Poles: 1E4 Two regions Range ~H(jω) 20log|H(jω)| Angle ...ssawyer/CircuitsFall2018_all/... · Electric Circuits HW09 solution Fall 2018 Problem 3) 2nd Order Filter Circuits R L

Electric Circuits HW09 solution Fall 2018

Problem 1) Sketching Bode Plots

Sketch the 'ideal' Bode plots of the magnitude (dB) and phase for the following transfer functions.

Note to grader: Students may also include0.1wc and 10wc which is fine.Give full credit if they show their work and madethe correct Bode chart regardless of how theyinterpreted the chart below.

1.1: H s( ) 10s

s 1 104

=

Zeros: 0 Poles: 1E4 Two regions

Range ~H(jω) 20log|H(jω)| Angle [degrees]

ω<1E4 -s1E-3 +20dB/decade -90

ω = 1E4 17dB -135

1E4<ω -10 20dB -180

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6110

5000

Ess

ssH

1.2:

s < 10, H(s) ~5E3/1E7= 5E-4, constant 10<s<5000, H(s) ~5E-3/s, -20dB/decade 5000<s<1E6, H(s) ~ 1/E-6, constant s<1E6, H(s) ~1/s, -20dB/decade

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Problem 2) Transfer functions, Bode plots, first order circuits

Determine the transfer functions in the following circuits

U1

OPAMP

+

-

OUTU2

OPAMP

+

-

OUTR1

1k

R2

4k

R32k

R4

8k

R5

100

L10.01

1

2

0

0

0Vin Vout

2.1 Draw the above circuit as a three stage network (put boxes around the stages for example andlabel Hn(2) n=1,2.3....etc). Indicate the tranfer function for each stage.

H1(s) = -R2/R1=-4 H2(s) =sL1/(R5+sL1)=s/(s+1E4) H3(s) = 1+R4/R3=1+4=5

H1(s)

H2(s)

H3(s)

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2.2 Determine the transfer functions, H(s) Vout(s)/Vin(s), for the circuit.

H(s)=H1(s)H2(s)H3(s) = -20 s/(s+1E4)

2.3. What is the gain of the circuit?

K = 20 2.4 What is the cutoff frequency?

ωc = 1E4 [rad/s]

2.5 Sketch an approximate Bode (dB-log) plot of the magnitude.

2.6 Sketch an approximate Bode (phase-log) plot of the phase.

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U1

uA741

+3

-2

V+7

V-4

OUT6

OS11

OS25

R2

4k

Vin

R42k

C131.2E-9

0

0

R5

50k U2

uA741

+3

-2

V+7

V-4

OUT6

OS11

OS25

R3

2kR62k

0

0

R7

500

L17.95E-3

1

2Vout

2.7. Draw the above circuit as a three stage network (put boxes around the stages for example andlabel Hn(2) n=1,2.3....etc). Indicate the tranfer function for each stage.

H1(s) – Low pass filter, 1

11

c

c

ssH

with 64192.31350

11

EEc

H2(s) – Non-inverting amplifier, 34

211

R

RsH

H3(s) – High pass filter, 2

1cs

ssH

with k

Ec 63395.7

5002

H4(s) – Non-inverting amplifier, 24

211

R

RsH

H1(s)

H4(s)

H3(s)

H2(s)

2.8 What is the transfer function of the circuit?

ks

s

ss

s

sKsHsHsHsHsH

cc

c

63641

6416

21

14321

not so great bandpass filter, low gain in passband

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2.9 Sketch an approximate Bode (db-log) of the magnitude.

2.10. Sketch an approximate Bode (phase-log) plot of the phase.

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Problem 3) 2nd Order Filter Circuits

R L

1E-4

1 2 C

1E-10

V1

3.1 What is the resonant frequency of the circuit?

The transfer function will take the form

22

2 21oss

sN

LCs

L

Rs

sNsH

giving ωo = sqrt(1/LC) = 1E7 [rad/s]

ω01 1 107

3.2. When R = 100.01 kΩ, R1 100.01kΩ L1 1 104H

3.2.1 What is the damping ratio, ζ?

α1

R1

2 L1

α1 5 108

1

s

ζ1

α1

ω01

ζ1 50.0051

s ζ 50 1=

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3.2.2. Sketch the Bode plot of the magnitude (dB-log) when Vout is the voltage across theinductor, H(s) = VL(s)/V1(s)

(Answer check: 20log(|H|) = -120dB at s = 1E4) (Answer check: 20log(|H|) = -40dB at s = 1E7)

14190001.12

2

EsEs

ssH

Zeros: 0 (double) Poles: 1E5,1E9

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3.2.3. Sketch the Bode plot of the magnitude (dB-log) when Vout is the voltage across the resistor H(s) = VR(s)/V1(s)

(Answer check: 20log(|H|) = -20dB at s = 1E4) (Answer check: 20log(|H|) = 0dB at s = 1E7)

14190001.1

90001.12 EsEs

sEsH

Zeros: 0 Poles: 1E5,1E9

3.3. When R = 2000Ω,

3.3.1. What is the damping ratio, ζ?

α = R/(2L) = 2000/(2*1E-4) = 1E7 [rad/s] δ = 1E7/1E7 ~ 1

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3.3.2. Sketch the Bode plot of the magnitude (dB-log) when Vout is the voltage across the inductor,H(s) = VL(s)/V1(s)

(Answer check: 20log(|H|) = -120dB at s = 1E4) (Answer check: 20log(|H|) = -6dB at s = 1E7, with correction)

141722

2

EsEs

ssH

Zeros: 0 (double) Poles: 1E7 (double The asymptotes are shown as the dashed red line. The correction to the straight line approximate at the resonant frequency is 20log(1/2δ) = -6dB

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3.3.3. Sketch the Bode plot of the magnitude (dB-log) when Vout is the voltage across the resistor, H(s) = VR(s)/V1(s)

(Answer check: 20log(|H|) = -54dB at s = 1E4) (Answer check: 20log(|H|) = 0dB at s = 1E7)

14172

722 EsEs

sEsH

Zeros: 0 Poles: 1E7 (double) Applying HW8 concepts to the overdamped circuit The asymptotes are shown as the dashed red line. Additionally, we locate the vertex as 0-20log(1/2δ)=0-(-6dB) = 6dB

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Problem 4) Design problems-Transfer functions

For the problem design specifications, determine a transfer function that meets the requirements. The answers may not be unique. You should provide a plot that verifies that your transfer function meets the specifications.

1. Design 1 a. Bandpass filter with a passband of 100 [rad/s] to 100E3 [rad/s] b. In the passband, the gain should be 0<gain<10dB c. The rolloff (slope) in the stopbands should have a magnitude of 40dB/decade

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1. Design 2 a. Narrow bandpass filter centered at f = 960MHz b. The gain at the center of the passband should be 20dB c. The 17dB points should be ~940MHz at ~980MHz

ωo = 2πf = 6.031E9 [rad/s] A narrow passband filter is an strongly underdamped circuit. In order to obtain to obtain a 20dB gain at the resonant frequency, a gain term of 100 is needed. The transfer function than takes the form

9031.62

2100

2 Ess

ssH

Testing a damping ratio of 0.05 gives α ~ 7.5E7

Zooming into the resonant frequency

Diagrams hererepresent adifferent problem.This one won't begraded.

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Design a filter that meets the specifications below. You need to pick values for any resistors, capacitors or inductors in your circuit. Simulate the circuit in PSpice to verify that your design meets specifications. Note, small deviations from the design specificationsare allowed and in a real circuit always exist, but they need to be small. Show calculations to justify your design (guessing in PSpice is not a design solution).

ω [rad/s] |H(s)| in dB

10 6

100 6

1000 6

1E4 3

1E5 -14

1E6 -34

The passband has a gain of 6dB, correspond to a gain term of 2. At 1E4, a -3dB point exists and after that there is a drop of -20dB/decade. The response is characteristic of a first order low pass filter with a cutoff frequency of 1E4 rad/s cascade with an 2 gain amplifier circuit. Using an RC circuit, pick appropriate values for R an C such that 1/RC = 1E4 and set the output voltage of the first order RC stage across the capacitor.

U4

OPAMP

+

-OUT

R7

100

0

R8

2k

R91k

0

VoutVin

C11E-6

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Design a filter that meets the specifications below. You need to pick values for any resistors, capacitors or inductors in your circuit. Simulate the circuit in PSpice to verify that your design meets specifications. Use the ideal amplifier component called OPAMP in your simulations. Note, small deviations from the design specifications are allowed and in a real circuit always exist, but they need to be small. Show calculations to justify your design (guessing in PSpice is not a design solution). (note: this problem has units Hz)

a. Highpass filter with a cutoff frequency of 159MHz b. In the specified frequency range, the gain must be >-3dB c. The rolloff in the stopband should be 80dB d. The circuit should contain a single unity gain (voltage follower) amplifier

stage (no other amplifiers should be in your circuit).

The circuit must be fourth order, which we can achieve by cascading two second order circuits with the unity gain opamp. In order to achieve a passband with only 3dB variation, we can set each second order circuit to have a cutoff frequency at -1.5dB. In [rad/s], the cutoff frequency is 1E9. For each second order RLC series circuit, we can choose a capacitor value of 1E-12 [F], the inductor then needs to be 1E-6 [H]. At ω = ωo, the transfer function of the second order circuit has a dB value of -1.5dB,

giving 5.12

1log20

. The damping ratio is then 0.594.

,

oL

R 22 R = 1.19[kΩ]

Frequency

1.0MHz 10MHz 100MHz 1.0GHz 10GHz 100GHz 1.0THzDB(V(L2:2))

-400

-300

-200

-100

-0

Note: The axis has scale log(f)

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Poles: 1E4 Two regions Range ~H(jω) 20log|H(jω)| Angle ...ssawyer/CircuitsFall2018_all/... · Electric Circuits HW09 solution Fall 2018 Problem 3) 2nd Order Filter Circuits R L