RevisionLecture3: FrequencyResponses - Imperial … Responses RevisionLecture3: FrequencyResponses •Frequency Responses •Plotting Frequency Responses •LF and HF Asymptotes •Corner

Revision Lecture 3:Frequency Responses

Revision Lecture 3:FrequencyResponses

• Frequency Responses

• Plotting FrequencyResponses

• LF and HF Asymptotes

• Corner frequencies (linearfactors)

• Sketching MagnitudeResponses (linear factors)

• Sketching Phase response

• Phase Response Example

• Filters

• Resonance

E1.1 Analysis of Circuits (2015-6186) Revision Lecture 3 – 1 / 10

Frequency Responses









• Filters

• Resonance


• Frequency response is the ratio of two voltage/current phasors in acircuit: output phasor ÷ input phasor.

◦ It is a complex number that depends on frequency.

◦ Calculate using nodal analysis.

• For a linear circuit, the frequency response is always a ratio of twopolynomials in jω having real coefficients that depend on thecomponent values.

Frequency Responses









• Filters

• Resonance


• Frequency response is the ratio of two voltage/current phasors in acircuit: output phasor ÷ input phasor.

◦ It is a complex number that depends on frequency.

◦ Calculate using nodal analysis.

• For a linear circuit, the frequency response is always a ratio of twopolynomials in jω having real coefficients that depend on thecomponent values.

YX = R+1/jωC

4R+1/jωC= jωRC+1

jω4RC+1

Plotting Frequency Responses









• Filters

• Resonance


• Plot the magnitude response and phase response as separate graphs.Use log scale for frequency and magnitude and linear scale for phase:this gives graphs that can be approximated by straight line segments.










• Filters

• Resonance



• If V2

V1

= A (jω)k = A× jk × ωk

◦ magnitude is a straight line with gradient k:

log∣

∣

∣

V2

V1

∣

∣

∣= log |A|+ k logω










• Filters

• Resonance



• If V2

V1



log∣

∣

∣

V2

V1

∣

∣


◦ phase is a constant k × π2 (+π if A < 0):

∠

(

V2

V1

)

= ∠A+ k∠j = ∠A+ k π2










• Filters

• Resonance



• If V2

V1



log∣

∣

∣

V2

V1

∣

∣



∠

(

V2

V1

)

= ∠A+ k∠j = ∠A+ k π2

• Measure magnitude using decibels = 20 log10|V2||V1| .










• Filters

• Resonance



• If V2

V1



log∣

∣

∣

V2

V1

∣

∣



∠

(

V2

V1

)

= ∠A+ k∠j = ∠A+ k π2


A gradient of k on log axes is the same as a gradient of 20k dB/decade(×10 in frequency) or 6k dB/octave ( ×2 in frequency).










• Filters

• Resonance



• If V2

V1



log∣

∣

∣

V2

V1

∣

∣



∠

(

V2

V1

)

= ∠A+ k∠j = ∠A+ k π2



YX =

1

jωC

R+ 1

jωC

= 1jωRC+1










• Filters

• Resonance



• If V2

V1



log∣

∣

∣

V2

V1

∣

∣



∠

(

V2

V1

)

= ∠A+ k∠j = ∠A+ k π2



0.1/RC 1/RC 10/RC-30

-20

-10

0

ω (rad/s)

|Gai

n| (

dB)

0.1/RC 1/RC 10/RC

-0.4

-0.2

0

ω (rad/s)

Pha

se/π

YX =

1

jωC

R+ 1

jωC

= 1jωRC+1=

1jωωc

+1where ωc =

1RC

LF and HF Asymptotes









• Filters

• Resonance


• Frequency response is always a ratio of two polynomials in jω withreal coefficients that depend on the component values.

◦ The terms with the lowest power of jω on top and bottom givesthe low-frequency asymptote

◦ The terms with the highest power of jω on top and bottomgives the high-frequency asymptote.










• Filters

• Resonance





Example: H(jω) = 60(jω)2+720(jω)

3(jω)3+165(jω)2+762(jω)+600

0.1 1 10 100 1000

-40

-20

0

ω (rad/s)0.1 1 10 100 1000

-0.5

0

0.5

ω (rad/s)

π










• Filters

• Resonance






3(jω)3+165(jω)2+762(jω)+600

0.1 1 10 100 1000

-40

-20

0

ω (rad/s)0.1 1 10 100 1000

-0.5

0

0.5

ω (rad/s)

πLF: H(jω) ≃ 1.2jω










• Filters

• Resonance






3(jω)3+165(jω)2+762(jω)+600

0.1 1 10 100 1000

-40

-20

0

ω (rad/s)0.1 1 10 100 1000

-0.5

0

0.5

ω (rad/s)

πLF: H(jω) ≃ 1.2jω

HF: H(jω) ≃ 20 (jω)−1

Corner frequencies (linear factors)









• Filters

• Resonance


• We can factorize the numerator and denominator into linear terms of

the form (ajω + b) ≃

{

b ω <∣

∣

ba

∣

∣

ajω ω >∣

∣

ba

∣

∣

.










• Filters

• Resonance




{

b ω <∣

∣

ba

∣

∣

ajω ω >∣

∣

ba

∣

∣

.

• At the corner frequency, ωc =∣

∣

ba

∣

∣ the slope of the magnituderesponse changes by ±1 ( ≡ ±20 dB/decade) because the linearterm introduces another factor of ω into the numerator ordenominator for ω > ωc.










• Filters

• Resonance




{

b ω <∣

∣

ba

∣

∣

ajω ω >∣

∣

ba

∣

∣

.


∣

ba

∣


• The phase changes by ±π2 because the linear term introduces

another factor of j into the numerator or denominator for ω > ωc.

◦ The phase change is gradual and takes place over the range0.1ωc to 10ωc (±π

2 spread over two decades in ω).










• Filters

• Resonance




{

b ω <∣

∣

ba

∣

∣

ajω ω >∣

∣

ba

∣

∣

.


∣

ba

∣






• When a and b are real and positive, it is often convenient to write

(ajω + b) = b(

jωωc

+ 1)

.










• Filters

• Resonance




{

b ω <∣

∣

ba

∣

∣

ajω ω >∣

∣

ba

∣

∣

.


∣

ba

∣






• When a and b are real and positive, it is often convenient to write

(ajω + b) = b(

jωωc

+ 1)

.

• The corner frequencies are the absolute values of the roots (valuesof jω) of the numerator and denominator polynomials.

Sketching Magnitude Responses (linear factors)









• Filters

• Resonance


1. Find corner frequencies: (a) factorize the numerator/denominatorpolynomials or (b) find their roots

2. Find LF and HF asymptotes. A (jω)k

has a slope of k; substitutethe value of ωc to get the value at first/last corner frequency.

3. At a corner frequency, the gradient of the magnitude responsechanges by ±1 (±20 dB/decade). + for numerator (top line) and −for denominator (bottom line).

4. |H(jω2)| =(

ω2

ω1

)k

|H(jω1)| if the gradient between them is k.










• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k


H(jω) = 1.2jω( jω

12+1)

( jω1+1)( jω

4+1)( jω

50+1)

0.1 1 10 100 1000

-40

-20

0

ω (rad/s)










• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k



12+1)

( jω1+1)( jω

4+1)( jω

50+1)

LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)

0.1 1 10 100 1000

-40

-20

0

ω (rad/s)










• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k



12+1)

( jω1+1)( jω

4+1)( jω

50+1)

LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)

|H(j4)| =(

41

)0× 1.2 = 1.2 0.1 1 10 100 1000

-40

-20

0

ω (rad/s)










• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k



12+1)

( jω1+1)( jω

4+1)( jω

50+1)

LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)

|H(j4)| =(

41

)0× 1.2 = 1.2

|H(j12)| =(

124

)−1× 1.2 = 0.4

0.1 1 10 100 1000

-40

-20

0

ω (rad/s)










• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k



12+1)

( jω1+1)( jω

4+1)( jω

50+1)

LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)

|H(j4)| =(

41

)0× 1.2 = 1.2

|H(j12)| =(

124

)−1× 1.2 = 0.4

0.1 1 10 100 1000

-40

-20

0

ω (rad/s)

|H(j50)| =(

5012

)0× 0.4 = 0.4 (−8 dB).










• Filters

• Resonance






4. |H(jω2)| =(

ω2

ω1

)k



12+1)

( jω1+1)( jω

4+1)( jω

50+1)

LF: 1.2jω ⇒|H(j1)| = 1.2 (1.6 dB)

|H(j4)| =(

41

)0× 1.2 = 1.2

|H(j12)| =(

124

)−1× 1.2 = 0.4

0.1 1 10 100 1000

-40

-20

0

ω (rad/s)

|H(j50)| =(

5012

)0× 0.4 = 0.4 (−8 dB). As a check: HF: 20 (jω)−1

Sketching Phase response









• Filters

• Resonance


• Phase response is more complicated than the magnitude response:(a) there are twice as many corner frquencies(b) negative factors add an extra π onto the phase(c) Phases wrap around: 2π ≡ 0.










• Filters

• Resonance



• A corner in the magnitude response at ωc gives rise to phaseresponse corners at 0.1ωc and 10ωc. The gradient change at eachcorner is ±π

4 radians per decade (⇒ gradient = m× π4 ).










• Filters

• Resonance





◦ For a factor (ajω + b) with ba > 0 (normally true) the first

change is in the same direction as for the magnitude response










• Filters

• Resonance







• a numerator factor has a positive gradient change atω = 0.1

∣

∣

ba

∣

∣ and negative at ω = 10∣

∣

ba

∣

∣.










• Filters

• Resonance








∣

∣

ba

∣


∣

ba

∣

∣.

◦ The signs of the gradient changes are reversed if ba < 0 and

reversed again for the denominator.










• Filters

• Resonance








∣

∣

ba

∣


∣

ba

∣

∣.



• LF asymptote = A (jω)k ⇒ LF phase = k π2 (+π if A < 0)










• Filters

• Resonance








∣

∣

ba

∣


∣

ba

∣

∣.



• LF asymptote = A (jω)k ⇒ LF phase = k π2 (+π if A < 0)

• When slope is mπ4 : ∠H(jω2) = mπ

4 × log10

(

ω2

ω1

)

+ ∠H(jω1)

Phase Response Example









• Filters

• Resonance


YX =

R+ 1

jωC

4R+ 1

jωC

= jωRC+14jωRC+1

0.1 1 10

-10

-5

0

ω RC0.1 1 10

-0.2

-0.1

0

ω RC

π










• Filters

• Resonance


YX =

R+ 1

jωC

4R+ 1

jωC

= jωRC+14jωRC+1 ⇒ ωD = 1

4RC = 0.25RC

0.1 1 10

-10

-5

0

ω RC0.1 1 10

-0.2

-0.1

0

ω RC

π

Phase Response: (all gradient changes ∆ = ±π4 /decade)

Denominator ωD = 0.25RC ⇒∆ = − at ω1 = 0.025

RC and + at ω3 = 2.5RC .










• Filters

• Resonance


YX =

R+ 1

jωC

4R+ 1

jωC


4RC = 0.25RC , ωN = 1

RC

0.1 1 10

-10

-5

0

ω RC0.1 1 10

-0.2

-0.1

0

ω RC

π




Numerator ωN = 1RC ⇒∆ = + at ω2 = 0.1

RC and − at ω4 = 10RC .










• Filters

• Resonance


YX =

R+ 1

jωC

4R+ 1

jωC


4RC = 0.25RC , ωN = 1

RC

0.1 1 10

-10

-5

0

ω RC0.1 1 10

-0.2

-0.1

0

ω RC

π

ω1

ω2

ω3

ω4















• Filters

• Resonance


YX =

R+ 1

jωC

4R+ 1

jωC


4RC = 0.25RC , ωN = 1

RC

0.1 1 10

-10

-5

0

ω RC0.1 1 10

-0.2

-0.1

0

ω RC

π

ω1

ω2

ω3

ω4






Corner ValuesLF asmptote: H(jω) = 1 ⇒ ∠H(jω1) = 0










• Filters

• Resonance


YX =

R+ 1

jωC

4R+ 1

jωC


4RC = 0.25RC , ωN = 1

RC

0.1 1 10

-10

-5

0

ω RC0.1 1 10

-0.2

-0.1

0

ω RC

π

ω1

ω2

ω3

ω4







∠H(jω2) = ∠H(jω1)−π4 log10

(

ω2

ω1

)

= −π4 log10 (4) = −0.151π










• Filters

• Resonance


YX =

R+ 1

jωC

4R+ 1

jωC


4RC = 0.25RC , ωN = 1

RC

0.1 1 10

-10

-5

0

ω RC0.1 1 10

-0.2

-0.1

0

ω RC

π

ω1

ω2

ω3

ω4







∠H(jω2) = ∠H(jω1)−π4 log10

(

ω2

ω1

)

= −π4 log10 (4) = −0.151π

∠H(jω3) = ∠H(jω2)− 0× log10

(

ω3

ω2

)

= −0.151π










• Filters

• Resonance


YX =

R+ 1

jωC

4R+ 1

jωC


4RC = 0.25RC , ωN = 1

RC

0.1 1 10

-10

-5

0

ω RC0.1 1 10

-0.2

-0.1

0

ω RC

π

ω1

ω2

ω3

ω4







∠H(jω2) = ∠H(jω1)−π4 log10

(

ω2

ω1

)

= −π4 log10 (4) = −0.151π

∠H(jω3) = ∠H(jω2)− 0× log10

(

ω3

ω2

)

= −0.151π

∠H(jω4) = ∠H(jω3) +π4 log10

(

ω4

ω3

)

= −0.151π + π4 log10 (4) = 0

Filters









• Filters

• Resonance


• Filter: a circuit designed to amplify some frequencies and/orattenuate others. Very widely used.

• The order of the filter is the highest power of jω in the denominatorof the frequency response.

• Often use op-amps to provide a low impedance output.

Filters









• Filters

• Resonance





YX = R

R+1/jωC

Filters









• Filters

• Resonance





YX = R

R+1/jωC= jωRC

jωRC+1

Filters









• Filters

• Resonance





YX = R

R+1/jωC= jωRC

jωRC+1 =jωRCjωa

+1

Filters









• Filters

• Resonance





0.1a a 10a

-30

-20

-10

0

ω

YX = R

R+1/jωC= jωRC

jωRC+1 =jωRCjωa

+1

Filters









• Filters

• Resonance





0.1a a 10a

-30

-20

-10

0

ω

YX = R

R+1/jωC= jωRC

jωRC+1 =jωRCjωa

+1

ZX =

(

1 + RB

RA

)

× jωRCjωa

+1

Resonance


• Resonant circuits have quadratic factors that cannot be factorized

◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)

◦ Resonant frequency: ω0 =√

ca determines the horizontal position

◦ Damping Factor: ζ = bω0

2c = b√4ac

determines the response shape

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)




2c = b√4ac


◦ Equivalently Quality Factor: Q ,ω×Stored EnergyPower Dissipation ≈ 1

2ζ = cbω0

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)




2c = b√4ac



2ζ = cbω0

• At ω = ω0, outer terms cancel (a (jω)2= −c): ⇒ H(jω) = jbω0 = 2jcζ

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)




2c = b√4ac



2ζ = cbω0


◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)




2c = b√4ac



2ζ = cbω0



XU =

1

jωC

R+jωL+ 1

jωC

= 1(jω)2LC+jωRC+1

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)




2c = b√4ac



2ζ = cbω0



XU =

1

jωC

R+jωL+ 1

jωC

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

2

√

CL , Q = ω0L

R = 12ζ

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)




2c = b√4ac



2ζ = cbω0



R = 5, 20, 60, 120

ζ = 140 ,

110 ,

310 ,

610

100 1k 10k-40

-20

0

20

ω (rad/s)

R=5

R=120

XU =

1

jωC

R+jωL+ 1

jωC

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

2

√

CL , Q = ω0L

R = 12ζ

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)




2c = b√4ac



2ζ = cbω0



R = 5, 20, 60, 120

ζ = 140 ,

110 ,

310 ,

610

Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,

56

100 1k 10k-40

-20

0

20

ω (rad/s)

R=5

R=120

XU =

1

jωC

R+jωL+ 1

jωC

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

2

√

CL , Q = ω0L

R = 12ζ

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)




2c = b√4ac



2ζ = cbω0



R = 5, 20, 60, 120

ζ = 140 ,

110 ,

310 ,

610

Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,

56

PeakGainCornerGain = 1

2ζ ≈ Q

100 1k 10k-40

-20

0

20

ω (rad/s)

R=5

R=120

XU =

1

jωC

R+jωL+ 1

jωC

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

2

√

CL , Q = ω0L

R = 12ζ

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)




2c = b√4ac



2ζ = cbω0


◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.◦ 3 dB bandwidth of peak ≃ 2ζω0 ≈ ω0

Q .

R = 5, 20, 60, 120

ζ = 140 ,

110 ,

310 ,

610

Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,

56


2ζ ≈ Q

100 1k 10k-40

-20

0

20

ω (rad/s)

R=5

R=120

XU =

1

jωC

R+jωL+ 1

jωC

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

2

√

CL , Q = ω0L

R = 12ζ

Resonance



◦ H(jω) = a (jω)2 + bjω + c = c

(

(

jωω0

)2

+ 2ζ(

jωω0

)

+ 1

)




2c = b√4ac



2ζ = cbω0


◦ |H(jω0)| = 2ζ times the straight line approximation at ω0.◦ 3 dB bandwidth of peak ≃ 2ζω0 ≈ ω0

Q . ∆phase = ±π over 2ζ decades

R = 5, 20, 60, 120

ζ = 140 ,

110 ,

310 ,

610

Q = |ZC(ω0) or ZL(ω0)|R = 20, 5, 53 ,

56


2ζ ≈ Q

100 1k 10k-40

-20

0

20

ω (rad/s)

R=5

R=120

XU =

1

jωC

R+jωL+ 1

jωC

= 1(jω)2LC+jωRC+1

ω0 =√

1LC , ζ = R

2

√

CL , Q = ω0L

R = 12ζ

Documents

RevisionLecture3: FrequencyResponses - Imperial … Responses RevisionLecture3: FrequencyResponses •Frequency Responses •Plotting Frequency Responses •LF and HF Asymptotes •Corner