ 3. **Integrate both sides over a closed, counter-clockwise contour $C$ that lies entirely within the ROC:** $$\oint_{C} X(z) z^{n-1} \, dz = \oint_{C} \left( \sum_{m=-\infty}^{\infty} x[m] z^{n-m-1} \right) \, dz$$ Since the power series converges uniformly within the ROC, we can swap the order of integration and summation: $$\oint_{C} X(z) z^{n-1} \, dz = \sum_{m=-\infty}^{\infty} x[m] \left( \oint_{C} z^{n-m-1} \, dz \right) \quad [44]$$ 4. **Evaluate the contour integral using Cauchy's Integral Theorem:** Let $z$ be represented in polar coordinates along the circular contour of radius $r$: $z = r e^{j\theta} \implies dz = j r e^{j\theta} d\theta = j z \, d\theta$. Evaluating $\oint_{C} z^{k} \, dz$ over $0 \le \theta \le 2\pi$: $$\oint_{C} z^k \, dz = \int_{0}^{2\pi} (r e^{j\theta})^k \left( j r e^{j\theta} \right) d\theta = j r^{k+1} \int_{0}^{2\pi} e^{j(k+1)\theta} \, d\theta$$ * **Case 1: If $k = -1$:** $$\oint_{C} z^{-1} \, dz = j r^{0} \int_{0}^{2\pi} 1 \, d\theta = j (2\pi) = 2\pi j \quad [45]$$ * **Case 2: If $k \neq -1$:** $$\oint_{C} z^k \, dz = j r^{k+1} \left[ \frac{e^{j(k+1)\theta}}{j(k+1)} \right]_{0}^{2\pi} = \frac{r^{k+1}}{k+1} \left( e^{j(k+1)2\pi} - 1 \right)$$ Since $k$ is an integer, $e^{j(k+1)2\pi} = 1$, making the integral evaluate to strictly **zero**. Thus, we establish the fundamental orthogonality relation of complex variables: $$\frac{1}{2\pi j} \oint_{C} z^{n-m-1} \, dz = \begin{cases} 1, & m = n \\ 0, & m \neq n \end{cases} \quad [45]$$ 5. **Substitute the orthogonality relation back into our integrated sum:** The summation on the right-hand side collapses completely, leaving only the term where $m = n$: $$\frac{1}{2\pi j} \oint_{C} X(z) z^{n-1} \, dz = \sum_{m=-\infty}^{\infty} x[m] \cdot \delta[m-n] = x[n]$$ $$x[n] = \frac{1}{2\pi j} \oint_{C} X(z) z^{n-1} \, dz \quad [Q.E.D.]$$ By Cauchy's Residue Theorem, this integral is simply the sum of residues of the integrand $X(z)z^{n-1}$ at all poles located inside the closed contour $C$: $$x[n] = \sum \left[ \text{Residues of } X(z) z^{n-1} \text{ at internal poles} \right] \quad [48]$$ --- ## 2. Inversion Method 1: Partial Fraction Expansion The **Partial Fraction Expansion** method is the absolute workhorse of Z-domain inversion. It mirrors the Laplace inversion process, with one critical difference: we expand **$\frac{X(z)}{z}$** instead of $X(z)$ directly. This step ensures that when we multiply back by $z$, every term has a $z$ in its numerator (e.g., $\frac{Az}{z-p}$), which maps directly to the standard exponential pair $A p^n u[n]$. ### Case A: Distinct (Simple) Poles If $X(z)$ is a proper rational fraction with distinct poles $p_1, p_2, \dots, p_N$, we write: $$\frac{X(z)}{z} = \frac{c_1}{z - p_1} + \frac{c_2}{z - p_2} + \dots + \frac{c_N}{z - p_N} \quad [54]$$ The residues $c_i$ are calculated using the cover-up method: $$c_i = \left. (z - p_i) \frac{X(z)}{z} \right|_{z = p_i} \quad [55]$$ Multiplying by $z$ yields the final Z-domain representation: $$X(z) = \sum_{i=1}^N \frac{c_i z}{z - p_i} \quad [56]$$ ### Case B: Multiple (Repeated) Poles If $\frac{X(z)}{z}$ contains a pole $p_j$ of multiplicity $k$, we must expand it using higher-order denominators: $$\frac{X(z)}{z} = \frac{c_{j1}}{z - p_j} + \frac{c_{j2}}{(z - p_j)^2} + \dots + \frac{c_{jk}}{(z - p_j)^k} + \sum \text{simple pole terms} \quad [55]$$ The residue for the highest-order term is: $$c_{jk} = \left. (z - p_j)^k \frac{X(z)}{z} \right|_{z = p_j} \quad [55]$$ The residues for the lower-order terms are evaluated using derivatives: $$c_{j(k-r)} = \frac{1}{r!} \left. \frac{d^r}{dz^r} \left[ (z - p_j)^k \frac{X(z)}{z} \right] \right|_{z = p_j} \quad [55]$$ --- ## 3. Inversion Method 2: Long Division (Power Series Expansion) The **Long Division Method** (also known as the Power Series Expansion) is used when we do not need a closed-form formula for $x[n]$ but instead want to evaluate the first few samples explicitly. It exploits the direct power series definition: $$X(z) = x[0] + x[1]z^{-1} + x[2]z^{-2} + x[3]z^{-3} + \dots \quad [42, 132]$$ By setting up a polynomial division, the coefficients of the quotients represent the discrete-time samples directly. > [!WARNING] > > **The Division Setup Trap (Causal vs. Anticausal):** > You cannot perform long division blindly. The direction of division depends entirely on the Region of Convergence (ROC)! > > 1. **For Causal Sequences ($|z| > p_{max}$):** We must divide by writing both the numerator and denominator in **descending powers of $z$** (or ascending powers of $z^{-1}$). This yields a quotient with negative exponents of $z$ ($z^{-1}, z^{-2}$), matching a right-sided sequence. > 2. **For Anticausal Sequences ($|z| < p_{min}$):** We must divide by writing both the numerator and denominator in **ascending powers of $z$** (or descending powers of $z^{-1}$). This yields a quotient with positive exponents of $z$ ($z^1, z^2$), matching a left-sided sequence. ``` Causal Division Setup (|z| > a): Anticausal Division Setup (|z| < a): q_0 + q_1 z⁻¹ + q_2 z⁻² q_0 z + q_1 z² + q_2 z³ ┌──────────────────────── ┌──────────────────────── z - a│ z -a + z│ z │ z - a │ z - (z²/a) └─────── └─────────── a ... (z²/a) ... ``` --- ## 4. Inversion Method 3: Cauchy's Residue Calculus Method For any rational fraction $X(z)$, the residue method calculates the time domain samples directly using contour residues. From our derivation, the inverse transform is evaluated as: $$x[n] = \sum \left[ \text{Residues of } X(z) z^{n-1} \text{ at internal poles} \right] \quad [48]$$ The calculation for residues depends on the multiplicity of the pole $p_i$: 1. **For a simple pole ($m=1$):** $$\text{Res}_{z = p_i} \left[ X(z) z^{n-1} \right] = \lim_{z \to p_i} (z - p_i) X(z) z^{n-1} \quad [49]$$ 2. **For a multiple pole of order $m$:** $$\text{Res}_{z = p_i} \left[ X(z) z^{n-1} \right] = \frac{1}{(m-1)!} \lim_{z \to p_i} \frac{d^{m-1}}{dz^{m-1}} \left[ (z - p_i)^m X(z) z^{n-1} \right] \quad [49]$$ > [!TIP] > > **Evaluating the Origin Pole ($z = 0$):** > When evaluating $X(z) z^{n-1}$, if $X(z)$ has a pole at the origin, or if $n-1 < 0$ (such as $n=0$ where $z^{n-1} = z^{-1}$), a pole is introduced at $z=0$. You **must** calculate the residue at $z=0$ for those specific samples, otherwise your $x[0]$ or left-sided values will be incorrect. --- ## 5. Inversion Method 4: Time-Shifting Property Method The time-shifting property is exceptionally useful for inverting functions written in terms of delay operators ($z^{-1}$). It bypasses partial fraction expansions of higher order. Recall the **Time-Shifting Property**: $$\mathcal{Z}\{x[n-k]\} = z^{-k} X(z) \quad [27, 110]$$ If we can factor out $z^{-k}$ from a function, we can invert the remaining term to $w[n]$ and then shift the result: $$X(z) = z^{-k} \cdot W(z) \implies x[n] = w[n-k] u[n-k] \quad [27]$$ --- ## 6. Exhaustive, Step-by-Step Solved Exam Numericals Let's solve the highest-yield numerical problems from the KUET Exam bank using these four methodologies. ### Problem 1: Partial Fractions on Complex Multiple Poles [PYQ 2016 - 5 Marks] > [!question] **KUET Exam Classic [PYQ 2016]** > > Determine the causal inverse Z-transform of the system function: > $$X(z) = \frac{1}{(1 + z^{-1})(1 - z^{-1})^2} \quad [298]$$ #### Step 1: Convert to positive powers of $z$ Multiply the numerator and denominator by $z^3$ to eliminate negative exponents: $$X(z) = \frac{z^3}{z^3 (1+z^{-1})(1-z^{-1})^2} = \frac{z^3}{(z+1)(z-1)^2}$$ #### Step 2: Set up the $\frac{X(z)}{z}$ expansion $$\frac{X(z)}{z} = \frac{z^2}{(z+1)(z-1)^2} = \frac{A}{z+1} + \frac{B}{z-1} + \frac{C}{(z-1)^2}$$ Here, we have a simple pole at $z = -1$ and a multiple pole of order 2 at $z = 1$. #### Step 3: Calculate the residues 1. **For simple pole $A$ at $z = -1$:** $$A = \left. (z+1) \frac{X(z)}{z} \right|_{z = -1} = \left. \frac{z^2}{(z-1)^2} \right|_{z = -1} = \frac{(-1)^2}{(-1 - 1)^2} = \frac{1}{4}$$ 2. **For the highest-order repeated pole term $C$ at $z = 1$:** $$C = \left. (z-1)^2 \frac{X(z)}{z} \right|_{z = 1} = \left. \frac{z^2}{z+1} \right|_{z = 1} = \frac{1^2}{1 + 1} = \frac{1}{2}$$ 3. **For the lower-order repeated pole term $B$ at $z = 1$ using derivative:** $$B = \frac{1}{1!} \left. \frac{d}{dz} \left[ (z-1)^2 \frac{X(z)}{z} \right] \right|_{z = 1} = \left. \frac{d}{dz} \left[ \frac{z^2}{z+1} \right] \right|_{z = 1}$$ Using the quotient rule: $$\frac{d}{dz}\left[ \frac{z^2}{z+1} \right] = \frac{(z+1)(2z) - z^2(1)}{(z+1)^2} = \frac{2z^2 + 2z - z^2}{(z+1)^2} = \frac{z^2 + 2z}{(z+1)^2}$$ Evaluating at $z = 1$: $$B = \frac{1^2 + 2(1)}{(1+1)^2} = \frac{3}{4}$$ #### Step 4: Reassemble $X(z)$ and invert Substitute the residues back into the $\frac{X(z)}{z}$ expression: $$\frac{X(z)}{z} = \frac{1/4}{z+1} + \frac{3/4}{z-1} + \frac{1/2}{(z-1)^2}$$ Multiply throughout by $z$: $$X(z) = \frac{1}{4} \frac{z}{z+1} + \frac{3}{4} \frac{z}{z-1} + \frac{1}{2} \frac{z}{(z-1)^2}$$ From our standard Z-transform lookup table: * $\frac{z}{z+1} \leftrightarrow (-1)^n u[n]$ * $\frac{z}{z-1} \leftrightarrow u[n]$ * $\frac{z}{(z-1)^2} \leftrightarrow n u[n] \quad [43, 119]$ Thus, the causal sequence is: $$\mathbf{x[n] = \left[ \frac{1}{4} (-1)^n + \frac{3}{4} + \frac{1}{2}n \right] u[n]} \quad [298]$$ --- ### Problem 2: Delay Cascade Inversion with Delay Elements [PYQ 2023 - 5 Marks] > [!question] **KUET Exam [PYQ 2023]** > > If $x[n]$ is causal, find the inverse Z-transform of the system function: > $$X(z) = \frac{1}{z(z-0.8)(z+0.4)} \quad [298]$$ Let's solve this using **Method 4 (Time-Shifting)** to highlight the power of delay operations. #### Step 1: Isolate the delay operator $z^{-1}$ Write $X(z)$ as: $$X(z) = z^{-1} \cdot \frac{1}{(z-0.8)(z+0.4)} = z^{-1} \cdot W(z)$$ where we define: $$W(z) = \frac{1}{(z-0.8)(z+0.4)}$$ #### Step 2: Expand $\frac{W(z)}{z}$ using partial fractions $$\frac{W(z)}{z} = \frac{1}{z(z-0.8)(z+0.4)} = \frac{A}{z} + \frac{B}{z-0.8} + \frac{C}{z+0.4}$$ Calculate residues: 1. **For $A$ at $z = 0$:** $$A = \left. \frac{1}{(z-0.8)(z+0.4)} \right|_{z=0} = \frac{1}{(-0.8)(0.4)} = \frac{1}{-0.32} = -\frac{25}{8}$$ 2. **For $B$ at $z = 0.8$:** $$B = \left. \frac{1}{z(z+0.4)} \right|_{z=0.8} = \frac{1}{0.8(1.2)} = \frac{1}{0.96} = \frac{25}{24}$$ 3. **For $C$ at $z = -0.4$:** $$C = \left. \frac{1}{z(z-0.8)} \right|_{z=-0.4} = \frac{1}{(-0.4)(-1.2)} = \frac{1}{0.48} = \frac{25}{12}$$ Substitute residues and multiply by $z$: $$W(z) = -\frac{25}{8} + \frac{25}{24} \frac{z}{z-0.8} + \frac{25}{12} \frac{z}{z+0.4}$$ #### Step 3: Invert $W(z)$ to the time domain $$w[n] = -\frac{25}{8} \delta[n] + \left[ \frac{25}{24} (0.8)^n + \frac{25}{12} (-0.4)^n \right] u[n]$$ #### Step 4: Apply the time delay $z^{-1}$ to obtain $x[n]$ Since $X(z) = z^{-1} W(z)$, the time-shifting property dictates that $x[n] = w[n-1] u[n-1]$: $$\mathbf{x[n] = -\frac{25}{8} \delta[n-1] + \left[ \frac{25}{24} (0.8)^{n-1} + \frac{25}{12} (-0.4)^{n-1} \right] u[n-1]} \quad [298]$$ *(Note: Evaluates exactly to $x[0] = 0$, $x[1] = 3.125$ A).* --- ### Problem 3: Multi-Domain ROC Boundary Solutions [PYQ 2017 - 10 Marks] > [!question] **KUET Exam Classic [PYQ 2017]** > > Determine the inverse Z-transform of the system function: > $$X(z) = \frac{z}{3z^2 - 4z + 1} \quad [298]$$ > under three different Region of Convergence (ROC) conditions: > (i) $|z| > 1$ (Causal) > (ii) $|z| < \frac{1}{3}$ (Anticausal) > (iii) $\frac{1}{3} < |z| < 1$ (Bilateral/Two-sided) #### Step 1: Perform Partial Fraction Expansion First, factor the denominator: $$3z^2 - 4z + 1 = 3(z - \frac{1}{3})(z - 1)$$ Set up the expansion of $\frac{X(z)}{z}$: $$\frac{X(z)}{z} = \frac{1}{3(z - 1/3)(z - 1)} = \frac{A}{z - 1} + \frac{B}{z - 1/3}$$ Calculate residues: 1. **For $A$ at $z = 1$:** $$A = \left. \frac{1}{3(z - 1/3)} \right|_{z=1} = \frac{1}{3(2/3)} = \frac{1}{2}$$ 2. **For $B$ at $z = 1/3$:** $$B = \left. \frac{1}{3(z - 1)} \right|_{z=1/3} = \frac{1}{3(-2/3)} = -\frac{1}{2}$$ Substitute residues and multiply by $z$: $$X(z) = \frac{1}{2} \frac{z}{z-1} - \frac{1}{2} \frac{z}{z-1/3}$$ #### Step 2: Evaluate under different ROCs ##### (i) For ROC $|z| > 1$ (Causal / Right-Sided Sequence) Since the ROC is exterior to the circle enclosing both poles ($z=1/3$ and $z=1$), the sequence is causal: * $\frac{z}{z-1} \leftrightarrow u[n]$ * $\frac{z}{z-1/3} \leftrightarrow \left(\frac{1}{3}\right)^n u[n]$ $$\mathbf{x[n] = \frac{1}{2} \left[ 1 - \left(\frac{1}{3}\right)^n \right] u[n]} \quad [298]$$ ##### (ii) For ROC $|z| < \frac{1}{3}$ (Anticausal / Left-Sided Sequence) Since the ROC is interior to the circle enclosing both poles, the sequence is anticausal: * $\frac{z}{z-1} \leftrightarrow -u[-n-1]$ * $\frac{z}{z-1/3} \leftrightarrow -\left(\frac{1}{3}\right)^n u[-n-1]$ $$x[n] = \frac{1}{2} (-u[-n-1]) - \frac{1}{2} \left( -\left(\frac{1}{3}\right)^n u[-n-1] \right)$$ $$\mathbf{x[n] = -\frac{1}{2} \left[ 1 - \left(\frac{1}{3}\right)^n \right] u[-n-1]} \quad [298]$$ ##### (iii) For ROC $\frac{1}{3} < |z| < 1$ (Bilateral / Two-Sided Sequence) Here, the ROC is a ring. The inner boundary is $|z| = 1/3$ (pole $z=1/3$ lies inside, representing a causal/right-sided component). The outer boundary is $|z| = 1$ (pole $z=1$ lies outside, representing an anticausal/left-sided component). * The pole at $z=1/3$ is right-sided: $\frac{z}{z-1/3} \leftrightarrow \left(\frac{1}{3}\right)^n u[n]$ * The pole at $z=1$ is left-sided: $\frac{z}{z-1} \leftrightarrow -u[-n-1]$ $$\mathbf{x[n] = -\frac{1}{2} u[-n-1] - \frac{1}{2} \left(\frac{1}{3}\right)^n u[n]} \quad [298]$$ --- ### Problem 4: Long Division Expansion (Dual Case Analysis) [PYQ 2017 - 8 Marks] > [!question] **KUET Exam [PYQ 2017]** > > Using long division, determine the inverse Z-transform of the function: > $$X(z) = \frac{1}{1 - \frac{3}{2}z^{-1} + \frac{1}{2}z^{-2}} \quad [135]$$ > when (a) ROC: $|z| > 1$, and (b) ROC: $|z| < \frac{1}{2}$. #### Case (a) ROC: $|z| > 1$ (Causal Sequence) Since the ROC is exterior to $|z|=1$, $x[n]$ is causal. We write the polynomials in descending powers of $z$ (ascending powers of $z^{-1}$): $$\text{Dividend: } 1, \quad \text{Divisor: } 1 - \frac{3}{2}z^{-1} + \frac{1}{2}z^{-2}$$ Let's perform the division step-by-step: 1. Divide first term $1$ by $1$: quotient is **$1$**. Multiply divisor: $1 - \frac{3}{2}z^{-1} + \frac{1}{2}z^{-2}$. Subtract from dividend to get remainder: $\frac{3}{2}z^{-1} - \frac{1}{2}z^{-2}$. 2. Divide $\frac{3}{2}z^{-1}$ by $1$: quotient is **$+\frac{3}{2}z^{-1}$**. Multiply divisor: $\frac{3}{2}z^{-1} - \frac{9}{4}z^{-2} + \frac{3}{4}z^{-3}$. Subtract to get remainder: $\frac{7}{4}z^{-2} - \frac{3}{4}z^{-3}$. 3. Divide $\frac{7}{4}z^{-2}$ by $1$: quotient is **$+\frac{7}{4}z^{-2}$**. Multiply divisor: $\frac{7}{4}z^{-2} - \frac{21}{8}z^{-3} + \frac{7}{8}z^{-4}$. Subtract to get remainder: $\frac{15}{8}z^{-3} - \frac{7}{8}z^{-4}$. 4. Divide $\frac{15}{8}z^{-3}$ by $1$: quotient is **$+\frac{15}{8}z^{-3}$**. The quotient series is: $$X(z) = 1 + \frac{3}{2}z^{-1} + \frac{7}{4}z^{-2} + \frac{15}{8}z^{-3} + \frac{31}{16}z^{-4} + \dots \quad [137]$$ Thus, taking the inverse Z-transform by matching coefficients: $$\mathbf{x[n] = \left\{ 1, \underline{\frac{3}{2}}, \frac{7}{4}, \frac{15}{8}, \frac{31}{16}, \dots \right\} \quad \text{for } n \ge 0} \quad [137]$$ #### Case (b) ROC: $|z| < \frac{1}{2}$ (Anticausal Sequence) Since the ROC is interior to $|z|=1/2$, $x[n]$ is anticausal. We write the polynomials in ascending powers of $z$ (descending powers of $z^{-1}$ to yield positive powers of $z$): $$\text{Divisor: } \frac{1}{2}z^{-2} - \frac{3}{2}z^{-1} + 1, \quad \text{Dividend: } 1$$ Perform the division: 1. Divide $1$ by $\frac{1}{2}z^{-2}$: quotient is **$2z^2$**. Multiply divisor: $1 - 3z + 2z^2$. Subtract to get remainder: $3z - 2z^2$. 2. Divide $3z$ by $\frac{1}{2}z^{-2}$: quotient is **$+6z^3$**. Multiply divisor: $3z - 9z^2 + 6z^3$. Subtract to get remainder: $7z^2 - 6z^3$. 3. Divide $7z^2$ by $\frac{1}{2}z^{-2}$: quotient is **$+14z^4$**. Multiply divisor: $7z^2 - 21z^3 + 14z^4$. Subtract to get remainder: $15z^3 - 14z^4$. 4. Divide $15z^3$ by $\frac{1}{2}z^{-2}$: quotient is **$+30z^5$**. The quotient series is: $$X(z) = 2z^2 + 6z^3 + 14z^4 + 30z^5 + 62z^6 + \dots \quad [138]$$ Taking the inverse Z-transform: $$\mathbf{x[n] = \left\{ \dots, 62, 30, 14, 6, 2, \underline{0}, 0 \right\} \quad \text{for } n \le 0} \quad [138]$$ --- ## 7. Common Mistakes That Cost Marks > [!warning] **Critical Exam Pitfalls** > > 1. **Blind Long Division Setup:** Setting up polynomial long division in descending order of $z$ for an anticausal sequence. Remember: causal ROC is divided in descending order of $z$, anticausal ROC is divided in ascending order of $z$. Mixing these up yields a diverging quotient series, costing you the full 8 marks. > 2. **Omitting the Residue at $z=0$:** Forgetting that when evaluating $\oint X(z) z^{n-1} \, dz$ at $n=0$, the integrand $X(z)z^{-1}$ has an additional pole at $z=0$. You **must** calculate this residue separately for $x[0]$, or your causal signal's initial value will be incorrect. > 3. **The Direct $X(z)$ Partial Fraction Trap:** Attempting to expand $X(z)$ directly instead of expanding $\frac{X(z)}{z}$ first. Decomposing $X(z)$ directly yields constant terms in the partial fractions (e.g., $\frac{A}{z-p}$) which transform to shifted sequences ($A p^{n-1}u[n-1]$) instead of standard exponential terms ($A p^n u[n]$), leading to tedious algebraic bookkeeping and high error rates. --- ## 8. Past Year Question (PYQ) Bank ### KUET Exam 2024 (Section B Q. 5a) * **Question:** Determine the inverse z-transform of the system function: $$H(z) = \frac{z^2 - 3z + 8}{(z-2)(z+2)(z+3)} \quad [298]$$ **(13 Marks)** * **Answer Plan:** 1. Check if $H(z)$ is proper. It has order 2 in numerator and 3 in denominator $\implies$ proper. 2. Set up the expansion of $\frac{H(z)}{z}$ with simple poles at $z=0, 2, -2, -3$. 3. Evaluate residues using the cover-up method: $A_0 = -2/3$, $A_1 = 3/20$, $A_2 = 9/4$, $A_3 = -26/15$. 4. Multiply by $z$ and take the inverse transform causal component term-by-term. ### KUET Exam 2023 (Section B Q. 6b) * **Question:** If $x[n]$ is causal, find the inverse z-transform of $X(z) = \frac{1}{z(z-0.8)(z+0.4)}$. **(05 Marks)** * **Answer Plan:** Solve using the time-shifting property: $X(z) = z^{-1} \cdot W(z)$. Find partial fractions of $\frac{W(z)}{z}$, invert $w[n]$, then shift by 1 sample: $x[n] = w[n-1]u[n-1]$. ### KUET Exam 2019 (Section B Q. 5c) * **Question:** Using the residue method, find the inverse Z-transform of $X(z) = \frac{1}{(z-0.2)(z-0.9)}$. **(05 Marks)** * **Answer Plan:** Formulate $X(z) z^{n-1} = \frac{z^{n-1}}{(z-0.2)(z-0.9)}$. Find residues for simple poles at $z=0.2$ and $z=0.9$. Sum residues to find $x[n]$ for $n \ge 1$, verifying that $x[0]=0$. --- ## 9. Self-Check Before Moving On - [ ] Can you explain why we expand $\frac{X(z)}{z}$ instead of $X(z)$ in partial fractions? - [ ] Do you know how to configure the polynomial order for causal vs. anticausal long division? - [ ] Can you derive Cauchy's residue integration formula from the forward transform definition? - [ ] Do you know when $X(z)z^{n-1}$ introduces a pole at the origin $z=0$? *** *Source: Signals and Systems Reference Book (K. Deergha Rao), Chapter 8, Sections 8.5 & 8.6.* --- [[10.04_Inverse_Z-Transform_Methods|← 10.04 Inverse Z-Transform Methods]] | [[10.00_Chapter_Map_-_Z-Transform_and_Discrete_Analysis|Chapter 10 Map]] | [[11.00_Chapter_Map_-_Discrete_Fourier_Transforms|Chapter 11 Map →]] # 10.05 Convolution, Correlation & Realization in Z-Domain Alright — let's tackle the final, highly scoring, and mathematically elegant module of Chapter 10: **Convolution, Correlation & Realization in the Z-Domain**! In continuous-time systems, we used the Laplace transform to bypass messy time-domain convolution integrals, converting them into simple algebraic multiplication in the complex $s$-plane. Discrete-time systems possess the exact same beautiful symmetry. By mapping our discrete-time signals to the complex $z$-plane, we can perform **linear convolution**, evaluate **signal cross-correlation**, solve **difference equations with non-zero initial conditions**, and construct **system realization block diagrams** with pure algebraic ease. Let's dive into the core proofs, detail how the unilateral transform tackles non-zero initial states, and secure maximum marks on your exam numericals! --- ## 1. Discrete Time-Domain Convolution vs. Z-Domain Multiplication In discrete-time linear time-invariant (LTI) systems, the output response $y[n]$ is mathematically defined as the **convolution sum** of the input sequence $x[n]$ and the system's impulse response $h[n]$: $$y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty} x[k] h[n - k]$$ Rather than sliding and summing arrays term-by-term in the time domain, the **Discrete Convolution Theorem** states that time-domain convolution maps directly to algebraic multiplication in the $z$-domain: $$\mathcal{Z}\{x[n] * h[n]\} = X(z) H(z)$$ where the resulting Region of Convergence (ROC) is at least the intersection of the individual ROCs: $\text{ROC}_y \supseteq \text{ROC}_x \cap \text{ROC}_h$. ### 1.1 Mathematically Rigorous Proof Let's prove this foundational theorem step-by-step to satisfy any 5-mark Section B theory question: 1. **State the definition of the Z-transform** of the convolved sequence $y[n] = x[n] * h[n]$: $$Y(z) = \sum_{n=-\infty}^{\infty} y[n] z^{-n} = \sum_{n=-\infty}^{\infty} \left[ \sum_{k=-\infty}^{\infty} x[k] h[n - k] \right] z^{-n}$$ 2. **Interchange the order of summation** {justified by the absolute convergence of the series within the common ROC}: $$Y(z) = \sum_{k=-\infty}^{\infty} x[k] \left[ \sum_{n=-\infty}^{\infty} h[n - k] z^{-n} \right]$$ 3. **Apply a change of variables** in the inner sum. Let $m = n - k$, which implies $n = m + k$. As $n \to \pm\infty$, the new index $m \to \pm\infty$: $$Y(z) = \sum_{k=-\infty}^{\infty} x[k] \left[ \sum_{m=-\infty}^{\infty} h[m] z^{-(m + k)} \right]$$ 4. **Split the complex exponential variable** $z^{-(m+k)} = z^{-m} z^{-k}$ and factor out terms independent of $m$: $$Y(z) = \sum_{k=-\infty}^{\infty} x[k] z^{-k} \left[ \sum_{m=-\infty}^{\infty} h[m] z^{-m} \right]$$ 5. **Identify the individual Z-transform sums**: * The inner sum is the Z-transform of $h[n]$: $\sum_{m=-\infty}^{\infty} h[m] z^{-m} = H(z)$. * The outer sum is the Z-transform of $x[n]$: $\sum_{k=-\infty}^{\infty} x[k] z^{-k} = X(z)$. 6. **Combine terms** to complete the proof: $$Y(z) = X(z) \cdot H(z) \quad [\text{Q.E.D.}]$$ --- ## 2. Discrete Time-Domain Correlation vs. Z-Domain Multiplication Signal **correlation** measures the similarity between two waveforms. * **Cross-correlation** $r_{x_1 x_2}[l]$ measures how closely a signal $x_1[n]$ resembles a shifted version of $x_2[n]$. * **Auto-correlation** $r_{xx}[l]$ compares a signal against a delayed version of itself to detect periodicities or hidden patterns. ### 2.1 The Z-Domain Correlation Theorem In the time domain, the cross-correlation sequence is defined as: $$r_{x_1 x_2}[l] = \sum_{n=-\infty}^{\infty} x_1[n] x_2[n - l]$$ Taking the Z-transform of both sides yields the **Correlation Theorem**: $$R_{x_1 x_2}(z) = X_1(z) X_2(z^{-1})$$ with an ROC of at least $\text{ROC}_{x_1} \cap \text{ROC}_{x_2^{-1}}$. > [!WARNING] > > **The Conjugate Complex Sequence Rule:** > If the signals are complex-valued, the cross-correlation incorporates a complex conjugate: > $$r_{x_1 x_2}[l] = \sum_{n=-\infty}^{\infty} x_1[n] x_2^*[n - l] \quad \longleftrightarrow \quad R_{x_1 x_2}(z) = X_1(z) X_2^*\left( \frac{1}{z^2} \right)$$ > For real-valued signals, this simplifies directly to the classic $X_1(z) X_2(z^{-1})$ formulation. ### 2.2 Proof of the Correlation Theorem 1. Write the Z-transform definition of the correlation sum: $$R_{x_1 x_2}(z) = \sum_{l=-\infty}^{\infty} r_{x_1 x_2}[l] z^{-l} = \sum_{l=-\infty}^{\infty} \left[ \sum_{n=-\infty}^{\infty} x_1[n] x_2[n - l] \right] z^{-l}$$ 2. Interchange the order of summation: $$R_{x_1 x_2}(z) = \sum_{n=-\infty}^{\infty} x_1[n] \left[ \sum_{l=-\infty}^{\infty} x_2[n - l] z^{-l} \right]$$ 3. Substitute $m = n - l$, meaning $l = n - m$. As $l \to \pm\infty$, the index $m \to \mp\infty$: $$R_{x_1 x_2}(z) = \sum_{n=-\infty}^{\infty} x_1[n] \left[ \sum_{m=-\infty}^{\infty} x_2[m] z^{-(n - m)} \right]$$ 4. Separate the powers of $z$: $z^{-(n - m)} = z^{-n} z^m$: $$R_{x_1 x_2}(z) = \sum_{n=-\infty}^{\infty} x_1[n] z^{-n} \left[ \sum_{m=-\infty}^{\infty} x_2[m] (z^{-1})^{-m} \right]$$ 5. Recognize the individual terms: * The outer sum is $X_1(z)$. * The inner sum matches the Z-transform definition of $x_2[n]$ evaluated at $z^{-1}$, which is $X_2(z^{-1})$. 6. Group terms to finish the proof: $$R_{x_1 x_2}(z) = X_1(z) X_2(z^{-1}) \quad [\text{Q.E.D.}]$$ --- ## 3. Unilateral Z-Transforms & Non-Zero Initial Conditions The standard bilateral Z-transform assumes a system is **initially relaxed** {carrying zero initial charge or current prior to input application}. However, to solve real physical systems characterized by difference equations with **non-zero initial conditions** (e.g., $y[-1] \neq 0$), we must employ the **Unilateral (One-Sided) Z-transform**: $$\mathcal{Z}_+\{y[n]\} = Y_+(z) = \sum_{n=0}^{\infty} y[n] z^{-n}$$ ### 3.1 The Time-Delay Property under Unilateral Constraints When we delay a unilateral sequence, initial condition parameters are introduced to account for past energy states: * **First-Order Delay:** $$\mathcal{Z}_+\{y[n-1]\} = z^{-1} Y_+(z) + y[-1]$$ *{Proof:* $\sum_{n=0}^{\infty} y[n-1]z^{-n} = y[-1] + \sum_{n=1}^{\infty} y[n-1]z^{-n} = y[-1] + z^{-1} \sum_{m=0}^{\infty} y[m]z^{-m} = z^{-1}Y_+(z) + y[-1]$ *}* * **Second-Order Delay:** $$\mathcal{Z}_+\{y[n-2]\} = z^{-2} Y_+(z) + z^{-1} y[-1] + y[-2]$$ ### 3.2 High-Yield Worked Case Study (Unilateral System Solution) > [!question] **KUET Past Year Exam Classic** > > Solve for the output response $y[n]$ of a discrete system described by the difference equation: > $$y[n] - \frac{1}{9} y[n-2] = x[n-1]$$ > driven by a step input $x[n] = u[n]$ with non-zero initial conditions $y[-1] = 1$ and $y[-2] = 0$. #### Step 1: Unilateral Transform with Initial Conditions Taking the unilateral Z-transform of both sides: $$\mathcal{Z}_+\{y[n]\} - \frac{1}{9} \mathcal{Z}_+\{y[n-2]\} = \mathcal{Z}_+\{x[n-1]\}$$ Substitute the time-delay expansion identities: $$\left[ Y_+(z) \right] - \frac{1}{9} \left[ z^{-2} Y_+(z) + z^{-1} y[-1] + y[-2] \right] = z^{-1} X_+(z) + x[-1]$$ Since $x[n] = u[n]$ is causal, $x[-1] = 0$. Inserting $y[-1] = 1$ and $y[-2] = 0$: $$Y_+(z) \left[ 1 - \frac{1}{9}z^{-2} \right] - \frac{1}{9} z^{-1} = z^{-1} X_+(z)$$ We know that $X_+(z) = \mathcal{Z}_+\{u[n]\} = \frac{1}{1 - z^{-1}}$: $$Y_+(z) \left[ 1 - \frac{1}{9}z^{-2} \right] = \frac{z^{-1}}{1 - z^{-1}} + \frac{1}{9} z^{-1}$$ Combine the right-hand terms over a common denominator: $$Y_+(z) \left[ 1 - \frac{1}{9}z^{-2} \right] = z^{-1} \left[ \frac{1}{1 - z^{-1}} + \frac{1}{9} \right] = z^{-1} \left[ \frac{9 + (1 - z^{-1})}{9(1 - z^{-1})} \right] = \frac{z^{-1}(10 - z^{-1})}{9(1 - z^{-1})}$$ #### Step 2: Isolate and Factorize $Y_+(z)$ Divide by the characteristic equation polynomial $1 - \frac{1}{9}z^{-2} = (1 - \frac{1}{3}z^{-1})(1 + \frac{1}{3}z^{-1})$: $$Y_+(z) = \frac{z^{-1}(10 - z^{-1})}{9 (1 - z^{-1})(1 - \frac{1}{3}z^{-1})(1 + \frac{1}{3}z^{-1})}$$ Multiply the numerator and denominator by $z^3$ to convert into positive powers of $z$: $$Y_+(z) = \frac{z^2(10z - 1)}{9 (z - 1)(z - \frac{1}{3})(z + \frac{1}{3})}$$ #### Step 3: Partial Fraction Expansion Expand $\frac{Y_+(z)}{z}$ to evaluate the residues: $$\frac{Y_+(z)}{z} = \frac{z(10z - 1)}{9 (z - 1)(z - \frac{1}{3})(z + \frac{1}{3})} = \frac{A}{z - 1} + \frac{B}{z - \frac{1}{3}} + \frac{C}{z + \frac{1}{3}}$$ * **Find $A$ (pole at $z = 1$) via the cover-up rule**: $$A = \left. \frac{z(10z - 1)}{9(z - \frac{1}{3})(z + \frac{1}{3})} \right|_{z = 1} = \frac{1(9)}{9(2/3)(4/3)} = \frac{9}{9 (8/9)} = \frac{9}{8} = 1.125$$ * **Find $B$ (pole at $z = \frac{1}{3}$)**: $$B = \left. \frac{z(10z - 1)}{9(z - 1)(z + \frac{1}{3})} \right|_{z = 1/3} = \frac{\frac{1}{3}(\frac{7}{3})}{9(-\frac{2}{3})(\frac{2}{3})} = \frac{\frac{7}{9}}{9(-\frac{4}{9})} = \frac{\frac{7}{9}}{-4} = -\frac{7}{36} \approx -0.194$$ * **Find $C$ (pole at $z = -\frac{1}{3}$)**: $$C = \left. \frac{z(10z - 1)}{9(z - 1)(z - \frac{1}{3})} \right|_{z = -1/3} = \frac{-\frac{1}{3}(-\frac{13}{3})}{9(-\frac{4}{3})(-\frac{2}{3})} = \frac{\frac{13}{9}}{9(\frac{8}{9})} = \frac{\frac{13}{9}}{8} = \frac{13}{72} \approx 0.181$$ Reconstruct $Y_+(z)$: $$Y_+(z) = \frac{9}{8} \left( \frac{z}{z - 1} \right) - \frac{7}{36} \left( \frac{z}{z - \frac{1}{3}} \right) + \frac{13}{72} \left( \frac{z}{z + \frac{1}{3}} \right)$$ Taking the inverse unilateral Z-transform: $$y[n] = \left[ 1.125 - 0.194 \left(\frac{1}{3}\right)^n + 0.181 \left(-\frac{1}{3}\right)^n \right] u[n]$$ This matches your physical initial state and represents a 100% rigorous, exam-scoring analytical solution! --- ## 4. High-Yield Solved "Exam Killers" Let's solve the five highest-yielding difference equations and array processing problems direct from past KUET examinations. ### Problem 4.1: Linear Convolution via Z-Transforms > [!question] **KUET Exam Section B [PYQ 2024, 2022, 2018, 2016]** > > Calculate the linear convolution of the sequences: > $$x_1[n] = \{4, \underline{-2}, 1\} \quad \text{and} \quad x_2[n] = \{\underline{1}, 1, 1, 1, 1, 1\}$$ > utilizing Z-transform properties. Note: The underline indicates the origin ($n = 0$) position. #### Step 1: Represent Signals in the Z-domain * For $x_1[n]$, the origin is at $-2$, meaning $x_1[-1] = 4$, $x_1[0] = -2$, and $x_1[1] = 1$: $$X_1(z) = 4z^1 - 2z^0 + 1z^{-1} = 4z - 2 + z^{-1}$$ * For $x_2[n]$, the origin is at the leftmost element, meaning it is a causal pulse of length 6: $$X_2(z) = 1 + z^{-1} + z^{-2} + z^{-3} + z^{-4} + z^{-5}$$ #### Step 2: Multiply the Transforms According to the Z-domain convolution property, the output transform is $Y(z) = X_1(z) X_2(z)$: $$Y(z) = (4z - 2 + z^{-1})(1 + z^{-1} + z^{-2} + z^{-3} + z^{-4} + z^{-5})$$ Let's expand this systematically by distributing each term of $X_1(z)$: 1. **Multiply by $4z$:** $$4z(1 + z^{-1} + z^{-2} + z^{-3} + z^{-4} + z^{-5}) = 4z + 4 + 4z^{-1} + 4z^{-2} + 4z^{-3} + 4z^{-4}$$ 2. **Multiply by $-2$:** $$-2(1 + z^{-1} + z^{-2} + z^{-3} + z^{-4} + z^{-5}) = -2 - 2z^{-1} - 2z^{-2} - 2z^{-3} - 2z^{-4} - 2z^{-5}$$ 3. **Multiply by $z^{-1}$:** $$z^{-1}(1 + z^{-1} + z^{-2} + z^{-3} + z^{-4} + z^{-5}) = z^{-1} + z^{-2} + z^{-3} + z^{-4} + z^{-5} + z^{-6}$$ #### Step 3: Group and Collect Like Powers Sum terms of matching degrees: * $z^1$ term: $4z$ * $z^0$ term: $4 - 2 = 2$ * $z^{-1}$ term: $4 - 2 + 1 = 3$ * $z^{-2}$ term: $4 - 2 + 1 = 3$ * $z^{-3}$ term: $4 - 2 + 1 = 3$ * $z^{-4}$ term: $4 - 2 + 1 = 3$ * $z^{-5}$ term: $-2 + 1 = -1$ * $z^{-6}$ term: $1$ Thus, the product polynomial is: $$Y(z) = 4z + 2 + 3z^{-1} + 3z^{-2} + 3z^{-3} + 3z^{-4} - z^{-5} + z^{-6}$$ #### Step 4: Inverse Transform to the Time Domain Taking the inverse transform of each term: $$y[n] = \{4, \underline{2}, 3, 3, 3, 3, -1, 1\}$$ where the underline indicates $y[0] = 2$ at the origin. --- ### Problem 4.2: Symmetrical Cross-Correlation via Z-Transforms > [!question] **KUET Exam Section B [PYQ 2025]** > > Find the cross-correlation sequence $r_{xy}[l]$ of the sequences: > $$x_1[n] = \{\underline{1}, 2, 3, 4\} \quad \text{and} \quad x_2[n] = \{\underline{2}, 3, 4, 5\}$$ > utilizing Z-transform properties. #### Step 1: Represent Signals in the Z-domain * For $x_1[n]$, the origin is at the leftmost element: $$X_1(z) = 1 + 2z^{-1} + 3z^{-2} + 4z^{-3}$$ * For $x_2[n]$, the origin is at the leftmost element: $$X_2(z) = 2 + 3z^{-1} + 4z^{-2} + 5z^{-3}$$ #### Step 2: Formulate the Correlation Equation According to the Correlation Theorem, $R_{x_1 x_2}(z) = X_1(z) X_2(z^{-1})$: $$X_2(z^{-1}) = 2 + 3z^1 + 4z^2 + 5z^3$$ Let's expand the product systematically: $$R_{x_1 x_2}(z) = (1 + 2z^{-1} + 3z^{-2} + 4z^{-3})(2 + 3z^1 + 4z^2 + 5z^3)$$ Let's multiply term by term: * $z^3$ term: $1 \cdot 5 = 5$ * $z^2$ term: $1 \cdot 4 + 2 \cdot 5 = 14$ * $z^1$ term: $1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 = 3 + 8 + 15 = 26$ * $z^0$ term: $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 = 2 + 6 + 12 + 20 = 40$ * $z^{-1}$ term: $2 \cdot 2 + 3 \cdot 3 + 4 \cdot 4 = 4 + 9 + 16 = 29$ * $z^{-2}$ term: $3 \cdot 2 + 4 \cdot 3 = 6 + 12 = 18$ * $z^{-3}$ term: $4 \cdot 2 = 8$ Thus, the correlation Z-transform is: $$R_{x_1 x_2}(z) = 5z^3 + 14z^2 + 26z^1 + 40 + 29z^{-1} + 18z^{-2} + 8z^{-3}$$ Taking the inverse Z-transform: $$r_{x_1 x_2}[l] = \{5, 14, 26, \underline{40}, 29, 18, 8\}$$ where the underline is at the zero-lag index ($l = 0$), where $r_{x_1 x_2}[0] = 40$. --- ### Problem 4.3: Impulse Response of a Second-Order System > [!question] **KUET Exam Section B [PYQ 2025]** > > Find the impulse response $h[n]$ of the system described by the difference equation: > $$y[n] - \frac{5}{6} y[n-1] + \frac{1}{6} y[n-2] = \frac{1}{3} x[n-1]$$ #### Step 1: Formulate the Transfer Function $H(z)$ Since the system is relaxed initially, taking the Z-transform of the difference equation: $$Y(z) - \frac{5}{6} z^{-1} Y(z) + \frac{1}{6} z^{-2} Y(z) = \frac{1}{3} z^{-1} X(z)$$ $$Y(z) \left[ 1 - \frac{5}{6} z^{-1} + \frac{1}{6} z^{-2} \right] = \frac{1}{3} z^{-1} X(z)$$ The transfer function is: $$H(z) = \frac{Y(z)}{X(z)} = \frac{\frac{1}{3} z^{-1}}{1 - \frac{5}{6} z^{-1} + \frac{1}{6} z^{-2}}$$ #### Step 2: Factorize the Denominator Find the roots of the quadratic equation $1 - \frac{5}{6}z^{-1} + \frac{1}{6}z^{-2} = 0$: $$1 - \frac{5}{6} z^{-1} + \frac{1}{6} z^{-2} = \left( 1 - \frac{1}{2} z^{-1} \right) \left( 1 - \frac{1}{3} z^{-1} \right)$$ Thus, $$H(z) = \frac{\frac{1}{3} z^{-1}}{\left( 1 - \frac{1}{2} z^{-1} \right) \left( 1 - \frac{1}{3} z^{-1} \right)}$$ #### Step 3: Partial Fraction Expansion Expand $H(z)$ into standard form: $$H(z) = \frac{A}{1 - \frac{1}{2} z^{-1}} + \frac{B}{1 - \frac{1}{3} z^{-1}}$$ Multiply both sides by the factored denominator: $$A \left( 1 - \frac{1}{3} z^{-1} \right) + B \left( 1 - \frac{1}{2} z^{-1} \right) = \frac{1}{3} z^{-1}$$ * To find $A$, evaluate at the pole $z^{-1} = 2$: $$A \left( 1 - \frac{2}{3} \right) = \frac{2}{3} \implies \frac{1}{3} A = \frac{2}{3} \implies A = 2$$ * To find $B$, evaluate at the pole $z^{-1} = 3$: $$B \left( 1 - \frac{3}{2} \right) = \frac{3}{3} \implies -\frac{1}{2} B = 1 \implies B = -2$$ Substitute back: $$H(z) = \frac{2}{1 - \frac{1}{2} z^{-1}} - \frac{2}{1 - \frac{1}{3} z^{-1}}$$ #### Step 4: Inverse Transform to the Time Domain Taking the inverse transform of each term for a causal LTI system: $$h[n] = \left[ 2 \left( \frac{1}{2} \right)^n - 2 \left( \frac{1}{3} \right)^n \right] u[n]$$ --- ### Problem 4.4: Complete System Response, Stability & Step Response > [!question] **KUET Exam Section B [PYQ 2021 - 16 Marks]** > > A discrete system is described by the difference equation: > $$y[n] + y[n-1] = x[n]$$ > with $y[n] = 0$ for $n < 0$. > (i) Determine the transfer function and discuss system stability. > (ii) Determine the impulse response $h[n]$. > (iii) Determine the output response when the input is a step sequence $x[n] = 10 u[n]$ (system is initially relaxed). #### Step 1: Transfer Function & Stability Analysis Taking the Z-transform of the difference equation: $$Y(z) + z^{-1} Y(z) = X(z) \implies Y(z) (1 + z^{-1}) = X(z)$$ The transfer function is: $$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1 + z^{-1}} = \frac{z}{z + 1}$$ * **Pole-Zero Map:** System has a zero at $z = 0$, and a pole at $z = -1$. * **Stability Evaluation:** The pole lies exactly on the unit circle ($|z| = 1$). Since there is a pole on the unit circle, the system is **marginally stable (oscillatory)** but not BIBO stable (because its impulse response is not absolutely summable). #### Step 2: Determine Impulse Response $h[n]$ Taking the inverse Z-transform of $H(z) = \frac{1}{1 + z^{-1}}$: $$h[n] = (-1)^n u[n]$$ This confirms the marginal, non-decaying oscillatory behavior. #### Step 3: Solve Step Response for $x[n] = 10 u[n]$ Given the input step $x[n] = 10 u[n] \leftrightarrow X(z) = \frac{10}{1 - z^{-1}} = \frac{10z}{z - 1}$: $$Y(z) = H(z) X(z) = \left( \frac{z}{z + 1} \right) \left( \frac{10z}{z - 1} \right) = \frac{10z^2}{(z - 1)(z + 1)}$$ Perform partial fraction expansion on $\frac{Y(z)}{z}$: $$\frac{Y(z)}{z} = \frac{10z}{(z - 1)(z + 1)} = \frac{A}{z - 1} + \frac{B}{z + 1}$$ * Find $A$ (pole at $z = 1$): $$A = \left. \frac{10z}{z + 1} \right|_{z = 1} = \frac{10}{2} = 5$$ * Find $B$ (pole at $z = -1$): $$B = \left. \frac{10z}{z - 1} \right|_{z = -1} = \frac{-10}{-2} = 5$$ Reassemble $Y(z)$: $$\frac{Y(z)}{z} = \frac{5}{z - 1} + \frac{5}{z + 1} \implies Y(z) = \frac{5}{1 - z^{-1}} + \frac{5}{1 + z^{-1}}$$ Taking the inverse Z-transform: $$y[n] = \left[ 5 + 5(-1)^n \right] u[n]$$ Let's check the first few terms of the sequence to confirm physical accuracy: * $y[0] = 5 + 5 = 10$ * $y[1] = 5 - 5 = 0$ * $y[2] = 5 + 5 = 10$ This oscillatory steady state perfectly tracks the marginally stable pole configuration! --- ### Problem 4.5: Deconvolution via Z-Transforms > [!question] **KUET Exam Section B [PYQ 2023 - 13 Marks]** > > A discrete LTI system has an impulse response $h[n] = \{1, 2, 3\}$ and output response $y[n] = \{1, 1, 2, -1, 3\}$. Determine the input sequence $x[n]$ using Z-transforms. #### Step 1: Represent Sequences in the Z-domain Since the leftmost elements are located at the origin: $$H(z) = 1 + 2z^{-1} + 3z^{-2}$$ $$Y(z) = 1 + z^{-1} + 2z^{-2} - z^{-3} + 3z^{-4}$$ #### Step 2: Formulate the Deconvolution Algebra Since $Y(z) = X(z) H(z)$, we can isolate the input transform algebraically: $$X(z) = \frac{Y(z)}{H(z)} = \frac{1 + z^{-1} + 2z^{-2} - z^{-3} + 3z^{-4}}{1 + 2z^{-1} + 3z^{-2}}$$ #### Step 3: Perform Polynomial Long Division Divide $Y(z)$ by $H(z)$ systematically in descending powers of $z^{-1}$ to yield the causal input sequence: ``` 1 - z⁻¹ + z⁻² ____________________________________ 1 + 2z⁻¹ + 3z⁻² | 1 + z⁻¹ + 2z⁻² - z⁻³ + 3z⁻⁴ 1 + 2z⁻¹ + 3z⁻² ___________________ - z⁻¹ - z⁻² - z⁻³ - z⁻¹ - 2z⁻² - 3z⁻³ _______________________ z⁻² + 2z⁻³ + 3z⁻⁴ z⁻² + 2z⁻³ + 3z⁻⁴ ___________________ 0 ``` Thus, the division terminates perfectly with zero remainder, proving: $$X(z) = 1 - z^{-1} + z^{-2}$$ #### Step 4: Inverse Z-Transform Taking the inverse transform of each term: $$x[n] = \{\underline{1}, -1, 1\}$$ where the underline is at the origin ($n = 0$), $x[0] = 1$. --- ## 5. s-Plane to z-Plane Realization Symmetries Just as in continuous state-space design, a discrete-time transfer function can be realized physically using three basic blocks: **adders**, **multipliers**, and **unit delays (registers)**. ``` Multiplier Block Unit Delay Block X(z) a X(z) z⁻¹ ───────o─────────> aX(z) ───────o─────────> z⁻¹X(z) ▲ │ │ a ▼ ``` We realize discrete-time systems using two standard canonical forms: ### 5.1 Direct Form I Realization * **Concept:** Implements the poles and zeros as two separate cascaded structures. * **Mechanism:** It first implements the feedforward zeros loop, followed by the feedback poles loop. * **Hardware Drawback:** Requires separate registers for both input delays and output delays, doubling memory usage. ### 5.2 Direct Form II Realization (Canonical) * **Concept:** Merges the input and output delay lines into a single, shared intermediate register network. * **Mechanism:** It defines an intermediate sequence $w[n]$, performing the feedback pole summation first, and then tapping the shared registers to compute the feedforward zeros. * **Hardware Advantage:** Minimizes the number of delay elements to exactly $N$ (the system order). --- ## 6. Common Mistakes That Cost Marks > [!WARNING] **Critical Exam Checkpoints** > > 1. **Bilateral vs. Unilateral Time-Shifting Slip:** Using the standard delay identity $x[n-k] \leftrightarrow z^{-k}X(z)$ when solving difference equations with non-zero initial conditions. Remember: for unilateral systems, you *must* expand the shift to include the initial condition terms ($y[-1], y[-2]$), or you will receive a $0/10$ grade penalty on the question. > 2. **Cross-Correlation Lag-Sign Trap:** Mixing up the cross-correlation transform formula as $X_1(z)X_2(z)$ instead of $X_1(z)X_2(z^{-1})$. Convolution corresponds to direct multiplication; correlation requires time-reversal of the second sequence, which flips $z$ to $z^{-1}$ in the complex plane. > 3. **Non-Zero Causal Inputs Shift Blunder:** Forgetting that if a difference equation includes delayed inputs like $x[n-1]$ and $x[n] = u[n]$, then $x[-1] = 0$. Keep track of your boundaries at the $t = 0^-$ switch transition! --- ## 7. Verbatim Past Year Question (PYQ) Bank > [!abstract] **Discrete Symmetries Exam Matrix** > > * **KUET 2025 Section B Q5c:** Find the cross-correlation sequence $r_{x_1x_2}(l)$ of the sequences $x_1(n) = (1, 2, 3, 4)$ and $x_2(n) = (2, 3, 4, 5)$. **(03 Marks)** > * **KUET 2025 Section B Q6b:** If $X(z) = 2 + 3z^{-1} + 4z^{-2}$, find the initial and final values of the corresponding sequence, $x(n)$. **(15 Marks)** > * **KUET 2024 Section B Q7b:** Find the convolution and correlation of the two sequences $x(n) = \{3, 1, 0, 1, 5\}$ and $h(n) = \{1, 2, 4, 1\}$. **(12 Marks)** > * **KUET 2023 Section B Q7a:** A system has an impulse response $h[n] = \{1, 2, 3\}$ and output response $y[n] = \{1, 1, 2, -1, 3\}$. Determine the input sequence $x[n]$ using Z-transforms. **(13 Marks)** > * **KUET 2022 Section B Q6b:** Determine the convolution and cross-correlation of the two sequences using Z-transform: $x(n) = \{1, 2, 3, 4\}$ and $y(n) = \{1, 1, 1, 1\}$. **(09 Marks)** > * **KUET 2021 Section B Q7a:** A system is described by the difference equation: $y(n) + y(n-1) = x(n)$ with $y(n) = 0$ for $n < 0$. Determine its transfer function, discuss system stability, find the impulse response $h(n)$, and solve for the step response when $x(n) = 10$ for $n \ge 0$. **(16 Marks)** --- ## 8. Interactive Self-Check Checklist - [ ] Prove the Z-domain Convolution Theorem using step-by-step index changes. - [ ] Prove the Z-domain Correlation Theorem and identify the conjugate requirements for complex signals. - [ ] Solve a second-order unilateral difference equation featuring non-zero $y[-1]$ and $y[-2]$ parameters. - [ ] Conduct polynomial deconvolution on a 4th-order output using long division. - [ ] Differentiate between Direct Form I and Direct Form II hardware resource requirements. *** [[10.04_Inverse_Z-Transform_Methods|← 10.04 Inverse Z-Transform Methods]] | [[10.00_Chapter_Map_-_Z-Transform_and_Discrete_Analysis|Chapter 10 Map]] | [[11.00_Chapter_Map_-_Discrete_Fourier_Transforms|Chapter 11 Map →]]