Properties of the Z-Transform

Chapter 23 defined the bilateral z-transform of a sequence as a power series in \(z^{-1}\), and established that the pair \(\big(X(z),\,\text{ROC}\big)\) — never \(X(z)\) alone — is what identifies a signal. That chapter also computed a handful of transforms directly from the sum, and by the end of it the method was beginning to look expensive. Evaluating \(\sum_n x[n]z^{-n}\) from scratch for \(n\,(0.4)^{n-3}\cos(\pi n/6)\,u[n-3]\) is not an attractive prospect.

It is also unnecessary. Almost every sequence an engineer meets is built from a few primitives by a small set of operations — delay, scaling by an exponential, multiplication by \(n\), convolution, accumulation — and each of those operations has a fixed, once-and-for-all effect on the transform. Learn the effect of the operations and you never evaluate the defining sum again; you decompose the signal, look up the primitives, and assemble.

This is exactly the strategy of Chapter 14 for the Fourier transform and Chapter 18 for the Laplace transform, and many of the results will look familiar in outline. The z-domain, however, adds a complication those chapters did not have in the same acute form. Because the transform converges only on an annulus, every property carries a second clause describing what happens to that annulus. A student who writes \(x[n-3] \leftrightarrow z^{-3}X(z)\) and stops has written half a theorem.

Linearity is immediate from the definition, because the defining sum is itself a linear operation on the sequence. If \(x_1[n] \leftrightarrow X_1(z)\) on \(R_1\) and \(x_2[n] \leftrightarrow X_2(z)\) on \(R_2\), then for any constants \(\alpha,\beta\),

The splitting of the sum into two sums is legitimate only where both converge, so the combination is certainly valid on \(R_1 \cap R_2\). That is the honest statement of the ROC clause, and it is stated as a containment rather than an equality for a reason.

The cancellation case is worth seeing once, because it is the mechanism by which infinite-length signals combine into finite-length ones. Take \(x_1[n] = (\tfrac12)^n u[n]\) and \(x_2[n] = (\tfrac12)^n u[n-1]\). Both are right-sided with a single pole at \(z = \tfrac12\), so both ROCs are \(|z| \gt \tfrac12\), and the intersection is \(|z| \gt \tfrac12\). But the two sequences agree at every index except \(n = 0\), so their difference is the single sample \(\delta[n]\) — a finite-length signal, whose transform is the constant \(1\), convergent for every \(z\) in the plane. The pole cancelled, and the ROC swelled from a punctured exterior to everything.

The opposite failure is also possible. If \(x_1[n] = 2^n u[n]\), whose ROC is \(|z| \gt 2\), is added to \(x_2[n] = (\tfrac12)^n u[-n-1]\), whose ROC is \(|z| \lt \tfrac12\), the two annuli do not overlap at all. The sum has no z-transform: there is no value of \(z\) at which the defining series converges. This is not a pathology invented for examinations — it is the discrete-time statement that a signal growing without bound in both directions of time cannot be represented.

Delay is the elementary operation of discrete-time systems — a shift register, a memory read, a sample of latency — so the transform of a shifted sequence is the property one uses most. The derivation is a change of summation index and nothing more.

Because the substitution simply relabels the terms, the set of \(|z|\) for which the series converges absolutely is unchanged — with one qualification. The factor \(z^{-n_0}\) is itself a pole at \(z=0\) when \(n_0 \gt 0\), and a pole at \(z=\infty\) when \(n_0 \lt 0\). If the original ROC included the point in question, the shift removes it.

For an infinite-length right-sided signal this never matters, because \(z=0\) was already excluded. It matters for finite-length signals, whose ROC is the whole plane apart from possibly \(0\) or \(\infty\). Delaying \(\delta[n]\) by three samples gives \(z^{-3}\), which is finite everywhere except the origin: the ROC lost exactly one point.

There is a trap here that catches more candidates than any other single item in this chapter, and it is worth naming. The property applies to \(x[n-n_0]\) — the whole signal shifted, argument and all. It does not apply when only part of the expression is shifted. The sequence \((\tfrac12)^n u[n-3]\) is not the shift of \((\tfrac12)^n u[n]\); the sequence \((\tfrac12)^{n-3}u[n-3]\) is. When you meet the first form, extract the constant factor first: \((\tfrac12)^n u[n-3] = (\tfrac12)^3 (\tfrac12)^{n-3}u[n-3]\). Worked Example 2 does this in full.

Finally, a version of the property that Chapter 25 will need. The unilateral transform \(X_u(z) = \sum_{n=0}^{\infty}x[n]z^{-n}\) ignores everything before \(n=0\), so shifting drags samples across the boundary at \(n=0\) and they must be accounted for explicitly:

Repeating the argument gives \(x[n-2] \leftrightarrow x[-2] + x[-1]z^{-1} + z^{-2}X_u(z)\), and so on. The stored past of the system appears as additive constants, which is exactly how a difference equation with non-zero initial conditions is solved. The bilateral transform, which sees all of time, has no such terms.

Multiplying a sequence by an exponential \(a^n\) — where \(a\) may be complex — is the discrete-time analogue of the Laplace shift \(e^{s_0 t}x(t) \leftrightarrow X(s-s_0)\) of Chapter 18. In the z-domain the effect is not a translation but a scaling of the variable, and the geometry that follows is worth dwelling on.

For the ROC, the series converges when \(z/a\) lies in the original ring \(r_1 \lt |z/a| \lt r_2\), that is when \(|a|r_1 \lt |z| \lt |a|r_2\). Both radii are multiplied by \(|a|\). Equivalently, and more usefully: every pole and zero of \(X\) at a location \(p\) moves to \(ap\).

Three special cases carry all the practical weight. When \(a\) is real and positive the pattern simply dilates radially — a stable pole can be pushed out through the unit circle, which is how a decaying signal is turned into a growing one. When \(a = -1\) the pattern is reflected through the origin, corresponding to the alternating multiplier \((-1)^n\); a lowpass system becomes a highpass one, a fact used constantly in filter design. When \(a = e^{j\Omega_0}\), so that \(|a|=1\), nothing is scaled at all and the entire pole–zero pattern is rotated by \(\Omega_0\) about the origin. The ROC is untouched, because \(|a|=1\).

Three further operations on the index have transforms that follow from the same style of substitution. Time reversal replaces \(n\) by \(-n\):

The ROC inverts with the variable: if \(X\) converged on \(r_1 \lt |z| \lt r_2\), then \(X(1/z)\) converges on \(1/r_2 \lt |z| \lt 1/r_1\). A right-sided signal becomes left-sided, an exterior ROC becomes an interior one, and each pole at \(p\) reappears at \(1/p\). A pole inside the unit circle is thrown outside it, which is precisely why reversing a stable causal signal in time produces something stable but anticausal — a point Chapter 25 will use when discussing inverse systems.

Time expansion — inserting \(k-1\) zeros between every pair of samples, the upsampling operation at the heart of multirate processing — is defined by \(x_{(k)}[n] = x[n/k]\) when \(n\) is an integer multiple of \(k\), and zero otherwise. Since only the indices \(n=km\) contribute,

with ROC \(R^{1/k}\): every radius is replaced by its \(k\)-th root, and every pole splits into \(k\) poles equally spaced on a circle. Note the contrast with continuous time, where compressing the axis divided the transform argument; here, because the inserted samples are zeros rather than interpolated values, the effect is a clean substitution.

Conjugation follows by conjugating the defining sum term by term, remembering that the conjugate of \(z^{-n}\) is \((z^*)^{-n}\):

The consequence is the one that matters. If \(x[n]\) is real then \(x^*[n]=x[n]\), so \(X(z) = X^*(z^*)\), which forces the coefficients of the rational function to be real and therefore forces every complex pole and zero to appear together with its conjugate. Every pole–zero plot in this book is symmetric about the real axis for exactly this reason, and it is why a resonance always costs two poles rather than one.

So far every property has been an operation on the index. This one is an operation on the transform, and it is the source of every repeated-pole transform pair you will ever need. Differentiate the defining series term by term with respect to \(z\) — legitimate inside the ROC, where a power series converges uniformly on compact subsets:

Apply it to the workhorse pair \(a^n u[n] \leftrightarrow z/(z-a)\) on \(|z| \gt |a|\). Differentiating the rational function by the quotient rule,

so \(n\,a^n u[n] \leftrightarrow a z^{-1}/(1-az^{-1})^2\) with ROC \(|z| \gt |a|\). Repeating the operation generates \(n^2 a^n u[n]\), \(n^3 a^n u[n]\) and the rest of the family, each with one more pole stacked at \(z=a\). Reading the correspondence backwards is how the inverse transform of a repeated pole is recognised: an \(m\)-fold pole at \(z=a\) in a causal transform always means a term of the form \(n^{m-1}a^n\) in the sequence, growing polynomially while the exponential decays.

This is the property the transform exists for. Chapter 7 established that an LTI system's output is the convolution of the input with the impulse response, \(y[n] = \sum_k x[k]h[n-k]\), and Chapter 8 built the system properties out of conditions on \(h[n]\). Convolution is exact and completely general, and it is also, for any input longer than a few samples, tedious. The z-transform removes the tedium entirely.

The middle step interchanges the two sums, which is permitted wherever both converge absolutely — that is, on the intersection of the two regions. The step after it is the time-shift property applied to the inner sum. Nothing else is involved: the whole proof is one exchange of order and one substitution.

A special case deserves its own name. Accumulation, the running sum \(y[n] = \sum_{k=-\infty}^{n}x[k]\), is the discrete-time counterpart of integration and appeared in Chapter 3 as the relation between the impulse and the step. It is not a new property at all, because a running sum is a convolution with the unit step: \(\sum_{k\le n}x[k] = \sum_k x[k]u[n-k] = x[n]*u[n]\). Since \(u[n] \leftrightarrow 1/(1-z^{-1})\) on \(|z| \gt 1\), the convolution property gives the answer immediately.

Two results extract a single sample of a signal from its transform without inverting anything. Both apply to causal signals only, which for these purposes means \(x[n]=0\) for \(n \lt 0\), and the restriction is not a technicality — the theorems are simply false without it.

The initial value theorem is little more than reading the series. For a causal signal the transform is a power series in \(z^{-1}\) with no positive powers at all:

because every term beyond the first carries a factor \(z^{-1}\) that vanishes as \(|z|\) grows. Continuing the idea, \(x[1] = \lim_{z\to\infty} z\big[X(z)-x[0]\big]\), and so on down the sequence — which is nothing other than long division of the numerator by the denominator, the power-series method of inversion from Chapter 23.

The final value theorem takes more care. Consider the transform of the first difference, \(x[n]-x[n-1]\), which by linearity and shifting is \((1-z^{-1})X(z)\). Write out that transform for a causal signal and let \(z\to 1\):

The last step is the telescoping of the sum: adding \(x[0]-x[-1]\), \(x[1]-x[0]\), \(x[2]-x[1]\) and so on up to \(N\) leaves only \(x[N]-x[-1] = x[N]\). The manipulation is valid only if the limit on the left actually exists, and that is the whole content of the precondition.

Collected in one place, the properties look like this. The ROC column is not decoration; treat any answer that omits it as incomplete.

The properties are only half the toolkit. The other half is a short list of primitive pairs, every one of which was derived in Chapter 23 or in this chapter, and from which everything else is assembled.

Notice the fourth and sixth rows against the third and fifth. The same algebraic expression serves two entirely different sequences, distinguished only by the ROC — the point Chapter 23 laboured, now visible as a table entry. Any problem that hands you \(X(z)\) without an ROC is under-specified, and the correct response is to give one answer for each admissible region.

Problems 1 to 3 exercise single properties; 4 to 6 combine several; 7 and 8 ask you to prove things. State the ROC with every transform — an answer without one is not an answer.