300-04 - Bode Plots - University of Victoria

Bode Plots

 Bode plots are semilog plots of the magnitude (in decibels) and
phase (in degrees) of a transfer function versus frequency.

 Bode plots contain the same information as the non-logarithmic
plots discussed in the previous chapter, but Bode plots are much
easier to construct.

 The transfer function can be written as

H  He j  H 

 In a Bode magnitude (amplitude) plot, the gain is plotted in
decibels (dB) versus frequency:

HdB  20 log10  H 

 In a Bode phase plot,  is plotted in degrees versus frequency.

 Both magnitude and phase plots are made on semilog graph paper.

1

Semilog Graph Paper

2

Sketching A Log Frequency Scale

1.25 3.1 6.3 

 1.6 2.5 4 8 200

0.5 1 25 10 20 50 100
decade

Note: There is no point =0 on a logarithmic frequency scale !

3

Bode Plots

 A transfer function may be written in terms of factors that have real and
imaginary parts. One such representation might be

H   K  j 1 1 j / z1  1 j21 / k   j / k 2 
1 j / p1  1 j2 2 / n
   j / 2  

n

 In this particular case, H() has seven different factors that can appear in

various combinations in a transfer function. These are:

1. A gain K

2. A pole (j)−1 or zero (j) at the origin

3. A simple pole 1/(1 + j/p1) or zero (1 + j/z1)
4. A quadratic pole 1/[1 + j22/n + (j/n) 2] or

zero [1 + j21/k + (j/k) 2].  is called the damping factor.

4

Example

Find Zi(), poles and zeroes, n and .

1  j   1  j   1 j  1  j 
     
Zi   5 j  2 j 1.5 3 2 1.5  K 1 j2 z1  z2 
 j 2  4 j  5 1  2
j2  2      j  2   j 
 5      n 
5 5 n 

K  3, z1  2, z2  1.5. Quadratic Pole: damping factor   2 , n  5
5

zeros at s  j  2 and 1.5 same as the answer on p. 13 of "Transfer Functions"

Note that, z1 and z2 in this example are based on the transfer function definition in the previous page. 5

Bode Plots

 In constructing a Bode plot, we plot each factor separately and
then combine them graphically.

 The factors can be considered one at a time and then combined
additively because of the logarithms involved.

 It is this mathematical convenience of the logarithm that makes
Bode plots a powerful engineering tool.

 Straight-line plots of the factors listed above are known as Bode
plots.

 Bode plots approximate the actual plots to a surprising degree of
accuracy.

6

Bode Plots

 A decade is an interval between two frequencies with a ratio of 10;

e.g., between 0 and 100, or between 10 Hz and 100 Hz. Thus,

±20 dB/decade means that the magnitude changes ± 20 dB
whenever the frequency changes tenfold or one decade.

H    K  j , slope  20 dB/decade
K =? H   1  K
HdB  20 log10  K   0 dB  K=1

If H    K  j 2 , slope  40 dB/decade

If H     K   slope  20 dB/decade

j 7

Bode Plots

 Constant term
For the gain K, the magnitude is 20 log10 (K) and the phase is 0◦; both are
constant with frequency.
If K is negative, the magnitude is 20 log10 (|K|) but the phase is ±180◦.

Bode plots for gain K 8

Bode Plots

 Zero at the origin

For the zero (jω) at the origin, the
magnitude is 20 log10(ω) and the phase
is 90◦.

The slope of the magnitude plot is 20
dB/decade, while the phase is 90◦ and
constant with frequency.

The Bode plots for the pole 1/(jω) are
similar except that the slope of the
magnitude plot is −20 dB/decade while
the phase is −90◦.

In general, for (jω)N, where  is an Bode plots for a zero (j) at

integer, the magnitude plot will have a the origin
slope of 20N dB/decade, while the phase
is 90N degrees.

9

 Simple Zero Bode Plots

H    1 j / z1
     tan1  / z1 
   0  0
     900
   z1   450

Corner frequency Phase change occurs over 2 decades

Bode plots of zero (1 + j/z1) 10

Bode Plots

 Simple Pole

The Bode plots for the pole 1/(1 + jω/p1) are similar to those of the zero
(1 + jω/z1), except that the corner frequency is at ω = p1, the magnitude
has a slope of −20 dB/decade, and the phase has a slope of −45◦ per
decade.

Magnitude Phase

HdB  20 log10 1 1  20 log10 1 j / p1 H    1 1 / p1

j / p1 j

  0  HdB  20 log10 1  0       tan1  / p1 
    HdB  20 log10  / p1
   0  0

  p1  HdB  20 log10 1 j1  20 log10 2      900

 3    z1   450

11

Bode Plots

Quadratic pole:

The significant peaking in the
neighborhood of the corner
frequency should be added to the
straight-line approximation if a
high level of accuracy is desired
(see later).  (zeta) is called the
damping factor.

Bode (magnitude) plot of quadratic pole [1 + j2/n − 2/n2]−1 12

Bode Plots

Quadratic pole:

n 4.81

If   0.2, the corrected straight-line

phase plot is between

n 4.81 and n 4.81

n 4.81 13

Phase change occurs over 2 decades

Bode (phase) plot of quadratic pole [1 + j2/n − 2/n2]−1

Bode Plots

 Quadratic zero:

For the quadratic zero [1 + j2/n − 2/n2], the plots of the quadratic pole

are inverted because the magnitude plot has a slope of (+)40 dB/decade while
the phase plot has a slope of (+)90◦ per decade.

14

Gain Plots for Complex Critical Frequencies

 In the gain responses, the low-frequency asymptotes are of
unity gain, 0dB.

 The high-frequency asymptotes are 40dB/decade.

 The asymptotes intersect at =0=n.
 The gain in the neighborhood of 0=n is a strong function

of the damping ratio . But generally speaking, the

straight-line gain is within 3dB of the actual gain for

0.3<<0.7.

15

Gain Plots for Complex Critical Frequencies

 When  falls outside this range (<0.3), the actual gain at
0 and at a few frequency points around 0 can be

computed to improve the gain plot.

 Alternatively, the amplitude peak can be estimated from

| H (max ) |dB over SL  20 log10 2 1 dB

1 2

and occurs at

max  0 1 2 2

16

Bode Plots

 There is another procedure for obtaining Bode plots that is faster
and perhaps more efficient than the one we have just discussed.

 It consists in realizing that zeros cause an increase in slope, while
poles cause a decrease.

 By starting with the low-frequency asymptote of the Bode plot,
moving along the frequency axis, and increasing or decreasing the
slope at each corner frequency, one can sketch the Bode plot
immediately from the transfer function without the effort of
making individual plots and adding them.

 This procedure can be used once you become proficient in the one
discussed here.

17

Bode straight-line approximations for magnitude and phase

18

Bode straight-line approximations for magnitude and phase

19

Example E.1

Construct the Bode plots for this circuit.

H  s   Vo (s)  R 1  R Ls 1  110s
Vi (s) R  sL  sC R Ls  LC s2 110s 1000
s2

110s 110s
 s2 110s 1000  (s 10)(s 100)

H     j  110 j  100
j
10  

We first put H() in the standard form by dividing out the poles and zeros.

H    (1  j 0.11 j 10 0)  0.11 j  (1  1 10)  (1  1 100)
10)(1 j
j j

20

Example E.1 1
0.11 j 1
H    (1  j 10)(1 j 10 0)  0.11 j  (1  10)  (1 j 100)
j

Magnitude (amplitude) plot:

AdB  HdB    20 log(0.11)  20 log j  20 log 1 j /10  20 log 1 j /100

21

Phase plot: Example E.2

H    H    (1  j 0.11 j 10 0)  1 j 0.11 j 10 0 e j1 1 2 
10)(1 j 10 1 j

1    90o zero @ origin
1     tan1  10 simple pole
2     tan1  100 simple pole

22

Example 1

Construct the Bode plots for the transfer function

H     j 200 j  10
j
 2

We first put H() in the standard form by dividing out the poles and zeros.

H    1 j 10 j j /10  H    

/ 21

H    1 j 10 j j /10

/2 1

    90  tan1  / 2  tan1  /10

HdB    20 log(10)  20 log j  20 log 1 j / 2  20 log 1 j /10

23

HdB    20 log 10  20 log j  20 log 1 j / 2  20 log 1 j /10

20

0
1 10 100

90 24

1 10 100
-90

    90  tan1  / 2  tan1  /10

25

26

Example 2

Construct the Bode plots for the transfer function

H  5 j  2
j  j 10

We first put H() in the standard form by dividing out the poles and zeros.

H  1 j / 2  H   
j 1 j /10

H   1 j / 2
j 1 j /10

    tan1  / 2  90  tan1  /10

HdB    20 log 1 j / 2  20 log j  20 log 1 j /10

27

HdB    20 log j  20 log 1 j / 2  20 log 1 j /10

20

0
1 10 100

90 28

1 10 100
-90

    90  tan1  / 2  tan1  /10

29

30

Example 3

Draw the Bode plots for H s  s2  s 1
60s 100

H     j 2 1  j  100  H     
 j60

H   1 j  1 j
100 1
100 1 j 2  3     j 2 j2    j 2
  n  n 
10 10  

    tan 1    tan 1  2  ; where  3 and n  10
 12 

H dB    20 log 1 j  40  20 log 1 j2    j 2
n  n 
 

31

HdB    20 log 1 j  40  20 log 1 j2 / n   j / n 2

where   3 and n  10

0

1 10

-40

90

1 10

-90

    tan 1    tan 1  2 / n 
 12 / n2 
 

32

Why is there such large disagreement ?
Note: Here =3

33

34

Example 3 cont’d

Note: A smarter way to solve this problem is to realize that the
quadratic pole can be split into two simple poles:

Hs  s2  s 1  (s s 1
60s 100 1.715)(s  58.28)

Recall 1 j2 /n  j /n 2  1 2 s /n  s /n 2  0

j s

n 2  2ns  s2  0  s1,2  n  n 2  n 2

Alternatively s2  as  b  0  s1,2  a / 2  a / 22  b

If   1 or a / 22  b, then a quadratic pole can be

split into two simple poles

35

Example (low )

Calculate the straight-line and the actual gain response of the RLC
circuit for v(t)  1V cost

ZEQ (s)   sC  1 1  sL
 sL  s2LC 1

sL

H (s)  s2LC 1  sL  sL / R
sL s2RLC  R s2LC 1 sL
R  LC  sL / R

s2  1

jL / R jL / R 0 L / R
1 LC2  jL / R 1 ( / 0 )2  jL / R H () SL    
H ()    L / R  1
LC 2
 RC

Compare: 1   2  j2    0  1, 2 0  L
 0    LC R
   0 

  L 0  L 1 L/C  0.011  0.3 36
2R 2R LC 2R

Example (low )

PSpice schematic editor

37

Example (low )

0 dB 1.0V

-20 dB 100mV

-40 dB 10mV

Straight-line Approximation

-60 dB 1.0mV 100KHz 1.0MHz 10MHz
16KHz

V(V2)

Frequency

| H (max ) | dB over SL 20 log10 2 1 dB

1 2

=20 log10 2  0.011 1 dB  33.5dB
1 0.0112
38

Example E.3

Calculate the dB value above the straight line.

Make straight-line (uncorrected )Bode plots.
1
Vo (s) R  sC 1 sRC 1 s4 102
Vi (s) sRC  s2 s4 102  s2102
H  s    1  1  LC  1 
sC
R  sL 

 1 s 25 10)2  simple zero at z=25
1 0.4(s 10)  (s
 quadratic pole

H    1  0.4( 1 j 25 10)2 Compare denominator with

j 10)  ( j [1 + j2/n − 2/n2]

 n  10 , 2  0.4,   0.2

| H (max ) |dB over SL  20 log10 2 1 dB  20 log10 2  0.2 1 dB  8.14 dB
1 0.22
1 2

max  n 1 2 2  10 1 2 2  10 1 2  0.22 rad/s=9.59 rad/s 39

Example E.3 Phase plot

Magnitude (amplitude) plot

40

Example 4

 Given the following Bode plot, obtain the transfer function H(s)

  0 : H    K  j 

40dB

  1: H    K  40dB  K  1020dB  100

At   1, 5, and 20,

changes of  20dB/decade occur

H    1  100 j  j / 20
j 1 j / 51

  j 1  104 j  j  20

j  5

H(s)   s  1  104 s  s  20

s  5

41

Example 5

 Given the following Bode plot, obtain the transfer function H(s)

At   0, H    0dB 1

H    1  1  j / 0.5
j  j /102
1

2 j  0.5
 0.01 j 1 j 102

200 j  0.5
  j 1 j 102

H(s)   200s  0.5
s 1s 102

42

Example 6

Find the numerical expression for H(s)

dB

At   0, H    0dB  H0 (s)  1

At   1, change of +20dB/decade  H1(s)  1 s

At   10, change of  20dB/decade  H 2 (s)  1 1 10  1
s 1 0.1s

At   100, change of  20dB/decade  H3 (s)  1 1  1 1
s 100 0.01s

H(s)  H0 (s)H1(s)H2 (s)H3(s)  1 1 s 0.01s  43

0.1s  1 