Signals and Systems - Electrical Engineering

596 C H A P T E R 10: Fourier Analysis of Discrete-Time Signals and Systems

Solution

Notice that if 1/αiis a zero ofHi(z), a pole at the conjugate reciprocalα∗iexists. To show that the

magnitude ofHi(ejω)is a constant for all frequencies, consider the magnitude-squared function

|Hi(ejω)|^2 =Hi(ejω)H∗(ejω)=K^2 i

(ejω− 1 /αi)(e−jω− 1 /α∗i)

(ejω−α∗i)(e−jω−αi)

=Ki^2

ejω(e−jω−αi)e−jω(ejω−α∗i)

αiαi∗(ejω−α∗i)(e−jω−αi)

=

Ki^2

|αi|^2

Thus, by lettingKi=|αi|, the above gives a unit magnitude. The cascade of theHi(z)gives a transfer

function of

H(z)=

∏

i

Hi(z)=

∏

i

|αi|

z− 1 /αi

z−αi

so that

H(ejω)=

∏

i

Hi(ejω)=

∏

i

|αi|

ejω− 1 /αi

ejω−αi

|H(ejω)|=

∏

i

|Hi(ejω)|= 1

∠H(ejω)=

∑

i

∠Hi(ejω)

which in turn gives

Y(ejω)=|X(ejω)|ej(∠X(e

jω)+∠H(ejω))

so that the magnitude of the output coincides with that of the input; however, the phase ofY(ejω)

is the sum of the phases ofX(ejω)andH(ejω). Thus, the all-pass system allows all frequency com-

ponents in the input to appear at the output with no change in the magnitude spectrum but with

a phase shift. n

10.3 Fourier Series of Discrete-Time Periodic Signals......................................

Like in the continuous-time domain, we are interested in finding the response of an LTI system to

a periodic signal. As in that case, we represent the periodic signal as a combination of complex

exponentials and use the eigenfunction property of LTI systems to find the response.

Notice that we are proceeding in the reverse order we followed in the continuous-time case: We are

considering the Fourier representation of periodic signals after that of aperiodic signals. Theoretically,

there is no reason why this cannot be done, but practically it has the advantage of ending with a