Frequency Response and Bode Plotsdspace.mit.edu/bitstream/handle/1721.1/78251/6-003...CT Frequency Response and Bode Plots March 9, 2010 Last Time Complex exponentials are eigenfunctions

6.003: Signals and Systems

CT Frequency Response and Bode Plots

March 9, 2010

(

Last Time

Complex exponentials are eigenfunctions of LTI systems.

H(s)es0t H(s0) es0t

H(s0) can be determined graphically using vectorial analysis. (s0 − z0)(s0 − z1)(s0 − z2) · · ·

H(s0) = K (s0 − p0)(s0 − p1)(s0 − p2) · · ·

z0 z0

s0 − z0 s0

s-planes0

Response of an LTI system to an eternal cosine is an eternal cosine:

same frequency, but scaled and shifted.

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

Frequency Response: H(s)|s←jω

|H(jω)|H(s) = s − z1 5

ω 5 s-plane

−5 0 5

∠H(jω)

5 σ

π/2−5

−5 5

−5 −π/2


|H(jω)|9 H(s) = 5

s − p1

ω 5 s-plane

−5 0 5

∠H(jω)

5 σ

π/2−5

−5 5

−5 −π/2


H(s) = 3 s − z1 |H

5(jω)|

s − p1

ω 5 s-plane

−5 0 5

∠H(jω)

5 σ

π/2−5

−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)|H(jω) = jω − z1 5

ω 5

−5 0 5

∠H(jω)

5 σ

π/2−5

−5 5

−5 −π/2



|H(jω)|H(jω) = jω − z1 5

ω 5 z1

−5 0 5

∠H(jω)

5 σ

π/2−5

−5 5

−5 −π/2



|H(jω)|H(jω) = jω − z1 5

ω

ω 5 z1

−5 0 5

∠H(jω)

5 σ

π/2−5

−5 5

−5 −π/2


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

H(s) = s − z1

log |H(jω)|

|H(jω)| z12 5

1 1

0 log ω

−5 0 5 −2 −1 0 1 2 z1

lim |H(jω)| = z1 ω→0

lim |H(jω)| = ω ω→∞

Asymptotic Behavior: Isolated Pole


H(s) = 9

ω 9

|H5(jω)|

s − p1 9

ω p15

−5 0 5

∠H(jω)

5 σ

π/2−5

−5 5

−5 −π/2


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

9 H(s) =

s − p1

log |H

9( /p

jω)| 1 |H(jω)| 0

5

−1

−2

−1

log ω

−5 0 5 −2 −1 0 1 2 p1

9lim |H(jω)| = ω→0 p1

9lim |H(jω)| = ω→∞ ω

Check Yourself

Compare log-log plots of the frequency-response magnitudes of

the following system functions:

H1(s) = 1 s + 1

and H2(s) = 1 s + 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:

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

s + 1 s + 10

log |H(jω)| 0 |H1(jω)|

−1 |H2(jω)|−1

−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) = 1 s + 1

and H2(s) = 1 s + 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


no scaling in either vertical or horizontal directions !

Asymptotic Behavior of More Complicated Systems

Constructing H(s0).

Q

H(s0) = K

∏

∏

q=1

P

p=1

(s0 − zq)

(s0 − pp)

s0 − z1

z1

← product of vectors for zeros

← product of vectors for poles

ω s-planes0

s0 − p1

σ p1

∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ ∣

∏ ∏

∏ ∏

∣ ∣

Asymptotic Behavior of More Complicated Systems

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

∣ Q ∣ Q ∣ (s0 − zq) ∣ ∣ s0 − zq

∣ ∣ ∣ ∣ ∣|H(s0)| = ∣ K

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

ω s-planes0

s 0−z 1

s0 −p1

σ z1 p1

∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ ∣

∏ ∏

∏ ∏

∣ ∣ ∣ ∣

∑ ∑

Bode Plot

The log of the magnitude is a sum of logs.

∣ Q ∣ Q ∣ (s0 − zq) ∣ ∣ s0 − zq

∣ ∣ ∣ ∣ ∣|H(s0)| = K

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

Q P ∣ ∣log |H(jω)| = log |K| + log ∣∣jω − zq

∣ − log ∣∣jω − pp

∣ q=1 p=1

log∣∣∣∣ jω

(jω + 1)(jω + 10)

∣∣∣∣

∣ ∣ ∣ ∣

∣ ∣ ∣ ∣

Bode Plot: Adding Instead of Multiplying

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

s + 10) 0

ω −1 10

s-plane −2

log |jω|

log ω −2 −1 0 1 2 3

0 ∣ 1 ∣ σ log ∣jω + 1 ∣ −10 10 −1 log ω

−2 −1 0 1 2 3 −1 ∣ 1 ∣

−10 log ∣jω + 10 ∣ −2 log ω

−2 −1 0 1 2 3

log∣∣∣∣ jω

(jω + 1)(jω + 10)

∣∣∣∣

∣ ∣ ∣ ∣

Bode Plot: Adding Instead of Multiplying ∣ ∣ ∣ ∣log ∣∣ jωjω + 1 ∣∣

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

0

ω −1 10

s-plane −2 log ω

−2 −1 0 1 2 3

σ −10 10

−1 log ∣ 1 ∣

−10 ∣jω + 10 ∣ −2 log ω

−2 −1 0 1 2 3

−2 −1 0 1 2 3

−1

−2 logω

log∣∣∣∣ 1jω + 10

∣∣∣∣

∣ ∣ Bode Plot: Adding Instead of Multiplying ∣ ∣

log ∣∣ (jω + 1)( jω

jω + 10) ∣∣

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

s + 10) −1

ω −2 10

s-plane −3 log ω

−2 −1 0 1 2 3

σ −10 10

−10

−2 −1 0 1 2 3

−1

−2 logω

log∣∣∣∣ 1jω + 10

∣∣∣∣

∣ ∣ Bode Plot: Adding Instead of Multiplying ∣ ∣

log ∣∣ (jω + 1)( jω

jω + 10) ∣∣

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

s + 10) −1

ω −2 10

s-plane −3 log ω

−2 −1 0 1 2 3

σ −10 10

−10


The angle response is simple at low and high frequencies.

|H(jω)|H(s) = s − z1 5

ω 5 s-plane

−5 0 5

∠H(jω)

5 σ

π/2−5

−5 5

−5 −π/2


Three straight lines provide a good approxmation versus log ω.

H(s) = s − z1

∠H(jω) ∠H(jω)

π/2

−5 5

−π/2

π 2

π 4

0 log ω

−2 −1 0 1 2 |z1|

lim ∠H(jω) = 0 ω→0

lim ∠H(jω) = π/2 ω→∞


The angle response is simple at low and high frequencies.

|H(jω)|9 H(s) = 5

s − p1

ω 5 s-plane

−5 0 5

∠H(jω) σ

π/2−5 5

−5 5

−5 −π/2


Three straight lines provide a good approxmation versus log ω.

9 H(s) =

s − p1

∠H(jω) ∠H(jω)

π/2

−π −5 5

−π−π/2

0

4

2 log ω

−2 −1 0 1 2 p1

lim ∠H(jω) = 0 ω→0

lim ∠H(jω) = −π/2ω→∞

∏

∏

∑ ∑

Bode Plot

The angle of a product is the sum of the angles. ⎛ ⎞ Q ⎜ (s0 − zq) ⎟⎜ ⎟ Q P ⎜ q=1 ⎟ ( ) ( ) ⎜ ⎟∠H(s0) = ∠ ⎜ K ⎟ = ∠K + ∠ s0 − zq − ∠ s0 − ppP ⎜ ⎟ ⎝ (s0 − pp) ⎠ q=1 p=1

p=1

ω

z1 p1

∠(s0 − z1) ∠(s0 − p1) σ

s-planes0

The angle of K can be 0 or π for systems described by linear differ

ential equations with constant, real-valued coefficients.

log∣∣∣∣ s

(s+ 1)(s+ 10)

∣∣∣∣Bode Plot

∠jω

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

s + 10) π/2

ω 0 10

s-plane −π/2 log ω

−2 −1 0 1 2 3 0 1∠σ

jω + 1−10 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

log∣∣∣∣ s

(s+ 1)(s+ 10)

∣∣∣∣Bode Plot

H(s) = s

(s + 1)(s + 10) π/2

ω 0 10

s-plane −π/2

σ −10 10

0 −10

−π/2

∠ jω jω + 1

log ω −2 −1 0 1 2 3

1∠ jω + 10

log ω −2 −1 0 1 2 3

log∣∣∣∣ s

(s+ 1)(s+ 10)

∣∣∣∣

−2 −1 0 1 2 3

0

−π/2 logω

∠ 1jω + 10

Bode Plot

∠ jω

(jω + 1)(jω + 10) H(s) = (s + 1)(

s

s + 10) π/2

ω 0 10


−2 −1 0 1 2 3

σ −10 10

−10

log∣∣∣∣ s

(s+ 1)(s+ 10)

∣∣∣∣

−2 −1 0 1 2 3

0

−π/2 logω

∠ 1jω + 10

Bode Plot

∠ jω

(jω + 1)(jω + 10) H(s) = (s + 1)(

s

s + 10) π/2

ω 0 10


−2 −1 0 1 2 3

σ −10 10

−10

∏ ∣ ∣

∏ ∣ ∣

∑ ∑ ∣ ∣ ∣ ∣

∑ ∑ ( ) ( )

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. Q P

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. Q P

∠H(jω) = ∠K + ∠ jω − zq − ∠ jω − pp q=1 p=1

Check Yourself

−1 0 1 2 3 4

−2

−3

−4 log ω

log |H(jω)|

Which corresponds to the Bode approximation above?

1. 1

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

s + 1 (s + 10)(s + 100)

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

s + 1 4.

s + 100 (s + 1)(s + 10)


Check Yourself

−1 0 1 2 3 4

−2

−3

−4 log ω

log |H(jω)|

Which corresponds to the Bode approximation above? 2

1. 1

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

s + 1 (s + 10)(s + 100)

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

s + 1 4.

s + 100 (s + 1)(s + 10)


ω [log scale]

ω [log scale]

log∣∣∣∣ s

(s+ 1)(s+ 10)

∣∣∣∣

Bode Plot: dB

log |H(jω)| 10s 0

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

ω −110

s-plane −2

−11

log ω −2 −1 0 1 2 3

σ ∠H(jω)−10 10 π/2

0−10

−π/2 log ω −2 −1 0 1 2 3

ω [log scale]

ω [log scale]

log∣∣∣∣ s

(s+ 1)(s+ 10)

∣∣∣∣

Bode Plot: dB

log |H(jω)| 10s 0

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

ω −1 10

s-plane −2

−11

ω [log scale] 0.01 0.1 1 10 100 1000

σ ∠H(jω)−10 10 π/2

0 −10

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

ω [log scale]

ω [log scale]

log∣∣∣∣ s

(s+ 1)(s+ 10)

∣∣∣∣

Bode Plot: dB

|H(jω)|[dB]= 20 log10 |H(jω)| 10s 0

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

ω −20 10

s-plane −40

−11

ω [log scale] 0.01 0.1 1 10 100 1000

σ ∠H(jω)−10 10 π/2

0 −10

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

ω [log scale]

ω [log scale]

log∣∣∣∣ s

(s+ 1)(s+ 10)

∣∣∣∣

Bode Plot: dB

|H(jω)|[dB]= 20 log10 |H(jω)| 10s 0

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

ω −20 20 dB/decade −20 dB/decade10

s-plane −40 ω [log scale]

0.01 0.1 1 10 100 1000

σ ∠H(jω)−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.

1

1 dB 3 dB

1 dB

H(jω) = jω + 1 20 log10 X0 X

√1 0 dB 2 ≈ 3 dB

−10 2 ≈ 6 dB 10 20 dB 100 40 dB

−20 ω [log scale] 0.1 1 10

0

|H(jω

)|[dB]

0.1 rad (6◦)∠H

(jω

)

−π/4

−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 1

2

−1 ( )2

3

−1 ( )2

4

Check Yourself

π

ω1. −1 −π

π

ω2. −1 1 −π

π

2 ω3. −1 −π

π

2 ω4. −1 −π

Check Yourself

π

1. −1 −π

π

2. −1 1 −π

π

3. −1

2

−π π

4. −1

2

−π

ω

ω if K < 0

ω

ω

Check Yourself

Could the phase plots of any of these systems be equal to

each other? [caution: this is a trick question] yes

−1

1

−1 1

2

−1 ( )2

3

−1 ( )2

4

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.

1 H(s) = ( )21 s s1 + +

Q ω0 ω0

s plane

ω0

−1 1 −2Q

log |H(jω)|

1 − 21 Q

)2 0

−1

√ ( )2−2 − 1 − 2

1 Q

log ω

ω

0−2 −1 0 1 2

√



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

s plane

ω0

−1 1 −2Q

log |H(jω)| ( )21 − 2

1 Q 0

−1

√ −2 log ω

− 1 − (

1 )2

ω02Q −2 −1 0 1 2

√

(



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

log |H(jω)|s plane

ω0 1 − (

1 )2

0

−1 −1 1 −2Q √ −2

log ω

1 )2

1 − ω0

2Q

− 2Q −2 −1 0 1 2

√

√



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

log |H(jω)|s plane

ω0 1 − (

1 )2

0

− 1

2Q

−1 −1

−2 log ω

1 − (

1 )2

ω0

2Q

− 2Q −2 −1 0 1 2

√

√



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

log |H(jω)|s plane

ω0 1 − (

1 )2

0

−1 −1

−2Q 1

−2 log ω

1 − (

1 )2

ω0

2Q

− 2Q −2 −1 0 1 2

Check Yourself

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

s ω0

plane

−1 −

1 2Q

√

1 − (

1 2Q

)2

−

√

1 − (

1 2Q

)2

H(s) = 1

1 + 1 Q s ω0

+

( s ω0

)2

−2 −1 0 1 2

0

−1

−2 log ω ω0

log |H(jω)|

Check Yourself

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

Analyze with vectors.

low frequencies high frequencies

ω/ω0 ω/ω0

− 1

2Q

σ/ω0 −1 −

1 2Q

1 2Q

σ/ω0−1

× 2 =1× 1 = 1 Q

Peak magnitude increases with Q !

1

√ (


As Q increases, the width of the peak narrows.

1 H(s) = ( )21 s s1 + +

Q ω0 ω0

s plane

ω0

−1 1 −2Q

log |H(jω)|

1 − 21 Q

)2 0

−1

√ ( )2−2 − 1 − 2

1 Q

log ω

ω

0−2 −1 0 1 2

√



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

s plane

ω0

−1 1 −2Q

log |H(jω)| ( )21 − 2

1 Q 0

−1

√ −2 log ω

− 1 − (

1 )2

ω02Q −2 −1 0 1 2

√

(



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

log |H(jω)|s plane

ω0 1 − (

21 )2

0

−1 −1 1 −2Q √ −2

log ω

1 )2

1 − ω0

Q

− 2Q −2 −1 0 1 2

√

√



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

log |H(jω)|s plane

ω0 1 − (

21 )2

0

− 1

2Q

−1 −1

−2 log ω

1 − (

1 )2

ω0

Q

− 2Q −2 −1 0 1 2

√

√



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

log |H(jω)|s plane

ω0 1 − (

21 )2

0Q

−1 −1

−2Q 1

−2 log ω

1 − (

1 )2

ω0− 2Q −2 −1 0 1 2

Check Yourself

Estimate the “3dB bandwidth” of the peak (assume Q > 3).

Let ωl (or ωh) represent the lowest (or highest) frequency for

which the magnitude is greater than the peak value divided by√2. The 3dB bandwidth is then ωh − ωl.

s ω0

plane

−1 −

1 2Q

√

1 − (

1 2Q

)2

−

√

1 − (

1 2Q

)2

−2 −1 0 1 2

0

−1

−2 log ω ω0

log |H(jω)|

Check Yourself

Estimate the “3dB bandwidth” of the peak (assume Q > 3).



ω/ω0 ω/ω0

−1 −

1 2Q

√2 1

2Q

1− 1

2Q

1

− 1

2Q

1 + 2Q

σ/ω0 σ/ω0−1

√ × 2 = 2 √

221 Q × 2 =

√

Q 2

Q

1 Bandwidth approximately

Q


As Q increases, the phase changes more abruptly with ω.

1 H(s) = ( )21 s s1 + +

Q ω0 ω0

s plane

ω0

−1

∠ |H(jω)|

0

−π/2

−π log ω

−2 −1 0 1 2 ω0

√ (

√



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

s plane

ω0

−1 1 −2Q

∠ |H(jω)|

1 − 21 Q

)2 0

−π/2

( )2−π − 1 − 21 Q

log ω

ω

0−2 −1 0 1 2

√



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

s plane

ω0

−1 1 −2Q

∠ |H(jω)| ( )21 − 2

1 Q 0

−π/2

√ −π log ω

− 1 − (

1 )2

ω02Q −2 −1 0 1 2

√

√



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

∠ |H(jω)|s plane

ω0 1 − (

1 )2

0

−π/2 −1 1 −2Q −π

log ω

1 − (

1 )2

ω0

2Q

− 2Q −2 −1 0 1 2

√

√



1 H(s) = ( )21 s s1 + +

Q ω0 ω0

∠ |H(jω)|s plane

ω0 1 − (

1 )2

0

− 1

2Q

−π/2 −1

−π log ω

1 − (

1 )2

ω0

2Q

− 2Q −2 −1 0 1 2

Check Yourself

Estimate change in phase that occurs over the 3dB bandwidth.

s ω0

plane

−1 −

1 2Q

√

1 − (

1 2Q

)2

−

√

1 − (

1 2Q

)2

H(s) = 1

1 + 1 Q s ω0

+

( s ω0

)2

−2 −1 0 1 2

0

−π/2

−π log ω ω0

∠ |H(jω)|

Check Yourself

Estimate change in phase that occurs over the 3dB bandwidth.



−1 −

1 2Q

π 2 − π

1− 1

2Q

ω/ω0 ω/ω0

− 1

2Q

11 + 2Q

σ/ω0 σ/ω0−1

π π π 3π 4 = 4 2 + 4 = 4

π Change in phase approximately 2 .

Summary

The frequency response of a system can be quickly determined using

Bode plots.

Bode plots are constructed from sections that correspond to single

poles and single zeros.

Responses for each section simply sum when plotted on logarithmic

coordinates.

MIT OpenCourseWarehttp://ocw.mit.edu

6.003 Signals and Systems Spring 2010

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

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Frequency Response and Bode Plotsdspace.mit.edu/bitstream/handle/1721.1/78251/6-003...CT Frequency Response and Bode Plots March 9, 2010 Last Time Complex exponentials are eigenfunctions

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