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We wish to plot both the magnitude and phase frequency response of the circuit.. The Bode plot plots the log magnitude and phase angle of the transfer function vs.. The dc gain term can

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:

Vout Vin

R

C

-1/RC

=

1/RC

=

RC 1

RC 1 Vin(s)

Vout(s)

= ) G(

p k

s-p

k /

s+

/

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

) ( ) (

p s

k s

G

−

= phase ∠G(s)= −∠(s− p)

To find the frequency response, let s=jω Then

(s−p) = ω2+ p2

and ∠(s−p)=arctan ω

−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

The contribution from a pole in dB is −20log( (s−p) )= −20log( ω2 + p2).

A pole at p=-1 has a magnitude response like the following

-30

-20

-10

0

Frequency (rad/sec)

-20 dB/decade p=1

The contribution from a zero in dB is +20log( (s−z) )= +20log( ω2 +z2)

A zero at z=-1 has a magnitude response like the following

0

20

40

Frequency (rad/sec)

z=1

+20 dB/decade

−p





 

 For negative, real poles or zeros (the ones we will see this semester in class),

∠(s−p)≅0o

for ω <p /10

≅90o

for ω >10p

=45o

for ω = p

The phase contribution from a pole at p=-1 is shown below

The phase contribution from a zero at z=-1 is shown below.

-270

-300

-330

-360

-390

Frequency (rad/sec)

z=1

+45 dB/decade

Here's the punchline:

for G(s)= k(s−z1)(s−z2) (s−z n)

(s− p1)(s−p2) (s− p m) ,

G(s)dB = (s−z)dB

zeros∑ − (s−p)dB

poles∑ +kdB

∠G(s)= ∠(s−z)

zeros∑ − ∠(s− p)

poles

∑

Going back to our low pass circuit,

-30

-60

-90

0

Frequency (rad/sec)

-45 deg/decade p=1

G(s) = 1/RC

s+1/RC = k

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 = kdB −(s+100)dB ∠G(s)= −∠(s+100)

An often useful alternative expression for the transfer function in terms of the dc gain is

G(s)=Ao(s/ z1−1)(s / z2−1) (s/ z n −1)

(s/ p1−1)(s / p2 −1) (s/ p m−1) ,

Ao =kz1z2 z n / p1p2 p m= dc gain

Now the Bode plot is assembled using

-50

0

50

Frequency (rad/sec)

k=100

p=100

total response

-30

-60

-90

0

Frequency (rad/sec)

-20dB/dec

-45 deg/dec

G(s)dB = (s/ z−1)dB

zeros∑ − (s / p−1)

poles∑ +AodB ∠G(s)= ∠(s/ z−1)

zeros∑ − ∠(s/ p−1)

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,Ao =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

0

50

Frequency (rad/sec)

-90

-180

0

p=1

p=20

Ao=14dB

-20dB/dec

-40dB/dec

-45deg/dec

-90deg/dec

-45deg/dec

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 re-labeling the dB axis to reflect the calculated gain in that region.

Phase Plot

two decades and centered at the pole frequency p Below 0.1p the angle

two decades and centered at the zero frequency z Below 0.1z the angle

3) Sum the individual responses to get the composite phase response.

Now consider a high pass circuit:

Vout Vin

R

C

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)= s

s+1/ RC, let 1/RC = 100 as before The Bode plot looks like this:

G(s)= s

s+1/ RC

k=1; kdB =0;

z=0; log(z)= −∞;

p= −1/ RC

-50

0

50

Frequency (rad/sec)

-270

-300

-330

-360

-390

let R1=9k then z=1000

R2=1k p=100

C= 1µF

-20

0

20

Frequency (rad/sec)

-50

0

50

Frequency (rad/sec)

Here are a few more examples of circuits and their Bode plots

Vout Vin

R1

C

R2 R1+R2(s+ 1

CR2)

C(R1+ R2))

s

1/CR2 +1)

1/C(R1+R2)+1)

R=R1+R2

C

Vout R1

C R2 1

ks(s−z2)

( s− p)2

then p=100, z1=0, and z2=1000, and k=0.1

-60

-40

-20

0

Frequency (rad/sec)

-270

-360

Frequency (rad/sec)