Chapter 1: Properties of the Z-Transform
Properties of Z-Transform
Overview
The Z-transform has many important properties, and by exploiting them we can derive the Z-transform of complex signals from known transforms. This chapter presents five properties—z-domain differentiation, scaling, time reversal, conjugation, and convolution—with rigorous proofs and worked examples.
The most fundamental properties, linearity and time shift, are covered in Introduction Chapter 2. For the definition of the Z-transform and the basic transform pairs, see Introduction Chapter 1.
Which Z-transform this chapter uses
The Introduction used the unilateral Z-transform (the sum over $n \geq 0$), which suits causal signals and difference equations with initial conditions. This chapter uses the bilateral Z-transform
$$X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}$$because properties such as time reversal are most naturally expressed in that form: $\mathcal{Z}\{x[-n]\} = X(1/z)$ holds only for the bilateral transform. The bilateral Z-transform and its region of convergence (ROC) are treated in detail in the next chapter.
The meaning of the complex variable $z$
Writing $z = re^{j\omega}$ in polar form, $|z| = r$ corresponds to the amplitude decay/growth rate and $\omega$ to the angular frequency. In particular, on the circle $|z| = 1$ (the unit circle), i.e. setting $z = e^{j\omega}$, the Z-transform reduces to the discrete-time Fourier transform (DTFT): $$X(e^{j\omega}) = \displaystyle\sum_{n=-\infty}^{\infty} x[n]\, e^{-j\omega n}$$ provided the unit circle lies within the region of convergence.
Note that the DTFT is a transform over continuous frequency for infinite-length signals, and differs from the discrete Fourier transform (DFT), which maps a finite-length $N$-point signal to $N$ discrete frequencies: $$X[k] = \displaystyle\sum_{n=0}^{N-1} x[n]\, e^{-j2\pi kn/N}$$ The DFT can be interpreted as the DTFT sampled at $N$ equally spaced frequency points.
1. z-Domain Differentiation
Theorem 1.1 (z-Domain Differentiation)
Multiplying a signal by $n$ gives
$$\mathcal{Z}\{n \cdot x[n]\} = -z \dfrac{dX(z)}{dz}$$The original ROC is contained in the ROC of the new transform. It is usually the same as that of $X(z)$, but it can be larger in degenerate cases where multiplying by $n$ makes terms cancel (for instance $x[n] = \delta[n]$ gives $n\,x[n] = 0$).
In general, applying it $k$ times gives
$$\mathcal{Z}\{n^k x[n]\} = \left(-z \dfrac{d}{dz}\right)^k X(z)$$Proof
Differentiating both sides of $X(z) = \displaystyle\sum_{n=-\infty}^{\infty} x[n] z^{-n}$ with respect to $z$ gives
$$\dfrac{dX(z)}{dz} = \displaystyle\sum_{n=-\infty}^{\infty} x[n] \cdot (-n) z^{-n-1} = -z^{-1} \displaystyle\sum_{n=-\infty}^{\infty} n \cdot x[n] \cdot z^{-n}$$Therefore,
$$-z \dfrac{dX(z)}{dz} = \displaystyle\sum_{n=-\infty}^{\infty} n \cdot x[n] \cdot z^{-n} = \mathcal{Z}\{n \cdot x[n]\}$$Example 1.1 (Z-transform of $n u[n]$)
Find $\mathcal{Z}\{n u[n]\}$.
Solution:
Since $X(z) = \mathcal{Z}\{u[n]\} = \dfrac{z}{z-1}$,
$$\begin{align} \dfrac{dX(z)}{dz} &= \dfrac{d}{dz}\left(\dfrac{z}{z-1}\right) = \dfrac{(z-1) - z}{(z-1)^2} = \dfrac{-1}{(z-1)^2} \end{align}$$Hence,
$$\mathcal{Z}\{n u[n]\} = -z \cdot \dfrac{-1}{(z-1)^2} = \dfrac{z}{(z-1)^2}, \quad |z| > 1$$Example 1.2 (Z-transform of $n a^n u[n]$)
Find $\mathcal{Z}\{n a^n u[n]\}$.
Solution:
Since $X(z) = \mathcal{Z}\{a^n u[n]\} = \dfrac{z}{z-a}$,
$$\dfrac{dX(z)}{dz} = \dfrac{(z-a) - z}{(z-a)^2} = \dfrac{-a}{(z-a)^2}$$Hence,
$$\mathcal{Z}\{n a^n u[n]\} = -z \cdot \dfrac{-a}{(z-a)^2} = \dfrac{az}{(z-a)^2}, \quad |z| > |a|$$Example 1.3 (Z-transform of $n^2 u[n]$)
Apply z-domain differentiation twice to find $\mathcal{Z}\{n^2 u[n]\}$.
Solution:
Let $Y(z) = \mathcal{Z}\{n u[n]\} = \dfrac{z}{(z-1)^2}$. Then
$$\dfrac{dY(z)}{dz} = \dfrac{(z-1)^2 - z \cdot 2(z-1)}{(z-1)^4} = \dfrac{(z-1) - 2z}{(z-1)^3} = \dfrac{-z-1}{(z-1)^3}$$Hence,
$$\mathcal{Z}\{n^2 u[n]\} = -z \cdot \dfrac{-z-1}{(z-1)^3} = \dfrac{z(z+1)}{(z-1)^3}, \quad |z| > 1$$2. Scaling
Theorem 1.2 (Scaling)
For a constant $a \neq 0$,
$$\mathcal{Z}\{a^n x[n]\} = X(z/a)$$That is, it equals $X(z)$ with $z$ replaced by $z/a$.
The ROC is $|a| \cdot \text{ROC}_X$, i.e. $\{z : |z/a| \in \text{ROC}_X\}$.
Proof
If the ROC of $X(z)$ is $R_1 < |z| < R_2$, then the ROC of $X(z/a)$ is $|a|R_1 < |z| < |a|R_2$.
Example 1.4
Find $\mathcal{Z}\{(0.5)^n u[n]\}$ using the scaling property.
Solution:
Since $\mathcal{Z}\{u[n]\} = \dfrac{z}{z-1}$ ($|z| > 1$), setting $a = 0.5$ gives
$$\mathcal{Z}\{(0.5)^n u[n]\} = \dfrac{z/0.5}{z/0.5 - 1} = \dfrac{2z}{2z - 1} = \dfrac{z}{z - 0.5}$$The ROC is $|z/0.5| > 1$, i.e. $|z| > 0.5$.
This agrees with the result of substituting $a = 0.5$ into $\mathcal{Z}\{a^n u[n]\} = \dfrac{z}{z-a}$.
Example 1.5 (Damped oscillation)
Find the Z-transform of $x[n] = r^n \cos(\omega_0 n) u[n]$ (with $0 < r < 1$).
Solution:
Since $\mathcal{Z}\{\cos(\omega_0 n) u[n]\} = \dfrac{z(z - \cos\omega_0)}{z^2 - 2z\cos\omega_0 + 1}$ ($|z| > 1$), the scaling property $z \to z/r$ gives
$$\begin{align} X(z) &= \dfrac{(z/r)((z/r) - \cos\omega_0)}{(z/r)^2 - 2(z/r)\cos\omega_0 + 1} \\ &= \dfrac{z(z - r\cos\omega_0)}{z^2 - 2rz\cos\omega_0 + r^2} \end{align}$$The ROC is $|z| > r$. The poles are at $z = re^{\pm j\omega_0}$, which lie inside the unit circle when $0 < r < 1$. If this sequence is regarded as the impulse response of a causal LTI system, that system is BIBO stable; stability is a property of the system, not of the signal.
3. Time Reversal
Theorem 1.3 (Time Reversal)
Reversing a signal in time gives
$$\mathcal{Z}\{x[-n]\} = X(z^{-1}) = X(1/z)$$The ROC is $1/\text{ROC}_X$, i.e. $\{z : 1/z \in \text{ROC}_X\}$.
Proof
Substituting $m = -n$,
$$\begin{align} \mathcal{Z}\{x[-n]\} &= \displaystyle\sum_{n=-\infty}^{\infty} x[-n] z^{-n} \\ &= \displaystyle\sum_{m=-\infty}^{\infty} x[m] z^{m} \\ &= \displaystyle\sum_{m=-\infty}^{\infty} x[m] (z^{-1})^{-m} \\ &= X(z^{-1}) \end{align}$$If the ROC of $X(z)$ is $R_1 < |z| < R_2$, then the ROC of $X(1/z)$ is $1/R_2 < |z| < 1/R_1$.
Example 1.6
Find the Z-transform of the time-reversed signal $x[-n] = a^{-n} u[-n]$ of $x[n] = a^n u[n]$.
Solution:
Since $X(z) = \dfrac{z}{z-a}$ ($|z| > |a|$),
$$\mathcal{Z}\{x[-n]\} = X(1/z) = \dfrac{1/z}{1/z - a} = \dfrac{1}{1 - az}$$Because the signal involves $u[-n]$, we can also verify by direct computation:
$$\mathcal{Z}\{a^{-n} u[-n]\} = \displaystyle\sum_{n=-\infty}^{0} a^{-n} z^{-n} = \displaystyle\sum_{m=0}^{\infty} (az)^{m} = \dfrac{1}{1 - az}, \quad |az| < 1$$The ROC is $|z| < |a|^{-1}$, which matches the reversal $|z| < 1/|a|$ of the original ROC $|z| > |a|$.
4. Conjugation
Theorem 1.4 (Conjugation)
For the complex conjugate of a signal,
$$\mathcal{Z}\{x^*[n]\} = X^*(z^*)$$The ROC is the same as that of $X(z)$.
Proof
This follows directly from the definition of the conjugate:
$$\begin{align} \mathcal{Z}\{x^*[n]\} &= \displaystyle\sum_{n} x^*[n] z^{-n} = \left( \displaystyle\sum_{n} x[n] (z^*)^{-n} \right)^* \\ &= \big(X(z^*)\big)^* = X^*(z^*) \end{align}$$The second equality uses $x^*[n] z^{-n} = \left(x[n] (z^*)^{-n}\right)^*$ (since $\left((z^*)^{-n}\right)^* = z^{-n}$).
Note: The case of real signals
When $x[n]$ is a real signal, $x^*[n] = x[n]$, so $X(z) = X^*(z^*)$ holds. That is, the Z-transform of a real signal has conjugate symmetry.
In particular, when $X(z)$ is a rational function, its poles and zeros must appear in conjugate pairs or lie on the real axis.
Example 1.7
For $x[n] = e^{j\omega_0 n} u[n]$, find the Z-transform of $x^*[n] = e^{-j\omega_0 n} u[n]$.
Solution:
Since $X(z) = \mathcal{Z}\{e^{j\omega_0 n} u[n]\} = \dfrac{z}{z - e^{j\omega_0}}$ ($|z| > 1$),
$$\mathcal{Z}\{e^{-j\omega_0 n} u[n]\} = X^*(z^*) = \dfrac{z^*}{z^* - e^{-j\omega_0}} \Bigg|_{z^* \to z} = \dfrac{z}{z - e^{-j\omega_0}}$$This also agrees with the result obtained from the scaling property with $a = e^{-j\omega_0}$.
5. Convolution
Theorem 1.5 (Convolution Theorem)
Let $X_1(z) = \mathcal{Z}\{x_1[n]\}$ and $X_2(z) = \mathcal{Z}\{x_2[n]\}$. Then
$$\mathcal{Z}\{x_1[n] * x_2[n]\} = X_1(z) \cdot X_2(z)$$The ROC contains at least $\text{ROC}_1 \cap \text{ROC}_2$.
Proof
Taking the Z-transform of the definition of convolution $y[n] = \displaystyle\sum_{k=-\infty}^{\infty} x_1[k] x_2[n-k]$ gives
$$\begin{align} Y(z) &= \displaystyle\sum_{n=-\infty}^{\infty} \left(\displaystyle\sum_{k=-\infty}^{\infty} x_1[k] x_2[n-k]\right) z^{-n} \end{align}$$For $z$ in the intersection of the two ROCs both series converge absolutely, so the order of summation may be interchanged. Substituting $m = n - k$,
$$\begin{align} &= \displaystyle\sum_{k=-\infty}^{\infty} x_1[k] \displaystyle\sum_{m=-\infty}^{\infty} x_2[m] z^{-(m+k)} \\ &= \left(\displaystyle\sum_{k=-\infty}^{\infty} x_1[k] z^{-k}\right) \left(\displaystyle\sum_{m=-\infty}^{\infty} x_2[m] z^{-m}\right) \\ &= X_1(z) \cdot X_2(z) \end{align}$$Example 1.8 (Output of an LTI system)
A system with impulse response $h[n] = (0.5)^n u[n]$ is driven by the input $x[n] = u[n]$. Find the output.
Solution:
Since $H(z) = \dfrac{z}{z-0.5}$ ($|z| > 0.5$) and $X(z) = \dfrac{z}{z-1}$ ($|z| > 1$),
$$Y(z) = H(z) X(z) = \dfrac{z}{z-0.5} \cdot \dfrac{z}{z-1} = \dfrac{z^2}{(z-0.5)(z-1)}$$Partial-fraction decomposition (studied in detail in Chapter 4). Each coefficient is found as a residue: $\left.\dfrac{z}{z-1}\right|_{z=0.5} = \dfrac{0.5}{-0.5} = -1$ and $\left.\dfrac{z}{z-0.5}\right|_{z=1} = \dfrac{1}{0.5} = 2$:
$$\dfrac{Y(z)}{z} = \dfrac{z}{(z-0.5)(z-1)} = \dfrac{-1}{z-0.5} + \dfrac{2}{z-1}$$ $$Y(z) = -\dfrac{z}{z-0.5} + \dfrac{2z}{z-1}$$Since the ROC $|z| > 1$ corresponds to a causal signal,
$$y[n] = \left[2 - (0.5)^n\right] u[n]$$Example 1.9 (Convolution of finite-length signals)
Find the convolution of $x_1[n] = \{1, 2, 3\}$ ($n = 0, 1, 2$) and $x_2[n] = \{1, 1\}$ ($n = 0, 1$) using the Z-transform.
Solution:
Since $X_1(z) = 1 + 2z^{-1} + 3z^{-2}$ and $X_2(z) = 1 + z^{-1}$,
$$\begin{align} Y(z) &= X_1(z) X_2(z) = (1 + 2z^{-1} + 3z^{-2})(1 + z^{-1}) \\ &= 1 + 3z^{-1} + 5z^{-2} + 3z^{-3} \end{align}$$Therefore, $y[n] = \{1, 3, 5, 3\}$ ($n = 0, 1, 2, 3$).
Since $x_1[n]$ and $x_2[n]$ are both finite-length causal signals, $Y(z)$ contains only negative powers of $z$, so the ROC is the entire $z$-plane except $z = 0$ (i.e. $|z| > 0$).
Summary Table of Z-Transform Properties
| Property | Time domain $x[n]$ | Z-domain $X(z)$ | ROC |
|---|---|---|---|
| Definition | $x[n]$ | $X(z)$ | $R_x$ |
| z-domain differentiation | $n x[n]$ | $-z \dfrac{dX(z)}{dz}$ | $R_x$ |
| Scaling | $a^n x[n]$ | $X(z/a)$ | $|a| R_x$ |
| Time reversal | $x[-n]$ | $X(1/z)$ | $1/R_x$ |
| Conjugation | $x^*[n]$ | $X^*(z^*)$ | $R_x$ |
| Convolution | $x_1[n] * x_2[n]$ | $X_1(z) X_2(z)$ | at least $R_1 \cap R_2$ |
Exercises
Exercise 1 (Linearity and z-Domain Differentiation)
Find the Z-transform of the following signals.
(a) $x[n] = 3^n u[n] + 2^n u[n]$
(b) $x[n] = (n+1) 2^n u[n]$
Solution
(a) By linearity,
$$X(z) = \dfrac{z}{z-3} + \dfrac{z}{z-2} = \dfrac{z(z-2) + z(z-3)}{(z-3)(z-2)} = \dfrac{2z^2 - 5z}{(z-3)(z-2)}, \quad |z| > 3$$(b) Since $(n+1)2^n = n \cdot 2^n + 2^n$, using linearity,
$$X(z) = \mathcal{Z}\{n \cdot 2^n u[n]\} + \mathcal{Z}\{2^n u[n]\} = \dfrac{2z}{(z-2)^2} + \dfrac{z}{z-2}$$ $$= \dfrac{2z + z(z-2)}{(z-2)^2} = \dfrac{z^2}{(z-2)^2}, \quad |z| > 2$$Exercise 2 (Time Shift and z-Domain Differentiation)
(a) Find $\mathcal{Z}\{u[n-3]\}$.
(b) Find $\mathcal{Z}\{n(n-1) a^n u[n]\}$. (Hint: apply z-domain differentiation twice.)
Solution
(a) Using $\mathcal{Z}\{u[n]\} = \dfrac{z}{z-1}$ and the time-shift property of the bilateral Z-transform used in this chapter (unlike the unilateral case in Introduction Chapter 2, no initial-value term appears),
$$\mathcal{Z}\{u[n-3]\} = z^{-3} \cdot \dfrac{z}{z-1} = \dfrac{z^{-2}}{z-1} = \dfrac{1}{z^2(z-1)}, \quad |z| > 1$$(b) Decompose as $n(n-1)a^n u[n] = n^2 a^n u[n] - n a^n u[n]$, or differentiate twice directly.
Let $F(z) = \mathcal{Z}\{a^n u[n]\} = \dfrac{z}{z-a}$ and $G(z) = \mathcal{Z}\{na^n u[n]\} = \dfrac{az}{(z-a)^2}$. Then
$$\dfrac{dG}{dz} = a \cdot \dfrac{(z-a)^2 - z \cdot 2(z-a)}{(z-a)^4} = a \cdot \dfrac{-(z+a)}{(z-a)^3}$$ $$\mathcal{Z}\{n^2 a^n u[n]\} = -z \cdot \dfrac{dG}{dz} = \dfrac{az(z+a)}{(z-a)^3}$$Therefore,
$$\mathcal{Z}\{n(n-1)a^n u[n]\} = \dfrac{az(z+a)}{(z-a)^3} - \dfrac{az}{(z-a)^2} = \dfrac{az(z+a) - az(z-a)}{(z-a)^3} = \dfrac{2a^2 z}{(z-a)^3}$$The ROC is $|z| > |a|$.
Exercise 3 (Scaling and Time Reversal)
(a) Find the Z-transform of $x[n] = r^n \sin(\omega_0 n) u[n]$ (with $0 < r < 1$).
(b) Find the Z-transform of $x[n] = -3^n u[-n-1]$.
Solution
(a) Apply the scaling $z \to z/r$ to $\mathcal{Z}\{\sin(\omega_0 n) u[n]\} = \dfrac{z\sin\omega_0}{z^2 - 2z\cos\omega_0 + 1}$ ($|z| > 1$):
$$X(z) = \dfrac{rz \sin\omega_0}{z^2 - 2rz\cos\omega_0 + r^2}, \quad |z| > r$$(b) From $\mathcal{Z}\{a^n u[n]\} = \dfrac{z}{z-a}$ ($|z| > |a|$) and the time-reversal property,
we have $\mathcal{Z}\{a^{-n} u[-n]\} = \dfrac{1}{1-az}$ ($|z| < 1/|a|$).
By direct computation, $-3^n u[-n-1]$ equals $-3^n$ for $n \leq -1$:
$$\mathcal{Z}\{-3^n u[-n-1]\} = -\displaystyle\sum_{n=-\infty}^{-1} 3^n z^{-n} = -\displaystyle\sum_{m=1}^{\infty} 3^{-m} z^{m} = -\displaystyle\sum_{m=1}^{\infty} \left(\dfrac{z}{3}\right)^m$$ $$= -\dfrac{z/3}{1 - z/3} = \dfrac{-z}{3 - z} = \dfrac{z}{z-3}, \quad |z| < 3$$Exercise 4 (Convolution)
Find the convolution $y[n] = x_1[n] * x_2[n]$ of $x_1[n] = (0.5)^n u[n]$ and $x_2[n] = (0.8)^n u[n]$.
Solution
Since $X_1(z) = \dfrac{z}{z-0.5}$ and $X_2(z) = \dfrac{z}{z-0.8}$,
$$Y(z) = \dfrac{z^2}{(z-0.5)(z-0.8)}$$Partial-fraction decomposition:
$$\dfrac{Y(z)}{z} = \dfrac{z}{(z-0.5)(z-0.8)} = \dfrac{A}{z-0.5} + \dfrac{B}{z-0.8}$$ $$\begin{align} A &= \dfrac{0.5}{0.5-0.8} = \dfrac{0.5}{-0.3} = -\dfrac{5}{3} \\ B &= \dfrac{0.8}{0.8-0.5} = \dfrac{0.8}{0.3} = \dfrac{8}{3} \end{align}$$ $$Y(z) = -\dfrac{5}{3} \cdot \dfrac{z}{z-0.5} + \dfrac{8}{3} \cdot \dfrac{z}{z-0.8}$$Since the ROC $|z| > 0.8$ corresponds to a causal signal,
$$y[n] = \dfrac{1}{3}\left[8(0.8)^n - 5(0.5)^n\right] u[n]$$Summary
- z-domain differentiation: multiplication by $n$ corresponds to $-z\dfrac{d}{dz}$
- Scaling: multiplication by $a^n$ corresponds to the substitution $z \to z/a$
- Time reversal: $n \to -n$ corresponds to $z \to 1/z$
- Conjugation: $x^*[n] \leftrightarrow X^*(z^*)$; real signals are conjugate symmetric
- Convolution: convolution in the time domain corresponds to multiplication in the Z-domain
- Together with linearity and time shift in Introduction Chapter 2, you can master all the basic properties of the Z-transform
- By combining these properties, you can efficiently derive the Z-transforms of complex signals
Frequently Asked Questions
What is z-domain differentiation?
The Z-transform of $n\cdot x[n]$ is given by $-z\,\dfrac{d}{dz}X(z)$. Applying this property repeatedly yields the transform of $n^k x[n]$ as well. It is useful for deriving the Z-transform of polynomial-weighted sequences such as the ramp signal $n\,u[n]$.
How is the time-reversal property expressed?
The Z-transform of $x[-n]$ is $X(1/z)$; that is, you simply replace $z$ with $1/z$. The ROC is also transformed to $1/\text{ROC}$. This property appears in the context of non-causal signals and the bilateral Z-transform.
What is the conjugation property?
The Z-transform of the conjugate $x^*[n]$ of a complex sequence $x[n]$ is $X^*(z^*)$. For real sequences ($x[n]=x^*[n]$), $X(z)=X^*(z^*)$ holds, which explains why poles and zeros of a real-coefficient transfer function appear as conjugate pairs.