Review of Analog Signal Processing Basics - Digital Signal Processing: Fundamentals and Applications - page 799

Solution:
Two signals hð s Þ and xð s Þ that are involved in the convolution integration are displayed in Figure B.12 . To evaluate
the convolution, the time-reversed signal xð s Þ and the shifted signal xðt s Þ are also plotted for reference. Figure
B.12 shows an overlap of hð s Þ and xðt s Þ. According to the overlapped (shaded) area, the lower limit and the
upper limit of the convolution integral are determined to be 0 and t , respectively. Hence,

e 10 s

Z t

y t ¼

t

5
10

e 10 s $ 5d s ¼

0

0

0:5e 100

¼0:5e 10t

Finally, the system response is found to be
y t ¼ 0:5u t 0:5e 10t u t

The solution is the same as that obtained using the Laplace transform method described in Example B.11.

B.3.3 Sinusoidal Steady-State Response
For linear analog systems, if the input to a system is a sinusoid of radian frequency u , the steady-state
response of the system will also be a sinusoid of the same frequency. Therefore, the transfer function,
which provides the relationship between a sinusoidal input and a sinusoidal output, is called the
steady-state transfer function. The steady-state transfer function is obtained from the Laplace transfer
function by substituting s ¼ ju , as shown in the following:
H ju ¼ HðsÞj s¼ju (B.35)

Thus we have a system relationship in a sinusoidal steady state as

YðjuÞ¼HðjuÞXðjuÞ (B.36)

Since HðjuÞ is a complex function, we may write it in the phasor form:

HðjuÞ¼AðuÞ : bðuÞ

(B.37)

where the quantity AðuÞ is the amplitude response of the system defined as

(B.38)
and the phase angle bðuÞ is the phase response of the system. The following example is presented to
illustrate the application.

AðuÞ¼jHðjuÞj

EXAMPLE B.15
Consider a linear system described by the differential equation shown in Example B.12, where xðtÞ and yðtÞ
designate the system input and system output, respectively. The transfer function has been derived as

H s ¼

10
s þ 10