Frequency Response and Bode Plots - DSpace@MIT: Home

Frequency Response of a High-Q System


As Q increases, the width of the peak narrows.


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


As Q increases, the width of the peak narrows.


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


As Q increases, the width of the peak narrows.


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
log ω0
−
1 − 12
2Q

−2 −1 0 1 2

Frequency Response of a High-Q System


As Q increases, the width of the peak narrows.


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
log ω0
−
1 − 12
2Q

−2 −1 0 1 2

Frequency Response of a High-Q System


As Q increases, the width of the peak narrows.


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
log ω0
−
1 − 12
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 log |H(jω)|
plane
1− 12 0
ω0 2Q

−1

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

Check Yourself


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

Analyze with vectors.

low frequencies high frequencies

ω/ω0 ω/ω0 1
1 + 2Q
1
1 − 2Q

−1 1 σ/ω0 −1 1 σ/ω0
2Q 2Q
− −

√ 1 × 2 = √ √ 1 × 2 = √
2 2Q 2 2 2Q 2

Q Q

1
Bandwidth approximately

Q

Frequency Response of a High-Q System


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


H(s) = 1

1 s s
2

Q ω0 ω0
1 + +


s plane ∠ |H(jω)|

ω0 0

−π/2
−1

−π log ω
−2 ω0
−1 0 1 2

Frequency Response of a High-Q System


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


H(s) = 1

1 s s
2

Q ω0 ω0
1 + +


s plane ∠ |H(jω)|

ω0
12 0
1− 2Q

−π/2

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

Frequency Response of a High-Q System


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


H(s) = 1

1 s s
2

Q ω0 ω0
1 + +


s plane ∠ |H(jω)|
ω0
1− 1 2
0

2Q

−π/2

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

Frequency Response of a High-Q System


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


H(s) = 1

1 s s
2

Q ω0 ω0
1 + +


s plane ∠ |H(jω)|
ω0
1− 12 0

2Q



−π/2


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

2Q

−2

Frequency Response of a High-Q System


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


H(s) = 1

1 s s
2

Q ω0 ω0
1 + +


s plane ∠ |H(jω)|
ω0
1− 12 0
2Q



−π/2

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

2Q

−2

Check Yourself


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

1 s2
H(s) = ω0

1+ 1 s +
Q ω0

s ∠ |H(jω)|
plane
12 0
ω0 2Q

1−

−π/2

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

Check Yourself


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

Analyze with vectors.

low frequencies high frequencies


ω/ω0 ω/ω0
1+ 1
1 2Q
2Q
1 −

−1 1 σ/ω0 −1 1 σ/ω0
2Q 2Q
− −

π − π = π π + π = 3π
2 4 4 2 4 4

Change in phase approximately π .

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 OpenCourseWare
http://ocw.mit.edu

6.003 Signals and Systems

Spring 2010

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