1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

9.3 • D IGITAL SIGNAL FILTERS 381

freq uency f s that is at least twice the highest input signal frequency to avoid
frequency fold-over, or aliasing. That is because the Fourier transform of a sam-

pled signal is periodic with period ws = }; , though we will not prove this here.

Aliasing prevents accurate recovery of the original signal from its samples.
Now it can be shown that the argument of the Fourier transform maps onto
the z,-plane unit circle via the formula

f
e = 21T-
fs'

where 8 is called the normalized frequency.

(9-37)

Therefore, the z,-transform evaluated on the unit circle is also periodic, except

with period 211'.

Definition 9.6 (Amplitude Response) The amplitude response A(fJ) is defined to

be the magnitude of the transfer function evaluated at the complex unit signal

z = e^18. The formula is

A(e) = IH(ei^9 )1 over the interval [O, 11']. (9-38) I

Proof Sinusoidal signals are linear combinations of ei^9 and e- •^0. To determine

the amplitude response we input x[n] = e^1 n.11, which has z-transform X(z) =

3[x[n]J = •-'e••.

The z,-transform of the output is Y(z) = X(z)H(z) = ,_• ... H(z), which can

be written as

(9-39)

It is possible to use a technique like partial fraction expansious to write (9-39)

in the form

Fo z

Y(z) = z--. 9 + Q(z) = Fo--. 9 + Q(z),

z -e' z - e•

(9-40)

where Fo = H(e^19 ).

For stable solutions, we have already mentioned (see Section 9.2.1) that the

poles of the transfer function must all lie inside the unit circle. Hence the terms

in the solution y[n] corresponding to the poles of Q(z) are all transient and decay

to zero as n-> oo. The steady state portion of the solution y[n] is the inverse of

the term Fo3[,_z.,.J and its magnitude is

(9-41)