1

Bode Diagram

Z. Aliyazicioglu

Electrical and Computer Engineering DepartmentCal Poly Pomona

ECE307-8

ECE 307-8 2

Bode Diagram

Real, First order Poles and Zeros

1

1

( )( )( )K s zH ss s p

+=

+

Bode diagram consists of two separate plotsThe amplitude of H(jω) varies with frequencyThe phase angle of H(jω) varies with frequency

Replace s with jω

The poles and zeros of H(s) are real and first order

1

1

( )( )( )

K j zH jj j p

ωωω ω

+=

+

Fist step put the expression for H(jω) in a standard form

11

11

1( )

1

jK zz

H jjp jp

ω

ωωω

+

= +

2

ECE 307-8 3

=Bode Diagram

Let Ko represent the constant quantity 10

1

K zKp

=

H(jω) in polar form

The amplitude value of H(jω)

0 1 01 1

1 1

11 1

1 1( ) ( 90 )

90 1 1

j jK Kz z

H jj jj jp p

ω ω

ω βω ωω β ω

+ ∠Ψ += = ∠Ψ −∠ −∠

∠ + ∠ +

The phase angle of H(jω)

01

1

1( )

1

jKz

H jjjp

ω

ωωω

+=

+

1 1( ) ( 90 )θ ω β= ∠Ψ −∠ −∠

11

1

tan ( )zω−Ψ = 1

11

tan ( )pωβ −=

Real, First order Poles and Zeros

ECE 307-8 4

Bode Diagram

The amplitude of value of H(jω) in decibel is

Some Results

Straight-Line Amplitude Plots

01

10 10 0 10 10 101 1

1

120 log 20log 20log 1 20log 20log 1

1dB

jKz j jA K

z pjp

ωω ωω

ωω

+= = + + − − +

+

100000.1005.6215

10000.803.1610

1000.006026

100.00401.4143

31.623010

10.00200.7071-3

AAdBAAdB

3

ECE 307-8 5

Bode Diagram

• The plot of 20 log10K0 is a horizontal straight line because of K0 is not a function of ω.

Straight-Line Amplitude Plots

1020log (2)6 dB

dBA =

=

Let say K0=2

20 log10K0 is positive for K0>1 20 log10K0 is zero for K0=1 20 log10K0 is negative for K0>1

ECE 307-8 6

Bode Diagram

The approximate plot of 20 log10|1+jω/z1| is two straight lines 1. For small values of ω, |1+jω/z1| is approximately 1

For large values of ω, |1+jω/z1| is approximately ω/z1

Straight-Line Amplitude Plots

101

20log 1 0dB 0j aszω ω+ → →

10 101 1

20log 1 20logj asz zω ω ω

+ → → ∞

On a log scale 20log |jω/z1| is straight line with a slope of 20dB/decade

This straight line intersection the 0dB axis at ω=z1 . This value of ω is called corner frequency

4

ECE 307-8 7

Bode Diagram

• Let’s assume that z1=100 1020log 1100jω

+

1020log 1 0 100100j asω ω+ = <

10 1020log 1 20log 20 / 100100 100j dB decade asω ω ω + → = >

>> w=0:10:10000;

>> a=20*log10(abs(1+j*w/100));

>> semilogx(w,a)

>> grid on

>> ylabel ('A_{dB}')

>> xlabel ('\omega (rad/s)')

ECE 307-8 8

Bode Diagram

The plot of -20 log10(ω) is q straight line having a slope of –20 dB/decade that intersect the 0 dB axis at ω=1

Straight-Line Amplitude Plots

5

ECE 307-8 9

Bode Diagram

The approximate plot of -20 log10|1+jω/p1| is two straight lines 1. For small values of ω, |1+jω/p1| is approximately 1

For large values of ω, |1+jω/p1| is approximately ω/p1

Straight-Line Amplitude Plots

101

20log 1 0 0j dB aspω ω− + → →

10 101 1

20log 1 20log , asjp pω ω ω

− + → − → ∞

On a log scale - 20log |jω/z1| is straight line with a slope of -20 dB/decade

This straight line intersection the 0 dB axis at ω=p1 . This value of ω is called corner frequency

ECE 307-8 10

Bode Diagram

• Let’s assume that p1=200 1020log 1200jω

− +

1020log 1 0 200200j asω ω− + = <

10 1020log 1 20log 20dB/decade 200200 200j asω ω ω − + → − = − >

>> w=0:10:10000;

>> a=-20*log10(abs(1+j*w/100));

>> semilogx(w,a)

>> grid on

>> ylabel ('A_{dB}')

>> xlabel ('\omega (rad/s)')

6

ECE 307-8 11

Example

Bode Diagram

1020log (2)

1020log 110jω

+

1020log ω−

1020log 1100jω

− +

10 10 10 10 10

2 11020 log 20log (2) 20log 1 20log 20log 1

10 1001100

dB

jj jA

j

ωω ωω

ωω

+= = + + − − +

+

ECE 307-8 12

MatLab

2

2 120 20010( )

1001100

ssH s

s s ss

+ + = =+ +

>> syms s

>> n=[0 20 200];

>> dn=[1 100 0];

>> g=tf(n,dn)

Transfer function:

20 s + 200

-----------

s^2 + 100 s

>> bode(g)

>> grid on

7

ECE 307-8 13

Bode Diagram

Example

L1

100 mH1 2

Vi1Vac0Vdc Vo

0

+

C1

10 mF

-

R1

11 ohms

2( ) 1

R sLH s Rs sL LC

=+ +

2110 110( )

110 1000 ( 10)( 100)s sH s

s s s s= =

+ + + +

0.11( )(1 )(1 )

10 100

jH jj j

ωω ω ω=+ +

10 10 10 1020log (0.11) 20log 20log 1 20log 110 100dBA j j jω ωω= + − + − +

Transfer function H(s) of the circuit

Writing H(jω)

In terms of dB

a. Bode Plotb. Find 20log10|H(jω)| at ω=50

rad/s and ω=1000 rad/s c. vi(t)=5cos(500t+150) find Vo(t)

ECE 307-8 14

1020log (0.11)

1020log jω

1020log 110j ω− +

1020log 1100j ω

− +

dBA

8

ECE 307-8 15

Matlab

>> w=1:10:10000;

>>h=(110*j*w)./((j*w+10).*(j*w+100));

>> a=20*log10(abs(h));

>> semilogx(w,a)

>> grid on

>> xlabel ('\omega (rad/s)')

>> ylabel ('A_{dB}')

11020log( 10)( 100)dB

jAj j

ωω ω

=+ +

ECE 307-8 16

Bode Diagram

>> dn=[1 110 1000];

>> n=[0 110 0];

>> g=tf(n,dn)

Transfer function:

110 s

------------------

s^2 + 110 s + 1000

>> bode (g)

>> grid on

9

ECE 307-8 17

Bode Diagram

b. 20log10|H(jω)| at ω=50 rad/s and ω=1000 rad/s

50

0.11 0.11( 50)( ) 0.9648 12.2550 50(1 )(1 ) (1 )(1 )10 100 10 100

j jH jj j j jω

ωω ω ω=

= = = ∠ −+ + + +

1000

0.11 0.11( 1000)( ) 0.1094 83.721000 1000(1 )(1 ) (1 )(1 )10 100 10 100

j jH jj j j jω

ωω ω ω=

= = = ∠ −+ + + +

10 1020log ( 50) 20log 0.9648 0.31 dBdBA H j= = = −

10 1020log ( 1000) 20log 0.0.1094 12.22 dBdBA H j= = = −

c. vi(t)=5cos(500t+150) find Vo(t) 500 /rad sω =

0.11( 500)( 500) 0.22 77.54500 500(1 )(1 )10 100

jH jj j

= = ∠ −+ +

0( ) 5 ( 500) cos(500 15 ( 500)

5(.22)cos(500 15 77.54 )1.1cos(500 62.54 )

v t H j t H j

tt

= + + ∠

= + −

= −

From Bode

500 12.5dBA dBω= = −

12.5 / 20( 500) 10 0.24H j −= =

ECE 307-8 18

Bode DiagramMore Accurate Amplitude Plots

1 10

10

20log 1 1

20log 23dB

dBA jω= = ± +

= ±

= ±

Amplitude value at the corner frequency ω=1 of H(jω)1020log 1dBA jω= ± +

0.5 10

10

20log 1 0.5

20log 5 / 41dB

dBA jω= = ± +

= ±

= ±

2 10

10

20log 1 2

20log 57dB

dBA jω= = ± +

= ±

= ±

10

ECE 307-8 19

Bode DiagramStraight-Line Phase Angle Plots

The Phase Angle of constant Ko is zero degreeThe phase angle of first order zero has two straight line

1. For values of ω= z1, 45 degree2. For values of ω>=10 z1, it is straight line 90 degree3. For values of ω<=0.1 z1, it is straight line 0 degree4. For values of ω<=0.1 z1<= ω , it is straight line straight line

having a slope of 45 degree/decadeThe phase angle of first order pole has two straight line

For values of ω= p1, -45 degree2. For values of ω>=10 p1, it is straight line -90 degree3. For values of ω<=0.1 p1, it is straight line 0 degree4. For values of ω<=0.1 p1<= ω , it is straight line straight line

having a slope of -45 degree/decadeThe phase angle of -20 log10(ω) is a straight line having a slope of

–90 degree/decade that intersect the 0 degree axis at ω=0

ECE 307-8 20

Straight-Line Phase Angle Plots2 1

10( )1

100

j

H jjj

ω

ωωω

+=

+

( )1 1 1( ) tan tan tan10 100

j ω ωθ ω ω− − − = − −

1tan10ω−

( )1tan ω−−

1tan100ω− −

11

ECE 307-8 21

Matlab

>> w=0.1:0.1:1000;>>h=2*(1+j*w/10)./((j*w).*(1+j*w/100));>> a=angle(h);>> deg=a*180/pi;>> semilogx(w,deg)>> grid on>> xlabel ('\omega (rad/s)')>>ylabel('\theta(\omega)')

ECE 307-8 22

Example

L1

100 mH1 2

Vi1Vac0Vdc Vo

0

+

C1

10 mF

-

R1

11 ohms

0.11( )(1 )(1 )

10 100

jH jj j

ωω ω ω=+ +

a. Straight-line phase angle plotb. Phase angle θ(ω) at ω=50 rad/s

ω=500 rad/s, and ω=1000 rad/s c. vi(t)=5cos(500t+150) find Vo(t)

1 1

1 2

( ) 90 tan tan10 100

90

j ω ωθ ω

β β

− − = − −

= − −

1 1

0.11 90( )

1 tan ( ) 1 tan ( )10 10 100 100

jH j

j j

ωω

ω ω ω ω− −

∠=

+ ∠ + ∠

1 10.11 50 50 50( 50) 90 tan ( ) tan ( ) 0.96 15.2550 50 10 1001 110 100

jH j

j j

− −= ∠ − ∠ −∠ == ∠ −+ +

1 10.11 500 500 500( 500) 90 tan ( ) tan ( ) 0.22 77.54500 500 10 1001 110 100

jH j

j j

− −= ∠ − ∠ −∠ == ∠ −+ +

12

ECE 307-8 23

Straight-line phase angle plot

90

1tan10ω− −

1tan100ω− −

1 1( ) 90 tan tan10 100

j ω ωθ ω − − = − −