Frequency Response and Bode Plots - DSpace@MIT: Home

6.003: Signals and Systems

CT Frequency Response and Bode Plots


March 9, 2010

Last Time


Complex exponentials are eigenfunctions of LTI systems.

es0t H(s) H(s0) es0t


H(s0) can be determined graphically using vectorial analysis.

H (s0 ) = K (s0 − z0)(s0 − z1)(s0 − z2) · · ·
(s0 − p0)(s0 − p1)(s0 − p2) · · ·

s0 s-plane

s0 − z0 s0

z0 z0

Response of an LTI system to an eternal cosine is an eternal cosine:
same frequency, but scaled and shifted.

cos(ω0t) H (s) |H(jω0)| cos ω0t + ∠H(jω0)

Frequency Response: H(s)|s←jω |H (j ω)| 5
5 5
H(s) = s − z1
0
ω ∠H (j ω)
5 s-plane π/2

−5

σ
−5 5

−5
−5 −π/2

Frequency Response: H(s)|s←jω

H (s) = s 9 |H (j ω)|
− p1 5

ω 0
5 s-plane ∠H (j ω)
π/2
−5 5
5
σ
−5 5

−5
−5 −π/2

Frequency Response: H(s)|s←jω

H (s) = 3 s − z1 |H (j ω)|
s − p1 5

ω 0
5 s-plane ∠H (j ω)
π/2
−5 5
5
σ
−5 5

−5
−5 −π/2

Poles and Zeros


Thinking about systems as collections of poles and zeros is an im­
portant design concept.
• simple: just a few numbers characterize entire system
• powerful: complete information about frequency response

Today: poles, zeros, frequency responses, and Bode plots.

Asymptotic Behavior: Isolated Zero


The magnitude response is simple at low and high frequencies.


H(jω) = jω − z1 |H (j ω)|
5

ω
5

−5 0 5

∠H (j ω)

−5 σ π/2
5

−5 5
−5 −π/2

Asymptotic Behavior: Isolated Zero


The magnitude response is simple at low and high frequencies.


H(jω) = jω − z1 |H (j ω)|
5

ω
5 z1

−5 0 5

∠H (j ω)

−5 σ π/2
5

−5 5
−5 −π/2

Asymptotic Behavior: Isolated Zero


The magnitude response is simple at low and high frequencies.


H(jω) = jω − z1 |H (j ω)|
5
ω
5 ω

z1

−5 0 5


∠H (j ω)


−5 σ π/2
5

−5 5
−5 −π/2

Asymptotic Behavior: Isolated Zero


Two asymptotes provide a good approxmation on log-log axes.


H(s) = s − z1


|H (j ω)| log |H (j ω)|

5
2
z1


1 1

−5 0 0 log ω
5 −2 z1
−1 0 1 2

lim |H (j ω)| = z1

ω→0

lim |H(jω)| = ω


ω→∞

Asymptotic Behavior: Isolated Pole


The magnitude response is simple at low and high frequencies.


H (s) = 9 9 |H (j ω)|
− p1 5
s ω

9

ω p1
5

−5 0 5

∠H (j ω)

−5 σ π/2
5

−5 5
−5 −π/2

Asymptotic Behavior: Isolated Pole


Two asymptotes provide a good approxmation on log-log axes.


H (s) = s 9
− p1

|H (j ω)|
log |H(jω)|

5
9/p1

0

−1
−1

−2 log ω
−5 0 5 −2 −1 2 p1
0 1

lim |H(jω)| = 9
p1
ω→0

lim |H (j ω)| =
9

ω→∞ ω

Check Yourself


Compare log-log plots of the frequency-response magnitudes of
the following system functions:

H1(s) = s 1 1 and H2(s) = s 1
+ + 10

The former can be transformed into the latter by

1. shifting horizontally
2. shifting and scaling horizontally
3. shifting both horizontally and vertically
4. shifting and scaling both horizontally and vertically
5. none of the above

Check Yourself


Compare log-log plots of the frequency-response magnitudes of the
following system functions:

11
H1(s) = s + 1 and H2(s) = s + 10

log |H(jω)| |H1(jω)|
0
−1
−1 |H2(jω)|

−2
log ω

−2 −1 0 1 2

Check Yourself


Compare log-log plots of the frequency-response magnitudes of
the following system functions:

H1(s) = s 1 1 and H2(s) = s 1
+ + 10

The former can be transformed into the latter by 3

1. shifting horizontally
2. shifting and scaling horizontally
3. shifting both horizontally and vertically
4. shifting and scaling both horizontally and vertically
5. none of the above

no scaling in either vertical or horizontal directions !

Asymptotic Behavior of More Complicated Systems


Constructing H(s0).


Q

(s0 − zq) ← product of vectors for zeros

← product of vectors for poles

H(s0) = K
q=1
P

(s0 − pp)

p=1

s0 ω
s-plane

s0 − z1 s0 − p1 σ
z1 p1

Asymptotic Behavior of More Complicated Systems


The magnitude of a product is the product of the magnitudes.


Q Q

q=1 (s0 − zq)
s0 − zq
P
|H(s0)| = K
= |K| q=1

(s0 − pp) P

p=1 s0 − pp

p=1

s0 ω s-plane

s0 − z1 s0 − p1

σ

z1 p1

Bode Plot


The log of the magnitude is a sum of logs.

Q Q

s0 − zq
q=1 (s0 − zq)

P q=1
|H(s0)| = K
= |K|
P
(s0 − pp)
s0 − pp
p=1
p=1

QP

log |H(jω)| = log |K| + log jω − zq − log jω − pp


q=1 p=1

Bode Plot: Adding Instead of Multiplying


H (s) = s log (jωl+og1|)jj(ωωj|ω
+ 10)
(s + 1)(s 0


+ 10)


s-plane ω
−1

10

−2
log ω

−2 −1 0 1 2 3


−10
0 log 1
σ jω + 1
10
log ω


−1


−2 −1 0 1 2 3


−1
log 1
−10 jω + 10

−2 log ω


−2 −1 0 1 2 3

Bode Plot: Adding Instead of Multiplying


(jloωg+j1ω)j(ω+jω1


log
10)
+
s
0

H (s) = (s + 1)(s
+ 10)

s-plane ω −1

10
−2


−2 −1 0 1 2 log ω
3

−10
σ
10


−1
log 1
−10 jω + 10

−2
log ω

−2 −1 0 1 2 3

Bode Plot: Adding Instead of Multiplying


jω

(jω + 1)(jω + 10)
log

H (s) = s −1
(s + 1)(s
+ 10)

s-plane ω −2

10
−3

−2 −1 0 1 2 log ω
3

−10
σ
10


−1 log 1
−10 jω + 10

−2 log ω

−2 −1 0 1 2 3

Bode Plot: Adding Instead of Multiplying


jω

(jω + 1)(jω + 10)
log

H (s) = s −1
(s + 1)(s
+ 10)

s-plane ω −2

10
−3

−2 −1 0 1 2 log ω
3

−10
σ
10


−1 log 1
−10 jω + 10

−2 log ω

−2 −1 0 1 2 3

Asymptotic Behavior: Isolated Zero


The angle response is simple at low and high frequencies.


H(s) = s − z1 |H (j ω)|
5

ω
5 s-plane

−5 0 5

∠H (j ω)

−5 σ π/2
5

−5 5
−5 −π/2

Asymptotic Behavior: Isolated Zero


Three straight lines provide a good approxmation versus log ω.


∠H (j ω) H(s) = s − z1
π/2
∠H (j ω)
−5 π
−π/2 2

π
54

0 log ω
−2 −1 0 1 2 |z1|

lim ∠H(jω) = 0


ω→0

lim ∠H(jω) = π/2


ω→∞

Asymptotic Behavior: Isolated Pole


The angle response is simple at low and high frequencies.


H (s) = s 9 |H (j ω)|
− p1 5

ω
5 s-plane

−5 0 5


∠H (j ω)


−5
σ π/2
5


−5 5
−5 −π/2

Asymptotic Behavior: Isolated Pole


Three straight lines provide a good approxmation versus log ω.


9
H(s) = s − p1

∠H (j ω) ∠H (j ω)

π/2 0

−5 5 − π
−π/2 4

− π log ω
2

−2 −1 0 1 2 p1

lim ∠H(jω) = 0


ω→0

lim ∠H(jω) = −π/2


ω→∞

Bode Plot


The angle of a⎛product is the ⎞sum of the angles.

Q

∠ ⎜⎜⎜⎜⎜⎝⎜ K (s0 − zq ) ⎟⎟⎠⎟⎟⎟⎟
=
∠H (s0 ) = − pp) Q ∠ s0 − zq P s0 − pp
q=1
∠K + −∠
P
q=1 p=1
(s0

p=1

s0 ω s-plane

∠(s0 − z1) ∠(s0 − p1) σ
z1 p1

The angle of K can be 0 or π for systems described by linear differ­
ential equations with constant, real-valued coefficients.

Bode Plot


H (s) = s log (s +∠1j)ω(ss
+ 10)
1)(s π/2

(s + + 10)

ω 0


s-plane
10


−π/2
log ω


−2 −1 0 1 2 3


0
1

jω +
−10
σ ∠ 1
10

−π/2

log ω


−2 −1 0 1 2 3


0
1

+
−10 ∠ jω 10

−π/2 log ω


−2 −1 0 1 2 3

Bode Plot


H (s) = (s + s + 10) log (s∠+jω1j)ω+(ss1+ 10)
1)(s π/2

s-plane ω 0
10
−π/2
−2 −1 0 1 2 log ω
3

σ
−10 10

0 ∠ jω 1 10
−10 +

−π/2 log ω

−2 −1 0 1 2 3

Bode Plot


H (s) = (s + s + 10) l∠og(jω(s++11)j)(ω(sjsω++1100))
1)(s π/2

ω 0


s-plane
10
−π/2

−2 −1 0 1 2
log ω
3

σ
−10 10

0 ∠ jω 1 10
−10 +

−π/2 log ω

−2 −1 0 1 2 3

Bode Plot


H (s) = (s + s + 10) l∠og(jω(s++11)j)(ω(sjsω++1100))
1)(s π/2

ω 0


s-plane
10
−π/2

−2 −1 0 1 2
log ω
3

σ
−10 10

0 ∠ jω 1 10
−10 +

−π/2 log ω

−2 −1 0 1 2 3

From Frequency Response to Bode Plot


The magnitude of H(jω) is a product of magnitudes.


Q

jω − zq


|H(jω)| = |K| q=1
P

jω − pp


p=1

The log of the magnitude is a sum of logs.

QP

log |H(jω)| = log |K| + log jω − zq − log jω − pp

q=1 p=1

The angle of H(jω) is a sum of angles.

QP

∠H(jω) = ∠K + ∠ jω − zq − ∠ jω − pp

q=1 p=1

Check Yourself


log |H(jω)|
−2

−3

−4 0 log ω
−1 1234

Which corresponds to the Bode approximation above?

1 s+1
1. (s + 1)(s + 10)(s + 100) 2. (s + 10)(s + 100)

(s + 10)(s + 100) s + 100
3. s + 1 4. (s + 1)(s + 10)

5. none of the above

Check Yourself


log |H(jω)|
−2

−3

−4 0 log ω
−1 1234

Which corresponds to the Bode approximation above? 2

1 s+1
1. (s + 1)(s + 10)(s + 100) 2. (s + 10)(s + 100)

(s + 10)(s + 100) s + 100
3. s + 1 4. (s + 1)(s + 10)

5. none of the above

Bode Plot: dB


H (s) = 10s log |H(jω)|

(s + 1)(s + 0


10)


ω −1
1 −1
10

s-plane

−2 ωlog[lωo
g scale]
−2 −1 0 1 2 3

−10 σ π/2 log (s∠+H1()j(sωs)+ 10)
10

0


−10


−π/2
ωlog[lωog scale]

−2 −1 0 1 2 3

Bode Plot: dB


H (s) = 10s log |H(jω)|

(s + 1)(s + 0


10)


s-plane ω −1
1 −1
10

−2 ω [log scale]
0.01 0.1 1 10 100 1000

−10
σ log (s∠+H1()j(sωs)+ 10)
10


π/2

0


−10


−π/2 ω [log scale]

0.01 0.1 1 10 100 1000

Bode Plot: dB


H (s) = 10s |H(jω)|[dB]= 20 log10 |H(jω)|

(s + 1)(s + 0


10)


ω −20
1 −1
10

s-plane

−40 ω [log scale]
0.01 0.1 1 10 100 1000

−10
σ log (s∠+H1()j(sωs)+ 10)
10


π/2

0


−10


−π/2 ω [log scale]

0.01 0.1 1 10 100 1000

Bode Plot: dB


H (s) = 10s 10)
|H(jω)|[dB]= 20 log10 |H(jω)|

(s + 1)(s + 0


s-plane ω −20 20 dB/decade −20 dB/decade
10
−40 ω [log scale]
0.01 0.1 1 10 100 1000

−10
σ log (s∠+H1()j(sωs)+ 10)
10


π/2

0


−10


−π/2 ω [log scale]

0.01 0.1 1 10 100 1000

Bode Plot: Accuracy


The straight-line approximations are surprisingly accurate.


0
H (j ω) = 1 1 X
20 log10 X

jω + √1

0 dB

|H (j ω)|[dB]
−10
1 dB 2 ≈ 3 dB
−20 3 dB 2 ≈ 6 dB

0.1 1 dB 10 20 dB


1 100 40 dB


ω [log scale]
10

0

∠H (j ω)
−π/4
0.1 rad
(6◦)

−π/2 ω [log scale]
0.01 0.1 1 10 100

Check Yourself


Could the phase plots of any of these systems be equal to
each other? [caution: this is a trick question]

−1 −1 1 ( )2 ( )2
1 2 −1 −1

3 4

Check Yourself


π

1. −π ω
−1 π ω
ω
2. ω
−1 1 −π

3. 2 π
−1 −π

4. 2 π
−1 −π

Check Yourself


π

1. −π ω
−1 π
ω
2. if K < 0
−1 1 −π
ω
3. 2 π
−1 −π ω

4. 2 π
−1 −π

Check Yourself


Could the phase plots of any of these systems be equal to
each other? [caution: this is a trick question] yes

( )2 ( )2

−1 −1 1 −1 −1

1234
phase of 2 could be same as phase of 3: depends on sign of K

Frequency Response of a High-Q System


The frequency-response magnitude of a high-Q system is peaked.


H(s) = 1

1 s s
2

Q ω0 ω0
1 + +


s plane log |H(jω)|

ω0
12 0
1− 2Q

−1

−1 − 1
2Q
1 2−2 ω
− 1− log ω0
2Q
−2 −1 0 1 2

Frequency Response of a High-Q System


The frequency-response magnitude of a high-Q system is peaked.


H(s) = 1

1 s s
2

Q ω0 ω0
1 + +


s plane log |H(jω)|
ω0
1− 1 2
0

2Q

−1

−1 − 1
2Q
−2 ω
12 log ω0
− 1− 2Q
−2 −1 0 1 2

Frequency Response of a High-Q System


The frequency-response magnitude of a high-Q system is peaked.


H(s) = 1

1 s s
2

Q ω0 ω0
1 + +


s plane log |H(jω)|

ω0
1− 12 0

2Q



−1


−1 − 1
2Q
−2
ω
−
1 − log ω0
12
−1 0 1 2

2Q

−2

Frequency Response of a High-Q System


The frequency-response magnitude of a high-Q system is peaked.


H(s) = 1

1 s s
2

Q ω0 ω0
1 + +


s plane log |H(jω)|

ω0
1− 12 0
2Q



−1

−1 − 1
2Q
−2 ω
−
1 − log ω0
12
−1 0 1 2

2Q

−2

Frequency Response of a High-Q System


The frequency-response magnitude of a high-Q system is peaked.


H(s) = 1

1 s s
2

Q ω0 ω0
1 + +


s plane log |H(jω)|

ω0
1− 12 0
2Q



−1

−1 − 1
2Q
−2 ω
−
1 − log ω0
12
−1 0 1 2

2Q

−2

Check Yourself


Find dependence of peak magnitude on Q (assume Q > 3).

H(s) = 1

1 + 1 s + s2
Q ω0 ω0

s log |H(jω)|
plane
1− 12 0
ω0 2Q

−1

−1 1
− 2Q
−2 log ω
12 2 ω0
− 1− 2Q −2 −1 0 1

Check Yourself


Find dependence of peak magnitude on Q (assume Q > 3).

Analyze with vectors.

low frequencies high frequencies

ω/ω0 ω/ω0

−1 1 σ/ω0 −1 1 σ/ω0
2Q 1×1=1 2Q
− −

1 × 2 = 1
2Q Q

Peak magnitude increases with Q !