EE202 Supplementary Materialsfor Self Study

• Circuit Analysis Using Complex Impedance

• Passive Filters and Frequency Response

Acknowledgment

• Dr. Furlani and Dr. Liu for lecture slides

• Ms. Colleen Bailey for homework and solution of complex impedance

• Textbook: Nilsson & Riedel, “Electric Circuits,” 8th edition

Steady-State Circuit Response to Sinusoidal Excitation - Analysis

Using Complex Impedance

Why Sinusoidal?US Power Grid 60Hz Sinusoidal

Household Power Line

Household Circuit Breaker Panel240V Central Air

120V Lighting, Plugs, etc.

Single FrequencySinusoidal Signal

)cos()( tVtv m

Sinusoidal Signal

• Amplitude– Peak-to-peak– Root-mean-square

• Frequency– Angular Frequency– Period

f 2

mV

mV2

2m

rms

VV

f

fT

1

Trigonometry Functions

sin)sin(

cos)cos(

cossinsincos)sin(

sinsincoscos)cos(

02

cos;10cos

12

sin;00sin

tdtt

tdtt

tdt

td

tdt

td

sin1

cos

cos1

sin

sincos

cossin

Appendix F

Other Periodic WaveformsFundamental and Harmonics

Resistor Only Circuit

• I=V/R, i(t)=v(t)/R

• Instantaneous Response

R-L Circuit

R

L

tLR

Ve

LR

Vi

tVRidt

diL

mtLRm

m

1

222

)/(

222

tan

)cos()cos(

)cos(

Transient Steady-state

Phase Shift

Time Delay or Phase Angle: t / T *2 or *360-degree

Phasor – Complex Number

Y

Imag(Z)

X

Real(Z)

Z

Real(Z)+j Imag(Z)

tan-1(Y/X)

Reference

Complex Number

||//1

)2^2^(||

||)sin(cos||

yImag(z) x,Real(z)

sin||,cos||

zez

yxsqrtz

ezjzz

zyzx

jyxz

j

j

Phasor Solution of R-L Circuit

LjR

eVeI

eeI

eVRdt

dL

tVRidt

diL

jmj

m

tjjm

tjm

m

I

II

)(

)cos(

Observations• Single Frequency for All Variables

• Phasor Solution of Diff Eq.– Algebraic equation– Extremely simple

• Phase– Delay between variables

• Physical Measurements– Real part of complex variables

v = Real{V}; i = Real{I}

tje

Resistor

RIV

Instantaneous Response

LjI

V

Inductor

CjIV

1

Capacitor

Impedance in Series

...321 ZZZZ eq Complex Impedance

Resistance, Reactance

=5000 rad/sec

Example

Apply ZL=jL, ZC=1/j C

Zab=90+j(160-40)=90+j120=sqrt(902+1202)exp{jtan-1(120/90)}=150 53.13 degree

I=750 30 deg / 150 53.13 deg = 5 -23.13 deg=5exp(-j23.13o)

Impedance in Parallel

...1111

321

ZZZZab

jBGZ

Y

YYYYab

1

...321

Complex Admittance

Conductance, Susceptance

=200000 rad/sec

Example

Apply ZL=jL, ZC=1/j C

Series: Use Z; Parallel: Use YY=0.2 36.87 deg; Z=5 -36.87 deg

V=IZ=40 -36.87 deg

Kirchhoff’s Laws

• Same

• Current at a Node– Addition of current vectors (phasors)

• Voltage Around a Loop or Mesh– Summation of voltage vectors (phasors)

Delta-T Transformation

cba

ba

cba

ac

cba

cb

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

3

2

1

3

133221

2

133221

1

133221

Z

ZZZZZZZ

Z

ZZZZZZZ

Z

ZZZZZZZ

c

b

a

Example

Delta-T Transformation

Series, Parallel, Series

Another Delta-T Transformation

Thevenin and Norton Transformation

Thevenin Equivalent Circuit

Norton Equivalent Circuits

Voltage dividerVo=36.12-j18.84 (V)

Find VTh

Vx=100-I*10, Vx=I*(120-j40)-10*Vx; solve Vx and IVTH=10Vx+I*120=784-j288 (V)

Find ZTh

Calculate Ia

Determine Vx

Calculate Ib

ZTh=VT/IT=91.2-j38.4 (Ohm)

Transformer

2221

2111

)(0

)(

IZLjRMIj

MIjILjRZV

L

ss

Lr

rab

ZLjR

MZ

ZLjRZ

22

22

11

Time differentiation replaced by j

AC Sine Wave, Ideal TransformerVoltage and Current

1

2

1

2

N

N

v

v

1

2

2

1

N

N

i

i

Power Conserved

Transformer

• Power Applications – Convert voltage

vout=(N2/N1) vin

• Signal Applications – Impedance transformation

Xab=(N1/N2)2 XL

– Match source impedance with load to maximize power delivered to load

Power Calculations

Frequency Response of Circuits• Analysis Over a Range of

Frequencies

• Amplifier Uniformity

• Filter Characteristics – Low pass filter– High pass filter– Bandpass filter– Equalizer

RC FiltersHigh Pass

Low Pass

Frequency Response

• High Pass

• Low Pass

Lj

RCin

o eCjR

R

VV

22 /1

1

1

Hj

RCin

o eCjR

Cj

VV

22 /1

1

1

1

Bode Plot

• Log10 (f)– Compress many orders of magnitude

• Vertical scale– Linear– log: 10log10(Vo/Vin)

Bode Plot

Appendix D, Appendix E

Summary• Sinusoidal, Steady-State Analysis

• Complex Impedance Z=jL

Z=1/jC

• All Circuit Analysis Methods Apply

• Analysis Power Systems

• Frequency Response of Circuits

Homework• Problem7.pdf

• Solution7.pdf

Frequency Response & Passive Filters

1. Filters2. Low Pass Filter3. High Bass Filter4. Band Pass Filter5. Band Stop Filter6. Series RLC Resonance4 Parallel RLC Resonance

90.7WSDL

Ocean City

90.3WHID

Salisbury

Frequency(MHz)

90.5WKHS Worton

91.3WMLU

Farmville

90.9WETA

Washington

91.1WHFCBel Air

91.5WBJC

Baltimore

Tuning a Radio

• Consider tuning in an FM radio station.

• What allows your radio to isolate one station from all of the adjacent stations?

Filters

• A filter is a frequency-selective circuit.

• Filters are designed to pass some frequencies and reject

others.

Frequency(MHz)90.9

WETAWashington

Different Kinds of Filters

• There are four basic kinds of filters:– Low-pass filter - Passes frequencies below a critical

frequency, called the cutoff frequency, and attenuates those above.

– High-pass filter - Passes frequencies above the critical frequency but rejects those below.

– Bandpass filter - Passes only frequencies in a narrow range between upper and lower cutoff frequencies.

– Band-reject filter - Rejects or stops frequencies in a narrow range but passes others.

Active and Passive Filters

• Filter circuits depend on the fact that the impedance of capacitors and inductors is a function of frequency

• There are numerous ways to construct filters, but there are two broad categories of filters:– Passive filters are composed of only passive

components (resistors, capacitors, inductors) and do not provide amplification.

– Active filters typically employ RC networks and amplifiers (opamps) with feedback and offer a number of advantages.

Impedance vs. Frequency

Calculate the impedance of a resistor, a capacitor and an inductor at the following frequencies.

1 L CZ j L Z j

C

f 100 Hz 1000 Hz 10,000 Hz

R 100 100 100

ZL j10 j100 j1000

ZC -j1000 -j100 -j10

RC Low-Pass Filter

A simple low pass filter can be constructed using a resistor and capacitor in series.

Transfer Function H()

VS

V ( )H ( )

V ( )o

Hv(ω) = Transfer function for Voltage

V SV ( ) H ( )V ( )

Amplitude of transfer function

Phase shift of transfer functionC V S

o

V

V

j j H jC V S

C V S

C V S

H

H

V e H e V e

V H V

H

Hv(ω) describes what the phase shift and amplitude scaling are.

A Transfer function H(ω) is the ratio of the output to the input Output( )H( )

Input( )

Different Kinds of FiltersIdeal frequency response of four types of filters:

a) lowpass b) highpass

c) bandpass d) bandstop

Gain

• Any circuit in which the output signal power is greater than the input signal– Power is referred to as an amplifier

• Any circuit in which the output signal power is less than the input signal power – Called an attenuator

Power gain is ratio of output power to input powerin

out

P

PAP

Voltage gain is ratio of output voltage to input voltagein

out

V

VAv

66

The Decibel

Bel is a logarithmic unit that represents a tenfold increase or decrease in power

in

out(bels) log

P

PAP 10

Because the bel is such a large unit, the decibel (dB) is often used

in

out(dB) log

P

PAP 1010

2

o(dB) 1010 logV

s

VA

V

For voltage

RC Low-Pass Filter

For this circuit, we want to compare the output (Vo) to the input (Vs):

v

v 2

11

( )1 1

1( )

1

Co s

C

o

s

o

s

j CH

j RCRj C

HRC

ZV V

R Z

V

V

V

V

RC Low-Pass Filter

The cutoff frequency is the frequency at which the output voltage amplitude is 70.7% of the input value (i.e., –3 dB).

2

co

co

So for our RC circuit:

1 1H( )

21

which implies:

or [Hz]2

o

s RC

RC

fRC

V

V

f (Hz)fco

actual filter output

passband reject-band

“ideal” filter output

cutoff frequency

o

s

V

V

0 dB

–3 dB

2

2o(dB) 10 1010 log 10 log .707 3dBV

s

VA

V

2

o(dB) 1010 logV

s

VA

V

Example

What is the cutoff frequency for this filter? Given:

8.2

0.0033

R k

C F

co

co

or [Hz]2

RC

fRC

co 5.88 kHzf

RL Low-Pass Filter

A low-pass filter can also be implemented with a resistor and inductor.

RL Low-Pass Filter

Comparing the output (Vo) to the input (Vs):

2

1

1

1

1

o sL

o

s

o

s

R

RLR j L jR

LR

V VR Z

V

V

V

V

RL Low-Pass Filter

The cutoff frequency for this circuit design is given by:

2

co

co

So for our RL circuit:

1 1

21

which implies:

or [Hz]2

o

s LR

R

L

Rf

L

V

V

f (Hz)fco

actual filter output

passband reject-band

“ideal” filter output

cutoff frequency

o

s

V

V

0 dB

–3 dB2

o(dB) 1010 logV

s

VA

V

EXAMPLE – RL Low Pass Filter

Design a series RL low-pass filter to filter out any noise above 10 Hz.

R and L cannot be specified independently to generate a value for fco = 10 Hz or co = 2fco. Therefore, let us choose L=100 mH. Then,

3(2 )(10)(100 10 ) 6.28coR L

2 2 22

20( )

400

RL

o s sRL

V V V

f(Hz) |Vs| |Vo|

1 1.0 0.995

10 1.0 0.707

60 1.0 0.164co co2

1 which implies: or [Hz]

21

o

s

R Rf

L LLR

V

V

FiltersNotice the placement of the elements in RC and RL low-pass filters.

What would result if the position of the elements were switched in each circuit?

RL low-pass filterRC low-pass filter

RC and RL High-Pass Filter Circuits

Switching elements results in a High-Pass Filter.

co co1

or [Hz]2 2

Rf f

RC L

f (Hz)fcoactual

passbandreject-band

“ideal”

cutoff frequency

o

s

V

V

0 dB

–3 dB

ExampleWhat resistor value R will produce a cutoff frequency of 3.4 kHz with a 0.047 mF capacitor? Is this a high-pass or low-pass filter?

co

co

1 [Hz]

2

1R=

2

fRC

C f

1004R

This is a High-Pass Filter

Bandpass Filter

A bandpass filter is designed to pass all frequencies within a band of frequencies, ω1 < ω0 < ω2

QL

RB 0

12

Bandwidth B of a Filter

B

Bandpass Filters

C/1LjR

R

V

V)(H

i

0

LC

10

Transfer function:Center frequency

Maximum occurs when 1/L C ( ) 1H

QL

RB 0

12

Bandwidth B of a Filter

B

0 1 2

Example – RLC Bandpass FiltersDesign a series RLC bandpass filter with cutoff frequencies f1=1kHz and f2 = 10 kHz.

Cutoff frequencies give us two equations but we have 3 parameters to choose. Thus, we need to select a value for either R, L, or C and use the equations to find other values. Here, we choose C=1μF.

0 1 2

0

22 6

0

2 1

3

2 6 2

(6280)(62800) 19,867rad/s

3162.28Hz21 1

2.533 mH2 (3162.28) (10 )

19,867rad/s0.3514

(2 *10000 2 *1000)rad/s

2.533(10 )143.24

(10 )(0.3514)

o

o

f

LC

Q

LR

CQ

f1=1kHz 1 = 2f1 = 6280 rad/sf2 = 10 kHz 2 = 2f2 = 62,800 rad/s

0 1 2

0

2 1

Q

0

1

LC

02

1L L L LR

Q Q C Q CQ LC

Bandstop FilterA bandstop filter is designed to stop or eliminate all frequencies within a band of frequencies, ω1 < ω0 < ω2

QL

RB 0

12

Bandwidth B of a Filter

B

Bandstop Filters

C/1LjR

C/1Lj

V

V)(H

i

0

LC

10

Transfer function:

Minimum occurs when 1/L C ( ) 0H

Center frequency

QL

RB 0

12

Bandwidth B of a Filter

B

Formulas for Band Pass and Band Stop Filters

82

LC

10

0 1 2

B

0

2 1

Q

2 1B

Series Resonance

Resonance is a condition in an RLC circuit in which the capacitive and inductive reactances are equal in magnitude, thereby resulting in a purely resistive impedance.

The series resonant circuit

Cj

1LjR

I

V)(HZ s

Input impedance:

C

1LjRZ

s/radLC

10

Resonant/center frequency:

Resonance occurs when imaginary part is 0

Resonance occurs when imaginary part is 0

1L

C

C

1LjRZ

At resonance:

1. The impedance is purely resistive, Z = R

2. The voltage and the current are in phase, pf=1

3. The magnitude of transfer function H(w) = Z(w) is minimum

4. The inductor voltage and capacitor voltage can be much more than the source voltage

s/radLC

10

Resonant/center frequency:Resonance occurs when imaginary part is 0

1L

C

C

1LjRZ

The current amplitude vs. frequency for the series resonant circuit

R

V

2

1)(P

2

m0

R4

V)(P)(P

2

m21

Maximum power:

LC

1

L2

R

L2

R

LC

1

L2

R

L2

R

2

2

2

1

Half power frequencies:

210

0

1rad/s

LC

Half of this power is obtained at 1 and 2

The “sharpness” of the resonance in a resonant circuit is measured quantitatively by the quality factor Q

BCRR

LQ 0

0

0 1

The quality factor of a resonant circuits is the ratio of its resonant frequency to its bandwidth

Quality Factor

0

2 1

Q

Series Resonance

QL

RB 0

12

Relation between Q and bandwidth B:

The higher the circuit Q, the smaller the bandwidth

Series Resonance

High Q circuit if, 10Q

2

B2

B

02

01

and half power frequency can be approximated as:

10Q

Example - Series Resonance

• The problem requires the formula for the frequency f.

• Only the inductance and capacitance matter.– 1/2 (0.25 H 10-7 F)1/2 = 1 kHz

100

10 V 250 mH

0.1 F

• Find the resonant frequency in the following circuit in Hz.

LCf

2

10

Series Resonant Circuit

Parallel Resonance

The parallel-resonant circuit

Lj

1Cj

R

1

V

I)(HY

Input admittance:

L

1Cj

R

1Y

s/radLC

10

Resonant frequency:

Resonance occurs when imaginary part is 0

Resonance occurs when imaginary part is 0

L

1Cj

R

1Y 1

CL

Parallel Resonance

Half power frequency:

RC

1B 12

L

RRC

BQ

0

00

LC

1

RC2

1

RC2

1

LC

1

RC2

1

RC2

1

2

2

2

1

Bandwidth B:

High Q circuit if, 10Q

2

B2

B

02

01

half power frequencies can be approximated as:

Homework Assignment

Chapter 11 of 8th Edition of Textbook

Problems 31, 36, 41

Filter Circuit Problems 1, 2

100 nF

1.8 k

VIN

0 V

VOUT

Prob. 1. For the filter circuit below

a. calculate the reactance of the capacitor at 10Hz, 100Hz, 1kHz, 10kHz and 100kHz.b. calculate the output voltage at each of these frequencies.c. calculate the cutoff frequency of this circuit.d. calculate VOUT at the break frequency.

e. plot a graph of output voltage against frequency on log graph paper.

Filter Analysis

VIN = 10 V0

154,15910100102

1

2

19fC

X C

915,15101001002

1

2

19fC

X C

591,11010010002

1

2

19fC

X C

1591010010102

1

2

193fC

X C

9.1510100101002

1

2

193fC

X C

Solution:i) calculate the reactance of the capacitor at 10Hz, 100Hz, 1kHz, 10kHz and 100kHz.At 10 Hz:

At 100 Hz:

At 1 kHz:

At 10 kHz:

At 100 kHz:

VVXR

XV IN

C

COUT 999.910

1591541800

1591542222

VVXR

XV IN

C

COUT 936.910

159151800

159152222

VVXR

XV IN

C

COUT 622.610

15911800

15912222

VVXR

XV IN

C

COUT 879.010

1591800

1592222

VVXR

XV IN

C

COUT 088.010

9.151800

9.152222

Hz

RCfb

19.8841010018002

12

1

9

V

VV INOUT

07.7

10707.02

1

ii) calculate the output voltage at each of these frequencies.At 10 Hz:

At 100 Hz:

At 1 kHz:

At 10 kHz:

At 100 kHz:

iii) calculate the break frequency of this circuit.

•calculate VOUT at the break frequency.

Frequency Response of Low Pass Filter

0

2

4

6

8

10

12

1 10 100 1000 10000 100000

Frequency (Hz)

Vo

ut

(V)

plot a graph of output voltage against frequency on log graph paper below.

Theoretical Break Frequency.

Filter Analysis

47 nF

3.3 k

VIN

0 V

VOUT

Prob. 2 Consider the filter circuit below:

a. calculate the reactance of the capacitor at 10Hz, 100Hz, 1kHz, 10kHz and 100kHz.b. calculate the output voltage at each of these frequencies.c. calculate the cutoff frequency of this circuit.d. calculate VOUT at the break frequency.

e. plot a graph of output voltage against frequency on log graph paper.

VIN = 10 V0

627,3381047102

1

2

19fC

X C

863,3310471002

1

2

19fC

X C

386,3104710002

1

2

19fC

X C

339104710102

1

2

193fC

X C

9.331047101002

1

2

193fC

X C

Solution :i) calculate the reactance of the capacitor at 10Hz, 100Hz, 1kHz, 10kHz and 100kHz.At 10 Hz:

At 100 Hz:

At 1 kHz:

At 10 kHz:

At 100 kHz:

VVXR

RV IN

C

OUT 097.0103386273300

33002222

VVXR

RV IN

C

OUT 969.010338633300

33002222

VVXR

RV IN

C

OUT 979.61033863300

33002222

VVXR

RV IN

C

OUT 947.9103393300

33002222

VVXR

RV IN

C

OUT 999.9109.333300

33002222

Hz

RCfb

14.1026104733002

12

1

9

V

VV INOUT

07.7

10707.02

1

ii) calculate the output voltage at each of these frequencies.At 10 Hz:

At 100 Hz:

At 1 kHz:

At 10 kHz:

At 100 kHz:

iii) calculate the break frequency of this circuit.

iv) calculate VOUT at the break frequency.

Frequency Response of High Pass Filter

0

2

4

6

8

10

12

1 10 100 1000 10000 100000

Frequency (Hz)

Vo

ut

(V)

Theoretical Break Frequency.

iv) plot a graph of output voltage against frequency on log graph paper below.

Example

What is the cutoff frequency for this filter? Given:

8.2

0.0033

R k

C F

co

co

or [Hz]2

RC

fRC

Example – RL Low Pass Filter

Design a series RL low-pass filter to filter out any noise above 10 Hz.

R and L cannot be specified independently to generate a value for wc. Therefore, let us choose L=100 mH. Then,

28.6)10100)(10)(2( 3 LR c

ii

LR

LR

o VVV2222 400

20)(

F(Hz) |V.| |Vo|

1 1.0 0.995

10 1.0 0.707

60 1.0 0.164

(1) Identify the following filter circuits as being low pass, high pass, band pass or band-stop (4pts).

Answers: (b) and (c) are high-pass; (a) and (d) are low-pass

oVsV oVsV

oVsV oVsV

(2) If the cut-off frequencies for each of the circuits is 1kHz and the resistance in each circuit is R=1000, find the values of L or C for each circuit (8pts).

3) Identify the following filter circuit as being low pass, high pass, band-pass or band-stop (2pts).

4) Assume C=1μF and the central frequency of this filter is 2MHz (i.e. 2e6 Hz).

(a) Determine the Inductance L (2pts).(b) Determine the bandwidth B if the Q of the

circuit is 100 (2pts).(c) Determine the filter angular cutoff frequencies 1 and

2 (2pts).