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).