0195136047.pdf

3.4 FREQUENCY RESPONSE 157

(a)

H(ω)

0

Passband

Ideal

Actual

ω ω

co

(b)

H(ω)

0

Passband

Ideal

Actual

ω ω

co

(c)

H(ω)

0

Passband

Ideal

Actual

ω ω

1 ω 2

(d)

H(ω)

0

Passband

Ideal

Actual

ω ω

1 ω 2

Passband

Figure 3.4.5Ideal response characteristics.(a)Low pass.(b)High pass.(c)Band pass.(d)Band reject.

H ̄(jω)=k

(

1 +jω (^01)
)(
1 +jω (^02)
)
···
(
1 +jω (^0) m
)
(
1 +jωp 1
)(
1 +jωp 2
)
···
(
1 +jωpn
) (3.4.6)
where− 01 ,− 02 , ...,− (^0) mare the zeros, and−p 1 ,−p 2 , ...,−pnare the poles of the network
function. Equation (3.4.6) is clearly the product of a constant and a group of terms having the form
( 1 +jω/ω 0 )or 1/( 1 +jω/ω 0 ). Each of these terms can be considered as an individual phasor,
andH(jω) ̄ has a magnitude given by the product of the individual magnitudes (or the sum of the
individual terms expressed in dB) and an angle given by the sum of the individual angles. Thus
the behavior of functions( 1 +jω/ω 0 )and 1/( 1 +jω/ω 0 )is to be clearly understood for the
construction of Bode plots.
Let us first considerH ̄ 1 (j ω)= 1 +jω/ω 0. Forω/ω 0 <<1 (i.e., low frequencies), the
magnitudeH 1 (ω)∼=1, orH 1 (ω)dB=20 log 1=0 dB. Thebreak frequencyisω=ω 0. At high
frequencies (i.e.,ω/ω 0 >>1), the function behaves as
H 1 (ω)∼=
ω
ω 0
or H 1 (ω)dB=20 log
ω
ω 0
atω/ω 0 =1,
H 1 (ω)dB=0dB
atω/ω 0 =10,
H 1 (ω)dB=20 log 10=20 dB
and atω/ω 0 =100,
H 1 (ω)dB=20 log 100=40 dB