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Lecture #21 EGR 272 – Circuit Theory II
Bode Plots
We have seen that determining the frequency response for 1st and 2nd order circuits involved a significant amount of work. Using the same methods for higher order circuits would become very difficult. A new method will be introduced here, called the Bode plot, which will allow us to form accurate “straight-line” approximations for the log-magnitude and phase responses quite easily for even high-order transfer functions. This technique will also show how various types of terms in a transfer function affect the log-magnitude and phase responses
Illustration - A Bode plot is used to make a good estimate of the actual response.
Read: Chapter 14 in Electric Circuits, 6th Edition by Nilsson
w(rad/s)
Actual log-magnitude response
Bode “straight-line” approximation
(w on a log scale)
20log|H(jw)|
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Lecture #21 EGR 272 – Circuit Theory II
Decibels
2 2 2
1 1 1
V I PNote that the quantities , , and are unitless quantities.
V I P
However, when scaled logs of the quantities are taken, the unit of decibels (dB), is assigned.
210
1
210
1
210
1
V20log log-magnitude (LM) of the voltage gain in dB
V
I20log log-magnitude (LM) of the current gain in dB
I
P10log log-magnitude (LM) of the power gain in dB
P
There are two types of Bode plots:
• The Bode straight-line approximation to the log-magnitude (LM) plot, LM versus w (with w on a log scale)
• The Bode straight-line approximation to the phase plot, (w) versus w (with w on a log scale)
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Lecture #21 EGR 272 – Circuit Theory II
Standard form for H(jw):
Before drawing a Bode plot, it is necessary to find H(jw) and put it in “standard form.”Show the “standard form” for H(jw) below:
1 2 N
1 2 M
K(s z )(s z ) (s z )H(s)
(s p )(s p ) (s p )
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Lecture #21 EGR 272 – Circuit Theory II
Example:
Find H(jw) for H(s) shown below and put H(jw) in “standard form.”
10(s)(s + 200)H(s)
(s 1000)
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Lecture #21 EGR 272 – Circuit Theory II
Show how the LM and phase of each term in 20log|H(jw)| is additive (or acts separately).
Drawing Bode plots:To draw a Bode plot for any H(s), we need to:1) Recognize the different types of terms that can occur in H(s) (or H(jw))2) Learn how to draw the log-magnitude and phase plots for each type of term.
The additive effect of terms in H(jw):
The reason that Bode plot approximations are used with the log-magnitude is due to the fact that this makes individual terms in the LM additive. The phase is also additive.
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Lecture #21 EGR 272 – Circuit Theory II
5 types of terms in H(jw)1) K (a constant)
2) (a zero) or (a pole)
3) jw (a zero) or 1/jw (a pole)
4)
5) Any of the terms raised to a positive integer power.
Each term is now examined in detail.
1
w1 j
w
2
2 2 20 0
2 20 0
2 w w 11 j - (a complex zero) or (a complex pole)
2 w ww w 1 j - w w
1
1w
1 j w
2
1
wFor example, 1 j (a double zero)
w
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Lecture #21 EGR 272 – Circuit Theory II
1. Constant term in H(jw)If H(jw) = K = K/0 Then LM = 20log(K) and (w) = 0 , so the LM and phase responses are:
LM (dB)
w 0
w 0o
(w)
1
20log(K)
10 100
Summary: A constant in H(jw):• Adds a constant value to the LM graph (shifts the entire graph up or down)• Has no effect on the phase
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Lecture #21 EGR 272 – Circuit Theory II
2. A) 1 + jw/w1 (a zero): The straight-line approximations are:2
-1
1 1 1
2
-1
1 1
w w wIf H(jw) 1 j 1 tan
w w w
w wThen LM 20log 1 and (w) tan
w w
To determine the LM and phase responses, consider 3 ranges for w:1) w << w1
2) w >> w1
3) w = w1
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Lecture #21 EGR 272 – Circuit Theory II
So the Bode approximations (LM and phase) for 1 + jw/w1 are shown below.
Summary: A 1 + jw/w1 (zero) term in H(jw): • Causes an upward break at w = w1 in the LM plot. There is a 0dB effect before the
break and a slope of +20dB/dec or +6dB/oct after the break.• Adds 90 to the phase plot over a 2 decade range beginning a decade before w1 and
ending a decade after w1 .
LM
w 0dB
= +20dB/dec
w
90o
(w)
= + 6dB/oct
20dB
w1
slope
0o 10w1
45o
w1 10w1 0.1w1
= +45 deg/dec slope
(for 2 decades)
Discuss the amount of error between the actual responses and the Bode approximations.
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Lecture #21 EGR 272 – Circuit Theory II
2. B) (a pole): The straight-line approximations are:
To determine the LM and phase responses, consider 3 ranges for w:1) w << w1
2) w >> w1
3) w = w1
-1
2 21
-1
11 1 1
-1
21
1
1 1 0 1 wIf H(jw) tan
w w1 j w w w1 tan 1 w
w w w
1 wThen LM 20log and (w) -tan
ww1
w
1
1w
1 j w
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Lecture #21 EGR 272 – Circuit Theory II
So the Bode approximations (LM and phase) for are shown below.
1
1w
1 j w
LM w 0dB
= -20dB/dec
w
-90o
(w)
= - 6dB/oct -20dB
w1
slope 0o 10w1
-45o
w1 10w1 0.1w1
= -45 deg/dec slope
(for 2 decades)
Summary: A 1 + jw/w1 (zero) term in H(jw): • Causes an downward break at w = w1 in the LM plot. There is a 0dB effect before
the break and a slope of -20dB/dec or -6dB/oct after the break.• Adds -90 to the phase plot over a 2 decade range beginning a decade before w1
and ending a decade after w1 .
Discuss the amount of error between the actual responses and the Bode approximations.
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Lecture #21 EGR 272 – Circuit Theory II
Example: Sketch the LM and phase plots for the following transfer function.
H(s) 10(s 100)
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Lecture #21 EGR 272 – Circuit Theory II
Example: Sketch the LM and phase plots for the following transfer function.
100H(s)
s 2000
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Lecture #21 EGR 272 – Circuit Theory II
Example: Sketch the LM and phase plots for the following transfer function.
200(s 50)H(s)
s 400
