0195136047.pdf

158 TIME-DEPENDENT CIRCUIT ANALYSIS

Since factors of 10 are linear increments on the logarithmic frequency scale, the plot of 20 log

(ω/ω 0 )versusωon a semilog paper is a straight line with a slope of+20 dB/decade or+ 6

dB/octave, as shown in Figure 3.4.6. Note that an octave represents a factor of 2 in frequency,

whereas a decade represents a factor of 10. The angle associated withH ̄ 1 (j ω)isθ 1 (ω) =

tan−^1 (ω/ω 0 ). At low frequencies(ω≤ 0. 1 ω 0 ), θ 1 ∼=0; for frequenciesω≥ 10 ω 0 ,θ 1 ∼=90°;

and atω=ω 0 ,θ 1 =+45°. This leads to the straight-line approximations forθ 1 (ω), as shown

in Figure 3.4.6. For 0. 1 ω 0 ≤ω≤ 10 ω 0 , the slope ofθ 1 (ω)versusωis approximately given by

+45°/decade (over two decades centered at the critical frequency) or+ 13 .5°/octave. Thus the

asymptotic Bode plot forH ̄ 1 (j ω)= 1 +jω/ω 0 is drawn in Figure 3.4.6. Figure 3.4.7 gives the

asymptotic Bode plot forH ̄ 2 (j ω)= 1 /( 1 +jω/ω 0 )with a similar kind of reasoning. The student

is encouraged to justify the plot. The dashed curves in Figures 3.4.6 and 3.4.7 indicate the exact

magnitude and phase responses.

The process of developing asymptotic Bode plots is then one of expressing the network

function in the form of Equation (3.4.6), locating the break frequencies, plotting the component

asymptotic lines, and adding these to get the resultant. The following example illustrates the

procedure.

0

− 90

− 45

45

90

− 20

20

0

40

θ 1 (ω), deg

Slope =

+20 dB/decade or

+6 dB/octave

Slope = + 45 ° /decade or

+13.5° /octave

0.1 ω 0 1 ω 0 10 ω 0 100 ω 0

ω

Asymptotic

Exact

ω

Asymptotic

Exact

ω, rad/s

H 1 (ω) dB Figure 3.4.6Asymptotic Bode plot forH ̄ 1 (j ω)=

1 +jω/ω 0.

EXAMPLE 3.4.1

Sketch the asymptotic Bode plot for

H(s)=

104 (s+ 50 )

s^2 + 510 s+ 5000