ECE113 DSP Homework 6: Z-Transform Tutorial for Proakis Problems

Introduction: Why Z-Transforms Matter in DSP

Digital signal processing (DSP) is the backbone of modern audio, image, and communication systems. Whether you're streaming music on Spotify, using noise cancellation on AirPods, or analyzing brain waves in an EEG, Z-transforms are the mathematical tool that lets engineers understand and design digital filters. In ECE113, homework 6 from Proakis (4th Edition) covers problems 3.14, 3.16, 3.18, 3.32, 3.35, 3.38, 3.40, 3.42, 3.51, and 5.20. This tutorial will guide you through the core concepts without giving away full solutions, using examples inspired by current trends like AI audio processing and crypto trading algorithms.

Problem 3.14 (d) and (g): Region of Convergence (ROC)

The Z-transform of a sequence x(n) is defined as X(z) = Σ x(n) z^{-n}. The region of convergence (ROC) is the set of z for which the sum converges. For finite-length sequences, the ROC is the entire z-plane except possibly 0 and ∞. For right-sided sequences, the ROC is outside a circle; for left-sided, inside. Problem 3.14 asks you to find the Z-transform and ROC for given sequences.

Tip: Always check the sequence type. For example, x(n) = (0.5)^n u(n) is right-sided → ROC: |z| > 0.5. For x(n) = -(0.5)^n u(-n-1), it's left-sided → ROC: |z| < 0.5. Think of ROC like the coverage area of a 5G cell tower: the signal is valid only within a certain region.

In part (g), you might have a sequence like x(n) = (1/3)^n u(n) + (1/2)^n u(-n-1). The ROC is the intersection of the individual ROCs: |z| > 1/3 and |z| < 1/2 → annular region 1/3 < |z| < 1/2. If the intersection is empty, the Z-transform does not exist.

Problem 3.16 (d): Inverse Z-Transform via Partial Fractions

Given X(z) = (1 + 2z^{-1} + z^{-2}) / (1 - 0.8z^{-1} + 0.64z^{-2}), find the inverse Z-transform. The denominator is (1 - 0.8z^{-1} + 0.64z^{-2}). Factor it into poles: z^2 - 0.8z + 0.64 = 0 → poles at z = 0.4 ± j0.6928 (magnitude 0.8). Use partial fraction expansion. Write X(z) = A + B/(1 - p1 z^{-1}) + C/(1 - p2 z^{-1}) where p1 and p2 are the poles. Then inverse transform each term using the table: p^n u(n) for causal ROC.

Trend analogy: Partial fractions are like decomposing a complex audio signal into individual frequency components – similar to how Spotify's EQ breaks music into bass, mid, and treble. Each pole corresponds to a resonant frequency.

Problem 3.18 (d): Z-Transform Properties

This problem likely uses the differentiation property: n x(n) ↔ -z dX(z)/dz. For example, if x(n) = n (0.5)^n u(n), then X(z) = -z d/dz [1/(1-0.5z^{-1})] = 0.5z^{-1}/(1-0.5z^{-1})^2. ROC: |z| > 0.5. Practice applying properties like time reversal, convolution, and initial value theorem – they save time.

Problem 3.32: System Function from Difference Equation

Given a difference equation like y(n) = 0.5y(n-1) + x(n) - 0.5x(n-1), take Z-transform both sides: Y(z) = 0.5z^{-1}Y(z) + X(z) - 0.5z^{-1}X(z). Then H(z) = Y(z)/X(z) = (1 - 0.5z^{-1})/(1 - 0.5z^{-1}) = 1. Wait, that cancels? That means the system is an all-pass filter? Actually check: numerator and denominator are identical → H(z)=1, so output equals input. But that's trivial. More interesting example: y(n) = 0.8y(n-1) + x(n) - 0.8x(n-1) gives H(z) = (1 - 0.8z^{-1})/(1 - 0.8z^{-1}) = 1 again. Hmm. Actually, for a first-order difference equation with coefficient symmetry, the system function cancels. So Problem 3.32 might involve a more complex equation. The key steps: (1) Z-transform, (2) solve for Y(z)/X(z), (3) simplify.

Problem 3.35 (c) and (g): Frequency Response from Transfer Function

The frequency response is H(e^{jω}) = H(z)|_{z=e^{jω}}. For part (c), note the hint: a missing comma between x(n-1) and x(n) likely means the sequence is x(n-1), x(n)? Actually, the hint says: “a ',' is obviously missing between x(n-1) and x(n).” So the sequence might be {x(n-1), x(n)}? That is ambiguous. Probably the problem originally had x(n) = ... with two terms separated by a comma. For part (g), the sequence x(n) does not have a Z-transform because it's not absolutely summable (e.g., a sinusoidal sequence). Instead, use the fundamental property: for an LTI system with transfer function H(z), the response to e^{jω0 n} is H(e^{jω0}) e^{jω0 n}. So the output is the same sinusoid scaled by the magnitude and shifted by the phase of H(e^{jω0}).

Trend analogy: This is like how a graphic equalizer in a music app (e.g., Apple Music's EQ) shapes the frequency response. Each slider corresponds to |H(e^{jω})| at that frequency. If you boost 60 Hz, the bass gets louder – that's the magnitude response.

Problem 3.38 (b): Stability and Causality

Given a pole-zero plot, determine if the system is stable and causal. A causal system has ROC outside the outermost pole. Stability requires the unit circle to be inside the ROC. So for a causal system, all poles must be inside the unit circle. For non-causal, the ROC must include the unit circle. Example: poles at 0.9 and 1.2. If causal, ROC: |z| > 1.2 → unit circle (|z|=1) is inside? Actually, |z| > 1.2 excludes the unit circle, so unstable. If anti-causal (left-sided), ROC: |z| < 0.9 → unit circle not included. If two-sided, ROC: 0.9 < |z| < 1.2 → includes unit circle → stable but not causal.

Problem 3.40: Convolution via Z-Transform

Compute the convolution of two sequences using Z-transform: y(n) = x(n) * h(n) ↔ Y(z) = X(z) H(z). Then inverse transform. For example, x(n) = u(n) - u(n-3) (finite length) and h(n) = (0.5)^n u(n). Compute X(z) = 1 + z^{-1} + z^{-2}, H(z) = 1/(1-0.5z^{-1}), multiply, then partial fractions to get y(n).

Problem 3.42: System Response to Initial Conditions

Given a difference equation with initial conditions, use one-sided Z-transform: Z{x(n-1)} = z^{-1}X(z) + x(-1). For example, y(n) - 0.5y(n-1) = x(n) with y(-1)=2 and x(n)=u(n). Then Y(z)(1 - 0.5z^{-1}) = X(z) + 0.5y(-1) = 1/(1-z^{-1}) + 1. Solve for Y(z) and inverse transform.

Problem 3.51: Realization of Digital Filters

This problem likely involves drawing block diagrams (direct form I, II, cascade, parallel) from a given transfer function. For H(z) = (1 + 0.5z^{-1})/(1 - 0.8z^{-1} + 0.64z^{-2}), the direct form II uses fewer delays. The key is to write the difference equation: y(n) = 0.8y(n-1) - 0.64y(n-2) + x(n) + 0.5x(n-1). Then draw the signal flow graph.

Problem 5.20: Discrete-Time Fourier Transform (DTFT) and Z-Transform Relationship

This problem connects DTFT and Z-transform. The DTFT is the Z-transform evaluated on the unit circle: X(e^{jω}) = X(z)|_{z=e^{jω}} provided the ROC includes the unit circle. For example, x(n) = a^n u(n), |a|<1, then DTFT is 1/(1-ae^{-jω}). If |a|>1, the DTFT does not exist because the sum diverges.

Conclusion: Practice Makes Perfect

DSP homework can be challenging, but breaking down each problem into steps – Z-transform, ROC, partial fractions, inverse transform – builds confidence. Use the Proakis textbook as your reference, and don't forget to check your ROC for uniqueness. For extra practice, try applying these concepts to real-world signals: the impulse response of a room (reverberation) is a causal system, and its Z-transform helps design audio filters. Good luck with ECE113 homework 6!