transfer function review overview of bode...

Bode Plots

H(s)x(t) y(t)

• Bode plots are standard method of plotting the magnitude andphase of H(s)

• Both plots use a logarithmic scale for the x-axis

• Frequency is in units of radians/second (rad/s)

• The phase is plotted on a linear scale in degrees

• Magnitude is plotted on a linear scale in decibels

HdB(jω) � 20 log10 |H(jω)|

J. McNames Portland State University ECE 222 Bode Plots Ver. 1.19 3

Overview of Bode Plots

• Transfer function review

• Piece-wise linear approximations

• First-order terms

• Second-order terms (complex poles & zeros)


Decibel Scales

It is important to become adept at translating between amplitude,|H(jω)|, and decibels, HdB(jω).

Amplitude (|H(jω)|) Decibels (20 log10 |H(jω)|)1 20 log10 1 =10 20 log10 10 =100 20 log10 100 =1000 20 log10 1000 =0.1 20 log10 0.1 =0.01 20 log10 0.01 =0.001 20 log10 0.001 =

12 20 log10

12 = -6.0206

2 20 log10 2 =√12 20 log10

√12 =


Transfer Function Review

H(s)x(t) y(t)

Recall that if H(s) is known and

x(t) = A cos(ωt + φ),

then we can find the steady-state solution for y(t):

yss(t) = A|H(jω)| cos (ωt + φ + ∠H(jω))


Example 1: Bode Plot

101

102

103

104

105

0

10

20

|H(j

ω)|

(dB

)

Active Lowpass RC Filter

101

102

103

104

105

100

120

140

160

180

∠ H

(jω

) (

degr

ees)

Frequency (rad/s)


Example 1: Bode Plots

vo(t)

-

+vs(t)

RL

20 nF

1 kΩ10 kΩ

1. Find the transfer function of the circuit shown above.

2. Generate the bode plot.


Example 1: MATLAB Code

w = logspace(1,5,500);H = -50e3./(j*w + 5e3);

subplot(2,1,1);h = semilogx(w,20*log10(abs(H)));set(h,’LineWidth’,1.4);ylabel(’|H(j\omega)| (dB)’);title(’Active Lowpass RC Filter’);set(gca,’Box’,’Off’);grid on;set(gca,’YLim’,[-5 25]);

subplot(2,1,2);h = semilogx(w,angle(H)*180/pi);set(h,’LineWidth’,1.4);ylabel(’\angle H(j\omega) (degrees)’);set(gca,’Box’,’Off’);grid on;set(gca,’YLim’,[85 185]);

xlabel(’Frequency (rad/s)’);


Example 1: Workspace


Example 2: Bode Plot

101

102

103

104

105

−2

0

2

4

6

8

|H(j

ω)|

(dB

)

Active Lead/Lag RC Filter

101

102

103

104

105

−180

−175

−170

−165

−160

∠ H

(jω

) (

degr

ees)

Frequency (rad/s)


Example 2: Bode Plots

-

+vs(t)

RL

1 kΩ1 kΩ

1 μF2 μF

1. Find the transfer function of the circuit shown above.

2. Generate the bode plot.


Example 2: MATLAB Code

w = logspace(1,5,500);H = -2*(j*w+500)./(j*w + 1000);

subplot(2,1,1);h = semilogx(w,20*log10(abs(H)));set(h,’LineWidth’,1.4);ylabel(’|H(j\omega)| (dB)’);title(’Active Lead/Lag RC Filter’);set(gca,’Box’,’Off’);grid on;set(gca,’YLim’,[-2 8]);

subplot(2,1,2);h = semilogx(w,angle(H)*180/pi);set(h,’LineWidth’,1.4);ylabel(’\angle H(j\omega) (degrees)’);set(gca,’Box’,’Off’);grid on;set(gca,’YLim’,[-180 -160]);

xlabel(’Frequency (rad/s)’);




Magnitude Components

Consider the expression for the transfer function magnitude:

|HdB(jω)| = 20 log10 |H(jω)|

= 20 log10

∣∣∣∣∣k s±�(1 − sz1

) . . . (1 − szm

)(1 − s

p1) . . . (1 − s

pn)

∣∣∣∣∣s=jω

= 20 log10 |k| · |jω|±�|1 − jω

z1| . . . |1 − jω

zm|

|1 − jωp1| . . . |1 − jω

pn|

= 20 log10 |k| ± � 20 log10 ω

+20 log10

∣∣∣∣1 − jω

z1

∣∣∣∣ + · · · + 20 log10

∣∣∣∣1 − jω

zm

∣∣∣∣−20 log10

∣∣∣∣1 − jω

p1

∣∣∣∣ − · · · − 20 log10

∣∣∣∣1 − jω

pn

∣∣∣∣


Bode Plot Approximations

• Until recently (late 1980’s) bode plots were drawn by hand

• There were many rules-of-thumb, tables, and template plots tohelp

• Today engineers primarily use MATLAB, or the equivalent

• Why discuss the old method of plotting by hand?

– It is still important to understand how the poles, zeros, andgain influence the Bode plot

– These ideas are used for transfer function synthesis, analogcircuit design, and control systems

• We will discuss simplified methods of generating Bode plots

• Based on asymptotic approximations


Magnitude Components Comments

|HdB(ω)| = 20 log10 |k| ± � 20 log10 ω

+20 log10

∣∣∣1 − jωz1

∣∣∣ + · · · + 20 log10

∣∣∣1 − jωzm

∣∣∣−20 log10

∣∣∣1 − jωp1

∣∣∣ − · · · − 20 log10

∣∣∣1 − jωpn

∣∣∣• Thus, |HdB(ω)| can be written as a sum of simple functions

• This is similar like using basis functions {δ(t),u(t),& r(t)} to writean expression for a piecewise linear signal

• We will use this approach to generate our piecewise linearapproximations of the bode plot

• Note that there are four types of components in this expression– Constant – Linear term– Zeros – Poles


Alternate Transfer Function Expressions

There are many equivalent expressions for transfer functions.

H(s) =N(s)D(s)

=bmsm + bm−1s

m−1 + · · · + b1s + b0

ansn + an−1sn−1 + · · · + a1s + a0

=bm

ans±� (s − z1)(s − z2) . . . (s − zm)

(s − p1)(s − p2) . . . (s − pn)

= k s±�

(1 − s

z1

)(1 − s

z2

). . .

(1 − s

zm

)(1 − s

p1

) (1 − s

p2

). . .

(1 − s

pn

)

• This last expression is called standard form

• The first step in making bode plots is to convert H(s) to standardform


Magnitude Components: Real Zeros

Consider two limiting conditions for a term containing a zero,20 log10

∣∣1 − jωz

∣∣First condition: ω � |z|

limωz →0

20 log10

∣∣1 − jωz

∣∣ = 0

Thus, if ω|z| � 1, then 20 log10

∣∣1 − jωz

∣∣ ≈ 0.

Second condition: ω � |z|lim

ωz →∞

20 log10

∣∣1 − jωz

∣∣ = 20 log10 | − jωz | = 20 log10 |ω| − 20 log10 |z|

Thus, if ω|z| � 1, then this term is linear (on a log scale) with a slope

of 20 dB per decade and an x-axis intercept at ω = |z|.


Magnitude Components: Constant

20

0

-20

-40

(dB)

(rad/sec)

40

|H(jω)|

ω

The constant term, 20 log10 |k|, is a straight line on the Bode plot.


Magnitude Components: Real Zeros Continued

20

0

-20

-40

(dB)

(rad/sec)

40

|H(jω)|

ω

Our piecewise approximation joins these two linear asymptoticapproximations at ω = |z|.Plot the piecewise approximation of the term 20 log10

∣∣1 − jωz

∣∣.


Magnitude Components: Linear Term

20

0

-20

-40

(dB)

(rad/sec)

40

|H(jω)|

ω

The linear term, ±� 20 log10 |ω|, is a line on the magnitude plot with aslope equal to ±� 20 dB per decade.

The x-axis intercept occurs at ω = 1 rad/s.

Plot the bode magnitude plots for H(s) = s, 1s ,s2, 1

s2 .


Magnitude Components: Real Poles Continued

10−2

10−1

100

101

102

−50

−40

−30

−20

−10

0

10 Bode Magnitude Real Pole: p = −1 rad/s

Mag

(dB

)

Frequency (rad/sec)

The approximation is least accurate at ω = |p|. The true magnitude is3 dB less than the approximation at this corner frequency.


Magnitude Components: Real Zeros Continued 2

10−2

10−1

100

101

102

−10

0

10

20

30

40

50 Bode Magnitude Real Zero: z = ±1 rad/s

Mag

(dB

)

Frequency (rad/sec)

The approximation is least accurate at ω = |z|. The true magnitude is3 dB higher than the approximation at this corner frequency.


Complex Poles & Zeros

• Complex poles and zeros require special attention

• Will discuss later

• You will not be expected to plot approximations with complexpoles or zeros on exams

• There are essentially 3 steps to generating piecewise linearapproximations of bode plots

1. Convert to standard form

2. Plot the components

3. Graphically add the components together


Magnitude Components: Real Poles

20

0

-20

-40

(dB)

(rad/sec)

40

|H(jω)|

ω

• Consider two limiting conditions for a term containing a pole,

−20 log10

∣∣∣1 − jωp

∣∣∣• This is just the negative of the expression for a zero

• The piecewise approximation is the mirror image of that for a zero


Example 3: Solution

10−1

100

101

102

103

104

105

−40

−20

0

20

40

60

Bode Magnitude Example 1

Mag

(dB

)

Frequency (rad/sec)


Example 3: Magnitude Components

Draw the piecewise approximation of the bode magnitude plot for

H(s) =(s + 10)(s + 100)2

10s2(s + 1000)


Example 4: Magnitude Components

Draw the piecewise approximation of the bode magnitude plot for

H(s) =1011s(s + 100)

(s + 10)(s + 1000)(s + 10, 000)2



20

0

-20

-40

(dB)

(rad/sec)

40

60

-60

101 102 103100 104 105

|H(jω)|

ω


Phase Components

H(jω) = k(jω)�(1 − jω

z1) . . . (1 − jω

zm)

(1 − jωp1

) . . . (1 − jωpn

)

Each of these terms can be expressed in polar form: a + jb = Aejθ.Note that (jω)� = ω�j� = ω�(ej π

2 )� = ω�ej π2 �

H(jω) = |k|ejηkπ (ω�ej�π2 )N1e

jθ1 . . . Nmejθm

D1ejφ1 . . . Dnejφn

= |k| |ω|�N1 . . . Nm

D1 . . . Dn×

exp(j(ηkπ + �π

2 + θ1 + · · · + θn − φ1 − · · · − φn))

where

ηk =

{0 k ≥ 01 k < 0



20

0

-20

-40

(dB)

(rad/sec)

40

60

-60

101 102 103100 104 105

|H(jω)|

ω


Phase Components: Comments

∠H(jω) = ηkπ + �π2 + θ1 + · · · + θn − φ1 − · · · − φn

= ηkπ + �π2 + ∠

(1 − jω

z1

)+ · · · + ∠

(1 − jω

zm

)

−∠(

1 − jω

p1

)− · · · − ∠

(1 − jω

pn

)

• Thus the phase of H(jω) is also a linear sum of the phases due toeach component

• We will consider each of the four components in turn– Constant – Linear term– Zeros – Poles


Example 4: Solution

100

102

104

106

−30

−20

−10

0

10

20

30

40

50

60

70

80 Bode Magnitude Example 2

Mag

(dB

)

Frequency (rad/sec)


Phase Components: Real Zeros

Consider three limiting conditions for a term containing a zero,∠1 − jω

z

First condition: ω � |z|limωz →0

∠(1 − jω

z

)= 0◦

Thus, if ω|z| � 1, then ∠

(1 − jω

z

) ≈ 0◦.

Second condition: ω = |z|∠

(1 − jω

z

)∣∣ω=|z| = ∠ (1 − jηz) = −ηz45◦

where ηz = sign(z)

Third condition: ω � |z|lim

ωz →∞

∠(1 − jω

z

)= ∠

(− jωz

)= −ηz90◦

Thus, if ω|z| � 1, then ∠

(1 − jω

z

) ≈ −ηz90◦.


Phase Components: Constant

0

(deg)

(rad/sec)

∠H(jω)

ω

180◦

90◦

−90◦

−180◦

The complex angle of the constant term, k, is either 0◦ if k > 0 or180◦ if k < 0.


Phase Components: Real Zeros Continued

0

(deg)

(rad/sec)

∠H(jω)

ω

180◦

90◦

−90◦

−180◦

Our piecewise approximation joins these three linear asymptoticapproximations at ω = 10−1|z| and ω = 10|z|.Plot the piecewise approximation of the term ∠

(1 − jω

z

). Assume

that z is in the left-hand plane (i.e. Re{z} < 0)


Phase Components: Linear Term

0

(deg)

(rad/sec)

∠H(jω)

ω

180◦

90◦

−90◦

−180◦

The linear term, ∠(jω)� = ∠j� = �× 90◦, is a constant multiple of 90◦

Plot the bode phase plots for H(s) = s, 1s ,s2, 1

s2 .


Phase Components: Real Zeros in Right Plane

10−2

10−1

100

101

102

−100

−90

−80

−70

−60

−50

−40

−30

−20

−10

0

10

Phas

e (d

eg)

Frequency (rad/sec)

The approximation is least accurate at ω = 0.1|z| and ω = 10|z|.


Phase Components: Real Zeros in Left Plane

10−2

10−1

100

101

102

−10

0

10

20

30

40

50

60

70

80

90

100

Phas

e (d

eg)

Frequency (rad/sec)

The approximation is least accurate at ω = 0.1|z| and ω = 10|z|.


Phase Components: Real Poles

Real poles in the left half plane have the same phase as real zeros inthe right half plane. We will only discuss poles in the left half planebecause only these systems are stable.

First condition: ω � |p|

limωp →0

−∠(1 − jω

p

)= −0◦

Second condition: ω = |p|

−∠(1 − jω

p

)∣∣∣ω=|p|

= −∠ (1 − sign(p) × j) = −∠ (1 + j) = −45◦

Third condition: ω � |p|

limωp →∞

−∠(1 − jω

p

)= −∠

(− jω

p

)= −∠j = −90◦


Phase Components: Real Zeros Continued 2

0

(deg)

(rad/sec)

∠H(jω)

ω

180◦

90◦

−90◦

−180◦

If the zero is in the right half plane (i.e. Re{z} > 0), then the phaseapproaches −90◦ asymptotically.

Plot the piecewise approximation of the term ∠(1 − jω

z

). Assume

that z is in the right half plane.


Example 5: Phase Components

0

(deg)

(rad/sec)101 102 103100 104 105

270

-270

∠H(jω)

180◦

90◦

−90◦

−180◦

Draw the piecewise approximation of the bode phase plot for

H(s) =(s + 10)(s + 100)2

10s2(s + 1000)


Phase Components: Real Poles

0

(deg)

(rad/sec)

∠H(jω)

ω

180◦

90◦

−90◦

−180◦

Plot the piecewise approximation of the term −∠(1 − jω

p

). Assume

that p is in the left-hand plane (i.e. Re{p} < 0)


Example 5: Solution

10−1

100

101

102

103

104

105

−150

−100

−50

0

50

Phas

e (d

eg)

Frequency (rad/sec)


Phase Components: Real Poles in Left Plane

10−2

10−1

100

101

102

−100

−90

−80

−70

−60

−50

−40

−30

−20

−10

0

10

Phas

e (d

eg)

Frequency (rad/sec)

The approximation is least accurate at ω = 0.1|p| and ω = 10|p|.


Example 7: Magnitude

20

0

-20

-40

(dB)

(rad/sec)

40

60

-60

101 102 103100 104 105

|H(jω)|

ω

Plot the magnitude of

H(s) = 10s + 10

s + 1000


Example 6: Phase Components

0

(deg)

(rad/sec)101 102 103100 104 105

270

-270

∠H(jω)

180◦

90◦

−90◦

−180◦

Draw the piecewise approximation of the bode phase plot for

H(s) =1011s(s + 100)

(s + 10)(s + 1000)(s + 10, 000)2


Example 7: Phase

0

(deg)

(rad/sec)101 102 103100 104 105

270

-270

∠H(jω)

180◦

90◦

−90◦

−180◦

Plot the phase of

H(s) = 10s + 10

s + 1000


Example 6: Solution

100

102

104

106

−150

−100

−50

0

50

100

Phas

e (d

eg)

Frequency (rad/sec)


Example 8: Phase

0

(deg)

(rad/sec)101 102 103100 104 105

270

-270

∠H(jω)

180◦

90◦

−90◦

−180◦

Plot the phase of

H(s) = −10s − 10

s + 1000


Example 7: Solution

10−1

100

101

102

103

104

105

−20

0

20

40 Bode Example 1

Mag

(dB

)

10−1

100

101

102

103

104

105

0

50

100

Phas

e (d

eg)

Frequency (rad/sec)


Example 8: Solution

10−1

100

101

102

103

104

105

−20

0

20

40 Bode Example 2

Mag

(dB

)

10−1

100

101

102

103

104

105

−150

−100

−50

0

Phas

e (d

eg)

Frequency (rad/sec)



20

0

-20

-40

(dB)

(rad/sec)

40

60

-60

101 102 103100 104 105

|H(jω)|

ω

Plot the magnitude of

H(s) = −10s − 10

s + 1000


Example 9: Phase

0

(deg)

(rad/sec)101 102 103100 104 105

270

-270

∠H(jω)

180◦

90◦

−90◦

−180◦


Example 9: Circuit Example

vs(t) vo(t)

-

+

100 mH 10 mF

11 Ω

Draw the straight-line approximations of the transfer function for thecircuit shown above. Hint: Recall from one of the Transfer FunctionsExamples

H(s) =RL s

s2 + RL s + 1

LC


Example 9: Solution

10−1

100

101

102

103

104

−40

−20

0

20 Bode Example 3

Mag

(dB

)

10−1

100

101

102

103

104

−100

−50

0

50

100

Phas

e (d

eg)

Frequency (rad/sec)



20

0

-20

-40

(dB)

(rad/sec)

40

60

-60

101 102 103100 104 105

|H(jω)|

ω


Complex Poles Magnitude

20 log10 |C(jω)| = 20 log10

∣∣∣∣∣1 −(

ω

ωn

)2

+jω

Qωn

∣∣∣∣∣−1

= −20 log10

√(1 − ω2

ω2n

)2

+(

ω

Qωn

)2

For ω � ωn,

20 log10 |C(jω)| ≈ −20 log10 |1| = 0 dB

For ω � ωn,

20 log10 |C(jω)| ≈ −20 log10

ω2

ω2n

= −40 log10

ω

ωndB

At these extremes, the behavior is identical to two real poles.


Complex Poles

Complex poles can be expressed in the following form:

C(s) =ω2

n

s2 + 2ζωns + ω2n

=1

1 + 2ζ sωn

+(

sωn

)2

• ωn is called the undamped natural frequency

• ζ (zeta) is called the damping ratio

• The poles are p1,2 = (−ζ ±√

ζ2 − 1) ωn

• If ζ ≥ 1, the poles are real

• If 0 < ζ < 1, the poles are complex

• If ζ = 0, the poles are imaginary: p1,2 = ±jωn

• If ζ < 0, the poles are in the right half plane (Re{p} > 0) and thesystem is unstable


Complex Poles Magnitude Continued

20 log10 |C(jω)| = −20 log10

√(1 − ω2

ω2n

)2

+(

ω

Qωn

)2

For ω = ωn,

20 log10 |C(jωn)| = −20 log10

1Q

= 20 log10 Q = QdB


Complex Poles Continued

The transfer function C(s) can also be expressed in the following form

C(s) =1

1 + 2ζ sωn

+(

sωn

)2 =1

1 + sQωn

+(

sωn

)2

where

Q � 12ζ

The meaning of Q, the Quality factor, will become clear in thefollowing slides.

C(jω) =1

1 −(

ωωn

)2

+ jωQωn


Complex Poles Phase

10−2

10−1

100

101

102

−180

−160

−140

−120

−100

−80

−60

−40

−20

0

Frequency (rad/sec)

Phas

e (d

eg)

Complex Poles

Q = 0.1Q = 0.5Q = 0.707Q = 1Q = 2Q = 10Q = 100


Complex Poles Magnitude

10−1

100

101

−60

−50

−40

−30

−20

−10

0

10

20

30

40

Frequency (rad/sec)

Mag

(dB

)

Complex Poles

Q = 0.1Q = 0.5Q = 0.707Q = 1Q = 2Q = 10Q = 100


Complex Zeros

• Left half plane:

– Inverted magnitude of complex poles

– Inverted phase of complex poles

• Right half plane:

– Inverted magnitude of complex poles

– Same phase of complex poles

• This is the same relationship real zeros had to real poles


Complex Poles Phase

∠C(jω) = ∠ 1

1 −(

ωωn

)2

+ jωQωn

For ω � ωn,∠C(jω) ≈ ∠1 = 0◦

For ω � ωn,

∠C(jω) ≈ ∠ 1− ω

Qωn

= ∠ − Qωn

ω= ∠ − 1 = −180◦

For ω = ωn,

∠C(jω) ≈ ∠ 1Qj

= ∠1j

= ∠ − j = −90◦

At the extremes, the behavior is identical to two real poles. At othervalues of ω near ωn, the behavior is more complicated.


Complex Poles Maximum

What is the frequency at which |C(jω)| is maximized?

C(jω) =1

1 +(

jωQωn

)+

(jωωn

)2

|C(jω)| =1√(

1 − ω2

ω2n

)2

+(

ωQωn

)2

• For high values of Q, the maximum of |C(jω)| > 1

• This is called peaking

• The largest Q before the onset of peaking is Q = 1√2≈ 0.707

• This curve is said to be maximally flat

• This is also called a Butterworth response

• In this case, |C(jωn)| = −3 dB and ωn is the cutoff frequency


Complex Zeros Magnitude

10−1

100

101

−40

−20

0

20

40

60

80

Frequency (rad/sec)

Mag

(dB

)

Complex Zeros

Q = 0.1Q = 0.5Q = 0.707Q = 1Q = 2Q = 10Q = 100


Complex Poles Maximum Continued

If Q > 0.707, the maximum magnitude and frequency are as follows:

ωr = ωn

√1 − 1

2Q2|C(jωr)| =

Q

1 − 14Q2

• ωr is called the resonant frequency or the damped naturalfrequency

• As Q → ∞, ωr → ωn

• For sufficiently large Q (say Q > 5)

– ωr ≈ ωn

– |C(jωr)| ≈ Q

• Peaked responses are useful in the synthesis of high-order filters

• Complex zeros (in the left half plane) have the inverted magnitudeand phase of complex poles


Complex Zeros Phase

10−2

10−1

100

101

102

0

20

40

60

80

100

120

140

160

180

Frequency (rad/sec)

Phas

e (d

eg)

Complex Zeros

Q = 0.1Q = 0.5Q = 0.707Q = 1Q = 2Q = 10Q = 100


Complex Poles Example Continued 2

102

103

104

105

106

−120

−100

−80

−60

−40

−20

0

20

40

Frequency (rad/sec)

Mag

(dB

)

Resonance Example

R = 5 ΩR = 50 ΩR = 707 ΩR = 1 kΩR = 5 kΩR = 50 kΩ


Complex Poles Example

vs(t) vo(t)

-

+

50 mH R

200 nF

Generate the bode plot for the circuit shown above.

H(s) =1

LC

s2 + RL s + 1

LC

=ω2

n

s2 + 2ζωns + ω2n

=ω2

n

s2 + ωn

Q s + ω2n



102

103

104

105

106

−180

−160

−140

−120

−100

−80

−60

−40

−20

0

Frequency (rad/sec)

Phas

e (d

eg)

Resonance Example




ωn =

√1

LC= 10 k rad/s

ζ =R

2L

√LC =

R

2

√CL = R × 0.001

Q =1√LC

L

R=

√L

C

1R

=500R

R = 5 Ω ζ = 0.005 Q = 100 Very Light Damping

R = 50 Ω ζ = 0.05 Q = 10 Light Damping

R = 707 Ω ζ = 1.41 Q = 0.707 Strong Damping

R = 1 kΩ ζ = 1 Q = 0.5 Critical Damping

R = 5 kΩ ζ = 5 Q = 0.1 Over Damping



0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Time (sec)

Out

put (

V)

Resonance Example



transfer function review overview of bode...

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