EE 205 Dr. A. Zidouri
Electric Circuits II
Frequency Selective Circuits (Filters)
Bode Plots: Complex Poles and Zeros
Lecture #41
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EE 205 Dr. A. Zidouri
The material to be covered in this lecture is as follows:
o
o
o
o
Bode Diagrams for Complex Poles and Zeros
Straight-Line Amplitude Plots
Correcting Straight-Line Amplitude Plots
Phase Angle Plots
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EE 205 Dr. A. Zidouri
After finishing this lecture you should be able to:
¾
¾
¾
¾
Use Bode Diagrams for Complex Poles and Zeros
Draw Straight-Line Approximation of the Amplitude Plot for a Pair of Complex Poles (Zeros)
Draw Straight-Line Approximation of the Phase Angle Plot
Correct the plots for a Pair of Complex Poles (Zeros)
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EE 205 Dr. A. Zidouri
Bode Diagrams for Complex Poles and Zeros
The complex poles and zeros of H(s) always appear in conjugate pairs
Always combine the conjugate pair into a single quadratic term
Once the rules for handling poles is understood, their application to zeros becomes apparent.
K
H ( s) =
( s +α − jβ )( s +α + jβ )
The product ( s + α − j β )( s + α + j β ) can be written as:
Let’s consider
2
( s + α ) + β 2 = s2 + 2α s + α 2 + β 2
(41-1)
(41-2)
or s 2 + 2α s + α 2 + β 2 = s 2 + 2ζωn s + ωn2
(41-3)
where ωn2 = α 2 + β 2
(41-4)
and ζωn = α
(41-5)
The term n is the corner frequency of the quadratic factor
The term ζ is the damping coefficient of the quadratic factor,
™ if ζ < 1 the roots are complex
™ if ζ > 1 the roots are real,
™ if ζ = 1 this is the critical value.
For real roots we treat them as we have seen in previous lecture (Lec#40).
ω
Assume
ζ < 1 then
H ( s) =
K
s 2 + 2ζωn s + ωn2
(41-5)
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EE 205 Dr. A. Zidouri
H ( s) =
K
ωn2
H ( jω ) =
Let
1
s ⎞+⎛ s
1+ 2ζ ⎛⎜
⎞
⎜
⎟
⎝ ωn ⎠
⎟
⎝ ωn ⎠
(41-6) in standard form will be
2
K0
(41-8)
1− ⎜⎜ ω2 ⎟⎟ + j ⎜⎜ 2ζω ⎟⎟
⎛
⎞
⎛
⎝ ωn ⎠
⎝
u = ω then
ωn
From which:
2
ωn
⎞
0
2
AdB = 20log10 H ( jw)
( 2ζ u )
and θ (ω ) = −β1 = − tan −1
K
1
ωn2 1+ ⎛ s ⎞2 + 2ζ ⎛ s ⎞
⎜
⎟
⎜
⎟
⎝ ωn ⎠
⎝ ωn ⎠
(41-7)
K0 = K2
where
ωn
⎠
H ( jω ) = 1− u K+ j 2ζ u
⎛
2
−20log10 ⎜⎜ ⎛⎜1− u 2 ⎞⎟ +
⎝
⎠
⎝
H ( s) =
1
2 ⎞2
⎟
⎟
⎠
2ζ u
1− u 2
(41-9) in polar form
H ( jω ) =
⎛
2
= 20log10 K0 − 20log10 ⎜⎜ ⎛⎜1− u 2 ⎞⎟ +
⎝
⎠
⎝
⎡
K0
(41-10)
1− u 2 + j 2ζ u ∠β1
( 2ζ u )
1
2 ⎞2
⎟
⎟
⎠
(41-11)
⎤
= −10log10 ⎢u 4 + 2u 2 ⎛⎜ 2ζ 2 −1⎞⎟ +1⎥ (41-12)
⎣
⎝
⎠
⎦
(41-13)
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EE 205 Dr. A. Zidouri
Approximate Amplitude Plots
⎡
⎤
AdB = 20log10 K0 −10log10 ⎢u 4 + 2u 2 ⎛⎜ 2ζ 2 −1⎞⎟ +1⎥
⎝
⎣
Because
⎠
⎦
u = ω therefore u → 0 as ω → 0 and u → ∞ as ω → ∞ thus:
ωn
⎡
⎤
−10log10 ⎢u 4 + 2u 2 ⎛⎜ 2ζ 2 −1⎞⎟ +1⎥ → 0
⎝
⎠
as
u →0
⎣
⎦
⎡
⎤
−10log10 ⎢u 4 + 2u 2 ⎛⎜ 2ζ 2 −1⎞⎟ +1⎥ → −40log10 u
⎝
⎠
⎣
⎦
as
u →∞
We conclude that the approximate amplitude plot
consists of two straight lines.
o For frequencies ω < ωn
the straight line
has a slope of 0dB
o For frequencies ω > ωn
the straight line
has a slope of -40dB/decade.
These two straight lines join on the 0dB axis at u=1 or
ω = ωn . Fig. 41-1 shows the straight line
approximation for a quadratic factor with ζ < 1 .
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EE 205 Dr. A. Zidouri
Correcting Straight-Line Amplitude Plots:
Correcting the straight-line amplitude plot for a pair of complex poles is not as easy as correcting
a first-order real pole, because the corrections depend on the damping coefficient ζ
The straight-line amplitude plot can be corrected by locating four points on the actual curve:
o Point 1 at ½ the corner frequency
ω
ωC
=
1
2
⎡
⎤
= ω = u then ⎢u 4 + 2u 2 ⎛⎜ 2ζ 2 −1⎞⎟ +1⎥ = 1 + 1 ⎛⎜ 2ζ 2 −1⎞⎟ +1 = ζ 2 + 5
ωn
⎝
⎠
⎠
8
⎣
⎦ 8 2⎝
AdB (ωn 2) = −10log10 ⎛⎜ ζ 2 + 0.5625 ⎞⎟
⎝
⎠
o Point 2 at peak amplitude frequency
ω p = ωn 1− 2ζ 2 = u , thus A (ω p ) = −10log10 ⎛⎜ 4ζ 2 (1 − ζ ) ⎞⎟
⎝
⎠
ω
o Point 3 at the corner frequency
= u =1
ωn
A (ωn ) = −10log10 4ζ or A (ωn ) = −20log10 2ζ
2
The amplitude peaks at
dB
2
dB
dB
o Point 4 at zero amplitude frequency
ω = ωn 2 ⎛⎜1− 2ζ 2 ⎞⎟ = 2ω p
0
⎝
⎠
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EE 205 Dr. A. Zidouri
AdB
AdB 3
20
2
3
2
1
1
4
0
0
-1
-20
-2
-3
-4
-5
-40
-6
1
10
-7 0
10
n/2
p
n
0
Fig. 41-2 Four points on the corrected amplitude plot for a pair of complex poles
n/2
n
0
2
10
(rad/s)
p
Fig. 41-4 the amplitude plot for Example 41-1
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EE 205 Dr. A. Zidouri
Example 41-1
Compute the transfer function for the circuit shown in Fig. 41-3
a) Find the corner frequency ωn ,
b) The value of K0 ,
c) The damping coefficient ζ
d) Make a straight line amplitude plot ranging from 10 to 500
rad/s.
e) Sketch the actual corrected amplitude in dB at
ωn
2
, ω p , ωn ,and
ω0 .
1Ω
Vi
Vo
Fig. 41-3 The circuit for Example 41-1
f) Describe the type of filter represented by the circuit in Fig. 41-3
Solution:
1
2500
LC
Transforming to s domain H ( s ) =
= 2
s 2 + R s + 1 s + 20s + 2500
L LC
2
a) from the expression of H(s), ωn = 2500 , therefore ωn = 50 rad s
2500 or 1
b) By definition K =
0
ωn2
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EE 205 Dr. A. Zidouri
c) The coefficient of s is
2ζωn therefore ζ = 20 = 0.20
2ωn
d) See Fig. 41-4
e) The actual amplitudes are
AdB (ωn 2) = −10log10 ( 0.6025) = 2.2dB
ω p = 50 0.92 = 47.96 rad s
AdB (ω p ) = −10log10 ( 0.16)( 0.96) = 8.14dB
AdB (ωn ) = −20log10 ( 0.4) = 7.96dB
ω0 = 2ω p = 67.82dB
( )
AdB ω0 = 0dB
Fig. 41-4 shows the corrected plot.
f) From the plot we see that this filter is a LPF, its cutoff frequency appears to be 55rad/s almost the
same as that predicted by the straight-line approximation.
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EE 205 Dr. A. Zidouri
Phase Angle Plots
¾ The phase angle is zero at zero frequency
¾ The phase angle is -90o at corner frequency
¾ The phase angle is -180o at large frequency as
ω →∞
For small values of ζ , the phase angle changes rapidly in the vicinity of the corner frequency.
The line tangent to the phase angle curve at -90o has a slope of -2.3/ ζ rad/decade or -132/ ζ
degree/decade and it intersects the 0o and -180o lines at u1 = 4.81−ζ and u2 = 4.81ζ respectively.
Fig. 41-5 depicts the straight-line approximation for ζ = 0.3 and shows the actual phase angle plot.
0.62 = 4.81−ζ
1.6 = 4.81ζ
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EE 205 Dr. A. Zidouri
Self Test 41:
The numerical expression for a transfer function is
H (s) =
9 Compute the corner frequency,
9 the damping coefficient,
9 frequencies when H jω is unity,
25×108
s 2 + 20 ×103 s + 25×108
( )
9 peak amplitude of H ( jω ) in dB,
9 frequency at which the peak occurs, and
9 amplitude of H jω at half the corner frequency.
( )
Answer:
9 ωC
= 50 krad s ,
9 ζ = 0.2 ,
9 0 , 67.82 krad s
9 8.14dB,
9 47.96 krad
9 2.20dB
s
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