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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!

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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.

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: $X(z)={\sum }_{m=-\infty }^{\infty }x[m]z^{-m}[44,303]$ where $z$ is a complex variable enclosing the origin within its ROC.

2. Multiply both sides of the forward transform equation by $z^{n-1}$: $X(z)z^{n-1}=\left({\sum }_{m=-\infty }^{\infty }x[m]z^{-m}\right)z^{n-1}$ $X(z)z^{n-1}={\sum }_{m=-\infty }^{\infty }x[m]z^{n-m-1}[44]$

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: ${\oint }_{C}X(z)z^{n-1}dz={\sum }_{m=-\infty }^{\infty }x[m]\left({\oint }_C{z}^{n-m-1}dz\right)[44]$

4. Evaluate the contour integral using Cauchy’s Integral Theorem: Let $z$ be represented in polar coordinates along the circular contour of radius $r$: $z=r{e}^{j\theta }⟹dz=jr{e}^{j\theta }d\theta =jzd\theta$. Evaluating ${\oint }_C{z}^{k}dz$ over $0\le \theta \le 2\pi$: ${\oint }_C{z}^{k}dz={\int }_0^{2\pi }(r{e}^{j\theta }{)}^k\left(jr{e}^{j\theta }\right)d\theta =j{r}^{k+1}{\int }_0^{2\pi }{e}^{j(k+1)\theta }d\theta$

   • Case 1: If $k=-1$: ${\oint }_C{z}^{-1}dz=j{r}^0{\int }_0^{2\pi }1d\theta =j(2\pi )=2\pi j[45]$
   • Case 2: If $k\ne -1$: ${\oint }_C{z}^{k}dz=j{r}^{k+1}{\left[\frac{e^{j(k+1)\theta }}{j(k+1)}\right]}_0^{2\pi }=\frac{r^{k+1}}{k+1}\left(e^{j(k+1)2\pi }-1\right)$ 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: $\frac{1}{2\pi j}{\oint }_C{z}^{n-m-1}dz=\{\begin{array}{ll}1,& m=n\\ 0,& m\ne n\end{array}[45]$

5. Substitute the orthogonality relation back into our integrated sum: The summation on the right-hand side collapses completely, leaving only the term where $m=n$: $\frac{1}{2\pi j}{\oint }_{C}X(z)z^{n-1}dz={\sum }_{m=-\infty }^{\infty }x[m]\cdot \delta [m-n]=x[n]$ $x[n]=\frac{1}{2\pi j}{\oint }_{C}X(z)z^{n-1}dz[Q.E.D.]$

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 $C$: $x[n]=\sum \left[\text{Residues of }X(z)z^{n-1}\text{ at internal poles}\right][48]$

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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: $\frac{X(z)}{z}=\frac{c_1}{z-p_1}+\frac{c_2}{z-p_2}+\cdots +\frac{c_N}{z-p_N}[54]$

The residues $c_i$ are calculated using the cover-up method: $c_i={(z-p_i)\frac{X(z)}{z}|}_{z=p_i}[55]$

Multiplying by $z$ yields the final Z-domain representation: $X(z)={\sum }_{i=1}^N\frac{c_{i}z}{z-p_i}[56]$

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: $\frac{X(z)}{z}=\frac{c_{j1}}{z-p_j}+\frac{c_{j2}}{(z-p_j{)}^2}+\cdots +\frac{c_{jk}}{(z-p_j{)}^k}+\sum \text{simple pole terms}[55]$

The residue for the highest-order term is: $c_{jk}={(z-p_j{)}^k\frac{X(z)}{z}|}_{z=p_j}[55]$

The residues for the lower-order terms are evaluated using derivatives: $c_{j(k-r)}=\frac{1}{r!}{\frac{d^r}{d{z}^r}\left[(z-p_j{)}^k\frac{X(z)}{z}\right]|}_{z=p_j}[55]$

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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: $X(z)=x[0]+x[1]z^{-1}+x[2]z^{-2}+x[3]z^{-3}+\dots [42,132]$

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) ...

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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: $x[n]=\sum \left[\text{Residues of }X(z)z^{n-1}\text{ at internal poles}\right][48]$

The calculation for residues depends on the multiplicity of the pole $p_i$:

1. For a simple pole ($m=1$): ${\text{Res}}_{z=p_i}\left[X(z)z^{n-1}\right]={\lim }_{z\to p_i}(z-p_i)X(z)z^{n-1}[49]$
2. For a multiple pole of order $m$: ${\text{Res}}_{z=p_i}\left[X(z)z^{n-1}\right]=\frac{1}{(m-1)!}{\lim }_{z\to p_i}\frac{d^{m-1}}{d{z}^{m-1}}\left[(z-p_i{)}^{m}X(z)z^{n-1}\right][49]$

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.

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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: $\mathcal{Z}\{x[n-k]\}=z^{-k}X(z)[27,110]$

If we can factor out $z^{-k}$ from a function, we can invert the remaining term to $w[n]$ and then shift the result: $X(z)=z^{-k}\cdot W(z)⟹x[n]=w[n-k]u[n-k][27]$

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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]

KUET Exam Classic [PYQ 2016]

Determine the causal inverse Z-transform of the system function: $X(z)=\frac{1}{(1+z^{-1})(1-z^{-1}{)}^2}[298]$

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 $z=-1$: $A={(z+1)\frac{X(z)}{z}|}_{z=-1}={\frac{z^2}{(z-1{)}^2}|}_{z=-1}=\frac{(-1{)}^2}{(-1-1{)}^2}=\frac{1}{4}$

2. For the highest-order repeated pole term $C$ at $z=1$: $C={(z-1{)}^2\frac{X(z)}{z}|}_{z=1}={\frac{z^2}{z+1}|}_{z=1}=\frac{1^2}{1+1}=\frac{1}{2}$

3. For the lower-order repeated pole term $B$ at $z=1$ using derivative: $B=\frac{1}{1!}{\frac{d}{dz}\left[(z-1{)}^2\frac{X(z)}{z}\right]|}_{z=1}={\frac{d}{dz}\left[\frac{z^2}{z+1}\right]|}_{z=1}$ 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:

• $\frac{z}{z+1}\leftrightarrow (-1{)}^{n}u[n]$
• $\frac{z}{z-1}\leftrightarrow u[n]$
• $\frac{z}{(z-1{)}^2}\leftrightarrow nu[n][43,119]$

Thus, the causal sequence is: $\mathbf{x}[\mathbf{n}]=\left[\frac{1}{4}(-1{)}^{\mathbf{n}}+\frac{3}{4}+\frac{1}{2}\mathbf{n}\right]\mathbf{u}[\mathbf{n}][298]$

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Problem 2: Delay Cascade Inversion with Delay Elements [PYQ 2023 - 5 Marks]

KUET Exam [PYQ 2023]

If $x[n]$ is causal, find the inverse Z-transform of the system function: $X(z)=\frac{1}{z(z-0.8)(z+0.4)}[298]$

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 $z=0$: $A={\frac{1}{(z-0.8)(z+0.4)}|}_{z=0}=\frac{1}{(-0.8)(0.4)}=\frac{1}{-0.32}=-\frac{25}{8}$
2. For $B$ at $z=0.8$: $B={\frac{1}{z(z+0.4)}|}_{z=0.8}=\frac{1}{0.8(1.2)}=\frac{1}{0.96}=\frac{25}{24}$
3. For $C$ at $z=-0.4$: $C={\frac{1}{z(z-0.8)}|}_{z=-0.4}=\frac{1}{(-0.4)(-1.2)}=\frac{1}{0.48}=\frac{25}{12}$

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 $x[n]=w[n-1]u[n-1]$: $\mathbf{x}[\mathbf{n}]=-\frac{25}{8}\delta [\mathbf{n}-1]+\left[\frac{25}{24}(0.8{)}^{\mathbf{n}-1}+\frac{25}{12}(-0.4{)}^{\mathbf{n}-1}\right]\mathbf{u}[\mathbf{n}-1][298]$

(Note: Evaluates exactly to $x[0]=0$, $x[1]=3.125$ A).

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Problem 3: Multi-Domain ROC Boundary Solutions [PYQ 2017 - 10 Marks]

KUET Exam Classic [PYQ 2017]

Determine the inverse Z-transform of the system function: $X(z)=\frac{z}{3z^2-4z+1}[298]$ 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 $z=1$: $A={\frac{1}{3(z-1/3)}|}_{z=1}=\frac{1}{3(2/3)}=\frac{1}{2}$
2. For $B$ at $z=1/3$: $B={\frac{1}{3(z-1)}|}_{z=1/3}=\frac{1}{3(-2/3)}=-\frac{1}{2}$

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:

• $\frac{z}{z-1}\leftrightarrow u[n]$
• $\frac{z}{z-1/3}\leftrightarrow {\left(\frac{1}{3}\right)}^{n}u[n]$

$\mathbf{x}[\mathbf{n}]=\frac{1}{2}\left[1-{\left(\frac{1}{3}\right)}^{\mathbf{n}}\right]\mathbf{u}[\mathbf{n}][298]$

(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:

• $\frac{z}{z-1}\leftrightarrow -u[-n-1]$
• $\frac{z}{z-1/3}\leftrightarrow -{\left(\frac{1}{3}\right)}^{n}u[-n-1]$

$x[n]=\frac{1}{2}(-u[-n-1])-\frac{1}{2}\left(-{\left(\frac{1}{3}\right)}^{n}u[-n-1]\right)$ $\mathbf{x}[\mathbf{n}]=-\frac{1}{2}\left[1-{\left(\frac{1}{3}\right)}^{\mathbf{n}}\right]\mathbf{u}[-\mathbf{n}-1][298]$

(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: $\frac{z}{z-1}\leftrightarrow -u[-n-1]$

$\mathbf{x}[\mathbf{n}]=-\frac{1}{2}\mathbf{u}[-\mathbf{n}-1]-\frac{1}{2}{\left(\frac{1}{3}\right)}^{\mathbf{n}}\mathbf{u}[\mathbf{n}][298]$

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Problem 4: Long Division Expansion (Dual Case Analysis) [PYQ 2017 - 8 Marks]

KUET Exam [PYQ 2017]

Using long division, determine the inverse Z-transform of the function: $X(z)=\frac{1}{1-\frac{3}{2}{z}^{-1}+\frac{1}{2}{z}^{-2}}[135]$ 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 $z^{-1}$): $\text{Dividend: }1,\text{Divisor: }1-\frac{3}{2}{z}^{-1}+\frac{1}{2}{z}^{-2}$

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: $X(z)=1+\frac{3}{2}{z}^{-1}+\frac{7}{4}{z}^{-2}+\frac{15}{8}{z}^{-3}+\frac{31}{16}{z}^{-4}+\dots [137]$

Thus, taking the inverse Z-transform by matching coefficients: $\mathbf{x}[\mathbf{n}]=\left\{1,\underline{\frac{3}{2}},\frac{7}{4},\frac{15}{8},\frac{31}{16},\dots \right\}\text{for }\mathbf{n}\ge 0[137]$

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 $z$): $\text{Divisor: }\frac{1}{2}{z}^{-2}-\frac{3}{2}{z}^{-1}+1,\text{Dividend: }1$

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: $X(z)=2z^2+6z^3+14z^4+30z^5+62z^6+\dots [138]$

Taking the inverse Z-transform: $\mathbf{x}[\mathbf{n}]=\left\{\dots ,62,30,14,6,2,\underline{0},0\right\}\text{for }\mathbf{n}\le 0[138]$

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7. Common Mistakes That Cost Marks

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.

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8. Past Year Question (PYQ) Bank

KUET Exam 2024 (Section B Q. 5a)

• Question: Determine the inverse z-transform of the system function: $H(z)=\frac{z^2-3z+8}{(z-2)(z+2)(z+3)}[298]$ (13 Marks)
• Answer Plan:
  1. Check if $H(z)$ is proper. It has order 2 in numerator and 3 in denominator $⟹$ 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 $x[0]=0$.

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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 $z=0$?

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Source: Signals and Systems Reference Book (K. Deergha Rao), Chapter 8, Sections 8.5 & 8.6.