← 10.01 Z-Transform: Definitions & Region of Convergence (ROC) | Chapter 10 Map | 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 with ROC , and with ROC , then: The Region of Convergence (ROC) is at least the intersection: {Where and are arbitrary scalar constants [8.21, 101].}
1.2 Mathematical Proof
Starting directly from the Bilateral Z-transform summation definition:
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]]*