Properties of the z-transform *

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Properties of the z-Transform

*

Nguyen Huu Phuong

This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 3.0

In this section many properties (can be considered as theorems) of the two-sided (bilateral) ztransform are presented :

• Linearity • Time shift

• Convolution in time

• Relation to Discrete Time Fourier Transform (DTFT) • Others

Not all the properties are considered in details .

In the following the z-transform pair is x (n) ↔ X (z) .

1 Linearity

Linearity can be expressed as

a1x1(n) + a2x2(n) ↔ a1X1(z) + a2X2(z) (1)

where a1, a2 are constants. The same form applies for more signals . So linearity means a linear combination of inputs gives the same linear combination of outputs .

For the ztransform and many other transforms (Laplace , Fourier . . .) linearity may be obvious but very important property . It allows us to nd the transform and the inverse transform when there is a combination of many terms .

Example 1

Find the ztransform of the causal cosine signal

x (n) = (cosnω0) u (n) Solution

Express x(n) in terms of complex exponentials : x (n) = (cosnω0) u (n) = 1 2e jnω_{u (n) +}1 2e −jnω_{u (n)} then

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X (z) = 1₂_1−e_jω01 _z−1 + 1 2 1 1−e−jω0z−1 = 1−z−1cosω0 1−2z−1_cosω 0+z−2

2 Time shift

First let's consider the z-transform of the unit sample (also called unit impulse) δ (n) and of its delay δ (n − n0): X (z) =P∞ n=0δ (n) z−n= [z−n]z=0= 1 X (z) =P∞ n=0δ (n − n0) z−n= [z−n]z=n0= z −n0

Thus a delay of n 0 samples corresponds to the factor z−n0 of the transform expression. By expressing the signal x(n) in terms of unit samples and applying the linearity we obtain the general result

x (n − n0) ↔ X (z) z−n0 (time delay) (2)

x (n + n0) ↔ X (z) z+n0 (time advance) (3)

Fig.4.2 Illustrates the relations . Because of this , we use the notation z−1 _{for unit delay and z for unit} advance in block diagram of systems (section 1.5.2).

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Figure 3: Time shift property

Example 2

Unit sample is just the subtraction of the unit step and its delay: u (n) − u (n − 1) = δ (n)

Find the z transform of (a) Causal unit sample x(n) = u(n) (b) Anticausal unit sample x(n) = -u(-n-1) Remember u(n) is also called right-sided signal , and -u(-n-1) or u(-n-1) left-sided signal , while a signal existing in both positive and negative n is called two-sided (or bilateral) signal. Solution

(a) Let's write

x (n) − x (n − 1) = u (n) − u (n − 1) = δ (n) Take the z transform of both sides , using the time shift property:

X (z) − z−1X (z) = 1 or X (z) = 1 1 − z−1 = z z − 1

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Figure 3

(b) For the anticausal signal we write

x (n) = x (n − 1) = −u (−n − 1) + u [− (n − 1) − 1] = u (−n) − u (−n − 1) = δ (−n) Remember δ (−n) is just δ (n) (see section 1.4.1), so transforming both sides gives

X (z) − z−1X (z) = 1 or X (z) = 1 1 − z−1 = z z − 1

Notice that the two signals in (a) and (b) have dierent expressions in time domain but their z-transforms are the same . However these two transforms have dierent region of convergences (ROC) (see Section 4.4).

Example 3

Find the z transform of causal rectangular pulse of N samples p (n) = 1 , 0 ≤ n ≤ N − 1

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Figure 3

Solution

We can apply directly denition (4.1) to nd the transform. Otherwise , we write the rectangular pulse as

p (n) = u (n) − u (n − N ) Take the transform , utilizing the delay property :

P (z) = Z [u (n)] − Z [u (n − N )]

= Z [u (n)] − z−NZ [u (n)] = 1 − z−N Z [u (n)]

= 1−z_1−z−N−1

3 Convolution in time

As for Fourier transform, the most important and powerful property (theorem) of the ztransform for ap-plication aspect is the convolution in time which states : The convolution of two time functions corresponds to normal multiplication of their z-transforms :

x1(n) ∗ x2(n) ↔ X1(z) X2(z) (4)

The most frequently convolution is between input signal x(n) and impulse response h(n) of the system :

x (n) ∗ h (n) ↔ X (z) H (z) (5)

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Figure 5: Convolution property

Output signal in time domain is given by

y (n) = x (n) ∗ h (n) and in z domain by Y (z) = X (z) H (z) (6) From this H (z) = Y (z) X (z) (7)

Thus the transfer function (or system function) of a system is the ratio of the ztransform of the output to the z-transform of the input .This point of view enables us to determine the system function and , throught the inverse transform , the impule response .

Proof :

The convolution property is demonstrated as follows .First let's write x (n) = x1(n) ∗ x2(n) =

∞ X

k=−∞

x1(k) x2(n − k) Take the ztransform

X (z) = ∞ X n=−∞ x (z) z−n= ∞ X n=−∞ " _∞ X k=−∞ x1(k) x2(n − k) # z−n Change the order of summation and use the delay property :

X (z) =P∞

k=−∞x1(k) P∞

n=−∞x2(n − k) z−n

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Example 4

Apply input sequence

x (n) = [1, 2, − 1, − 2, 1, 2] to system whose impulse response is

h (n) = [0, 1, 2] Find the output signal by way of convolution .

Solution

Take the ztransform of x(n) and h(n) :

X (z) = 1 + 2z−1− z−2_{− 2z}−3_{+ z}−4_{+ 2z}−5

H (z) = 0 + z−1+ 2z−2 The transform of the output is

Y (z) = H (z) X (z)

= z−1+ 4z−2+ 3z−3− 4z−4_{− 3z}−5_{+ 4z}−6_{+ 4z}−7 The coecients of X(z) constitute the signal y(n)

y (n) = [0, 1, 4, 3, − 4, − 3, 4, 4]

If we convolve directly x(n) with h(n), e.g. by graphical method , we'll get the same result .

4 Some other properties

There are many other properties of z-transform, below is some of them. (a) Time reversal

x (−n) ↔ X z−1 (8) Proof : Z [x (−n)] = ∞ X n=−∞ x (−n) z−n= ∞ X k=−∞ x (k) z−1k = X z−1 For example u (n) → 1 1 − z−1 ⇒ u (−n) ↔ 1 1 − z (b) Scaling by discrete exponential

anx (n) ↔ X a−1z (9)

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X2 .

(d) Dierentiation in z-domain

nx (n) ↔ −zdX (z)

dz (11)

Proof:

By dierentiating both sides of the denition (4.3), we get dX(z) dz = P∞ n=−∞x (n) (−n) z−n−1= −z−1 P∞ n=−∞[nx (n)] z−n = −z−1Z [nx (n)] Which is another form of the property stated above .

For example let's nd the z-transform of the signal

X (n) = nanu (n) To apply the dierentiation we write

x (n) = nx1(n) wherer x1(n) = anu (n) The z-transform of x1(n) (table 4.1) is

X1(z) = 1 1 − az−1 Thus X (z) = −zdX1(z) dz = az−1 (1 − az−1₎2 (e) Complex conjugation

x∗(n) ↔ (X (z∗))∗ (12)

(f) Initial value

x (0) = lim

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(g) Final value

lim

n→∞x (n) = limz→−1((z − 1) X (z)) (14)

The meaning of this property is similar to above but this case we would like to know the nal value of x(n) .

5 Relation to Discrete-Time Fourier Transform (DTFT)

With respect to discrete time signals and systems , the ztransform is related to Fourier transform the same way as the Laplace transform is to the Fourier transform for continuous-time signals and systems. For this we make the replacement

z = ejω (15)

into the deniton(4.3) of the z-transform and get X (ω) = ∞ X n=−∞ x (n) e−jωn (signal) H (ω) = ∞ X n=−∞ h (n) e−jnω (system)

which are just the Fourier transform we know. Remember that X( ω) and H( ω) are spectra and ω is the digital angular frequency having unit of radian /sample , and that both X( ω) and H( ω) are periodic with a period of 2π . We conclude

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Figure 15: Along the unit circle, the z-transform is just the Fourier transform

H (ω) = [H (z)]_z=ejω (16)

Relation between X( ω) and X(z) is similar . Henceforth, when discussing about the Fourier transform and the z-transform we just mention either X( ω) and X(z) , or H( ω) and H(z) as representative

The magnitude and phase of z with respect to ωare |z| = |ejω_{| = 1}

Φz= ∠|ejω| = ω

Thus the Fourier transform is just the ztransform when z is conned on the unit circle (Fig 4.5).When z moves along this circle the frequence ω changes. As X( ω) and H( ω) are 2π - periodic we need to consider them only in a period of 2 π, usually in the interval [- π , π] or [0 , 2 π].