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The Frequency Domain Representation<br />

of LTI Systems 75<br />

In general, the frequency response H(e jω )isacomplex function of<br />

ω. The magnitude |H(e jω )| of H(e jω )iscalled the magnitude (or gain)<br />

response function, and the angle ̸ H(e jω )iscalled the phase response<br />

function as we shall see below.<br />

3.3.2 RESPONSE TO SINUSOIDAL SEQUENCES<br />

Let x(n) =A cos(ω 0 n + θ 0 )beaninput to an LTI system h(n). Then<br />

from (3.17) we can show that the response y(n) isanother sinusoid of the<br />

same frequency ω 0 , with amplitude gained by |H(e jω0 )| and phase shifted<br />

by ̸ H(e jω0 ), that is,<br />

y(n) =A|H(e jω0 )| cos(ω 0 n + θ 0 + ̸ H(e jω0 )) (3.18)<br />

This response is called the steady-state response, denoted by y ss (n). It<br />

can be extended to a linear combination of sinusoidal sequences.<br />

∑<br />

A k cos(ω k n + θ k ) −→ H(e jω ) −→ ∑ k A k|H(e jω k<br />

)|<br />

k<br />

cos(ω k n + θ k + ̸ H(e jω k<br />

))<br />

3.3.3 RESPONSE TO ARBITRARY SEQUENCES<br />

Finally, (3.17) can be generalized to arbitrary absolutely summable sequences.<br />

Let X(e jω )=F[x(n)] and Y (e jω )=F[y(n)]; then using the<br />

convolution property (3.11), we have<br />

Y (e jω )=H(e jω ) X(e jω ) (3.19)<br />

Therefore an LTI system can be represented in the frequency domain by<br />

X(e jω ) −→ H(e jω ) −→ Y (e jω )=H(e jω ) X(e jω )<br />

The output y(n) isthen computed from Y (e jω ) using the inverse<br />

discrete-time Fourier transform (3.2). This requires an integral operation,<br />

which is not a convenient operation in MATLAB. As we shall see in<br />

Chapter 4, there is an alternate approach to the computation of output to<br />

arbitrary inputs using the z-transform and partial fraction expansion. In<br />

this chapter we will concentrate on computing the steady-state response.<br />

□ EXAMPLE 3.13 Determine the frequency response H(e jω )ofasystem characterized by h(n) =<br />

(0.9) n u(n). Plot the magnitude and the phase responses.<br />

Solution Using (3.16),<br />

∞∑<br />

∞∑<br />

H(e jω )= h(n)e −jωn = (0.9) n e −jωn<br />

=<br />

−∞<br />

∞∑<br />

(0.9e −jω ) n 1<br />

=<br />

1 − 0.9e −jω<br />

0<br />

0<br />

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).<br />

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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