11: Frequency Responses 11: Frequency Responses • Frequency Response • Sine Wave Response • Logarithmic axes • Logs of Powers + • Straight Line Approximations • Plot Magnitude Response • Low and High Frequency Asymptotes • Phase Approximation + • Plot Phase Response + • RCR Circuit • Summary E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 1 / 12
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11: Frequency Responses
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 1 / 12
Frequency Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 2 / 12
If x(t) is a sine wave, then y(t) will also be asine wave but with a different amplitude andphase shift. X is an input phasor and Y is theoutput phasor.
Frequency Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 2 / 12
If x(t) is a sine wave, then y(t) will also be asine wave but with a different amplitude andphase shift. X is an input phasor and Y is theoutput phasor.
The gain of the circuit is YX =
1/jωC
R+1/jωC= 1
jωRC+1
Frequency Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 2 / 12
If x(t) is a sine wave, then y(t) will also be asine wave but with a different amplitude andphase shift. X is an input phasor and Y is theoutput phasor.
The gain of the circuit is YX =
1/jωC
R+1/jωC= 1
jωRC+1
This is a complex function of ω so we plot separate graphs for:
Frequency Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 2 / 12
If x(t) is a sine wave, then y(t) will also be asine wave but with a different amplitude andphase shift. X is an input phasor and Y is theoutput phasor.
The gain of the circuit is YX =
1/jωC
R+1/jωC= 1
jωRC+1
This is a complex function of ω so we plot separate graphs for:
Magnitude:∣
∣
YX
∣
∣ = 1|jωRC+1| =
1√1+(ωRC)2
Frequency Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 2 / 12
If x(t) is a sine wave, then y(t) will also be asine wave but with a different amplitude andphase shift. X is an input phasor and Y is theoutput phasor.
The gain of the circuit is YX =
1/jωC
R+1/jωC= 1
jωRC+1
This is a complex function of ω so we plot separate graphs for:
Magnitude:∣
∣
YX
∣
∣ = 1|jωRC+1| =
1√1+(ωRC)2
Magnitude Response
Frequency Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 2 / 12
If x(t) is a sine wave, then y(t) will also be asine wave but with a different amplitude andphase shift. X is an input phasor and Y is theoutput phasor.
The gain of the circuit is YX =
1/jωC
R+1/jωC= 1
jωRC+1
This is a complex function of ω so we plot separate graphs for:
Magnitude:∣
∣
YX
∣
∣ = 1|jωRC+1| =
1√1+(ωRC)2
Phase Shift: ∠(
YX
)
= −∠ (jωRC + 1) = − arctan(
ωRC1
)
Magnitude Response
Frequency Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 2 / 12
If x(t) is a sine wave, then y(t) will also be asine wave but with a different amplitude andphase shift. X is an input phasor and Y is theoutput phasor.
The gain of the circuit is YX =
1/jωC
R+1/jωC= 1
jωRC+1
This is a complex function of ω so we plot separate graphs for:
Magnitude:∣
∣
YX
∣
∣ = 1|jωRC+1| =
1√1+(ωRC)2
Phase Shift: ∠(
YX
)
= −∠ (jωRC + 1) = − arctan(
ωRC1
)
Magnitude Response Phase Response
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
0 100 200 300 400 5000
0.5
1
ω (rad/s)
|Y/X
|
0 100 200 300 400 500
-80
-60
-40
-20
0
ω (rad/s)
Pha
se (
°)
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
0 0.5 1
-0.4
-0.2
0X
YX-Y
ω=50
Real
Imag
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
0 100 200 300 400 5000
0.5
1
ω (rad/s)
|Y/X
|
0 100 200 300 400 500
-80
-60
-40
-20
0
ω (rad/s)
Pha
se (
°)
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
0 0.5 1
-0.4
-0.2
0X
YX-Y
ω=50
Real
Imag
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
0 20 40 60 80 100 120-1
-0.5
0
0.5
1
x
y
time (ms)
x=bl
ue, y
=re
d
w = 50 rad/s, Gain = 0.89, Phase = -27°
0 100 200 300 400 5000
0.5
1
ω (rad/s)
|Y/X
|
0 100 200 300 400 500
-80
-60
-40
-20
0
ω (rad/s)P
hase
(°)
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
0 100 200 300 400 5000
0.5
1
ω (rad/s)
|Y/X
|
0 100 200 300 400 500
-80
-60
-40
-20
0
ω (rad/s)
Pha
se (
°)
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
0 0.5 1
-0.4
-0.2
0X
YX-Y
ω=100
Real
Imag
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
0 100 200 300 400 5000
0.5
1
ω (rad/s)
|Y/X
|
0 100 200 300 400 500
-80
-60
-40
-20
0
ω (rad/s)
Pha
se (
°)
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
0 0.5 1
-0.4
-0.2
0X
YX-Y
ω=100
Real
Imag
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
0 20 40 60 80 100 120-1
-0.5
0
0.5
1
x
y
time (ms)
x=bl
ue, y
=re
d
w = 100 rad/s, Gain = 0.71, Phase = -45°
0 100 200 300 400 5000
0.5
1
ω (rad/s)
|Y/X
|
0 100 200 300 400 500
-80
-60
-40
-20
0
ω (rad/s)P
hase
(°)
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
0 100 200 300 400 5000
0.5
1
ω (rad/s)
|Y/X
|
0 100 200 300 400 500
-80
-60
-40
-20
0
ω (rad/s)
Pha
se (
°)
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
0 0.5 1
-0.4
-0.2
0X
YX-Y
ω=300
Real
Imag
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
0 100 200 300 400 5000
0.5
1
ω (rad/s)
|Y/X
|
0 100 200 300 400 500
-80
-60
-40
-20
0
ω (rad/s)
Pha
se (
°)
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
0 0.5 1
-0.4
-0.2
0X
YX-Y
ω=300
Real
Imag
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
0 20 40 60 80 100 120-1
-0.5
0
0.5
1x
y
time (ms)
x=bl
ue, y
=re
d
w = 300 rad/s, Gain = 0.32, Phase = -72°
0 100 200 300 400 5000
0.5
1
ω (rad/s)
|Y/X
|
0 100 200 300 400 500
-80
-60
-40
-20
0
ω (rad/s)
Pha
se (
°)
Sine Wave Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 3 / 12
RC = 10ms
YX = 1
jωRC+1 = 10.01jω+1
0 0.5 1
-0.4
-0.2
0X
YX-Y
ω=300
Real
Imag
ω = 50 ⇒ YX = 0.89∠− 27◦
ω = 100 ⇒ YX = 0.71∠− 45◦
ω = 300 ⇒ YX = 0.32∠− 72◦
0 20 40 60 80 100 120-1
-0.5
0
0.5
1x
y
time (ms)
x=bl
ue, y
=re
d
w = 300 rad/s, Gain = 0.32, Phase = -72°
0 100 200 300 400 5000
0.5
1
ω (rad/s)
|Y/X
|
0 100 200 300 400 500
-80
-60
-40
-20
0
ω (rad/s)
Pha
se (
°)The output, y(t), lags the input, x(t), by up to 90◦.
Logarithmic axes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 4 / 12
We usually use logarithmic axes for frequency and gain (but not phase)because % differences are more significant than absolute differences.E.g. 5 kHz versus 5.005 kHz is less significant than 10Hz versus 15Hzeven though both differences equal 5Hz.
Logarithmic axes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 4 / 12
We usually use logarithmic axes for frequency and gain (but not phase)because % differences are more significant than absolute differences.E.g. 5 kHz versus 5.005 kHz is less significant than 10Hz versus 15Hzeven though both differences equal 5Hz.
Logarithmic axes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 4 / 12
We usually use logarithmic axes for frequency and gain (but not phase)because % differences are more significant than absolute differences.E.g. 5 kHz versus 5.005 kHz is less significant than 10Hz versus 15Hzeven though both differences equal 5Hz.
Logarithmic axes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 4 / 12
We usually use logarithmic axes for frequency and gain (but not phase)because % differences are more significant than absolute differences.E.g. 5 kHz versus 5.005 kHz is less significant than 10Hz versus 15Hzeven though both differences equal 5Hz.
Note that 0 does not
exist on a log axis and so
the starting point of the
axis is arbitrary.
Logarithmic axes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 4 / 12
We usually use logarithmic axes for frequency and gain (but not phase)because % differences are more significant than absolute differences.E.g. 5 kHz versus 5.005 kHz is less significant than 10Hz versus 15Hzeven though both differences equal 5Hz.
Logarithmic voltage ratios are specified in decibels (dB) = 20 log10|V2||V1| .
Note that 0 does not
exist on a log axis and so
the starting point of the
axis is arbitrary.
Logarithmic axes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 4 / 12
We usually use logarithmic axes for frequency and gain (but not phase)because % differences are more significant than absolute differences.E.g. 5 kHz versus 5.005 kHz is less significant than 10Hz versus 15Hzeven though both differences equal 5Hz.
Logarithmic voltage ratios are specified in decibels (dB) = 20 log10|V2||V1| .
Common voltage ratios:
|V2||V1|
1
dB 0
Note that 0 does not
exist on a log axis and so
the starting point of the
axis is arbitrary.
Logarithmic axes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 4 / 12
We usually use logarithmic axes for frequency and gain (but not phase)because % differences are more significant than absolute differences.E.g. 5 kHz versus 5.005 kHz is less significant than 10Hz versus 15Hzeven though both differences equal 5Hz.
Logarithmic voltage ratios are specified in decibels (dB) = 20 log10|V2||V1| .
Common voltage ratios:
|V2||V1|
0.1 1 10 100
dB −20 0 20 40
Note that 0 does not
exist on a log axis and so
the starting point of the
axis is arbitrary.
Logarithmic axes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 4 / 12
We usually use logarithmic axes for frequency and gain (but not phase)because % differences are more significant than absolute differences.E.g. 5 kHz versus 5.005 kHz is less significant than 10Hz versus 15Hzeven though both differences equal 5Hz.
Logarithmic voltage ratios are specified in decibels (dB) = 20 log10|V2||V1| .
Common voltage ratios:
|V2||V1|
0.1 0.5 1 2 10 100
dB −20 -6 0 6 20 40
Note that 0 does not
exist on a log axis and so
the starting point of the
axis is arbitrary.
Logarithmic axes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 4 / 12
We usually use logarithmic axes for frequency and gain (but not phase)because % differences are more significant than absolute differences.E.g. 5 kHz versus 5.005 kHz is less significant than 10Hz versus 15Hzeven though both differences equal 5Hz.
Logarithmic voltage ratios are specified in decibels (dB) = 20 log10|V2||V1| .
Common voltage ratios:
|V2||V1|
0.1 0.5√
0.5 1√
2 2 10 100
dB −20 -6 -3 0 3 6 20 40
Note that 0 does not
exist on a log axis and so
the starting point of the
axis is arbitrary.
Logarithmic axes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 4 / 12
We usually use logarithmic axes for frequency and gain (but not phase)because % differences are more significant than absolute differences.E.g. 5 kHz versus 5.005 kHz is less significant than 10Hz versus 15Hzeven though both differences equal 5Hz.
Logarithmic voltage ratios are specified in decibels (dB) = 20 log10|V2||V1| .
Common voltage ratios:
|V2||V1|
0.1 0.5√
0.5 1√
2 2 10 100
dB −20 -6 -3 0 3 6 20 40
Note that 0 does not
exist on a log axis and so
the starting point of the
axis is arbitrary.
Note: P ∝ V 2 ⇒ decibel power ratios are given by 10 log10P2
P1
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logω
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logω
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)The phase is constant ∀ω.
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)The phase is constant ∀ω.
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.c only affects the line’s vertical position.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)The phase is constant ∀ω.
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.c only affects the line’s vertical position.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)The phase is constant ∀ω.
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.c only affects the line’s vertical position.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)The phase is constant ∀ω.
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.c only affects the line’s vertical position.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)The phase is constant ∀ω.If c > 0, phase = 90◦× magnitude slope.
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.c only affects the line’s vertical position.
If |H| is measured in decibels, a slope of ris called 6r dB/octave or 20r dB/decade.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)The phase is constant ∀ω.If c > 0, phase = 90◦× magnitude slope.
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.c only affects the line’s vertical position.
If |H| is measured in decibels, a slope of ris called 6r dB/octave or 20r dB/decade.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)The phase is constant ∀ω.If c > 0, phase = 90◦× magnitude slope.Negative c adds ±180◦ to the phase.
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
Suppose we plot the magnitude and phase of H = c (jω)r
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.c only affects the line’s vertical position.
If |H| is measured in decibels, a slope of ris called 6r dB/octave or 20r dB/decade.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)The phase is constant ∀ω.If c > 0, phase = 90◦× magnitude slope.Negative c adds ±180◦ to the phase.
Note: Phase angles are modulo 360◦, i.e.+180◦ ≡ −180◦ and 450◦ ≡ 90◦.
Logs of Powers +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 5 / 12
H = c (jω)r
has a straight-line magnitude graph and a constant phase.
Magnitude (log-log graph):|H| = cωr ⇒ log |H| = log |c|+r logωThis is a straight line with a slope of r.c only affects the line’s vertical position.
If |H| is measured in decibels, a slope of ris called 6r dB/octave or 20r dB/decade.
Phase (log-lin graph):∠H = ∠jr + ∠c = r × π
2 (+π if c < 0)The phase is constant ∀ω.If c > 0, phase = 90◦× magnitude slope.Negative c adds ±180◦ to the phase.
Note: Phase angles are modulo 360◦, i.e.+180◦ ≡ −180◦ and 450◦ ≡ 90◦.
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ): H(jω) ≈ 1
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ): H(jω) ≈ 1
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ): H(jω) ≈ 1
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC
Approximate the magnitude responseas two straight lines
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ): H(jω) ≈ 1⇒ |H(jω)| ≈ 1
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC
Approximate the magnitude responseas two straight lines
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ): H(jω) ≈ 1⇒ |H(jω)| ≈ 1
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ |H(jω)| ≈ 1RCω−1
Approximate the magnitude responseas two straight lines
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ): H(jω) ≈ 1⇒ |H(jω)| ≈ 1
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ |H(jω)| ≈ 1RCω−1
Approximate the magnitude responseas two straight lines
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ): H(jω) ≈ 1⇒ |H(jω)| ≈ 1
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ |H(jω)| ≈ 1RCω−1
Approximate the magnitude responseas two straight lines intersecting at thecorner frequency, ωc =
1RC .
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ): H(jω) ≈ 1⇒ |H(jω)| ≈ 1
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ |H(jω)| ≈ 1RCω−1
Approximate the magnitude responseas two straight lines intersecting at thecorner frequency, ωc =
1RC .
At the corner frequency:
(a) the gradient changes by −1 (= −6 dB/octave = −20 dB/decade).
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ): H(jω) ≈ 1⇒ |H(jω)| ≈ 1
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ |H(jω)| ≈ 1RCω−1
Approximate the magnitude responseas two straight lines intersecting at thecorner frequency, ωc =
1RC .
At the corner frequency:
(a) the gradient changes by −1 (= −6 dB/octave = −20 dB/decade).
(b) |H(jωc)| =∣
∣
∣
11+j
∣
∣
∣= 1√
2= −3 dB (worst-case error).
Straight Line Approximations
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 6 / 12
Key idea: (ajω + b) ≈{
ajω for |aω| ≫ |b|b for |aω| ≪ |b|
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ): H(jω) ≈ 1⇒ |H(jω)| ≈ 1
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ |H(jω)| ≈ 1RCω−1
Approximate the magnitude responseas two straight lines intersecting at thecorner frequency, ωc =
1RC .
At the corner frequency:
(a) the gradient changes by −1 (= −6 dB/octave = −20 dB/decade).
(b) |H(jωc)| =∣
∣
∣
11+j
∣
∣
∣= 1√
2= −3 dB (worst-case error).
A linear factor (ajω + b) has a corner frequency of ωc =∣
∣
ba
∣
∣.
Plot Magnitude Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 7 / 12
The gain of a linear circuit is always a rational polynomial in jω and iscalled the transfer function of the circuit. For example:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600
Plot Magnitude Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 7 / 12
The gain of a linear circuit is always a rational polynomial in jω and iscalled the transfer function of the circuit. For example:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomials
Plot Magnitude Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 7 / 12
The gain of a linear circuit is always a rational polynomial in jω and iscalled the transfer function of the circuit. For example:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: Sort corner freqs: 1, 4, 12, 50
Plot Magnitude Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 7 / 12
The gain of a linear circuit is always a rational polynomial in jω and iscalled the transfer function of the circuit. For example:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: Sort corner freqs: 1, 4, 12, 50Step 3: For ω < 1 all linear factors equaltheir constant terms:|H| ≈ 20ω×12
1×4×50 = 1.2ω1.
Plot Magnitude Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 7 / 12
The gain of a linear circuit is always a rational polynomial in jω and iscalled the transfer function of the circuit. For example:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: Sort corner freqs: 1, 4, 12, 50Step 3: For ω < 1 all linear factors equaltheir constant terms:|H| ≈ 20ω×12
1×4×50 = 1.2ω1.
Step 4: For 1 < ω < 4, the factor (jω + 1) ≈ jω so|H| ≈ 20ω×12
ω×4×50 = 1.2ω0
Plot Magnitude Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 7 / 12
The gain of a linear circuit is always a rational polynomial in jω and iscalled the transfer function of the circuit. For example:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: Sort corner freqs: 1, 4, 12, 50Step 3: For ω < 1 all linear factors equaltheir constant terms:|H| ≈ 20ω×12
1×4×50 = 1.2ω1.
Step 4: For 1 < ω < 4, the factor (jω + 1) ≈ jω so|H| ≈ 20ω×12
ω×4×50 = 1.2ω0 = +1.58 dB.
Plot Magnitude Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 7 / 12
The gain of a linear circuit is always a rational polynomial in jω and iscalled the transfer function of the circuit. For example:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: Sort corner freqs: 1, 4, 12, 50Step 3: For ω < 1 all linear factors equaltheir constant terms:|H| ≈ 20ω×12
1×4×50 = 1.2ω1.
Step 4: For 1 < ω < 4, the factor (jω + 1) ≈ jω so|H| ≈ 20ω×12
ω×4×50 = 1.2ω0 = +1.58 dB.
Step 5: For 4 < ω < 12, |H| ≈ 20ω×12ω×ω×50 = 4.8ω−1.
Plot Magnitude Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 7 / 12
The gain of a linear circuit is always a rational polynomial in jω and iscalled the transfer function of the circuit. For example:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: Sort corner freqs: 1, 4, 12, 50Step 3: For ω < 1 all linear factors equaltheir constant terms:|H| ≈ 20ω×12
1×4×50 = 1.2ω1.
Step 4: For 1 < ω < 4, the factor (jω + 1) ≈ jω so|H| ≈ 20ω×12
ω×4×50 = 1.2ω0 = +1.58 dB.
Step 5: For 4 < ω < 12, |H| ≈ 20ω×12ω×ω×50 = 4.8ω−1.
Step 6: For 12 < ω < 50, |H| ≈ 20ω×ωω×ω×50 = 0.4ω0 = −7.96 dB.
Plot Magnitude Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 7 / 12
The gain of a linear circuit is always a rational polynomial in jω and iscalled the transfer function of the circuit. For example:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: Sort corner freqs: 1, 4, 12, 50Step 3: For ω < 1 all linear factors equaltheir constant terms:|H| ≈ 20ω×12
1×4×50 = 1.2ω1.
Step 4: For 1 < ω < 4, the factor (jω + 1) ≈ jω so|H| ≈ 20ω×12
ω×4×50 = 1.2ω0 = +1.58 dB.
Step 5: For 4 < ω < 12, |H| ≈ 20ω×12ω×ω×50 = 4.8ω−1.
Step 6: For 12 < ω < 50, |H| ≈ 20ω×ωω×ω×50 = 0.4ω0 = −7.96 dB.
Step 7: For ω > 50, |H| ≈ 20ω×ωω×ω×ω = 20ω−1.
Plot Magnitude Response
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 7 / 12
The gain of a linear circuit is always a rational polynomial in jω and iscalled the transfer function of the circuit. For example:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: Sort corner freqs: 1, 4, 12, 50Step 3: For ω < 1 all linear factors equaltheir constant terms:|H| ≈ 20ω×12
1×4×50 = 1.2ω1.
Step 4: For 1 < ω < 4, the factor (jω + 1) ≈ jω so|H| ≈ 20ω×12
ω×4×50 = 1.2ω0 = +1.58 dB.
Step 5: For 4 < ω < 12, |H| ≈ 20ω×12ω×ω×50 = 4.8ω−1.
Step 6: For 12 < ω < 50, |H| ≈ 20ω×ωω×ω×50 = 0.4ω0 = −7.96 dB.
Step 7: For ω > 50, |H| ≈ 20ω×ωω×ω×ω = 20ω−1.
At each corner frequency, the graph is continuous but its gradient changesabruptly by +1 (numerator factor) or −1 (denominator factor).
Low and High Frequency Asymptotes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 8 / 12
You can find the low and high frequency asymptotes without factorizing:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Low and High Frequency Asymptotes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 8 / 12
You can find the low and high frequency asymptotes without factorizing:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Low and High Frequency Asymptotes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 8 / 12
You can find the low and high frequency asymptotes without factorizing:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Low Frequency Asymptote:
Low and High Frequency Asymptotes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 8 / 12
You can find the low and high frequency asymptotes without factorizing:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Low Frequency Asymptote:
From factors: HLF(jω) =20jω(12)(1)(4)(50) = 1.2jω
Low and High Frequency Asymptotes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 8 / 12
You can find the low and high frequency asymptotes without factorizing:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Low Frequency Asymptote:
From factors: HLF(jω) =20jω(12)(1)(4)(50) = 1.2jω
Lowest power of jω on top and bottom: H (jω) ≃ 720(jω)600 = 1.2jω
Low and High Frequency Asymptotes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 8 / 12
You can find the low and high frequency asymptotes without factorizing:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Low Frequency Asymptote:
From factors: HLF(jω) =20jω(12)(1)(4)(50) = 1.2jω
Lowest power of jω on top and bottom: H (jω) ≃ 720(jω)600 = 1.2jω
High Frequency Asymptote:
Low and High Frequency Asymptotes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 8 / 12
You can find the low and high frequency asymptotes without factorizing:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Low Frequency Asymptote:
From factors: HLF(jω) =20jω(12)(1)(4)(50) = 1.2jω
Lowest power of jω on top and bottom: H (jω) ≃ 720(jω)600 = 1.2jω
High Frequency Asymptote:
From factors: HHF(jω) =20jω(jω)
(jω)(jω)(jω) = 20 (jω)−1
Low and High Frequency Asymptotes
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 8 / 12
You can find the low and high frequency asymptotes without factorizing:
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Low Frequency Asymptote:
From factors: HLF(jω) =20jω(12)(1)(4)(50) = 1.2jω
Lowest power of jω on top and bottom: H (jω) ≃ 720(jω)600 = 1.2jω
High Frequency Asymptote:
From factors: HHF(jω) =20jω(jω)
(jω)(jω)(jω) = 20 (jω)−1
Highest power of jω on top and bottom: H (jω) ≃ 60(jω)2
3(jω)3= 20 (jω)
−1
Phase Approximation +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 9 / 12
Gain: H(jω) = 1jωRC+1
Phase Approximation +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 9 / 12
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ):
H(jω) ≈ 1
Phase Approximation +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 9 / 12
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ):
H(jω) ≈ 1⇒ ∠1 = 0
Phase Approximation +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 9 / 12
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ):
H(jω) ≈ 1⇒ ∠1 = 0
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC
Phase Approximation +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 9 / 12
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ):
H(jω) ≈ 1⇒ ∠1 = 0
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ ∠j−1 = −π2
Phase Approximation +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 9 / 12
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ):
H(jω) ≈ 1⇒ ∠1 = 0
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ ∠j−1 = −π2
Approximate the phase response asthree straight lines.
Phase Approximation +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 9 / 12
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ):
H(jω) ≈ 1⇒ ∠1 = 0
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ ∠j−1 = −π2
Approximate the phase response asthree straight lines.
By chance, they intersect close to0.1ωc and 10ωc where ωc =
1RC .
Phase Approximation +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 9 / 12
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ):
H(jω) ≈ 1⇒ ∠1 = 0
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ ∠j−1 = −π2
Approximate the phase response asthree straight lines.
By chance, they intersect close to0.1ωc and 10ωc where ωc =
1RC .
Between 0.1ωc and 10ωc the phase changes by −π2 over two decades.
This gives a gradient = −π4 radians/decade.
Phase Approximation +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 9 / 12
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ):
H(jω) ≈ 1⇒ ∠1 = 0
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ ∠j−1 = −π2
Approximate the phase response asthree straight lines.
By chance, they intersect close to0.1ωc and 10ωc where ωc =
1RC .
Between 0.1ωc and 10ωc the phase changes by −π2 over two decades.
This gives a gradient = −π4 radians/decade.
(ajω + b) in denominator⇒ ∆gradient = ∓π
4 /decade at ω = 10∓1∣
∣
ba
∣
∣.
Phase Approximation +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 9 / 12
Gain: H(jω) = 1jωRC+1
Low frequencies (ω ≪ 1RC ):
H(jω) ≈ 1⇒ ∠1 = 0
High frequencies (ω ≫ 1RC ): H(jω) ≈ 1
jωRC⇒ ∠j−1 = −π2
Approximate the phase response asthree straight lines.
By chance, they intersect close to0.1ωc and 10ωc where ωc =
1RC .
Between 0.1ωc and 10ωc the phase changes by −π2 over two decades.
This gives a gradient = −π4 radians/decade.
(ajω + b) in denominator⇒ ∆gradient = ∓π
4 /decade at ω = 10∓1∣
∣
ba
∣
∣.
The sign of ∆gradient is reversed for (a) numerator factors and (b) ba < 0.
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Step 6: At ω = 0.1 gradient becomes −π4 rad/decade. φ is still π
2 .
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Step 6: At ω = 0.1 gradient becomes −π4 rad/decade. φ is still π
2 .Step 7: At ω = 0.4, φ = π
2 − 0.6π4 = 0.35π. New gradient is −π
2 .
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Step 6: At ω = 0.1 gradient becomes −π4 rad/decade. φ is still π
2 .Step 7: At ω = 0.4, φ = π
2 − 0.6π4 = 0.35π. New gradient is −π
2 .Step 8: At ω = 1.2, φ = 0.35π − 0.48π
2 = 0.11π. New gradient is −π4 .
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Step 6: At ω = 0.1 gradient becomes −π4 rad/decade. φ is still π
2 .Step 7: At ω = 0.4, φ = π
2 − 0.6π4 = 0.35π. New gradient is −π
2 .Step 8: At ω = 1.2, φ = 0.35π − 0.48π
2 = 0.11π. New gradient is −π4 .
Steps 9-13: Repeat for each gradient change.
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Step 6: At ω = 0.1 gradient becomes −π4 rad/decade. φ is still π
2 .Step 7: At ω = 0.4, φ = π
2 − 0.6π4 = 0.35π. New gradient is −π
2 .Step 8: At ω = 1.2, φ = 0.35π − 0.48π
2 = 0.11π. New gradient is −π4 .
Steps 9-13: Repeat for each gradient change.
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Step 6: At ω = 0.1 gradient becomes −π4 rad/decade. φ is still π
2 .Step 7: At ω = 0.4, φ = π
2 − 0.6π4 = 0.35π. New gradient is −π
2 .Step 8: At ω = 1.2, φ = 0.35π − 0.48π
2 = 0.11π. New gradient is −π4 .
Steps 9-13: Repeat for each gradient change.
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Step 6: At ω = 0.1 gradient becomes −π4 rad/decade. φ is still π
2 .Step 7: At ω = 0.4, φ = π
2 − 0.6π4 = 0.35π. New gradient is −π
2 .Step 8: At ω = 1.2, φ = 0.35π − 0.48π
2 = 0.11π. New gradient is −π4 .
Steps 9-13: Repeat for each gradient change.
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Step 6: At ω = 0.1 gradient becomes −π4 rad/decade. φ is still π
2 .Step 7: At ω = 0.4, φ = π
2 − 0.6π4 = 0.35π. New gradient is −π
2 .Step 8: At ω = 1.2, φ = 0.35π − 0.48π
2 = 0.11π. New gradient is −π4 .
Steps 9-13: Repeat for each gradient change. Final gradient is always 0.
Plot Phase Response +
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 10 / 12
H(jω) = 60(jω)2+720(jω)
3(jω)3+165(jω)2+762(jω)+600= 20jω(jω+12)
(jω+1)(jω+4)(jω+50)
Step 1: Factorize the polynomialsStep 2: List corner freqs: ± = num/denωc = {1−, 4−, 12+, 50−}
Step 3: Gradient changes at 10∓1ωc.Sign depends on num/den and sgn
(
ba
)
:.1−, 10+; .4−, 40+; 1.2+, 120−; 5−, 500+
Step 4: Put in ascending order and calculate gaps as log10ω2
ω1
decades:
.1− (.6) .4− (.48) 1.2+ (.62) 5− (.3) 10+ (.6) 40+ (.48) 120− (.62) 500+.
Step 5: Find phase of LF asymptote: ∠1.2jω = +π2 .
Step 6: At ω = 0.1 gradient becomes −π4 rad/decade. φ is still π
2 .Step 7: At ω = 0.4, φ = π
2 − 0.6π4 = 0.35π. New gradient is −π
2 .Step 8: At ω = 1.2, φ = 0.35π − 0.48π
2 = 0.11π. New gradient is −π4 .
Steps 9-13: Repeat for each gradient change. Final gradient is always 0.
At 0.1 and 10 times each corner frequency, the graph is continuous but itsgradient changes abruptly by ±π
4 rad/decade.
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
Corner freqs: 0.25RC
−, 1RC
+
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
Corner freqs: 0.25RC
−, 1RC
+LF Asymptote: H(jω) = 1
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
Corner freqs: 0.25RC
−, 1RC
+LF Asymptote: H(jω) = 1
Magnitude Response:
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
Corner freqs: 0.25RC
−, 1RC
+LF Asymptote: H(jω) = 1
Magnitude Response:Gradient Changes: −20 dB/dec at ω = 0.25
RC and +20 at ω = 1RC
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
Corner freqs: 0.25RC
−, 1RC
+LF Asymptote: H(jω) = 1
Magnitude Response:Gradient Changes: −20 dB/dec at ω = 0.25
RC and +20 at ω = 1RC
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
Corner freqs: 0.25RC
−, 1RC
+LF Asymptote: H(jω) = 1
Magnitude Response:Gradient Changes: −20 dB/dec at ω = 0.25
RC and +20 at ω = 1RC
Line equations: H(jω) = (a) 1, (b) 14jωRC , (c) jωRC
4jωRC = 0.25
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
Corner freqs: 0.25RC
−, 1RC
+LF Asymptote: H(jω) = 1
Magnitude Response:Gradient Changes: −20 dB/dec at ω = 0.25
RC and +20 at ω = 1RC
Line equations: H(jω) = (a) 1, (b) 14jωRC , (c) jωRC
4jωRC = 0.25
Phase Response:
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
Corner freqs: 0.25RC
−, 1RC
+LF Asymptote: H(jω) = 1
Magnitude Response:Gradient Changes: −20 dB/dec at ω = 0.25
RC and +20 at ω = 1RC
Line equations: H(jω) = (a) 1, (b) 14jωRC , (c) jωRC
4jωRC = 0.25
Phase Response:LF asymptote: φ = ∠1 = 0
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
Corner freqs: 0.25RC
−, 1RC
+LF Asymptote: H(jω) = 1
Magnitude Response:Gradient Changes: −20 dB/dec at ω = 0.25
RC and +20 at ω = 1RC
Line equations: H(jω) = (a) 1, (b) 14jωRC , (c) jωRC
4jωRC = 0.25
Phase Response:LF asymptote: φ = ∠1 = 0
Gradient changes of ±π4 /decade at: ω = 0.025
RC
−, 0.1RC
+, 2.5RC
+, 10RC
−.
RCR Circuit
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 11 / 12
YX =
R+ 1
jωC
3R+R+ 1
jωC
= jωRC+14jωRC+1
Corner freqs: 0.25RC
−, 1RC
+LF Asymptote: H(jω) = 1
Magnitude Response:Gradient Changes: −20 dB/dec at ω = 0.25
RC and +20 at ω = 1RC
Line equations: H(jω) = (a) 1, (b) 14jωRC , (c) jωRC
4jωRC = 0.25
Phase Response:LF asymptote: φ = ∠1 = 0
Gradient changes of ±π4 /decade at: ω = 0.025
RC
−, 0.1RC
+, 2.5RC
+, 10RC
−.
At ω = 0.1RC , φ = 0− π
4 log100.1
0.025 = −π4 × 0.602 = −0.15π
Summary
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 12 / 12
• Frequency response: magnitude and phase of YX as a function of ω
◦ Only applies to sine waves
Summary
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 12 / 12
• Frequency response: magnitude and phase of YX as a function of ω
◦ Only applies to sine waves
◦ Use log axes for frequency and gain but linear for phase
⊲ Decibels = 20 log10V2
V1
= 10 log10P2
P1
Summary
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 12 / 12
• Frequency response: magnitude and phase of YX as a function of ω
◦ Only applies to sine waves
◦ Use log axes for frequency and gain but linear for phase
⊲ Decibels = 20 log10V2
V1
= 10 log10P2
P1
• Linear factor (ajω + b) gives corner frequency at ω =∣
∣
ba
∣
∣.
◦ Magnitude plot gradient changes by ±20 dB/decade@ω =∣
∣
ba
∣
∣.
Summary
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 12 / 12
• Frequency response: magnitude and phase of YX as a function of ω
◦ Only applies to sine waves
◦ Use log axes for frequency and gain but linear for phase
⊲ Decibels = 20 log10V2
V1
= 10 log10P2
P1
• Linear factor (ajω + b) gives corner frequency at ω =∣
∣
ba
∣
∣.
◦ Magnitude plot gradient changes by ±20 dB/decade@ω =∣
∣
ba
∣
∣.
◦ Phase gradient changes in two places by:
⊲ ±π4 rad/decade@ω = 0.1×
∣
∣
ba
∣
∣
⊲ ∓π4 rad/decade@ω = 10×
∣
∣
ba
∣
∣
Summary
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 12 / 12
• Frequency response: magnitude and phase of YX as a function of ω
◦ Only applies to sine waves
◦ Use log axes for frequency and gain but linear for phase
⊲ Decibels = 20 log10V2
V1
= 10 log10P2
P1
• Linear factor (ajω + b) gives corner frequency at ω =∣
∣
ba
∣
∣.
◦ Magnitude plot gradient changes by ±20 dB/decade@ω =∣
∣
ba
∣
∣.
◦ Phase gradient changes in two places by:
⊲ ±π4 rad/decade@ω = 0.1×
∣
∣
ba
∣
∣
⊲ ∓π4 rad/decade@ω = 10×
∣
∣
ba
∣
∣
• LF/HF asymptotes: keep only the terms with the lowest/highest powerof jω in numerator and denominator polynomials
Summary
11: Frequency Responses
• Frequency Response
• Sine Wave Response
• Logarithmic axes
• Logs of Powers +
• Straight LineApproximations
• Plot Magnitude Response
• Low and High FrequencyAsymptotes
• Phase Approximation +
• Plot Phase Response +
• RCR Circuit
• Summary
E1.1 Analysis of Circuits (2018-10340) Frequency Responses: 11 – 12 / 12
• Frequency response: magnitude and phase of YX as a function of ω
◦ Only applies to sine waves
◦ Use log axes for frequency and gain but linear for phase
⊲ Decibels = 20 log10V2
V1
= 10 log10P2
P1
• Linear factor (ajω + b) gives corner frequency at ω =∣
∣
ba
∣
∣.
◦ Magnitude plot gradient changes by ±20 dB/decade@ω =∣
∣
ba
∣
∣.
◦ Phase gradient changes in two places by:
⊲ ±π4 rad/decade@ω = 0.1×
∣
∣
ba
∣
∣
⊲ ∓π4 rad/decade@ω = 10×
∣
∣
ba
∣
∣
• LF/HF asymptotes: keep only the terms with the lowest/highest powerof jω in numerator and denominator polynomials
For further details see Hayt Ch 16 or Irwin Ch 12.
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