Review of Analog Signal Processing Basics - Digital Signal Processing: Fundamentals and Applications - page 800

a. Derive the steady-state transfer function.

b. Derive the amplitude response and phase response.

c. If the input is given as a sinusoid, that is, xðtÞ¼5 sin ð10t þ 30 ÞuðtÞ, find the steady-state response y ss ðtÞ.

Solution:
a. By substituting s ¼ ju into the transfer function in terms of a suitable form, we get the steady-state transfer
function as

H ju ¼

1 1

10 þ 1 ¼

s

ju
10 þ 1

b. The amplitude response and phase response are found to be

A u ¼

1
u
10

r

2

þ1

b u ¼ : tan 1 u
10

c. When u ¼ 10 rad/sec, the input sinusoid can be written in terms of the phasor form as
X j10 ¼ 5 : 30

For the amplitude and phase of the steady-state transfer function at u ¼ 10, we have

A 10 ¼

1
10
10

s ¼ 0:7071

2

þ1

b 10 ¼tan 1 10
10

¼45

Hence, we yield

H j10 ¼ 0:7071 : 45

Using Equation (B.36) , the system output in phasor form is obtained as

Y j10 ¼ H j10 X j10 ¼ 1:4141 : 45

5 : 30

Y j10 ¼ 3:5355 : 15

Converting the output in phasor form back to the time domain results in the steady-state system output:
y ss t ¼ 3:5355 sin 10t 15

u t