5.2 SOME [PROPERTIES OF THE](#page-11-0) z-TRANSFORM

← Back to LINEAR SYSTEMS AND SIGNALS Overview

5.2 SOME PROPERTIES OF THE z**-TRANSFORM**

The z-transform properties are useful in the derivation of z-transforms of many functions and also in the solution of linear difference equations with constant coefficients. Here we consider a few important properties of the z-transform.

In our discussion, the variable n appearing in signals, such as x[n] and y[n], may or may not stand for time. However, in most applications of our interest, n is proportional to time. For this reason, we shall loosely refer to the variable n as time.

5.2-1 Time-Shifting Properties

In the following discussion of the shift property, we deal with shifted signals x[n]u[n], x[n−k]u[n− k], x[n−k]u[n], and x[n+k]u[n]. Unless we physically understand the meaning of such shifts, our understanding of the shift property remains mechanical rather than intuitive or heuristic. For this reason, using a hypothetical signal x[n], we have illustrated various shifted signals for k = 1 in Fig. 5.4.

RIGHT SHIFT (DELAY) If

x[n]u[n] ⇐⇒ X[z]

then

$$x[n-1]u[n-1] \Longleftrightarrow \frac{1}{z}X[z] \tag{5.12}$$

In general,

$$x[n-m]u[n-m] \Longleftrightarrow \frac{1}{z^m}X[z]$$

\n(5.13)

Moreover,

$$x[n-1]u[n] \Longleftrightarrow \frac{1}{z}X[z] + x[-1]$$

\n(5.14)

Repeated application of this property yields

$$x[n-2]u[n] \Longleftrightarrow \frac{1}{z} \left[ \frac{1}{z} X[z] + x[-1] \right] + x[-2] = \frac{1}{z^2} X[z] + \frac{1}{z} x[-1] + x[-2]$$

In general, for integer value of m,

$$x[n-m]u[n] \Longleftrightarrow z^{-m}X[z] + z^{-m}\sum_{n=1}^{m}x[-n]z^{n}$$

\n(5.15)

A look at Eqs. (5.12) and (5.14) shows that they are identical except for the extra term x[−1] in Eq. (5.14). We see from Figs. 5.4c and 5.4d that x[n − 1]u[n] is the same as x[n − 1]u[n − 1] plus x[−1]δ[n]. Hence, the difference between their transforms is x[−1].

5.2 Some Properties of the z-Transform 503

Proof. For the integer value of m,

$$\mathcal{Z}{x[n-m]u[n-m]} = \sum_{n=0}^{\infty} x[n-m]u[n-m]z^{-n}$$

Recall that x[n − m]u[n − m] = 0 for n < m so that the limits on the summation on the right-hand side can be taken from n = m to ∞. Therefore,

$$\mathcal{Z}\{x[n-m]u[n-m]\} = \sum_{n=m}^{\infty} x[n-m]z^{-n}$$ $$= \sum_{r=0}^{\infty} x[r]z^{-(r+m)}$$ $$= \frac{1}{z^m} \sum_{r=0}^{\infty} x[r]z^{-r} = \frac{1}{z^m} X[z]$$

To prove Eq. (5.15), we have

$$\mathcal{Z}\{x[n-m]u[n]\} = \sum_{n=0}^{\infty} x[n-m]z^{-n} = \sum_{r=-m}^{\infty} x[r]z^{-(r+m)}$$ $$= z^{-m} \left[ \sum_{r=-m}^{-1} x[r]z^{-r} + \sum_{r=0}^{\infty} x[r]z^{-r} \right]$$ $$= z^{-m} \sum_{n=1}^{m} x[-n]z^{n} + z^{-m}X[z]$$

LEFT SHIFT (ADVANCE) If

$$x[n]u[n] \Longleftrightarrow X[z]$$

then

$$x[n+1]u[n] \Longleftrightarrow zX[z] - zx[0]$$

Repeated application of this property yields

$$x[n+2]u[n] \Longleftrightarrow z\{z(X[z] - zx[0]) - x[1]\} = z^2 X[z] - z^2 x[0] - zx[1]$$

and for the integer value of m,

$$x[n+m]u[n] \Longleftrightarrow z^m X[z] - z^m \sum_{n=0}^{m-1} x[n]z^{-n}$$

\n(5.16)

504 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM

Proof. By definition,

$$\mathcal{Z}\{x[n+m]u[n]\} = \sum_{n=0}^{\infty} x[n+m]z^{-n}$$ $$= \sum_{r=m}^{\infty} x[r]z^{-(r-m)}$$ $$= z^m \sum_{r=m}^{\infty} x[r]z^{-r}$$ $$= z^m \left[ \sum_{r=0}^{\infty} x[r]z^{-r} - \sum_{r=0}^{m-1} x[r]z^{-r} \right]$$ $$= z^m X[z] - z^m \sum_{r=0}^{m-1} x[r]z^{-r}$$

EXAMPLE 5.4 z**-Transform Using the Right-Shift Property**

The signal x[n] can be expressed as a product of n and a gate pulse u[n] − u[n − 6]. Therefore,

$$x[n] = n\{u[n] - u[n-6]\} = nu[n] - nu[n-6]$$

We cannot find the z-transform of nu[n − 6] directly by using the right-shift property [Eq. (5.13)]. So we rearrange it in terms of (n−6)u[n−6] as follows:

$$x[n] = nu[n] - (n - 6 + 6)u[n - 6]$$

= $nu[n] - (n - 6)u[n - 6] - 6u[n - 6]$

We can now find the z-transform of the bracketed term by using the right-shift property [Eq. (5.13)]. Because u[n] ⇐⇒ z/(z−1),

$$u[n-6] \Longleftrightarrow \frac{1}{z^6} \frac{z}{z-1} = \frac{1}{z^5(z-1)}$$

Also, because nu[n] ⇐⇒ z/(z−1)2,

$$(n-6)u[n-6] \Longleftrightarrow \frac{1}{z^6} \frac{z}{(z-1)^2} = \frac{1}{z^5(z-1)^2}$$

Therefore,

$$X[z] = \frac{z}{(z-1)^2} - \frac{1}{z^5(z-1)^2} - \frac{6}{z^5(z-1)} = \frac{z^6 - 6z + 5}{z^5(z-1)^2}$$

DR ILL 5.5 z**-Transform Using the Right-Shift Property**

Using only the fact that u[n] ⇐⇒ z/(z−1) and the right-shift property [Eq. (5.13)], find the z-transforms of the signals in Figs. 5.2 and 5.3.

ANSWERS

See Ex. 5.2d and Drill 5.1a.

5.2-2 z**-Domain Scaling Property (Multiplication by** γ ⁿ)

Scaling in the z-domain is equivalent to multiplying a time-domain signal by an exponential. That is, if

$$x[n]u[n] \Longleftrightarrow X[z]$$

then

$$\gamma^n x[n]u[n] \Longleftrightarrow X\left[\frac{z}{\gamma}\right] \tag{5.17}$$

Proof.

$$\mathcal{Z}\{\gamma^{n}x[n]u[n]\} = \sum_{n=0}^{\infty} \gamma^{n}x[n]z^{-n} = \sum_{n=0}^{\infty} x[n]\left(\frac{z}{\gamma}\right)^{-n} = X\left[\frac{z}{\gamma}\right]$$

DR ILL 5.6 Using the z**-Domain Scaling Property**

Use Eq. (5.17) to derive pairs 6 and 8 in Table 5.1 from pairs 2 and 3, respectively.

5.2-3 z**-Domain Differentiation Property (Multiplication by** n**)**

Multiplying a signal by n in the time domain produces differentiation in the z-domain. That is, if

$$x[n]u[n] \Longleftrightarrow X[z]$$

then

$$nx[n]u[n] \Longleftrightarrow -z\frac{d}{dz}X[z] \tag{5.18}$$

Proof.

$$-z\frac{d}{dz}X[z] = -z\frac{d}{dz}\sum_{n=0}^{\infty}x[n]z^{-n}$$ $$= -z\sum_{n=0}^{\infty} -nx[n]z^{-n-1}$$ $$= \sum_{n=0}^{\infty}nx[n]z^{-n} = \mathcal{Z}\{nx[n]u[n]\}$$

DR ILL 5.7 Using the z**-Domain Differentiation Property**

Use Eq. (5.18) to derive pairs 3 and 4 in Table 5.1 from pair 2. Similarly, derive pairs 8 and 9 from pair 6.

5.2-4 Time-Reversal Property

If

$$x[n] \Longleftrightarrow X[z]$$

then†

$$x[-n] \Longleftrightarrow X[1/z]$$

Proof.

$$\mathcal{Z}{x[-n]} = \sum_{n=-\infty}^{\infty} x[-n]z^{-n}$$ $$x^*[-n] \Longleftrightarrow X^*[1/z^*]$$

^† For complex signal ^x[n], the time-reversal property is modified as follows:

Changing the sign of the dummy variable n yields

$$\mathcal{Z}{x[-n]} = \sum_{n=-\infty}^{\infty} x[n]z^n$$ $$= \sum_{n=-\infty}^{\infty} x[n](1/z)^{-n}$$ $$= X[1/z]$$

The region of convergence is also inverted; that is, if the ROC of x[n] is |z| > |γ |, then the ROC of x[−n] is |z| < 1/|γ |.

DR ILL 5.8 Using the Time-Reversal Property

Use the time-reversal property and pair 2 in Table 5.1 to show that u[−n] ⇐⇒ −1/(z−1) with the ROC |z| < 1.

5.2-5 Convolution Property

The time-convolution property states that if‡

$$x_1[n] \Longleftrightarrow X_1[z]$$

and $x_2[n] \Longleftrightarrow X_2[z]$ ,

then (time convolution)

$$x_1[n] * x_2[n] \Longleftrightarrow X_1[z]X_2[z] \tag{5.19}$$

Proof. This property applies to causal as well as noncausal sequences. We shall prove it for the more general case of noncausal sequences, where the convolution sum ranges from −∞ to ∞.

We have

$$\mathcal{Z}\{x_1[n] * x_2[n]\} = \mathcal{Z}\left[\sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m]\right]$$ $$= \sum_{n=-\infty}^{\infty} z^{-n} \sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m]$$

‡ There is also the frequency-convolution property, which states that

$$x_1[n]x_2[n] \Longleftrightarrow \frac{1}{2\pi j} \oint X_1[u]X_2\left[\frac{z}{u}\right]u^{-1} du$$

508 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM

Interchanging the order of summation, we have

$$\mathcal{Z}[x_1[n]*x_2[n]] = \sum_{m=-\infty}^{\infty} x_1[m] \sum_{n=-\infty}^{\infty} x_2[n-m]z^{-n}$$ $$= \sum_{m=-\infty}^{\infty} x_1[m] \sum_{r=-\infty}^{\infty} x_2[r]z^{-(r+m)}$$ $$= \sum_{m=-\infty}^{\infty} x_1[m]z^{-m} \sum_{r=-\infty}^{\infty} x_2[r]z^{-r}$$ $$= X_1[z]X_2[z]$$

LTID SYSTEM RESPONSE

It is interesting to apply the time-convolution property to the LTID input–output equation y[n] = x[n] ∗ h[n]. Since h[n] ⇐⇒ H[z], it follows from Eq. (5.19) that

$$Y[z] = X[z]H[z] \tag{5.20}$$

DR ILL 5.9 Using the Convolution Property

Use the time-convolution property and appropriate pairs in Table 5.1 to show that u[n] ∗ u[n−1] = nu[n].

INITIAL AND FINAL VALUES

For a causal x[n], the initial value theorem states that

$$x[0] = \lim_{z \to \infty} X[z]$$

This result follows immediately from Eq. (5.7).

If (z−1)X[z] has no poles outside the unit circle, then the final value theorem states that

$$\lim_{N \to \infty} x[N] = \lim_{z \to 1} (z - 1)X[z]$$

This can be shown from the fact that

$$x[n] - x[n-1] \Longleftrightarrow \left\{1 - \frac{1}{z}\right\} X[z] = \frac{(z-1)X[z]}{z}$$

and

$$\frac{(z-1)X[z]}{z} = \sum_{n=-\infty}^{\infty} \{x[n] - x[n-1]\}z^{-n}$$ $$\lim_{z \to 1} \frac{(z-1)X[z]}{z} = \lim_{z \to 1} (z-1)X[z] = \lim_{z \to 1} \lim_{N \to \infty} \sum_{n=-\infty}^{N} \{x[n] - x[n-1]\}z^{-n} = \lim_{N \to \infty} x[N]$$

All these properties of the z-transform are listed in Table 5.2.

Operation	x[n]	X[z]
Addition	x1[n] +x2[n]	X1[z] +X2[z]
Scalar multiplication	ax[n]	aX[z]
Right shifting	x[n−m]u[n−m]	1 zm X[z]
	x[n−m]u[n]	“m 1 1 n zm X[z] + x[−n]z zm n=1
	x[n−1]u[n]	1 X[z] +x[−1] z
	x[n−2]u[n]	1 1 z2 X[z] + x[−1] +x[−2] z
	x[n−3]u[n]	1 1 1 z3 X[z] + z2 x[−1] + x[−2] +x[−3] z
Left shifting	x[n+m]u[n]	m ”−1 −n zmX[z] −zm x[n]z
	x[n+1]u[n]	n=0 zX[z] −zx[0]
	x[n+2]u[n]	z2X[z] −z2x[0] −zx[1]
	x[n+3]u[n]	z3X[z] −z3x[0] −z2x[1] −zx[2]
Multiplication by γ n	γ nx[n]u[n]	z ! X γ
Multiplication by n	nx[n]u[n]	d −z X[z] dz
Time reversal	x[−n]	X[1/z]
Time convolution	x1[n] ∗ x2[n]	X1[z]X2[z]
Initial value	x[0]	lim X[z] z→∞
Final value	lim x[N] N→∞	lim (z−1)X[z] Poles of (z −1)X[z] z→1
		inside the unit circle

TABLE 5.2 z-Transform Properties

and

5.3 z**-TRANSFORM SOLUTION OF LINEAR DIFFERENCE EQUATIONS**

The time-shifting (left-shift or right-shift) property has set the stage for solving linear difference equations with constant coefficients. As in the case of the Laplace transform with differential equations, the z-transform converts difference equations into algebraic equations that are readily solved to find the solution in the z domain. Taking the inverse z-transform of the z-domain solution yields the desired time-domain solution. The following examples demonstrate the procedure.

EXAMPLE 5.5 z**-Transform Solution of a Linear Difference Equation**

Solve

$$y[n+2] - 5y[n+1] + 6y[n] = 3x[n+1] + 5x[n]$$

if the initial conditions are y[−1] = 11/6, y[−2] = 37/36, and the input x[n] = (2)−nu[n].

As we shall see, difference equations can be solved by using the right-shift or the leftshift property. Because the difference equation here is in advance form, the use of the left-shift property in Eq. (5.16) may seem appropriate for its solution. Unfortunately, this left-shift property requires a knowledge of auxiliary conditions y[0], y[1], … , y[N − 1] rather than of the initial conditions y[−1], y[−2], … , y[−n], which are generally given. This difficulty can be overcome by expressing the difference equation in delay form (obtained by replacing n with n−2) and then using the right-shift property.† The resulting delay-form difference equation is

$$y[n] - 5y[n-1] + 6y[n-2] = 3x[n-1] + 5x[n-2]$$

\n(5.21)

We now use the right-shift property to take the z-transform of this equation. But before proceeding, we must be clear about the meaning of a term like y[n − 1] here. Does it mean y[n − 1]u[n − 1] or y[n − 1]u[n]? In any equation, we must have some time reference n = 0, and every term is referenced from this instant. Hence, y[n−k] means y[n−k]u[n]. Remember also that although we are considering the situation for n ≥ 0, y[n] is present even before n = 0 (in the form of initial conditions). Now

$$y[n]u[n] \Longleftrightarrow Y[z]$$

\n

$$y[n-1]u[n] \Longleftrightarrow \frac{1}{z}Y[z] + y[-1] = \frac{1}{z}Y[z] + \frac{11}{6}$$

\n

$$y[n-2]u[n] \Longleftrightarrow \frac{1}{z^2}Y[z] + \frac{1}{z}y[-1] + y[-2] = \frac{1}{z^2}Y[z] + \frac{11}{6z} + \frac{37}{36}$$

Noting that for causal input x[n],

$$x[-1] = x[-2] = \cdot \cdot \cdot = x[-n] = 0$$

^† Another approach is to find ^y[0], ^y[1], ^y[2], … , ^y[^{N − 1}] from ^y[−1], ^y[−2], … , ^y[−n] iteratively, as in Sec. 3.5-1, and then apply the left-shift property to the advance-form difference equation.

We obtain

$$x[n] = (2)^{-n}u[n] = (2^{-1})^{n}u[n] = (0.5)^{n}u[n] \Longleftrightarrow \frac{z}{z - 0.5}$$ $$x[n-1]u[n] \Longleftrightarrow \frac{1}{z}X[z] + x[-1] = \frac{1}{z}\frac{z}{z-0.5} + 0 = \frac{1}{z-0.5}$$ $$x[n-2]u[n] \Longleftrightarrow \frac{1}{z^2}X[z] + \frac{1}{z}x[-1] + x[-2] = \frac{1}{z^2}X[z] + 0 + 0 = \frac{1}{z(z-0.5)}$$

In general,

$$x[n - r]u[n] \Longleftrightarrow \frac{1}{z^r}X[z]$$

Taking the z-transform of Eq. (5.21) and substituting the foregoing results, we obtain

$$Y[z] - 5\left[\frac{1}{z}Y[z] + \frac{11}{6}\right] + 6\left[\frac{1}{z^2}Y[z] + \frac{11}{6z} + \frac{37}{36}\right] = \frac{3}{z - 0.5} + \frac{5}{z(z - 0.5)}$$

or

^{1− 5} z + 6 z2 ^Y[z] − ^{3− 11} z ^{= 3 z}−0.5 ⁺ 5 ^z(z−0.5) (5.22)

from which we obtain

$$(z2 - 5z + 6)Y[z] = \frac{z(3z2 - 9.5z + 10.5)}{(z - 0.5)}$$

so that

$$Y[z] = \frac{z(3z^2 - 9.5z + 10.5)}{(z - 0.5)(z^2 - 5z + 6)}$$

and

$$\frac{Y[z]}{z} = \frac{3z^2 - 9.5z + 10.5}{(z - 0.5)(z - 2)(z - 3)} = \frac{(26/15)}{z - 0.5} - \frac{(7/3)}{z - 2} + \frac{(18/5)}{z - 3}$$

Therefore,

$$Y[z] = \frac{26}{15} \left(\frac{z}{z - 0.5}\right) - \frac{7}{3} \left(\frac{z}{z - 2}\right) + \frac{18}{5} \left(\frac{z}{z - 3}\right)$$ $$y[n] = \left[\frac{26}{15} (0.5)^n - \frac{7}{3} (2)^n + \frac{18}{5} (3)^n\right] u[n] \tag{5.23}$$

and

This example demonstrates the ease with which linear difference equations with constant coefficients can be solved by the z-transform. This method is general: it can be used to solve a single difference equation or a set of simultaneous difference equations of any order as long as the equations are linear with constant coefficients.

Comment.

Sometimes, instead of initial conditions y[−1], y[−2], … , y[−n], auxiliary conditions y[0], y[1], … , y[N − 1] are given to solve a difference equation. In this case, the equation can be solved by expressing it in the advance form and then using the left-shift property (see Drill 5.11).

DR ILL 5.10 z**-Transform Solution of a Linear Difference Equation**

Solve the following equation if the initial conditions y[−1] = 2, y[−2] = 0, and the input x[n] = u[n]:

$$y[n+2] - \frac{5}{6}y[n+1] + \frac{1}{6}y[n] = 5x[n+1] - x[n]$$

ANSWER

^y[n] = ¹²−15 ¹ 2 n ^{+ 14} 3 1 3 ⁿ u[n]

DR ILL 5.11 Difference Equation Solution Using y[0], y[1], … , y[N −1]

Solve the following equation if the auxiliary conditions are y[0] = 1, y[1] = 2, and the input x[n] = u[n]:

y[n] +3y[n−1] +2y[n−2] = x[n−1] +3x[n−2]

ANSWER

^y[n] = ^{2 3 +}2(−1)^{n − 5 3} (−2)ⁿ u[n]

ZERO-INPUT AND ZERO-STATE COMPONENTS

In Ex. 5.5 we found the total solution of the difference equation. It is relatively easy to separate the solution into zero-input and zero-state components. All we have to do is to separate the response into terms arising from the input and terms arising from initial conditions (IC). We can separate the response in Eq. (5.22) as follows:

$$\left(1 - \frac{5}{z} + \frac{6}{z^2}\right)Y[z] - \underbrace{\left(3 - \frac{11}{z}\right)}_{\text{IC terms}} = \underbrace{\frac{3}{z - 0.5} + \frac{5}{z(z - 0.5)}}_{\text{input terms}}$$

Therefore,

$$\left(1 - \frac{5}{z} + \frac{6}{z^2}\right)Y[z] = \underbrace{\left(3 - \frac{11}{z}\right)}_{\text{IC terms}} + \underbrace{\frac{(3z + 5)}{z(z - 0.5)}}_{\text{input terms}}$$

Multiplying both sides by z² yields

$$(z2 - 5z + 6)Y[z] = \underbrace{z(3z - 11)}_{\text{IC terms}} + \underbrace{\frac{z(3z + 5)}{z - 0.5}}_{\text{input terms}}$$

and

$$Y[z] = \underbrace{\frac{z(3z-11)}{z^2 - 5z + 6}}_{\text{zero-input response}} + \underbrace{\frac{z(3z+5)}{(z-0.5)(z^2 - 5z + 6)}}_{\text{zero-state response}}$$

We expand both terms on the right-hand side into modified partial fractions to yield

$$Y[z] = \underbrace{\left[5\left(\frac{z}{z-2}\right) - 2\left(\frac{z}{z-3}\right)\right]}_{\text{zero-input response}} + \underbrace{\left[\frac{26}{15}\left(\frac{z}{z-0.5}\right) - \frac{22}{3}\left(\frac{z}{z-2}\right) + \frac{28}{5}\left(\frac{z}{z-3}\right)\right]}_{\text{zero-state response}}$$

and

$$y[n] = \underbrace{(5(2)^n - 2(3)^n) u[n]}_{\text{zero-input response}} + \underbrace{\left(\frac{26}{15}(0.5)^n - \frac{22}{3}(2)^n + \frac{28}{5}(3)^n\right) u[n]}_{\text{zero-state response}}$$

=

$$\left[-\frac{7}{3}(2)^n + \frac{18}{5}(3)^n + \frac{26}{15}(0.5)^n\right] u[n]$$

which agrees with the result in Eq. (5.23).

DR ILL 5.12 Separating Zero-Input and Zero-State Responses

Solve

$$y[n+2] - \frac{5}{6}y[n+1] + \frac{1}{6}y[n] = 5x[n+1] - x[n]$$

if the initial conditions are y[−1] = 2, y[−2] = 0, and the input x[n] = u[n]. Separate the response into zero-input and zero-state responses.

ANSWER

$$y[n] = \underbrace{\left(3\left(\frac{1}{2}\right)^n - \frac{4}{3}\left(\frac{1}{3}\right)^n\right)u[n]}_{\text{zero-input response}} + \underbrace{\left(12 - 18\left(\frac{1}{2}\right)^n + 6\left(\frac{1}{3}\right)^n\right)u[n]}_{\text{zero-state response}}$$

\n

$$= \left[12 - 15\left(\frac{1}{2}\right)^n + \frac{14}{3}\left(\frac{1}{3}\right)^n\right]u[n]$$

5.3-1 Zero-State Response of LTID Systems: The Transfer Function

Consider an Nth-order LTID system specified by the difference equation

$$Q[E]y[n] = P[E]x[n]$$

or

$$(EN + a1EN-1 + \dots + aN-1E + aN)y[n]$$

= (b₀E^N + b₁E^N-1 + \dots + b_N-1E + b_N)x[n]

or

$$y[n+N] + a_1y[n+N-1] + \cdots + a_{N-1}y[n+1] + a_Ny[n]$$

= $b_0x[n+N] + \cdots + b_{N-1}x[n+1] + b_Nx[n]$ (5.24)

We now derive the general expression for the zero-state response: that is, the system response to input x[n] when all the initial conditions y[−1] = y[−2]=···= y[−N] = 0 (zero state). The input x[n] is assumed to be causal so that x[−1] = x[−2]=···= x[−N] = 0.

Equation (5.24) can be expressed in delay form as

$$y[n] + a_1y[n-1] + \dots + a_Ny[n-N] = b_0x[n] + b_1x[n-1] + \dots + b_Nx[n-N]$$

(5.25)

Because y[−r] = x[−r] = 0 for r = 1, 2,…,N,

$$y[n-m]u[n] \Longleftrightarrow \frac{1}{z^m}Y[z]$$ $$x[n-m]u[n] \Longleftrightarrow \frac{1}{z^m}X[z] \qquad m=1,2,\ldots,N$$

Now the z-transform of Eq. (5.25) is given by

$$\left(1+\frac{a_1}{z}+\frac{a_2}{z^2}+\cdots+\frac{a_N}{z^N}\right)Y[z] = \left(b_0+\frac{b_1}{z}+\frac{b_2}{z^2}+\cdots+\frac{b_N}{z^N}\right)X[z]$$

Multiplication of both sides by z^N yields

$$(zN + a1zN-1 + \dots + aN-1z + aN)Y[z]$$

= (b₀z^N + b₁z^N-1 + \dots + b_N-1z + b_N)X[z]

Therefore,

$$Y[z] = \left(\frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N}\right) X[z]$$

= $\frac{P[z]}{Q[z]} X[z]$

We have shown in Eq. (5.20) that Y[z] = X[z]H[z]. Hence, it follows that

$$H[z] = \frac{P[z]}{Q[z]} = \frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N}$$

(5.26)

As in the case of LTIC systems, this result leads to an alternative definition of the LTID system transfer function as the ratio of Y[z] to X[z] (assuming all initial conditions zero).

$$H[z] \equiv \frac{Y[z]}{X[z]} = \frac{\mathcal{Z}[zero-state response]}{\mathcal{Z}[input]}$$

ALTERNATE INTERPRETATION OF THE z-TRANSFORM

So far we have treated the z-transform as a machine that converts linear difference equations into algebraic equations. There is no physical understanding of how this is accomplished or what it means. We now discuss more intuitive interpretation and meaning of the z-transform.

In Ch. 3, Eq. (3.38), we showed that the LTID system response to an everlasting exponential zⁿ is H[z]zⁿ. If we could express every discrete-time signal as a linear combination of everlasting exponentials of the form zⁿ, we could readily obtain the system response to any input. For example, if

$$x[n] = \sum_{k=1}^{K} X[z_k] z_k^n$$

\n(5.27)

the response of an LTID system to this input is given by

$$y[n] = \sum_{k=1}^{K} X[z_k]H[z_k]z_k^n$$

Unfortunately, a very small class of signals can be expressed in the form of Eq. (5.27). However, we can express almost all signals of practical utility as a sum of everlasting exponentials over a continuum of values of z. This is precisely what the z-transform in Eq. (5.2) does.

$$x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz$$

\n(5.28)

Invoking the linearity property of the z-transform, we can find the system response y[n] to input x[n] in Eq. (5.28) as†

$$y[n] = \frac{1}{2\pi j} \oint X[z]H[z]z^{n-1} dz = \mathcal{Z}^{-1}{X[z]H[z]}$$

Clearly,

Y[z] = X[z]H[z]

This viewpoint of finding the response of LTID system is illustrated in Fig. 5.6a. Just as in continuous-time systems, we can model discrete-time systems in the transformed manner by

^† In computing ^y[n], the contour along which the integration is performed is modified to consider the ROC of X[z] as well as H[z]. We ignore this consideration in this intuitive discussion.

Figure 5.6 The transformed representation of an LTID system.

representing all signals by their z-transforms and all system components (or elements) by their transfer functions, as shown in Fig. 5.6b.

The result Y[z] = H[z]X[z] greatly facilitates derivation of the system response to a given input. We shall demonstrate this assertion by an example.

EXAMPLE 5.6 Transfer Function to Find the Zero-State Response

Find the response y[n] of an LTID system described by the difference equation

$$y[n+2] + y[n+1] + 0.16y[n] = x[n+1] + 0.32x[n]$$

or

$$(E2 + E + 0.16)y[n] = (E + 0.32)x[n]$$

for the input x[n] = (−2)−nu[n] and with all the initial conditions zero (system in the zero state).

From the difference equation, we find

$$H[z] = \frac{P[z]}{Q[z]} = \frac{z + 0.32}{z^2 + z + 0.16}$$

For the input x[n] = (−2)−nu[n]=[(−2)−¹] nu(n) = (−0.5)nu[n],

$$X[z] = \frac{z}{z+0.5}$$

and

$$Y[z] = X[z]H[z] = \frac{z(z+0.32)}{(z^2 + z + 0.16)(z+0.5)}$$

Therefore,

$$\frac{Y[z]}{z} = \frac{(z+0.32)}{(z^2+z+0.16)(z+0.5)} = \frac{(z+0.32)}{(z+0.2)(z+0.8)(z+0.5)}$$ $$= \frac{2/3}{z+0.2} - \frac{8/3}{z+0.8} + \frac{2}{z+0.5}$$

so that

$$Y[z] = \frac{2}{3} \left( \frac{z}{z+0.2} \right) - \frac{8}{3} \left( \frac{z}{z+0.8} \right) + 2 \left( \frac{z}{z+0.5} \right)$$

and

$$y[n] = \left[\frac{2}{3}(-0.2)^n - \frac{8}{3}(-0.8)^n + 2(-0.5)^n\right]u[n]$$

EXAMPLE 5.7 Transfer Function of a Unit Delay

Show that the transfer function of a unit delay is 1/z.

If the input to the unit delay is x[n]u[n], then its output (Fig. 5.7) is given by

y[n] = x[n−1]u[n−1]

The z-transform of this equation yields [see Eq. (5.12)]

$$Y[z] = \frac{1}{z}X[z] = H[z]X[z]$$

It follows that the transfer function of the unit delay is

$$H[z] = \frac{1}{z}$$

x[n]u[n] X[z] x[n - 1]u[n - 1] ^Y[z] ^X[z] ¹ z 1 z Figure 5.7 Ideal unit delay and its transfer function.

DR ILL 5.13 Transfer Function to Find Zero-State Response and Difference Equation

A discrete-time system is described by the following transfer function:

$$H[z] = \frac{z - 0.5}{(z + 0.5)(z - 1)}$$

• (a) Find the system response to input x[n] = 3−(n+1) u[n] if all initial conditions are zero.
• (b) Write the difference equation relating the output y[n] to input x[n] for this system.

ANSWERS

• (a) ^y[n] = ¹ 3 1 ^{2 −}0.8(−0.5)^{n +}0.3 ¹ 3 ⁿ u[n]
• (b) y[n+2] −0.5y[n+1] −0.5y[n] = x[n+1] −0.5x[n]

5.3-2 Stability

Equation (5.26) shows that the denominator of H[z] is Q[z], which is apparently identical to the characteristic polynomial Q[γ ] defined in Ch. 3. Does this mean that the denominator of H[z] is the characteristic polynomial of the system? This may or may not be the case: if P[z] and Q[z] in Eq. (5.26) have any common factors, they cancel out, and the effective denominator of H[z] is not necessarily equal to Q[z]. Recall also that the system transfer function H[z], like h[n], is defined in terms of measurements at the external terminals. Consequently, H[z] and h[n] are both external descriptions of the system. In contrast, the characteristic polynomial Q[z] is an internal description. Clearly, we can determine only external stability, that is, BIBO stability, from H[z]. If all the poles of H[z] are within the unit circle, all the terms in h[n] are decaying exponentials, and as shown in Sec. 3.9, h[n] is absolutely summable. Consequently, the system is BIBO-stable. Otherwise the system is BIBO-unstable.

If P[z] and Q[z] do not have common factors, then the denominator of H[z] is identical to Q[z]. † The poles of H[z] are the characteristic roots of the system. We can now determine internal stability. The internal stability criterion in Sec. 3.9-2 can be restated in terms of the poles of H[z], as follows.

• 1. An LTID system is asymptotically stable if and only if all the poles of its transfer function H[z] are within the unit circle. The poles may be repeated or simple.
• 2. An LTID system is unstable if and only if either one or both of the following conditions exist: (i) at least one pole of H[z] is outside the unit circle; (ii) there are repeated poles of H[z] on the unit circle.

^† There is no way of determining whether any common factors in ^P[z] and ^Q[z] were canceled out. This is because in our derivation of H[z], we generally get the final result after the cancellations have been effected. When we use internal description of the system to derive Q[z], however, we find pure Q[z] unaffected by any common factor in P[z].

3. An LTID system is marginally stable if and only if there are no poles of H[z] outside the unit circle, and there are some simple poles on the unit circle.

DR ILL 5.14 Transfer Function to Determine Stability

Show that an accumulator whose impulse response is h[n] = u[n] is marginally stable but BIBO-unstable.

5.3-3 Inverse Systems

If H[z] is the transfer function of a system S, then Si, its inverse system, has a transfer function Hi[z] given by

$$H_i[z] = \frac{1}{H[z]}$$

This follows from the fact the inverse system Sⁱ undoes the operation of S. Hence, if H[z] is placed in cascade with Hi[z], the transfer function of the composite system (identity system) is unity. For example, an accumulator whose transfer function is H[z] = z/(z − 1) and a backward difference system whose transfer function is Hi[z] = (z−1)/z are inverse of each other. Similarly if

$$H[z] = \frac{z - 0.4}{z - 0.7}$$

its inverse system transfer function is

$$H_i[z] = \frac{z - 0.7}{z - 0.4}$$

as required by the property H[z]Hi[z] = 1. Hence, it follows that

$$h[n] * h_i[n] = \delta[n]$$

DR ILL 5.15 Inverse Systems

Find the impulse responses of an accumulator and a first-order backward difference system. Show that the convolution of the two impulse responses yields δ[n].

5.4 SYSTEM REALIZATION

Because of the similarity between LTIC and LTID systems, conventions for block diagrams and rules of interconnection for LTID are identical to those for continuous-time (LTIC) systems. It is not necessary to rederive these relationships. We shall merely restate them to refresh the reader’s memory.

The block diagram representations of the basic operations, such as an adder, a scalar multiplier, unit delay, and pickoff points, is shown in Fig. 3.13. In our development, the unit delay, which is represented by a box marked D in Fig. 3.13, will be represented by its transfer function 1/z. All the signals will also be represented in terms of their z-transforms. Thus, the input and the output will be labeled X[z] and Y[z], respectively.

When two systems with transfer functions H1[z] and H2[z] are connected in cascade (as in Fig. 4.18b), the transfer function of the composite system is H1[z]H2[z]. If the same two systems are connected in parallel (as in Fig. 4.18c), the transfer function of the composite

Figure 5.8 Realization of an Nth-order causal LTID system transfer function by using (a) DFI, (b) canonic direct (DFII), and (c) the transpose form of DFII.

system is H1[z] + H2[z]. For a feedback system (as in Fig. 4.18d), the transfer function is G[z]/(1+G[z]H[z]).

We now consider a systematic method for realization (or simulation) of an arbitrary Nth-order LTID transfer function. Since realization is basically a synthesis problem, there is no unique way of realizing a system. A given transfer function can be realized in many different ways. We present here the two forms of direct realization. Each of these forms can be executed in several other ways, such as cascade and parallel. Furthermore, a system can be realized by the transposed version of any known realization of that system. This artifice doubles the number of system realizations. A transfer function H[z] can be realized by using time delays along with adders and multipliers.

We shall consider a realization of a general Nth-order causal LTID system, whose transfer function is given by

$$H[z] = \frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N}$$

(5.29)

This equation is identical to the transfer function of a general Nth-order proper LTIC system given in Eq. (4.36). The only difference is that the variable z in the former is replaced by the variable s in the latter. Hence, the procedure for realizing an LTID transfer function is identical to that for the LTIC transfer function with the basic element 1/s (integrator) replaced by the element 1/z (unit delay). The reader is encouraged to follow the steps in Sec. 4.6 and rederive the results for the LTID transfer function in Eq. (5.29). Here we shall merely reproduce the realizations from Sec. 4.6 with integrators (1/s) replaced by unit delays (1/z).

The direct form I (DFI) is shown in Fig. 5.8a, the canonic direct form (DFII) is shown in Fig. 5.8b and the transpose of canonic direct is shown in Fig. 5.8c. The DFII and its transpose are canonic because they require N delays, which is the minimum number needed to implement the Nth-order LTID transfer function in Eq. (5.29). In contrast, the form DFI is a noncanonic because it generally requires 2N delays. The DFII realization in Fig. 5.8b is also called a canonic direct form.

EXAMPLE 5.8 Canonical Realizations of Transfer Functions

Find the canonic direct and the transposed canonic direct realizations of the following transfer functions: (a) 2 z+5 , (b) 4z+28 ^z+¹ , (c) ^z z+7 , and (d) 4z+28 z² +6z+5 .

All four of these transfer functions are special cases of H[z] in Eq. (5.29). (a)

$$H[z] = \frac{2}{z+5}$$

For this case, the transfer function is of the first order (N = 1); therefore, we need only one delay for its realization. The feedback and feedforward coefficients are

$$a_1 = 5$$

and $b_0 = 0$ , $b_1 = 2$

522 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM

We use Fig. 5.8 as our model and reduce it to the case of N = 1. Figure 5.9a shows the canonic direct (DFII) form, and Fig. 5.9b its transpose. The two realizations are almost the same. The minor difference is that in the DFII form, the gain 2 is provided at the output, and in the transpose, the same gain is provided at the input.

Figure 5.9 Realization of transfer function 2/(z+5): (a) canonic direct form and (b) its transpose.

In a similar way, we realize the remaining transfer functions. (b)

$$H[z] = \frac{4z + 28}{z + 1}$$

In this case also, the transfer function is of the first order (N = 1); therefore, we need only one delay for its realization. The feedback and feedforward coefficients are

$$a_1 = 1$$

and $b_0 = 4$ , $b_1 = 28$

Figure 5.10 illustrates the canonic direct and its transpose for this case.†

Figure 5.10 Realization of (4z +28)/(z +1): (a) canonic direct form and (b) its transpose.

$$H[z] = \frac{4z + 28}{z+1} = 4 + \frac{24}{z+1}$$

Hence, this transfer function can also be realized as two transfer functions in parallel.

^† Transfer functions with ^{N = M} may also be expressed as a sum of a constant and a strictly proper transfer function. For example,

$$H[z] = \frac{z}{z+7}$$

Here N = 1 and b⁰ = 1,b¹ = 0 and a¹ = 7. Figure 5.11 shows the direct and the transposed realizations. Observe that the realizations are almost alike.

Figure 5.11 Realization of z/(z +7): (a) canonic direct form and (b) its transpose.