0195136047.pdf

3.4 FREQUENCY RESPONSE 159

Solution

H(s)=

100 ( 1 +s/ 50 )

( 1 +s/ 10 )( 1 +s/ 500 )

or H(jω) ̄ =

100 ( 1 +jω/ 50 )

( 1 +jω/ 10 )( 1 +jω/ 500 )

The break frequencies are 10 and 500 rad/s for the denominator and 50 rad/s for the numerator.

The component straight-line segments are drawn as shown in Figure E3.4.1. Note that the effect

of the constant multiplier 100 inH ̄(jω)is to add a constant value, 20 log 100=40 dB, as marked

in the Bode plot. The resultant asymptotic magnitudeH(ω) and angleθ(ω)characteristics are

indicated by the dashed lines.

H(ω), dB

0

− 60

− 90

− 30

+ 60

+ 30

+ 90

− 30

− 20

− 10

20

0

40

30

10

θ(ω), deg

0.1 1 10 100 1000

Angle of

1

( 1 + jω /10)

10,000

ω

ω

Angular frequency, rad/s (logarithmic scale)

Magnitude of

1

( 1 + jω /500)

Angle of

1

( 1 + jω /500)

Magnitude of

Magnitude of (1 + jω /50)

Angle of (1 + jω /50)

1

( 1 + jω /10)

Angle θ (ω)

H(ω) dB

40 dB

Figure E3.4.1Asymptotic Bode plot.

As seen from Example 3.4.1, the frequency response in terms of the asymptotic Bode plot is
obtained with far less computation than needed for the exact characteristics. With little additional
effort, by correcting errors at a few frequencies within the asymptotic plot, one can get a sufficiently
accurate result for most engineering purposes. While we have consideredH ̄(jω)of the type
given by Equation (3.4.6) with only simple poles and zeros, two additional cases need further
consideration: