Modern Control Engineering

402 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method

The amplitude ratio of the output to the input is

while the phase angle fis

Thus, for the input x(t)=Xsinvt, the steady-state output yss(t)can be obtained from Equation

(7–5) as follows:

(7–6)

From Equation (7–6), it can be seen that for small v, the amplitude of the steady-state output

yss(t)is almost equal to Ktimes the amplitude of the input. The phase shift of the output is small

for small v. For large v, the amplitude of the output is small and almost inversely proportional to

v. The phase shift approaches –90° as vapproaches infinity. This is a phase-lag network.

EXAMPLE 7–2 Consider the network given by

Determine whether this network is a lead network or lag network.

For the sinusoidal input x(t)=Xsinvt, the steady-state output yss(t)can be found as follows:

Since

we have

and

Thus the steady-state output is

From this expression, we find that if then Thus, if

then the network is a lead network. If then the network is a lag network.

Presenting Frequency-Response Characteristics in Graphical Forms. The

sinusoidal transfer function, a complex function of the frequency v, is characterized by

its magnitude and phase angle, with frequency as the parameter. There are three

commonly used representations of sinusoidal transfer functions:

T 16 T 2 ,

T 17 T 2 , tan-^1 T 1 v-tan-^1 T 2 v 7 0. T 17 T 2 ,

yss(t)=

XT 221 +T^21 v^2

T 121 +T^22 v^2

sin Avt+tan-^1 T 1 v-tan-^1 T 2 vB

f= /G(jv)=tan-^1 T 1 v-tan-^1 T 2 v

@G(jv)@ =

T 221 +T^21 v^2

T 121 +T^22 v^2

G(jv)=

jv+

1

T 1

jv+

1

T 2

=

T 2 A 1 +T 1 jvB

T 1 A 1 +T 2 jvB

G(s)=

s+

1

T 1

s+

1

T 2

yss(t)=

XK

21 +T^2 v^2

sin Avt-tan-^1 TvB

f=/G(jv)=-tan-^1 Tv

@G(jv)@=

K

21 +T^2 v^2

Openmirrors.com