Bode Diagramzaliyazici/ece307/Bode plot.pdf · 2 ECE 307-8 3 Bode Diagram= Let K o represent the...
Transcript of Bode Diagramzaliyazici/ece307/Bode plot.pdf · 2 ECE 307-8 3 Bode Diagram= Let K o represent the...
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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
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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
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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
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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)')
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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
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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
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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
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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
− −= ∠ − ∠ −∠ == ∠ −+ +
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ECE 307-8 23
Straight-line phase angle plot
90
1tan10ω− −
1tan100ω− −
1 1( ) 90 tan tan10 100
j ω ωθ ω − − = − −