Questions: Z-Transform Properties and Inverse Techniques

Question 1

A digital filter is described by the difference equation y[n] = 0.8·y[n−1] + 0.2·x[n]. Why is taking the Z-transform of this equation useful?

• A. The Z-transform converts the discrete-time sequence into a continuous-time signal that is easier to analyze
• B. The time-shift property converts the delay y[n−1] into z⁻¹·Y(z), turning the difference equation into an algebraic equation: Y(z) = 0.8z⁻¹Y(z) + 0.2X(z), which can be solved for the transfer function H(z) = Y(z)/X(z) by algebra
• C. The Z-transform eliminates the need to specify initial conditions, allowing any solution to be found
• D. The convolution property converts the sum y[n] + x[n] into a product, simplifying the equation

Answer: (B)

The time-shift property — that a delay of k samples multiplies the Z-transform by z⁻ᵏ — is what makes difference equations tractable. The difference equation y[n] = 0.8·y[n−1] + 0.2·x[n] involves a relation between a sequence and a delayed version of itself: in the time domain, this is a recursive definition. After Z-transforming, it becomes the algebraic equation Y(z)(1 − 0.8z⁻¹) = 0.2X(z), giving H(z) = 0.2/(1 − 0.8z⁻¹). This is a simple rational function that can be analyzed for poles, stability, and frequency response — all operations that would be far harder on the original recursive equation.

Question 2

You have X(z) = 1/(1 − 0.5z⁻¹). The same rational function can correspond to different time-domain sequences. What determines which sequence it represents?

• A. The numerator polynomial — a numerator of 1 always gives a causal sequence
• B. The magnitude of the pole — poles with |a| < 1 always give causal sequences
• C. The region of convergence (ROC) — ROC |z| > 0.5 gives x[n] = (0.5)ⁿu[n] (causal), while ROC |z| < 0.5 gives x[n] = −(0.5)ⁿu[−n−1] (anti-causal)
• D. The sign of the pole — positive poles always produce causal sequences

Answer: (C)

The region of convergence uniquely determines which sequence corresponds to a given Z-transform. For the pole at z = 0.5, two sequences have the same rational form: the causal sequence x[n] = (0.5)ⁿu[n] converges for |z| > 0.5, and the anti-causal sequence x[n] = −(0.5)ⁿu[−n−1] converges for |z| < 0.5. Without specifying the ROC, you cannot determine which sequence X(z) represents. In practice, causal sequences have ROCs outside the outermost pole, and stable causal sequences have ROCs that include the unit circle. The ROC is not a mathematical formality — it is the information that resolves the ambiguity.

Question 3

Convolution of two sequences in the time domain corresponds to multiplication of their Z-transforms, which is why filtering (a convolution operation) becomes a simple multiplication in the Z-domain.

• T. True
• F. False

Answer: True

True. The convolution property states that if Y(z) = Z{y[n]} and X(z) = Z{x[n]}, then the Z-transform of the convolution y[n] * x[n] is Y(z)·X(z). This means that applying a filter h[n] to an input x[n] — which requires computing the convolution sum Σh[k]x[n−k] in the time domain — becomes the multiplication H(z)·X(z) in the Z-domain. This is the key to the Z-transform workflow: transform the filter and input, multiply, then invert. The multiplication is simple algebra; it's the inversion that requires partial fractions or long division.

Question 4

The Region of Convergence (ROC) is a purely mathematical technicality with no physical significance; for practical engineering purposes, most Z-transforms correspond to a unique time-domain sequence regardless of ROC.

• T. True
• F. False

Answer: False

False. The ROC has direct physical significance. For a causal, stable discrete-time system, the ROC must include the unit circle |z| = 1 — this is the condition for BIBO stability, because the system's frequency response is the Z-transform evaluated on the unit circle. Two sequences with identical rational Z-transforms but different ROCs represent physically different systems: one stable and causal, one stable and anti-causal, or one unstable. When you specify a system's poles (e.g., a pole at z = 0.8), knowing the ROC is |z| > 0.8 tells you the system is causal and stable; ROC |z| < 0.8 would tell you it's anti-causal. Ignoring the ROC leads to ambiguity in both analysis (what sequence does this represent?) and design (is this filter stable?).

Question 5

Explain why the region of convergence (ROC) is essential when performing the inverse Z-transform. Use a specific example with a simple pole to illustrate the ambiguity that the ROC resolves.

Model answer: A rational Z-transform X(z) does not uniquely identify a time-domain sequence — the ROC specifies which of several possible sequences is intended. Consider X(z) = 1/(1 − 2z⁻¹), which has a pole at z = 2. Two sequences have this Z-transform: (1) x₁[n] = 2ⁿu[n], which is the causal (right-sided) sequence with ROC |z| > 2; and (2) x₂[n] = −2ⁿu[−n−1], which is the anti-causal (left-sided) sequence with ROC |z| < 2. Both sequences produce exactly the same rational function X(z), but they are completely different time-domain signals — x₁ grows without bound for positive n, while x₂ grows without bound for negative n. Without the ROC, inverse Z-transform is undefined. In practice, specifying that the system is causal restricts the ROC to the exterior of the outermost pole, uniquely selecting the causal sequence.

The ROC is the bridge between the algebraic Z-domain and the physical time domain. It encodes information about the signal's direction in time (causal vs. anti-causal vs. two-sided) that the rational function alone cannot express. The practical implication: whenever you design or analyze a filter as H(z), implicitly you're also choosing an ROC (usually the causal one), and this choice determines stability, realizability, and physical meaning.