Bode plots for ratio of first/second order factors

Problem: Draw the Bode plots for

G(s) =s + 3

(s + 2)(s2 + 2s + 25)

Solution: We first convert G(s) showing eachterm normalized to a low-frequency gain ofunity. The second order term is normalizedby factoring ω2

n, forming

s2

ω2n

+ 2ζs

ωn+ 1

Thus

G(s) =3

2 × 25

(s3 + 1)

(s2 + 1)( s2

25 + 225s + 1)

The Bode log-magnitude diagram is the sumof the individual first and second terms of G(s).

The low frequency value for G(s) is 350, or

−24.44dB, which is found by letting s → 0.We see that the break frequencies are at 2,3 and 5, and set the extent of our plot fromω = 0.01rad/s to ω = 100rad/s.

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Table below: Bode magnitude diagram slopes

Start: Start: Start:pole zero atat -2 at -3 ωn = 5

Frequency 0.01 2 3 5(rad/s)

pole at -2 0 -20 -20 -20

zero at -3 0 0 20 20

ωn = 5 0 0 0 -40

Total slope 0 -20 0 -40(dB/dec)

The Bode plot starts at −24.44dB and con-

tinue until the first break frequency at 2rad/s,

yielding -20dB/decade slope downwards un-

til the next break frequency at 3rad/s, which

causes +20dB/decade slope upwards, which

when added to the previous -20dB, gives a net

slope of 0. At ωn = 5, the second order term

initiates a -40dB downward slope which con-

tinues to infinity.

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The correction to the log-magnitude curve due

to the underdamped second order term is found

by plotting a point −20 log2ζ above the asymp-

tote at the natural frequency ωn = 5. Note

ζ = 0.2, the correction is 7.9dB.

Figure above: Bode log-magnitude plot for

G(s) = s+3(s+2)(s2+2s+25)

; (a) components; (b)

composite

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We now turn to the phase plot. The first or-

der pole at −2 yields a phase angle that starts

at 0 and end at −90 via a −45◦ starting a

decade below the break frequency and ending

at a decade above the break frequency.

The first order zero at −3 yields a phase angle

that starts at 0 and end at +90 via a +45◦

starting a decade below the break frequency

and ending at a decade above the break fre-

quency.

Table below: Bode phase diagram slopes.

Start: Start: Start: End: End: End:pole zero pole zeroat -2 at -3 ωn = 5 at -2 at 3 ωn = 5

Frequency 0.2 0.3 0.5 20 30 50(rad/s)

pole at -2 -45 -45 -45 0

zero at -3 45 45 45 0

ωn = 5 -90 -90 -90 0

Total slope -45 0 -90 -45 -90 0(deg/dec)

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The second order poles yield a phase angle that

starts at 0 and end at −180 via a −90◦ start-

ing a decade below the break frequency (i,e.

natural frequency at ωn = 5) and ending at a

decade above the break frequency.

Figure above: Bode phase plot for G(s) =s+3

(s+2)(s2+2s+25); (a) components; (b) com-

posite

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Frequency response using Matlab

We can use Matlab to make Bode plots using

bode(G), where G(s) = numgdeng

, and G is an LTI

transfer function object.

Problem: Draw the Bode plots for

G(s) =s + 3

(s + 2)(s2 + 2s + 25)

using Matlab.

≫ numg=[1 3];

≫ deng=conv([1 2],[1 2 25]);

≫ G=tf(numg,deng);

≫ bode(G);

≫ grid on;

≫ title(’open loop frequency response’)

≫ [mag,phase,w] = bode(G);

≫ points=[20*log10(mag(:,:))’, phase(:,:)’,w]

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−80

−60

−40

−20

0

Mag

nitu

de (

dB)

10−1

100

101

102

−180

−135

−90

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

Figure above; The Bode plots for

G(s) =s + 3

(s + 2)(s2 + 2s + 25).

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We can also use Matlab to make polar plots

using nyquist(G), where G(s) = numgdeng

, and G is

an LTI transfer function object.

Problem: Draw the polar plots for

G(s) =s + 3

(s + 2)(s2 + 2s + 25)

using Matlab.

≫ numg=[1 3];

≫ deng=conv([1 2],[1 2 25]);

≫ G=tf(numg,deng);

≫ nyquist(G);

≫ grid on;

≫ title(’polar plots’)

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−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.20 dB

−20 dB

−10 dB−6 dB−4 dB−2 dB

20 dB

10 dB 6 dB4 dB 2 dB

polar plots

Real Axis

Imag

inar

y A

xis

Figure above; The polar plots for

G(s) =s + 3

(s + 2)(s2 + 2s + 25).

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Problem: Sketch the Bode plots and polar

plots by hand and verify the results using Mat-

lab for

G(s) =s + 20

(s + 2)(s + 7)(s + 50).

≫ numg=[1 20];

≫ deng=conv([1 2],[1 7]);

≫ deng=conv(deng,[1 50]);

≫ G=tf(numg,deng);

≫ bode(G);

≫ grid on;

≫ title(’open loop frequency response’)

≫ [mag,phase,w] = bode(G);

≫ points=[20*log10(mag(:,:))’, phase(:,:)’,w]

10

−120

−100

−80

−60

−40

−20

Mag

nitu

de (

dB)

10−1

100

101

102

103

−180

−135

−90

−45

0

Pha

se (

deg)

open loop frequency response

Frequency (rad/sec)

Figure above; The Bode plots for

G(s) =s + 20

(s + 2)(s + 7)(s + 50).

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Problem: Sketch the Bode plots by hand and

verify the results using Matlab for

G(s) =s + 20

(s + 2)(s + 7)(s + 50).

≫ numg=[1 20];

≫ deng=conv([1 2],[1 7]);

≫ deng=conv(deng,[1 50]);

≫ G=tf(numg,deng);

≫ nyquist(G);

≫ grid on;

≫ title(’polar plots’)

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−1 −0.8 −0.6 −0.4 −0.2 0 0.2−0.02

−0.015

−0.01

−0.005

0

0.005

0.01

0.015

0.020 dB −20 dB−10 dB−6 dB−4 dB−2 dB20 dB 10 dB 6 dB4 dB2 dB

polar plots

Real Axis

Imag

inar

y A

xis

Figure above; The polar plots for

G(s) =s + 20

(s + 2)(s + 7)(s + 50).

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