Transfer Function Review Overview of Bode...
Transcript of Transfer Function Review Overview of Bode...
Bode Plots
H(s)x(t) y(t)
• Bode plots are standard method of plotting the magnitude andphase of H(s)
• Both plots use a logarithmic scale for the x-axis
• Frequency is in units of radians/second (rad/s)
• The phase is plotted on a linear scale in degrees
• Magnitude is plotted on a linear scale in decibels
HdB(jω) � 20 log10 |H(jω)|
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Overview of Bode Plots
• Transfer function review
• Piece-wise linear approximations
• First-order terms
• Second-order terms (complex poles & zeros)
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Decibel Scales
It is important to become adept at translating between amplitude,|H(jω)|, and decibels, HdB(jω).
Amplitude (|H(jω)|) Decibels (20 log10 |H(jω)|)1 20 log10 1 =10 20 log10 10 =100 20 log10 100 =1000 20 log10 1000 =0.1 20 log10 0.1 =0.01 20 log10 0.01 =0.001 20 log10 0.001 =
12 20 log10
12 = -6.0206
2 20 log10 2 =√12 20 log10
√12 =
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Transfer Function Review
H(s)x(t) y(t)
Recall that if H(s) is known and
x(t) = A cos(ωt + φ),
then we can find the steady-state solution for y(t):
yss(t) = A|H(jω)| cos (ωt + φ + ∠H(jω))
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Example 1: Bode Plot
101
102
103
104
105
0
10
20
|H(j
ω)|
(dB
)
Active Lowpass RC Filter
101
102
103
104
105
100
120
140
160
180
∠ H
(jω
) (
degr
ees)
Frequency (rad/s)
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Example 1: Bode Plots
vo(t)
-
+vs(t)
RL
20 nF
1 kΩ10 kΩ
1. Find the transfer function of the circuit shown above.
2. Generate the bode plot.
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 5
Example 1: MATLAB Code
w = logspace(1,5,500);H = -50e3./(j*w + 5e3);
subplot(2,1,1);h = semilogx(w,20*log10(abs(H)));set(h,’LineWidth’,1.4);ylabel(’|H(j\omega)| (dB)’);title(’Active Lowpass RC Filter’);set(gca,’Box’,’Off’);grid on;set(gca,’YLim’,[-5 25]);
subplot(2,1,2);h = semilogx(w,angle(H)*180/pi);set(h,’LineWidth’,1.4);ylabel(’\angle H(j\omega) (degrees)’);set(gca,’Box’,’Off’);grid on;set(gca,’YLim’,[85 185]);
xlabel(’Frequency (rad/s)’);
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 8
Example 1: Workspace
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Example 2: Bode Plot
101
102
103
104
105
−2
0
2
4
6
8
|H(j
ω)|
(dB
)
Active Lead/Lag RC Filter
101
102
103
104
105
−180
−175
−170
−165
−160
∠ H
(jω
) (
degr
ees)
Frequency (rad/s)
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Example 2: Bode Plots
-
+vs(t)
RL
1 kΩ1 kΩ
1 μF2 μF
1. Find the transfer function of the circuit shown above.
2. Generate the bode plot.
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 9
Example 2: MATLAB Code
w = logspace(1,5,500);H = -2*(j*w+500)./(j*w + 1000);
subplot(2,1,1);h = semilogx(w,20*log10(abs(H)));set(h,’LineWidth’,1.4);ylabel(’|H(j\omega)| (dB)’);title(’Active Lead/Lag RC Filter’);set(gca,’Box’,’Off’);grid on;set(gca,’YLim’,[-2 8]);
subplot(2,1,2);h = semilogx(w,angle(H)*180/pi);set(h,’LineWidth’,1.4);ylabel(’\angle H(j\omega) (degrees)’);set(gca,’Box’,’Off’);grid on;set(gca,’YLim’,[-180 -160]);
xlabel(’Frequency (rad/s)’);
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 12
Example 2: Workspace
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 10
Magnitude Components
Consider the expression for the transfer function magnitude:
|HdB(jω)| = 20 log10 |H(jω)|
= 20 log10
∣∣∣∣∣k s±�(1 − sz1
) . . . (1 − szm
)(1 − s
p1) . . . (1 − s
pn)
∣∣∣∣∣s=jω
= 20 log10 |k| · |jω|±�|1 − jω
z1| . . . |1 − jω
zm|
|1 − jωp1| . . . |1 − jω
pn|
= 20 log10 |k| ± � 20 log10 ω
+20 log10
∣∣∣∣1 − jω
z1
∣∣∣∣ + · · · + 20 log10
∣∣∣∣1 − jω
zm
∣∣∣∣−20 log10
∣∣∣∣1 − jω
p1
∣∣∣∣ − · · · − 20 log10
∣∣∣∣1 − jω
pn
∣∣∣∣
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 15
Bode Plot Approximations
• Until recently (late 1980’s) bode plots were drawn by hand
• There were many rules-of-thumb, tables, and template plots tohelp
• Today engineers primarily use MATLAB, or the equivalent
• Why discuss the old method of plotting by hand?
– It is still important to understand how the poles, zeros, andgain influence the Bode plot
– These ideas are used for transfer function synthesis, analogcircuit design, and control systems
• We will discuss simplified methods of generating Bode plots
• Based on asymptotic approximations
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Magnitude Components Comments
|HdB(ω)| = 20 log10 |k| ± � 20 log10 ω
+20 log10
∣∣∣1 − jωz1
∣∣∣ + · · · + 20 log10
∣∣∣1 − jωzm
∣∣∣−20 log10
∣∣∣1 − jωp1
∣∣∣ − · · · − 20 log10
∣∣∣1 − jωpn
∣∣∣• Thus, |HdB(ω)| can be written as a sum of simple functions
• This is similar like using basis functions {δ(t),u(t),& r(t)} to writean expression for a piecewise linear signal
• We will use this approach to generate our piecewise linearapproximations of the bode plot
• Note that there are four types of components in this expression– Constant – Linear term– Zeros – Poles
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Alternate Transfer Function Expressions
There are many equivalent expressions for transfer functions.
H(s) =N(s)D(s)
=bmsm + bm−1s
m−1 + · · · + b1s + b0
ansn + an−1sn−1 + · · · + a1s + a0
=bm
ans±� (s − z1)(s − z2) . . . (s − zm)
(s − p1)(s − p2) . . . (s − pn)
= k s±�
(1 − s
z1
)(1 − s
z2
). . .
(1 − s
zm
)(1 − s
p1
) (1 − s
p2
). . .
(1 − s
pn
)
• This last expression is called standard form
• The first step in making bode plots is to convert H(s) to standardform
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Magnitude Components: Real Zeros
Consider two limiting conditions for a term containing a zero,20 log10
∣∣1 − jωz
∣∣First condition: ω � |z|
limωz →0
20 log10
∣∣1 − jωz
∣∣ = 0
Thus, if ω|z| � 1, then 20 log10
∣∣1 − jωz
∣∣ ≈ 0.
Second condition: ω � |z|lim
ωz →∞
20 log10
∣∣1 − jωz
∣∣ = 20 log10 | − jωz | = 20 log10 |ω| − 20 log10 |z|
Thus, if ω|z| � 1, then this term is linear (on a log scale) with a slope
of 20 dB per decade and an x-axis intercept at ω = |z|.
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Magnitude Components: Constant
20
0
-20
-40
(dB)
(rad/sec)
40
|H(jω)|
ω
The constant term, 20 log10 |k|, is a straight line on the Bode plot.
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Magnitude Components: Real Zeros Continued
20
0
-20
-40
(dB)
(rad/sec)
40
|H(jω)|
ω
Our piecewise approximation joins these two linear asymptoticapproximations at ω = |z|.Plot the piecewise approximation of the term 20 log10
∣∣1 − jωz
∣∣.
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Magnitude Components: Linear Term
20
0
-20
-40
(dB)
(rad/sec)
40
|H(jω)|
ω
The linear term, ±� 20 log10 |ω|, is a line on the magnitude plot with aslope equal to ±� 20 dB per decade.
The x-axis intercept occurs at ω = 1 rad/s.
Plot the bode magnitude plots for H(s) = s, 1s ,s2, 1
s2 .
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Magnitude Components: Real Poles Continued
10−2
10−1
100
101
102
−50
−40
−30
−20
−10
0
10 Bode Magnitude Real Pole: p = −1 rad/s
Mag
(dB
)
Frequency (rad/sec)
The approximation is least accurate at ω = |p|. The true magnitude is3 dB less than the approximation at this corner frequency.
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Magnitude Components: Real Zeros Continued 2
10−2
10−1
100
101
102
−10
0
10
20
30
40
50 Bode Magnitude Real Zero: z = ±1 rad/s
Mag
(dB
)
Frequency (rad/sec)
The approximation is least accurate at ω = |z|. The true magnitude is3 dB higher than the approximation at this corner frequency.
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Complex Poles & Zeros
• Complex poles and zeros require special attention
• Will discuss later
• You will not be expected to plot approximations with complexpoles or zeros on exams
• There are essentially 3 steps to generating piecewise linearapproximations of bode plots
1. Convert to standard form
2. Plot the components
3. Graphically add the components together
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Magnitude Components: Real Poles
20
0
-20
-40
(dB)
(rad/sec)
40
|H(jω)|
ω
• Consider two limiting conditions for a term containing a pole,
−20 log10
∣∣∣1 − jωp
∣∣∣• This is just the negative of the expression for a zero
• The piecewise approximation is the mirror image of that for a zero
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Example 3: Solution
10−1
100
101
102
103
104
105
−40
−20
0
20
40
60
Bode Magnitude Example 1
Mag
(dB
)
Frequency (rad/sec)
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Example 3: Magnitude Components
Draw the piecewise approximation of the bode magnitude plot for
H(s) =(s + 10)(s + 100)2
10s2(s + 1000)
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 25
Example 4: Magnitude Components
Draw the piecewise approximation of the bode magnitude plot for
H(s) =1011s(s + 100)
(s + 10)(s + 1000)(s + 10, 000)2
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 28
Example 3: Workspace
20
0
-20
-40
(dB)
(rad/sec)
40
60
-60
101 102 103100 104 105
|H(jω)|
ω
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Phase Components
H(jω) = k(jω)�(1 − jω
z1) . . . (1 − jω
zm)
(1 − jωp1
) . . . (1 − jωpn
)
Each of these terms can be expressed in polar form: a + jb = Aejθ.Note that (jω)� = ω�j� = ω�(ej π
2 )� = ω�ej π2 �
H(jω) = |k|ejηkπ (ω�ej�π2 )N1e
jθ1 . . . Nmejθm
D1ejφ1 . . . Dnejφn
= |k| |ω|�N1 . . . Nm
D1 . . . Dn×
exp(j(ηkπ + �π
2 + θ1 + · · · + θn − φ1 − · · · − φn))
where
ηk =
{0 k ≥ 01 k < 0
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 31
Example 4: Workspace
20
0
-20
-40
(dB)
(rad/sec)
40
60
-60
101 102 103100 104 105
|H(jω)|
ω
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Phase Components: Comments
∠H(jω) = ηkπ + �π2 + θ1 + · · · + θn − φ1 − · · · − φn
= ηkπ + �π2 + ∠
(1 − jω
z1
)+ · · · + ∠
(1 − jω
zm
)
−∠(
1 − jω
p1
)− · · · − ∠
(1 − jω
pn
)
• Thus the phase of H(jω) is also a linear sum of the phases due toeach component
• We will consider each of the four components in turn– Constant – Linear term– Zeros – Poles
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Example 4: Solution
100
102
104
106
−30
−20
−10
0
10
20
30
40
50
60
70
80 Bode Magnitude Example 2
Mag
(dB
)
Frequency (rad/sec)
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 30
Phase Components: Real Zeros
Consider three limiting conditions for a term containing a zero,∠1 − jω
z
First condition: ω � |z|limωz →0
∠(1 − jω
z
)= 0◦
Thus, if ω|z| � 1, then ∠
(1 − jω
z
) ≈ 0◦.
Second condition: ω = |z|∠
(1 − jω
z
)∣∣ω=|z| = ∠ (1 − jηz) = −ηz45◦
where ηz = sign(z)
Third condition: ω � |z|lim
ωz →∞
∠(1 − jω
z
)= ∠
(− jωz
)= −ηz90◦
Thus, if ω|z| � 1, then ∠
(1 − jω
z
) ≈ −ηz90◦.
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 35
Phase Components: Constant
0
(deg)
(rad/sec)
∠H(jω)
ω
180◦
90◦
−90◦
−180◦
The complex angle of the constant term, k, is either 0◦ if k > 0 or180◦ if k < 0.
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 33
Phase Components: Real Zeros Continued
0
(deg)
(rad/sec)
∠H(jω)
ω
180◦
90◦
−90◦
−180◦
Our piecewise approximation joins these three linear asymptoticapproximations at ω = 10−1|z| and ω = 10|z|.Plot the piecewise approximation of the term ∠
(1 − jω
z
). Assume
that z is in the left-hand plane (i.e. Re{z} < 0)
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 36
Phase Components: Linear Term
0
(deg)
(rad/sec)
∠H(jω)
ω
180◦
90◦
−90◦
−180◦
The linear term, ∠(jω)� = ∠j� = �× 90◦, is a constant multiple of 90◦
Plot the bode phase plots for H(s) = s, 1s ,s2, 1
s2 .
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 34
Phase Components: Real Zeros in Right Plane
10−2
10−1
100
101
102
−100
−90
−80
−70
−60
−50
−40
−30
−20
−10
0
10
Phas
e (d
eg)
Frequency (rad/sec)
The approximation is least accurate at ω = 0.1|z| and ω = 10|z|.
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 39
Phase Components: Real Zeros in Left Plane
10−2
10−1
100
101
102
−10
0
10
20
30
40
50
60
70
80
90
100
Phas
e (d
eg)
Frequency (rad/sec)
The approximation is least accurate at ω = 0.1|z| and ω = 10|z|.
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 37
Phase Components: Real Poles
Real poles in the left half plane have the same phase as real zeros inthe right half plane. We will only discuss poles in the left half planebecause only these systems are stable.
First condition: ω � |p|
limωp →0
−∠(1 − jω
p
)= −0◦
Second condition: ω = |p|
−∠(1 − jω
p
)∣∣∣ω=|p|
= −∠ (1 − sign(p) × j) = −∠ (1 + j) = −45◦
Third condition: ω � |p|
limωp →∞
−∠(1 − jω
p
)= −∠
(− jω
p
)= −∠j = −90◦
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 40
Phase Components: Real Zeros Continued 2
0
(deg)
(rad/sec)
∠H(jω)
ω
180◦
90◦
−90◦
−180◦
If the zero is in the right half plane (i.e. Re{z} > 0), then the phaseapproaches −90◦ asymptotically.
Plot the piecewise approximation of the term ∠(1 − jω
z
). Assume
that z is in the right half plane.
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 38
Example 5: Phase Components
0
(deg)
(rad/sec)101 102 103100 104 105
270
-270
∠H(jω)
180◦
90◦
−90◦
−180◦
Draw the piecewise approximation of the bode phase plot for
H(s) =(s + 10)(s + 100)2
10s2(s + 1000)
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 43
Phase Components: Real Poles
0
(deg)
(rad/sec)
∠H(jω)
ω
180◦
90◦
−90◦
−180◦
Plot the piecewise approximation of the term −∠(1 − jω
p
). Assume
that p is in the left-hand plane (i.e. Re{p} < 0)
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 41
Example 5: Solution
10−1
100
101
102
103
104
105
−150
−100
−50
0
50
Phas
e (d
eg)
Frequency (rad/sec)
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 44
Phase Components: Real Poles in Left Plane
10−2
10−1
100
101
102
−100
−90
−80
−70
−60
−50
−40
−30
−20
−10
0
10
Phas
e (d
eg)
Frequency (rad/sec)
The approximation is least accurate at ω = 0.1|p| and ω = 10|p|.
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 42
Example 7: Magnitude
20
0
-20
-40
(dB)
(rad/sec)
40
60
-60
101 102 103100 104 105
|H(jω)|
ω
Plot the magnitude of
H(s) = 10s + 10
s + 1000
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 47
Example 6: Phase Components
0
(deg)
(rad/sec)101 102 103100 104 105
270
-270
∠H(jω)
180◦
90◦
−90◦
−180◦
Draw the piecewise approximation of the bode phase plot for
H(s) =1011s(s + 100)
(s + 10)(s + 1000)(s + 10, 000)2
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 45
Example 7: Phase
0
(deg)
(rad/sec)101 102 103100 104 105
270
-270
∠H(jω)
180◦
90◦
−90◦
−180◦
Plot the phase of
H(s) = 10s + 10
s + 1000
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 48
Example 6: Solution
100
102
104
106
−150
−100
−50
0
50
100
Phas
e (d
eg)
Frequency (rad/sec)
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 46
Example 8: Phase
0
(deg)
(rad/sec)101 102 103100 104 105
270
-270
∠H(jω)
180◦
90◦
−90◦
−180◦
Plot the phase of
H(s) = −10s − 10
s + 1000
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 51
Example 7: Solution
10−1
100
101
102
103
104
105
−20
0
20
40 Bode Example 1
Mag
(dB
)
10−1
100
101
102
103
104
105
0
50
100
Phas
e (d
eg)
Frequency (rad/sec)
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 49
Example 8: Solution
10−1
100
101
102
103
104
105
−20
0
20
40 Bode Example 2
Mag
(dB
)
10−1
100
101
102
103
104
105
−150
−100
−50
0
Phas
e (d
eg)
Frequency (rad/sec)
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 52
Example 8: Magnitude
20
0
-20
-40
(dB)
(rad/sec)
40
60
-60
101 102 103100 104 105
|H(jω)|
ω
Plot the magnitude of
H(s) = −10s − 10
s + 1000
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 50
Example 9: Phase
0
(deg)
(rad/sec)101 102 103100 104 105
270
-270
∠H(jω)
180◦
90◦
−90◦
−180◦
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 55
Example 9: Circuit Example
vs(t) vo(t)
-
+
100 mH 10 mF
11 Ω
Draw the straight-line approximations of the transfer function for thecircuit shown above. Hint: Recall from one of the Transfer FunctionsExamples
H(s) =RL s
s2 + RL s + 1
LC
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 53
Example 9: Solution
10−1
100
101
102
103
104
−40
−20
0
20 Bode Example 3
Mag
(dB
)
10−1
100
101
102
103
104
−100
−50
0
50
100
Phas
e (d
eg)
Frequency (rad/sec)
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 56
Example 9: Magnitude
20
0
-20
-40
(dB)
(rad/sec)
40
60
-60
101 102 103100 104 105
|H(jω)|
ω
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 54
Complex Poles Magnitude
20 log10 |C(jω)| = 20 log10
∣∣∣∣∣1 −(
ω
ωn
)2
+jω
Qωn
∣∣∣∣∣−1
= −20 log10
√(1 − ω2
ω2n
)2
+(
ω
Qωn
)2
For ω � ωn,
20 log10 |C(jω)| ≈ −20 log10 |1| = 0 dB
For ω � ωn,
20 log10 |C(jω)| ≈ −20 log10
ω2
ω2n
= −40 log10
ω
ωndB
At these extremes, the behavior is identical to two real poles.
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 59
Complex Poles
Complex poles can be expressed in the following form:
C(s) =ω2
n
s2 + 2ζωns + ω2n
=1
1 + 2ζ sωn
+(
sωn
)2
• ωn is called the undamped natural frequency
• ζ (zeta) is called the damping ratio
• The poles are p1,2 = (−ζ ±√
ζ2 − 1) ωn
• If ζ ≥ 1, the poles are real
• If 0 < ζ < 1, the poles are complex
• If ζ = 0, the poles are imaginary: p1,2 = ±jωn
• If ζ < 0, the poles are in the right half plane (Re{p} > 0) and thesystem is unstable
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 57
Complex Poles Magnitude Continued
20 log10 |C(jω)| = −20 log10
√(1 − ω2
ω2n
)2
+(
ω
Qωn
)2
For ω = ωn,
20 log10 |C(jωn)| = −20 log10
1Q
= 20 log10 Q = QdB
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 60
Complex Poles Continued
The transfer function C(s) can also be expressed in the following form
C(s) =1
1 + 2ζ sωn
+(
sωn
)2 =1
1 + sQωn
+(
sωn
)2
where
Q � 12ζ
The meaning of Q, the Quality factor, will become clear in thefollowing slides.
C(jω) =1
1 −(
ωωn
)2
+ jωQωn
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 58
Complex Poles Phase
10−2
10−1
100
101
102
−180
−160
−140
−120
−100
−80
−60
−40
−20
0
Frequency (rad/sec)
Phas
e (d
eg)
Complex Poles
Q = 0.1Q = 0.5Q = 0.707Q = 1Q = 2Q = 10Q = 100
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 63
Complex Poles Magnitude
10−1
100
101
−60
−50
−40
−30
−20
−10
0
10
20
30
40
Frequency (rad/sec)
Mag
(dB
)
Complex Poles
Q = 0.1Q = 0.5Q = 0.707Q = 1Q = 2Q = 10Q = 100
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 61
Complex Zeros
• Left half plane:
– Inverted magnitude of complex poles
– Inverted phase of complex poles
• Right half plane:
– Inverted magnitude of complex poles
– Same phase of complex poles
• This is the same relationship real zeros had to real poles
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 64
Complex Poles Phase
∠C(jω) = ∠ 1
1 −(
ωωn
)2
+ jωQωn
For ω � ωn,∠C(jω) ≈ ∠1 = 0◦
For ω � ωn,
∠C(jω) ≈ ∠ 1− ω
Qωn
= ∠ − Qωn
ω= ∠ − 1 = −180◦
For ω = ωn,
∠C(jω) ≈ ∠ 1Qj
= ∠1j
= ∠ − j = −90◦
At the extremes, the behavior is identical to two real poles. At othervalues of ω near ωn, the behavior is more complicated.
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 62
Complex Poles Maximum
What is the frequency at which |C(jω)| is maximized?
C(jω) =1
1 +(
jωQωn
)+
(jωωn
)2
|C(jω)| =1√(
1 − ω2
ω2n
)2
+(
ωQωn
)2
• For high values of Q, the maximum of |C(jω)| > 1
• This is called peaking
• The largest Q before the onset of peaking is Q = 1√2≈ 0.707
• This curve is said to be maximally flat
• This is also called a Butterworth response
• In this case, |C(jωn)| = −3 dB and ωn is the cutoff frequency
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 67
Complex Zeros Magnitude
10−1
100
101
−40
−20
0
20
40
60
80
Frequency (rad/sec)
Mag
(dB
)
Complex Zeros
Q = 0.1Q = 0.5Q = 0.707Q = 1Q = 2Q = 10Q = 100
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 65
Complex Poles Maximum Continued
If Q > 0.707, the maximum magnitude and frequency are as follows:
ωr = ωn
√1 − 1
2Q2|C(jωr)| =
Q
1 − 14Q2
• ωr is called the resonant frequency or the damped naturalfrequency
• As Q → ∞, ωr → ωn
• For sufficiently large Q (say Q > 5)
– ωr ≈ ωn
– |C(jωr)| ≈ Q
• Peaked responses are useful in the synthesis of high-order filters
• Complex zeros (in the left half plane) have the inverted magnitudeand phase of complex poles
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 68
Complex Zeros Phase
10−2
10−1
100
101
102
0
20
40
60
80
100
120
140
160
180
Frequency (rad/sec)
Phas
e (d
eg)
Complex Zeros
Q = 0.1Q = 0.5Q = 0.707Q = 1Q = 2Q = 10Q = 100
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 66
Complex Poles Example Continued 2
102
103
104
105
106
−120
−100
−80
−60
−40
−20
0
20
40
Frequency (rad/sec)
Mag
(dB
)
Resonance Example
R = 5 ΩR = 50 ΩR = 707 ΩR = 1 kΩR = 5 kΩR = 50 kΩ
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 71
Complex Poles Example
vs(t) vo(t)
-
+
50 mH R
200 nF
Generate the bode plot for the circuit shown above.
H(s) =1
LC
s2 + RL s + 1
LC
=ω2
n
s2 + 2ζωns + ω2n
=ω2
n
s2 + ωn
Q s + ω2n
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 69
Complex Poles Example Continued 3
102
103
104
105
106
−180
−160
−140
−120
−100
−80
−60
−40
−20
0
Frequency (rad/sec)
Phas
e (d
eg)
Resonance Example
R = 5 ΩR = 50 ΩR = 707 ΩR = 1 kΩR = 5 kΩR = 50 kΩ
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 72
Complex Poles Example Continued 1
ωn =
√1
LC= 10 k rad/s
ζ =R
2L
√LC =
R
2
√CL = R × 0.001
Q =1√LC
L
R=
√L
C
1R
=500R
R = 5 Ω ζ = 0.005 Q = 100 Very Light Damping
R = 50 Ω ζ = 0.05 Q = 10 Light Damping
R = 707 Ω ζ = 1.41 Q = 0.707 Strong Damping
R = 1 kΩ ζ = 1 Q = 0.5 Critical Damping
R = 5 kΩ ζ = 5 Q = 0.1 Over Damping
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 70
Complex Poles Example Continued 4
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time (sec)
Out
put (
V)
Resonance Example
R = 5 ΩR = 50 ΩR = 707 ΩR = 1 kΩR = 5 kΩR = 50 kΩ
J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 73