Short notes

Z-Transform Short Notes

When you implement a first-order IIR digital filter on a DSP processor — difference equation y[n] = 0.9y[n−1] + 0.1x[n] — the Z-transform of this equation is Y(z) = 0.9z^(−1)Y(z) + 0.1X(z), giving transfer function H(z) = 0.1/(1 − 0.9z^(−1)) with a single pole at z = 0.9. The location of that pole inside the unit circle is what guarantees the filter is stable — and reading pole locations from H(z) is the entire point of Z-transform analysis.

ECE, EI

How it works

The Z-transform is defined as X(z) = Σ x[n]z^(−n) summed over all n, where z is a complex variable. The ROC is the set of |z| values for which the sum converges absolutely. For the causal sequence x[n] = aⁿu[n], X(z) = 1/(1−az^(−1)) = z/(z−a) with ROC |z| > |a|. Key pairs: δ[n] ↔ 1 (all z), u[n] ↔ z/(z−1) (|z| > 1), aⁿu[n] ↔ z/(z−a) (|z| > |a|). Poles of X(z) cannot lie inside the ROC; the ROC is always a ring or half-plane in the z-domain.

Key points to remember

For a causal stable discrete system, all poles of H(z) must lie strictly inside the unit circle |z| = 1. The Z-transform and DTFT are related: substitute z = e^(jω) into X(z) to get X(e^jω) — this is valid only when the ROC includes the unit circle. Inverse Z-transform methods include partial fractions (most common in exams), long division (gives coefficients directly), and contour integration. Initial value theorem: x[0] = lim(z→∞) X(z). The convolution property is Y(z) = X(z)H(z), mirroring the Laplace transform. A right-sided sequence has ROC outside a circle; a left-sided sequence has ROC inside a circle.

Exam tip

The examiner always asks you to find the inverse Z-transform using partial fractions and specify the ROC — state whether the sequence is causal or anti-causal based on the ROC, because that determines the final time-domain answer.

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