Chapter 15 – Frequency Response and Bode Plots

Report
Chapter 15
EGR 272 – Circuit Theory II
1
Read: Ch. 15, Sect. 1-2; and Appendices D&E in Electric Circuits, 9th Edition
by Nilsson
Bode Plots
We have determined the frequency response for several 1st and 2nd order circuits.
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 logmagnitude and phase responses.
Illustration - A Bode plot is used to make a good estimate of the actual response.
20log|H(jw)|
Bode “straight-line” approximation
Actual log-magnitude response
(w on a log scale)
w(rad/s)
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Decibels
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V2
I2
P2
Note that the quantities
,
, and
are unitless quantities.
V1
I1
P1
However, when scaled logs of the quantities are taken, the unit of decibels (dB),
is assigned.
V2
20log10
 log-magnitude (LM) of the voltage gain in dB
V1
20log10
I2
 log-magnitude (LM) of the current gain in dB
I1
10log10
P2
 log-magnitude (LM) of the power gain in dB
P1
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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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:
H(s) 
K(s  z1 )(s  z2 )  (s  z N )
(s  p1 )(s  p2 )  (s  pM )
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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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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.
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.
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5 types of terms in H(jw)
1) K (a constant)
2)
w
1j
w1
(a zero)
3) jw (a zero) or
or
1
w
1  j
w1
(a pole)
1/jw (a pole)
2 w
w2
4) 1  j
(a complex zero)
2
2
w0
w0
or
1
2 w
w2
1  j
2
2
w0
w0
5) Any of the terms raised to a positive integer power.
For example,

w
1  j 
w1 

2
(a double zero)
Each term is now examined in detail.
(a complex pole)
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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:
(w)
LM (dB)
20log(K)
0o
0
1
10
100
w
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
w
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2. A) 1 + jw/w1 (a zero): The straight-line approximations are:
If H(jw)  1  j
w

w1
2
w
w
1    tan -1  
 w1 
 w1 
2 



w 
-1  w 

Then LM  20log 1    and  (w)  tan  

w1  
w1 




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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The Bode approximations (LM and phase) for 1 + jw/w1 are shown below.
(w)
LM
90o
20dB
slope
= + 6dB/oct 45o
= +20dB/dec
0dB
w
w1
10w1
0o
0.1w1
w1
slope
= +45 deg/dec
(for 2 decades)
w
10w1
Discuss the amount of error between the actual responses and the Bode
approximations.
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 . The slope is +45 deg/dec or
+13.5 deg/oct.
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Chapter 15
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1
2B)
w
1  j
w1
If H(jw) 
1
w
1  j
w1
(a pole): The straight-line approximations are:

10
2
w
-1  w 
1    tan  
 w1 
 w1 

w
  tan -1  
2
 w1 
w
1   
 w1 
1






1
-1  w 
Then LM  20log 
and

(w)

-tan

 
2
 w1 

w 
 1    
 w1  

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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1
The Bode approximations (LM and phase) for
LM
0dB
w1
10w1
(w)
0.1w1
0o
w
slope
= -20dB/dec
o
-45
= - 6dB/oct
-20dB
are shown below.
w
1  j
w1
w1
10w1
w
slope
= -45 deg/dec
(for 2 decades)
-90o
Discuss the amount of error between the actual responses and the Bode
approximations.
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 . The slope is -45 deg/dec or -13.5 deg/oct.
Chapter 15
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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Example: Sketch the LM and phase plots for the following transfer function.
100
H(s) 
s  2000
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LM: Working with slopes on Bode plots:
• If the LM has a slope of 20 dB/dec, the change in LM between two
frequencies can be calculated using:
 w final 
LM  20  log 

 w initial 
Example: Determine the value of LM2 below.
LM
40 dB
-20 dB/dec
LM2
45
725
w (rad/s)
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Phase: Working with slopes on Bode plots:
• If the phase has a slope of 45 deg/dec, the change in phase (in degrees)
between two frequencies can be calculated using:
 w final 
 Phase 45 log

w
 initial 
Example: Determine the value of 2 below.
 (deg)
90
-45 deg/dec
2
150
5200
w (rad/s)
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Example: Sketch the LM and phase plots for the following transfer function.
200(s  50)
H(s) 
s  400
(Solution provided on the following slide)
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Chapter 15
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200(s  50)
H(s) 
s  400
Example: Sketch the Bode LM and phase plots for:
Solution:
First find H(jw) and put H(jw) into "standard form"
200(jw  50) 25(1 + jw/50)
H(jw) 
=
(standard form)
jw  400
1 + jw/400
Log-magnitude:
The LM will start level since there are no jw terms and will begin at LM = 20log(25) = 28 dB
The LM will break up at w = 50 due to the zero.
The pole at w = 400 will cancel the effect of the zero causing the graph to level out.
Since 50 to 400 is three octaves and the slope is 6 dB/oct, the graph increases by 18 dB from 28 to 46 dB.
Phase:
The phase plots begins at 0 degrees since there are no jw terms.
The phase begins increasing with a slope of 45 /dec at w = 5 (a decade before the zero).
The phase levels out at w = 40 (a decade before the pole) as the decreasing slope due to the pole cancels the effect of the zero.
The phase begins decreasing at w = 500 (a decade after the zero) as the zero stops increasing.
The phase levels out at w = 4000 (a decade after the pole) as the pole stops decreasing.
Since a zero causes a net increase of 90 in phase and a pole causes a net decrease of 90 in phase, the graph will end at 0 .
(w)
LM
46 dB
o
90
slope
= +20dB/dec
28 dB
50
400
w
o
0
slope
= - 45 deg/dec
5
40
500
4000
w
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Example: Sketch the LM and phase plots on the 4-cycle semi-log graph paper
shown below for the following transfer function. (Pass out 2 sheets of graph
paper.)
2,000,000(
s  2,000)
H(s) 
(s  8,000)(s 50,000)
Log-magnitude (LM) plot:
Chapter 15
EGR 272 – Circuit Theory II
Example: (continued)
2,000,000(
s  2,000)
H(s) 
(s  8,000)(s 50,000)
Phase plot:
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3A) jw term in H(jw)
If H(jw) = jw = w/90
Then LM = 20log(w) and (w) = 90
Calculate 20log(w) for several values of w to
show that the graph is a straight line for all
frequency with a slope of 20dB/dec (or 6dB/oct).
The LM and phase for H(jw) = jw are shown below.
w
0.1
1
10
100
1,000
10,000
100,000
20log(w)
(w)
LM
40 dB
slope
= +20dB/dec
= + 6dB/oct
20 dB
0 dB
1
10
100
w
90o
0o
1
10
100
• So a jw (zero) term in H(jw) adds an upward slope of +20dB/dec (or
+6dB/oct) to the LM plot.
• And a jw (zero) term in H(jw) adds a constant 90 to the phase plot.
w
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3B) 1/(jw) term in H(jw)
If H(jw) = 1/(jw) = (1/w)/-90
Then LM = 20log(1/w) = -20log(w) and (w) = -90
A few calculations could easily show that the graph of 20log(1/w)is a
straight line for all frequency with a slope of -20dB/dec (or -6dB/oct).
The LM and phase for H(jw) = 1/(jw) are shown below.
(w)
LM
0 dB
-20 dB
10
100
w
o
10
100
0
slope
= -20dB/dec
= - 6dB/oct
-90o
-40 dB
• So a 1/(jw) (pole) term in H(jw) adds a downward slope of -20dB/dec (or 6dB/oct) to the LM plot.
• And a 1/(jw) (pole) term in H(jw) adds a constant -90 to the phase plot.
w
Chapter 15
EGR 272 – Circuit Theory II
Example: Sketch the LM and phase plots for the following transfer function.
200(s)
H(s) 
s  5000
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Chapter 15
EGR 272 – Circuit Theory II
Example: Sketch the LM and phase plots for the following transfer function.
500(s  500)
H(s) 
(s)(s  8000)
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Chapter 15
EGR 272 – Circuit Theory II
Calculation of exact points to check Bode Plots
Evaluating H(jw) at a particular value of w is helpful to check Bode Plots. An
example is shown below.
Example:
5000(s + 200)
Suppose H(s) =
(s + 4000)
w 

250 1 + j

200


so H(jw) =
w 

1
+
j


4000


Evaluating this function at w = 4000 rad/s yields:
4000 

250 1 + j
 250 1 + j20 
200

 =
H(jw) =
= 3539.9542.1
4000 

1 + j
1
+
j


4000 

so  (4000) = 42.1
LM(4000) = 20log(3539.95) = 70.98 dB
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Chapter 15
EGR 272 – Circuit Theory II
Example:
Evaluate the H(jw) on the last page at w = 500 rad/s and w = 8000 rad/s.
Compare the values with the Bode plots. Do they appear to be correct?
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Chapter 15
EGR 272 – Circuit Theory II
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5. Poles and zeros raised to an integer power in H(s)
In the last class it was demonstrated that terms in H(jw) are additive.
Therefore, a double terms (such as a pole or zero that is squared) simply acts
like two terms, a triple term acts like three terms, etc.
Illustration: Show that (1 + jw/w1)N results in the following responses:
LM plot:
Has a 0dB contribution before its break frequency
Will increase at a rate of 20NdB/dec after the break
There will be an error of 3NdB at the break between the Bode straight-line
approximation and the exact LM
Phase plot:
Has a 0 degree contribution until 1 decade before its break frequency
Will increase at a rate of 45Ndeg/dec for two decades (from 0.1w1 to 10w1).
The total final phase contribution will be 90N degrees.
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So (1 + jw/w1)N results in the following responses:
w
LM
90No
20N dB
0 dB
slope
= +20NdB/dec
= + 6NdB/oct
w1
10w1
w
45No
0o
0.1w1
w1
slope
= +45N deg/dec
(for 2 decades)
w
10w1
Chapter 15
EGR 272 – Circuit Theory II
Example: Sketch the LM plot for the following transfer function.
2x1012s2
H(s) 
(s  800)2 (s  2,000)2
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Chapter 15
EGR 272 – Circuit Theory II
Generating LM and phase plots using Excel and MATLAB:
Refer to the handout entitled “Frequency Response” which includes detailed
examples of creating LM and phase plots using Excel and MATLAB.
Generating LM and phase plots using PSPICE:
Refer to the handout entitled “PSPICE Example: Frequency Response (LogMagnitude and Phase)” which includes a detailed example of creating LM and
phase plots using PSPICE.
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Bode Plots - Recall that there are 5 types of terms in H(s). The first four have
already been covered.
5. Complex terms in H(s)
A second order term in H(s) with complex roots has the general form s2 +
2s + wo2 . After factoring out the constant has the wo2 , the corresponding
term in H(jw) is:
2 w
w2
1  j
(a 2nd-order pole or zero with complex roots)
2
2
w0
w0
Key point: The complex term above is identical to a double real
pole or a double real zero in the Bode straight-line approximation,
but they differ in the exact curves.
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Complex peaks and damping ratio
Recall that  (zeta) = damping ratio where  

wo
2 w
w2
2 w
w2
then H(jw)  1  j
is equal to H(jw)  1  j
2
2
2
w0
w0
w0
w0
and at the break frequency (w  w o ) this yields
H(jw)  j2
so LM  20log(2 ) which yields a peak whose size depends on the value of 
Exercise: Complete the table below showing the value of the complex peak
(for a complex zero)

1*
0.5
0.2
0.1
0.05
20log(2) = Value of complex peak
*  = 1 is not complex. It corresponds to a double-zero.
Chapter 15
EGR 272 – Circuit Theory II
The following graphs from the text (Electric Circuits, 9th Edition, by Nilsson)
below show the Bode straight-line approximation and the exact curve for
different values of  .
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EGR 272 – Circuit Theory II
Example: Sketch the LM responses for H1(s), which has a double zero, and
H2(s), which has a complex zero.
104
H1 (s) 
(s  100)2
104
H2 (s)  2
(s  20s  104 )
A) Find H1(jw) and sketch the LM response
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Chapter 15
EGR 272 – Circuit Theory II
Example: Sketch the LM responses for H1(s), which has a double zero, and
H2(s), which has a complex zero.
104
H1 (s) 
(s  100)2
104
H2 (s)  2
(s  20s  104 )
B) Find H2(jw) and sketch the LM response
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EGR 272 – Circuit Theory II
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Filter order
Unfortunately, we can’t build ideal filters. However, the higher the order of a
filter, the more closely it will approximate an ideal filter.
The order of a filter is equal to the degree of the denominator of H(s).
(Of course, H(s) must also have the correct form.)
LM
Ideal LPF
slope = 
H(s) =
K
(s + w C )
K
4th-order LPF
H(s) =
(s + w C )4
slope = -80dB/dec
3rd-order LPF
slope = -60dB/dec
H(s) =
2nd-order LPF
slope = -40dB/dec
K
(s + w C )3
H(s) =
K
(s + w C )2
K
1st-order LPF
H(s) =
(s + w C )
slope = -20dB/dec
wC
w
Chapter 15
EGR 272 – Circuit Theory II
Recognizing filter types from the transfer function
Low-pass filter (LPF) - Discuss the form of LM, H(jw), and H(s).
High-pass filter (HPF) - Discuss the form of LM, H(jw), and H(s).
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EGR 272 – Circuit Theory II
Band-pass filter (BPF) - Discuss the form of LM, H(jw), and H(s).
Band-stop filter (BSF) - Discuss the form of LM, H(jw), and H(s). Also
define a “notch” filter.
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EGR 272 – Circuit Theory II
Practical Filter Examples: Discuss a stereo tuner, 5-band equalizer, 60 Hz
interference block.
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Active and passive filters
Passive filter - a filter constructed using passive circuit elements (R’s, L’s, and C’s)
Active filter - a filter constructed using active circuit elements (primarily op amps)
Analysis of active filters
Active filters are fairly easy to analyze. Recall that a resistive inverting
amplifier circuit has the following relationship:
R2
ACL =
_
Vin
R1
Vo
+
Vo
R
=- 2
Vin
R1
Replacing the resistors with impedances yields a transfer function, H(s):
Z2
Vin(s)
Z1
_
+
H(s) =
Vo(s)
Vo ( s)
Z
=- 2
Vin ( s)
Z1
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Example: Find H(s) =Vo(s)/Vin(s) for the circuit below and sketch the LM
response.
10mF
200 W
_
Vin
Vo
10mF
+
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EGR 272 – Circuit Theory II
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Impedance and Frequency Scaling
Filter tables are commonly available that list component values for various
types of filters. However, the tables could not possibly list component values
for all possible cutoff frequencies. Typically, filter tables will list cutoff
frequencies of perhaps 1 Hz or 1 kHz. The user can then scale the cutoff
frequencies to the desired values using frequency scaling. Similarly,
component values may be too large or too small (for example, a filter might be
shown using 1 ohm resistors or 1F capacitors) so the components need to be
scaled using impedance scaling.
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Impedance Scaling – a procedure used to change the impedance of a circuit
without affecting its frequency response. Consider the series RLC circuit
shown below.
R
Z(s)
sL
1
sC
Note that Z(s)  R  sL 
R'  K Z  R
Show that to scale the impedance by a factor KZ :
L'  K Z  L
C' 
C
KZ
1
sC
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Frequency Scaling – a procedure used to scale all break frequencies in the
frequency response of a circuit. The frequency response is scaled without
affecting the circuit impedance (the impedance is the same at the original and
new break frequencies). Consider the series RLC circuit shown below.
+
Vi(s)
sL
1
sC
+
R
_
Vo(s)
_
As seen in an earlier class, H(s) is described below.
H(s) 
Vo (s)

Vi (s)
s2
R
  s
L
1
R
  s 
LC
L
Note that H(s) described above is the transfer function of a band-pass filter.
The LM is shown on the following page.
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LM of the original band-pass filter (center frequency = wo):
|H(jw)|
|H(jwo)|
w
wo
LM of the frequency-scaled band-pass filter (center frequency = KFwo):
|H(jw/KF)|
|H(jwo)|
w
wo
KFwo
Chapter 15
EGR 272 – Circuit Theory II
Show that to scale the frequency by a factor KF :
Development :
45
R'  R
L
L' 
KF
1
C
sC
C' 
-j
KF
Z(jw)  R  jwL 
wC
w
and now substituting
in place of w as indicatedon thepreviousgraph yields :
KF
If Z(s)  R  sL 
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Example: Impedance and frequency scaling.
A) Determine H(s) = Vo(s)/Vi(s)
B) Sketch the LM response
+
Vi(s)
_
+
1
1F
Vo(s)
_
Chapter 15
EGR 272 – Circuit Theory II
Example: (continued)
C) A cutoff frequency of 1 rad/s and component values of 1 ohm and 1 Farad
are not very useful. Scale the frequency so that the cutoff frequency is 500
rad/s. Also use a 1 kW resistor instead of a 1 W resistor.
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Chapter 15
EGR 272 – Circuit Theory II
Example: (continued)
D) Draw the new circuit. Determine the new H(s) and verify that the LM
response has been shifted as planned.
E) Calculate the circuit impedance at the break frequency for the new circuit
and compare it to the circuit impedance at the break frequency for the
original circuit.
48

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