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Page 1: Bode Diagram 1

Bode Diagram (1)

Hany FerdinandoDept. of Electrical Engineering

Petra Christian University

Page 2: Bode Diagram 1

General Overview

This section discusses the steady-state response of sinusoidal input

The frequency response analysis uses Bode diagram

Students also learn how to plot Bode diagram

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What is frequency response?

G(s)X(s) Y(s)

If x(t) = X sin If x(t) = X sin t then y(t) = Y sin (t then y(t) = Y sin (t + t + ))

The sinusoidal transfer function G(jThe sinusoidal transfer function G(j) is a ) is a complex quantity and can be represented complex quantity and can be represented by the magnitude and phase angle with by the magnitude and phase angle with

frequency as a parameterfrequency as a parameter

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Presenting the freq. resp.

Bode Diagram or logarithmic plot (this section)

Nyquist plot or polar plot Log-magnitude versus phase plot

Matlab can be used to plot both Bode diagram and Nyquist plot

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Preparation

Bode diagram uses the open loop transfer function

The plot is a pair of magnitude and phase plots The representation of logarithmic

magnitude is 20 log |G(j)| in dB

The main advantage: multiplication is converted into addition

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Basic factor of G(j)

Gain K Derivative and integral factors (j)±1

First-order factors (1+j)±1

Quadratic factors [1+2(jn)+(jn)2]±1

One must pay attention to the corner frequency and the form of the equation must be fitted to

above

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Gain K It is real part only no phase angle The log-magnitude is a straight line at

20 log (K) If K > 1, then the magnitude is positive If K < 1, then the magnitude is negative

Varying K only influences the log-magnitude plot, the phase angle remains the same

Slope is 0 at corner frequency 0 rad/s

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Integral (j)-1

It has only imaginary part Log-magnitude = -20 log () Phase angle = 90o (constant) Slope is -20 dB/decade at corner

frequency =1 rad/s

)log(201

log201

log20

j

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Derivative (j)

It has only imaginary part Log-magnitude: 20 log () Phase angle: 90o (constant) Slope is 20 dB/decade at corner

frequency =1 rad/s

)log(20log20log20 j

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First order (1+j)±1 (1) Integral:

Corner frequency is at =1/T Slope is -20 dB/decade Phase angle is -45o at corner

frequency Derivative:

Corner frequency is at =1/T Slope is 20 dB/decade Phase angle is 45o at corner frequency

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First order (1+j)±1 (2)

-30

-25

-20

-15

-10

-5

0

Mag

nitu

de (

dB)

10-2

10-1

100

101

-90

-45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

21

1

j

45o at =0,5 rad/s

Slope: 20dB/dec

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Quadratic factors (1) Integral:

Corner frequency is at =n Slope is – 40 dB/decade Phase angle is -90o at corner frequency

Derivative: Corner frequency is at =n

Slope is 40 dB/decade Phase angle is 90o at corner frequency

Resonant freq: 221 nr

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Quadratic factors (2)12

2

)(

21

jj

n = √2

-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)

90o at corner freq. n = √2

Slope = 40dB/dec

Page 14: Bode Diagram 1

Example:

)2)(2(

)3(10)(

2

ssss

ssG

Draw Bode diagram for the following transfer function:

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Answer:

)2))((2)((

)3(10)(

2

jjjj

jsG

Substitute the s with j! We got

Make it to the standard form… Proot it!!!

122

)(1

2)(

13

5.7)(

2

jjjj

j

sG

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Answer:

122

)(1

2)(

13

5.7)(

2

jjjj

j

sG

13

j12

122

)(

jj1

12

j1)( j

from

We got

7.5

With = 0.35

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Answer:

7.5 is gain Log-magnitude = 20 log (7.5) Phase = 0o

The

Slope = 20 dB/decade Phase = 45o at = 3 rad/s

13

j

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Answer

The Slope = -20 dB/decade Phase = -90o (constant)

The

Slope = -20 dB/decade Phase = - 45o at = 2 rad/s

1)( j

1

12

j

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Answer:

The

Slope = - 40 dB/decade Phase = -90o at = √2 rad/s

12

122

)(

jj

The next step is to combine all magnitudes and phases respectively,

then add all of them to from a sketch of Bode diagram

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Next…

The Bode diagram has been discussed here, the next topic is Phase and Gain

Margin. Several important points will be discussed as well

Please prepare yourself by reading the book!


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