Why Can the Same Z-Transform Belong to Two Different Signals?

Studyloaf Team•October 10, 2026•Last verified October 10, 2026•7 min read

Because the algebraic expression is only half of a z-transform; the region of convergence (ROC) is the other half. The expression 1/(1 − az⁻¹) is the transform of the right-sided signal aⁿu[n] when |z| > |a|, and of the left-sided signal −aⁿu[−n−1] when |z| < |a|. With poles at 0.5 and 2, one expression has three possible ROCs and three different signals, and only the ring 0.5 < |z| < 2 gives a stable one.

What you'll learn

  • Why the ROC, not the formula, decides which signal you have
  • A numerical check showing the same formula summing two different series
  • A worked example with poles at 0.5 and 2: all three ROCs, causality and stability
  • The inverse transform for the stable case, value by value
  • The handful of ROC rules that answer most GATE questions

Conventions used here

The two-sided z-transform X(z) = Σ x[n]z⁻ⁿ over all integers n, with u[n] the unit step (1 for n ≥ 0, else 0). The ROC is the set of z for which that sum converges absolutely. A system is stable (bounded-input bounded-output) when its impulse response is absolutely summable.

Why does the ROC carry the information?

For x[n] = aⁿu[n] the sum is Σ (az⁻¹)ⁿ over n ≥ 0, a geometric series that converges only when |az⁻¹| < 1, that is |z| > |a|, and then equals 1/(1 − az⁻¹). For x[n] = −aⁿu[−n−1] the sum runs over n ≤ −1; substituting m = −n gives −Σ (a⁻¹z)ᵐ over m ≥ 1, which converges only when |z| < |a| and also equals 1/(1 − az⁻¹). Same formula, opposite regions. Neither signal has a transform at all on the other's region, so the formula plus its ROC identifies the signal uniquely.

A numerical check

Take a = 0.5. At z = 1, inside |z| > 0.5, summing 0.5ⁿ for n from 0 to 200 gives 2, which is 1/(1 − 0.5). At z = 0.25, outside that region, the same series blows up: its partial sums reach 63 after six terms, 2047 after eleven and over two million after twenty-one. But the left-sided signal −0.5ⁿu[−n−1] does converge at z = 0.25, to −1, and 1/(1 − 0.5/0.25) is also −1.

Worked example: X(z) = 1/(1 − 0.5z⁻¹) + 1/(1 − 2z⁻¹)

The three possible regions of convergence for poles at 0.5 and 2Three z-planes, each with poles marked at z equals 0.5 and z equals 2 and the unit circle dashed. Left: the shaded region is outside the circle of radius 2, giving a causal but unstable signal. Middle: the shaded ring between radius 0.5 and radius 2 contains the unit circle, giving a two-sided stable signal. Right: the shaded disc inside radius 0.5 gives an anticausal unstable signal.|z| > 2causalunstable0.5 < |z| < 2two-sidedstable|z| < 0.5anticausalunstable
The three ROCs for poles at z = 0.5 and z = 2, with the unit circle dashed. Only the middle one, the ring, contains the unit circle.
One expression, three signals. Each sum was checked numerically against the formula at a point inside its ROC.
ROCSignal x[n]Causal?Stable?
|z| > 20.5ⁿu[n] + 2ⁿu[n]yesno (2ⁿ grows)
0.5 < |z| < 20.5ⁿu[n] − 2ⁿu[−n−1]no (two-sided)yes
|z| < 0.5−0.5ⁿu[−n−1] − 2ⁿu[−n−1]no (anticausal)no (0.5ⁿ grows as n → −∞)
  • Each pole contributes one term, and the ROC decides whether that term is right-sided or left-sided.
  • A term is right-sided if the ROC lies outside its pole, left-sided if the ROC lies inside it.
  • For |z| > 2 the ROC is outside both poles, so both terms are right-sided.
  • For the ring it is outside 0.5 but inside 2, so the 0.5 term is right-sided and the 2 term left-sided.
  • For |z| < 0.5 it is inside both, so both terms are left-sided.

What does the stable signal look like?

The stable signal for the ring-shaped region of convergenceStem plot of x of n for n from minus 6 to 6. For n of 0 and above, x of n equals 0.5 to the n: 1, 0.5, 0.25 and so on. For negative n, x of n equals minus 2 to the n: minus 0.5 at n equals minus 1, minus 0.25 at minus 2, decaying towards zero. Both sides decay, so the signal is absolutely summable, with total 3.-6-4-20246-0.500.51nx[n]x[0] = 1x[−1] = −0.50.5ⁿ for n ≥ 0−2ⁿ for n < 0
The signal for the ring ROC, x[n] = 0.5ⁿu[n] − 2ⁿu[−n−1], for n = −6 to 6. Both sides decay towards zero.

For n ≥ 0 the signal is 0.5ⁿ: 1, 0.5, 0.25, 0.125 and so on. For n < 0 it is −2ⁿ: −0.5 at n = −1, −0.25 at n = −2, −0.125 at n = −3. Both halves shrink as you move away from n = 0, so the total of |x[n]| is finite: 2 from the right side plus 1 from the left side, which is 3. That absolute summability is exactly why this ROC, the only one containing the unit circle |z| = 1, is the stable one. It is also the only one of the three signals with a DTFT, because the DTFT is the z-transform evaluated on the unit circle.

The ROC rules worth memorising

  • The ROC never contains a pole, and its boundaries are circles through poles.
  • A right-sided signal has an ROC outside its outermost pole; a left-sided signal, inside its innermost pole; a two-sided signal, a ring between two poles.
  • A finite-length signal converges everywhere except possibly at z = 0 or z = ∞, so its ROC is essentially the whole plane.
  • A causal signal or system has an ROC of the form |z| > r, extending to infinity.
  • Stable means the ROC contains the unit circle.
  • So a causal system is stable exactly when every pole lies inside the unit circle.

Common mistakes

  • Inverting X(z) without asking which ROC is meant, and always writing right-sided terms.
  • Assuming every z-transform belongs to a causal signal.
  • Testing stability by looking at the poles alone, without the ROC: poles outside the unit circle are fine for a stable but non-causal signal.
  • Putting a pole inside the ROC. The ROC can only touch a pole on its boundary.
  • Forgetting the minus sign in the left-sided pair: 1/(1 − az⁻¹) goes with −aⁿu[−n−1], not +aⁿu[−n−1].

Check your understanding

  • Q1. X(z) = 1/(1 − 0.8z⁻¹) belongs to a causal signal. What is its ROC, and is the signal stable? Answer: |z| > 0.8, which contains the unit circle, so yes, it is stable.
  • Q2. X(z) = 1/(1 − 1.25z⁻¹) must belong to a stable signal. Which signal is it? Answer: x[n] = −1.25ⁿu[−n−1], with ROC |z| < 1.25; x[−1] = −0.8, x[−2] = −0.64, and Σ|x[n]| = 4.
  • Q3. Can a system with poles at 0.5 and 2 be both causal and stable? Answer: No. Causal needs |z| > 2, and that region excludes the unit circle.

Related

Official source

Last verified

The GATE 2027 EC syllabus lists the DTFT, DFT and z-transform under discrete-time signals in Section 2, Networks, Signals and Systems (checked against the official PDF from IIT Madras on 10 October 2026). Every series in this post was summed numerically in code and compared with the closed form at a point inside its ROC.

ecegateengineering-fundamentals