Bode Plot

ECE307-8

Bode Diagram

Z. Aliyazicioglu Electrical and Computer Engineering Department Cal Poly Pomona

Bode Diagram Bode diagram consists of two separate plots The amplitude of H(jω) varies with frequency The phase angle of H(jω) varies with frequency Real, First order Poles and Zeros The poles and zeros of H(s) are real and first order H (s ) =

K (s + z1) s(s + p1)

Replace s with jω H ( jω ) =

K ( jω + z1) jω ( jω + p1)

Fist step put the expression for H(jω) in a standard form  jω  K z1  1 +  z1   H ( jω ) =  jω  p1 jω  1 +  p1  

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Bode Diagram = Real, First order Poles and Zeros

K0 =

Let Ko represent the constant quantity

K z1 p1

H(jω) in polar form K0 1 + H ( jω ) =

jω ∠Ψ1 z1

K0 1 +

jω jω ∠90 1 + ∠β1 p1

=

D

The amplitude value of H(jω)

The phase angle of H(jω)

jω z1

jω jω 1 + p1

(∠Ψ1 − ∠90D − ∠β1)

K0 1 +

jω z1

jω 1 +

jω p1

H ( jω ) =

θ (ω ) = (∠Ψ1 − ∠90D − ∠β1) ω

Ψ1 = tan−1( ) z1

ω β1 = tan−1( ) p1

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Bode Diagram Straight-Line Amplitude Plots The amplitude of value of H(jω) in decibel is K0 1 + AdB = 20 log10

jω z1

jω ω 1+ p1

= 20log10 K0 + 20log10 1 +

jω jω − 20log10 ω − 20log10 1 + z1 p1

Some Results AdB

A

AdB

A

-3

0.7071

20

10.00

0

1

30

31.62

3

1.414

40

100.00

6

2

60

1000.00

10

3.16

80

10000.

15

5.62

100

100000.

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Bode Diagram Straight-Line Amplitude Plots • The plot of 20 log10K0 is a horizontal straight line because of K0 is not a function of ω. 20 log10K0 is positive for K0>1 20 log10K0 is zero for K0=1 20 log10K0 is negative for K0>1

Let say K0=2 AdB = 20log10 (2) = 6 dB

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Bode Diagram Straight-Line Amplitude Plots 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 20log10 1 +

jω → 0 dB as ω → 0 z1

For large values of ω, |1+jω/z1| is approximately ω/z1 20log10 1 +

ω  jω → 20log10   as ω → ∞ z1  z1 

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 ECE 307-8 6

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Bode Diagram • Let’s assume that z1=100 20log10 1 +

20log10 1 +

20log10 1 +

jω 100

jω = 0 as ω < 100 100

jω  ω  → 20log10   = 20 dB / decade as ω > 100 100  100 

>> w=0:10:10000; >> a=20*log10(abs(1+j*w/100)); >> semilogx(w,a) >> grid on >> ylabel ('A_{dB}') >> xlabel ('\omega (rad/s)')

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Bode Diagram Straight-Line Amplitude Plots 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

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Bode Diagram Straight-Line Amplitude Plots 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 −20log10 1 +

jω → 0dB as ω → 0 p1

For large values of ω, |1+jω/p1| is approximately ω/p1 −20log10 1 +

ω  jω → −20log10   , p1  p1 

as 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 9

Bode Diagram • Let’s assume that p1=200 −20log10 1 + −20log10 1 +

−20log10 1 +

jω 200

jω = 0 as ω < 200 200

jω  ω  → −20log10   = −20 dB/decade as ω > 200 200  200 

>> w=0:10:10000; >> a=-20*log10(abs(1+j*w/100)); >> semilogx(w,a) >> grid on >> ylabel ('A_{dB}') >> xlabel ('\omega (rad/s)')

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Bode Diagram Example jω jω jω 10 = 20log10 (2) + 20log10 1 + − 20log10 ω − 20log10 1 + jω 10 100 ω 1+ 100 2 1+

AdB = 20 log10

20log10 1 +

jω 10

20log10 (2)

−20log10 1 +

jω 100

−20log10 ω

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MatLab s   2 1+ 20s + 200 10   H (s ) = = s  s 2 + 100s  s 1 +   100  >> 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

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Bode Diagram L1 1

Example

C1 2

100 mH

10 mF

Vi R1

1Vac 0Vdc

11 ohms

+

a. Bode Plot Vo b. Find 20log10|H(jω)| at ω=50 rad/s and ω=1000 rad/s - c. v (t)=5cos(500t+150) find V (t) i o

0

Transfer function H(s) of the circuit H (s ) =

R s L R 1 s2 + s + L LC

Writing H(jω)

H (s ) =

110s 110s = s 2 + 110s + 1000 (s + 10)(s + 100) 0.11 jω

H ( jω ) = (1 + j

ω

10

)(1 + j

ω 100

)

In terms of dB AdB = 20log10 (0.11) + 20log10 jω − 20log10 1 + j

ω 10

− 20log10 1 + j

ω 100 ECE 307-8 13

20log10 jω

AdB 20log10 (0.11)

−20log10 1 + j

ω 10

−20log10 1 + j

ω 100

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Matlab AdB = 20log

110 jω ( jω + 10)( jω + 100)

>> 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}')

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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

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Bode Diagram b. 20log10|H(jω)| at ω=50 rad/s and ω=1000 rad/s 0.11 jω

H ( jω ) = (1 + j

ω

10

)(1 + j

ω 100

= )

ω =50

0.11( j 50) = 0.9648∠ − 12.25D 50 50 (1 + j )(1 + j ) 10 100

AdB = 20log10 H ( j 50) = 20log10 0.9648 = −0.31 dB 0.11 jω

H ( jω ) = (1 + j

ω

10

)(1 + j

ω 100

= )

ω =1000

0.11( j 1000) = 0.1094∠ − 83.72D 1000 1000 (1 + j )(1 + j ) 10 100

AdB = 20log10 H ( j1000) = 20log10 0.0.1094 = −12.22 dB

c. vi(t)=5cos(500t+150) find Vo(t) H ( j 500) =

ω = 500 rad / s

0.11( j 500) = 0.22∠ − 77.54D 500 500 (1 + j )(1 + j ) 10 100

v0 (t ) = 5 H ( j 500) cos(500t + 15D + ∠H ( j 500) = 5(.22)cos(500t + 15D − 77.54D )

From Bode AdB

ω =500

= −12.5 dB

H ( j 500) = 10 −12.5 / 20 = 0.24

= 1.1cos(500t − 62.54D ) ECE 307-8 17

Bode Diagram More Accurate Amplitude Plots Amplitude value at the corner frequency ω=1 of H(jω) AdB = ± 20log10 1 + jω AdB

ω =1

= ± 20log10 1 + j 1

AdB

ω =0.5

= ± 20log10 1 + j 0.5

AdB

ω =2

= ± 20log10 1 + j 2

= ± 20log10 2

= ± 20log10 5 / 4

= ± 20log10 5

= ± 3 dB

= ± 1dB

= ± 7 dB

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Bode Diagram Straight-Line Phase Angle Plots The Phase Angle of constant Ko is zero degree The phase angle of first order zero has two straight line 1. For values of ω= z1, 45 degree 2. For values of ω>=10 z1, it is straight line 90 degree 3. For values of ω<=0.1 z1, it is straight line 0 degree 4. For values of ω<=0.1 z1<= ω , it is straight line straight line having a slope of 45 degree/decade The phase angle of first order pole has two straight line For values of ω= p1, -45 degree 2. For values of ω>=10 p1, it is straight line -90 degree 3. For values of ω<=0.1 p1, it is straight line 0 degree 4. For values of ω<=0.1 p1<= ω , it is straight line straight line having a slope of -45 degree/decade The 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 19

jω 10 H ( jω ) = jω jω 1 + 100 2 1+

Straight-Line Phase Angle Plots

ω  tan−1    10 

 ω  − tan−1    100 

ω  −1 −1  ω   − tan (ω ) − tan  100   10   

θ ( jω ) = tan−1 

− tan−1 (ω )

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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)')

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L1

C1

1

2

Example

100 mH

10 mF

Vi R1

1Vac 0Vdc

11 ohms

+

a. Straight-line phase angle plot Vo b. Phase angle θ(ω) at ω=50 rad/s ω=500 rad/s, and ω=1000 rad/s - c. v (t)=5cos(500t+150) find V (t) i o

0

0.11 jω

H ( jω ) = (1 + j

ω

10

)(1 + j

ω 100

0.11 jω ∠90D

H ( jω ) = )

1+ j

ω 10

∠ tan−1(

ω 10

) 1+ j

ω 100

∠ tan−1(

ω 100

)

ω  −1  ω   − tan  100   10   

θ ( jω ) = 90D − tan−1 

= 90D − β1 − β 2 H ( j 50) =

H ( j 500) =

0.11 j 50 50 50 ) == 0.96∠ − 15.25D ∠90D − ∠ tan−1( ) − ∠ tan−1( 50 50 10 100 1+ j 1+ j 10 100 0.11 j 500 500 500 ) − ∠ tan−1( ) == 0.22∠ − 77.54D ∠90D − ∠ tan−1( 500 500 10 100 1+ j 1+ j 10 100 ECE 307-8 22

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Straight-line phase angle plot

90D

 ω  − tan−1    100 

ω  −1  ω   − tan  100   10   

θ ( jω ) = 90D − tan−1 

ω  − tan−1    10 

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