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Bode Plots
The Bode plot is a quick method to graphically evaluate the frequency dependence of
a circuit, once we know its “transfer function” G(s) (or equivalently G(ω)). Using simple
rules we can construct both a magnitude response and a phase response, describing
the behavior of the circuit in the sinusoidal steady state.
Let’s start with an example. Consider a simple low pass circuit:
R
Vin

Vout
G(s ) =

C

Vout(s)
1/RC
k
=
=
Vin(s)
s+1/RC s-p
k = 1/RC
p = -1/RC

The transfer function G(s) has a single “pole” p and a low frequency gain (when s is
small) of k/-p. We wish to plot both the magnitude and phase frequency response of
the circuit.
magnitude
phase

G( s) =

k

( s − p)
∠G(s) = −∠(s − p)

To find the frequency response, let s=jω. Then

and

(s − p) = ω 2 + p2
ω
∠(s − p) = arctan
.
 − p

The Bode plot plots the log magnitude and phase angle of the transfer function vs.
frequency, which is also on a log scale. Why log magnitude? Basically it is easier to
add graphically than to multiply or divide. The contribution from a single “pole” or
“zero” is expressed as an asymptotic approximation, and then the overall response is
found by simply summing the individual responses. What do the asymptotic
representations of (s − p) and ∠(s − p) look like?
Magnitude Response:

for ω << p, (s − p) ≅ p = constant
for ω >> p, (s − p) ≅ ω

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DLD 01/24/01

EE 216
Spring 2001

handout # 1

The contribution from a pole in dB is −20log( (s − p) ) = −20log( ω 2 + p2 ).
A pole at p=-1 has a magnitude response like the following.
p=1
0
-20 dB/decade
-10
-20
-30 -2
10

10

-1

0

10
Frequency (rad/sec)

10

1

10

2

2

2

The contribution from a zero in dB is +20log( (s − z) ) = +20log( ω + z ) .
A zero at z=-1 has a magnitude response like the following.
z=1

40

+20 dB/decade
20

0 -2
10

10

-1

0

10
Frequency (rad/sec)

10

1

10

2

ω
∠(s − p) = arctan
 − p
For negative, real poles or zeros (the ones we will see this semester in class),

Phase Response:

∠(s − p) ≅ 0 for ω < p /10
o

≅ 90 for ω > 10p
o

= 45o for ω = p
The phase contribution from a pole at p=-1 is shown below.

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DLD 01/24/01


EE 216
Spring 2001

handout # 1
p=1

0
-30

-45 deg/decade
-60
-90
10

-2

10

-1

0

10
Frequency (rad/sec)

10

1

10

2

The phase contribution from a zero at z=-1 is shown below.
z=1
-270
-300
+45 dB/decade

-330
-360
-390
10

-2

10

-1

0

10
Frequency (rad/sec)

Here's the punchline:
for G(s) =

G(s) dB =

10

1

10

2

k(s − z1)(s − z 2 )...(s − zn )
,
(s − p1)(s − p2 )...(s − pm )

∑ (s − z)

zeros

dB

− ∑ (s − p) dB + k dB
poles

∠G(s) = ∑ ∠(s − z) − ∑ ∠(s − p)
zeros

poles

Going back to our low pass circuit,

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DLD 01/24/01

EE 216
Spring 2001

handout # 1

1/RC
k
=
s +1/RC s - p
let's let R=1K and C=10 µF, so that k = p = -100. We can construct the Bode plot as
follows.
G(s) dB = k dB − (s + 100) dB
G(s) =

∠G(s) = −∠(s + 100)

50
k=100
0

total response
-20dB/dec
p=100

-50
10

0

10

1

2

10
Frequency (rad/sec)

10

3

10

4

0
-30
-45 deg/dec
-60
-90
10

0

10

1

2

10
Frequency (rad/sec)

10

3

10

4

An often useful alternative expression for the transfer function in terms of the dc gain
is
G(s) =

A o (s / z1 − 1)(s / z 2 − 1)...(s / z n − 1)
,
(s / p1 − 1)(s / p2 − 1)...(s / pm − 1)

Ao = kz1z2 ...z n / p1 p2...pm = dc gain.
Now the Bode plot is assembled using

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DLD 01/24/01


EE 216

Spring 2001

handout # 1
G(s) dB =

∑ (s / z − 1)

zeros

∠G(s) =

dB

−

∑ (s / p − 1) + A

o dB

poles

∑ ∠(s / z − 1) − ∑ ∠(s / p − 1)

zeros

poles

This has the advantage that all of the finite poles and zeros 'break' from the 0 dB axis.
The dc gain term can be incorporated either as a horizontal line representing the
appropriate gain, or by simply re-labeling the gain axis to take into account the gain at

dc. In this case, the poles and zeros all 'break' from the AodB axis. As an example,
consider a two pole transfer function specified by
G(s) =

100
,
(s + 1)(s + 20)

p1 = -1, p2 = -20, k = 100,A o = 5, Ao dB = 14
The Bode magnitude plot consists of a horizontal line at +14 dB, from which the
contrubution from the two poles at s=1 and s=20 break. When adding up the
responses we consider the line at +14 dB to be zero. Only after the composite curve
is drawn do we then think of the absolute gain being +14 dB at low frequencies. The
Bode plot looks like the following.
50
Ao=14dB
0

-20dB/dec

p=1

p=20
-40dB/dec

-50 -1
10

10

0

Frequency (rad/sec)

10

1

10

2

0
-45deg/dec
-90

-90deg/dec
-45deg/dec

-180
10

-1

10

0

Frequency (rad/sec)
-5-

10

1

10

2

DLD 01/24/01


EE 216
Spring 2001

handout # 1

If there are poles or zeros at zero frequency, then the dc gain is 0 (-∞ dB) and the first
representation of G(s) is perhaps the most convenient.
To summarize the rules for generating Bode plots:
Magnitude Plot
1) For each pole, draw a straight line with slope –20 dB/decade (-6 dB/octave)
intersecting the 0 dB axis at the pole frequency. This is the contribution for
that pole at frequencies greater than the pole frequency.
2) For each zero, draw a straight line with slope +20 dB/decade (+6 dB/octave)
intersecting the 0 dB axis at the zero frequency. This is the contribution for
that zero at frequencies greater than the zero frequency.
3) Sum the resulting partial plots to get the overall magnitude response
4) Set the vertical scale by choosing a convenient flat-band region and relabeling the dB axis to reflect the calculated gain in that region.
Phase Plot

1) For each pole, draw the angle response curve with slope –45o/decade over
two decades and centered at the pole frequency p. Below 0.1p the angle
response is 0o, and above 10p the angle response is –90o.
2) For each zero, draw the angle response curve with slope +45o/decade over
two decades and centered at the zero frequency z. Below 0.1z the angle
response is 0o, and above 10z the angle response is +90o.
3) Sum the individual responses to get the composite phase response.

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DLD 01/24/01


EE 216
Spring 2001

handout # 1

Now consider a high pass circuit:
s
s + 1/ RC
k = 1; k dB = 0;
z = 0; log(z) = −∞;
p = −1/ RC

C

G(s) =

Vin

Vout
R

How do we handle the zero at ω=0? The zero location is at -∞ on the log(ω) axis.
Similarly, the dc gain AodB is -∞. For this case we plot the magnitude response of the
zero as a line with slope +20 dB/decade with no break point. The angle response is a
constant +90o since for any finite frequency the zero will contribute +90o of phase.
We proceed as usual for the magnitude and angle plots, except for the absolute gain
of the magnitude plot. To fix the absolute gain we choose a convenient frequency,
typically in a region where the gain is flat, and evaluate the transfer function at that
point. We can adjust the axis labels to include this offset value.
For the high pass circuit, G(s) =
looks like this:

s
, let 1/RC = 100 as before. The Bode plot
s + 1/ RC

50

0

-50
10

0

10

1

2

10

3

0

10

1

2

10

3

10
Frequency (rad/sec)

-270
-300
-330
-360
-390

10

10
Frequency (rad/sec)
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DLD 01/24/01


EE 216
Spring 2001

handout # 1

Here are a few more examples of circuits and their Bode plots.
R1
Vin
Vout
C

s
R2
1
(
+ 1)
(s +
)
1/CR2
R1
R2
CR2

+
G(s) =
=
1
s
(s +
)
(
+ 1)
C(R1+ R2)
1/C(R1 + R2)

R2

let R1=9k
R2=1k
C= 1µF

then z=1000
p=100

20
0
-20
10

0

10

1

2

10
Frequency (rad/sec)

10

3

10

4

50
0
-50
10

0

10

1

2

10
Frequency (rad/sec)

-8-

10

3

10

4

DLD 01/24/01


EE 216
Spring 2001

handout # 1

C
Vin

R1
Vout

1
R=R1+R2

C
R2

G(s) =

If R1=1k R2=9k C=1µF
then p=100, z1=0, and z2=1000, and k=0.1

ks(s − z2 )
( s − p) 2

0
-20
-40
-60 0
10

10

1

2

3

2

3

10
10
Frequency (rad/sec)

10

4

10

5

-270
-360

10

0

10

1

10
10
Frequency (rad/sec)

-9-

10

4

10

5

DLD 01/24/01