PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

DTFT, DFT, ZT, and FFT 483

c. For the causal signal f(n), if f(n) = 0 for n < 0, then the exterior of the circle

defi ned by the pole with the largest magnitude constitutes the ROC.

d. For the noncausal signal f(n), if f(n) ≠ 0 for n < 0 , the ROC is inside the circle

defi ned by the pole with the smallest magnitude.

R.5.69 The ROC of a stable, LTI system includes the unit circle since that fact would

imply that the impulse response is absolutely summable.

R.5.70 A summary of the most-often used ZT pairs is given in Table 5.2.

R.5.71 The evaluation of the inverse ZT is given by the following equation:

fn Fzz dzn

c

()
∫ ()^1

where the symbol c∫ denotes an integration over the close contour c that includes

all the poles of F(z).

The contour integral can be evaluated by using the Cauchy’s residue theorem,

resulting in

fn()
∑ [residues ofFzz()n^1 at the poles insidec]

R.5.72 Observe that the inverse ZT as defi ned in R.5.71 involves a diffi cult computational

process, and thereby it is not practical and therefore seldom used. Simpler methods

based on conversion transform tables, including partial-fraction expansion, and

long division are the preferred methods.

Of course, it is far simpler to employ the MATLAB symbolic toolbox to evaluate

the ZT and the inverse ZT of arbitrary functions, as illustrated in the next sections.

Recall that the system impulse response h(n) of an LTI system is critical to evalu-

ate the response of any arbitrary sequence. Note that the system impulse response

can be evaluated by taking the inverse ZT of the system transfer function H(z).

TABLE 5.2

Table of ZT Pairs

Discrete Time Signals Bidirectional Relation ZT

f[n] F(z)

δ[n] 1

δ[n − n 0 ] zz zn (^0) for 0 and 
u[n] 1/(1 − z−^1 ) for |z| > 1
αnu(n) 1/(1 − αz−^1 ) for |z| > |α|
−αnu[−n − 1] 1/1 − αz−^1 for |z| < |α|
nαnu[n] αz−^1 /(1 − αz−^1 )^2 for |z| > |α|
−nαnu[−n − 1] αz−^1 /(1 − αz−^1 )^2 for |z| < |α|
αncos(ω 0 n)u[n]
1()
{1 [2 ( )] }
1
122





 

n nz
zz
cos z
cos
0
0

αnsin(ω 0 n)u[n]

 

n
n
z
zz
sin z
cos
()
{1 [2 ] }
0 1
0 12
2