Transcript EE202 Powerpoint Slides

EE202 Supplementary Materials
for 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 Panel
240V Central Air
120V Lighting, Plugs, etc.
Single Frequency
Sinusoidal Signal
v(t )  Vm cos(t   )
Sinusoidal Signal
• Amplitude
– Peak-to-peak
– Root-mean-square
• Frequency
– Angular Frequency
– Period
Vm
2  Vm
Vrms 
Vm
2
f
  2f
T
1
f
Trigonometry Functions
sin 0  0; sin

cos0  1; cos
2
1

0
2
cos(   )  cos cos  sin  sin 
sin(   )  cos sin   sin  cos
cos( )  cos
sin( )   sin 
d sin t
  cost
dt
d cost
  sin t
dt
1
sin

t

dt


cost


Appendix F
 cost  dt 
1

sin t
Other Periodic Waveforms
Fundamental and Harmonics
Resistor Only Circuit
• I=V/R, i(t)=v(t)/R
• Instantaneous Response
R-L Circuit
di
 Ri  Vm cos(t   )
dt
 Vm
i
cos(   )e ( R / L ) t 
R 2   2 L2
L
  tan1
R
L
Transient
Vm
R  L
2
2
2
cos(t     )
Steady-state
Phase Shift
Time Delay or Phase Angle: t / T *2 or *360-degree
Phasor – Complex Number
Z
Real(Z)+j Imag(Z)
Y
Imag(Z)
tan-1(Y/X)
X
Real(Z)
Reference
Complex Number
z  x  jy
x | z | cos , y | z | sin 
Real(z)  x, Imag(z) y
z | z | (cos  j sin  ) | z | e j
| z | sqrt( x ^ 2  y ^ 2)
1 / z  e  j / | z |
Phasor Solution of R-L Circuit
di
 Ri  Vm cos(t   )
dt
dI
L  RI  Vm e j (t  )
dt
I  I me j e jt
L
I me
j
Vm e j

R  jL
Observations
• Single Frequency for All Variables
e j t
• 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}
Resistor
V
R
I
Instantaneous Response
Inductor
V
 jL
I
Capacitor
V
1

I jC
Impedance in Series
Z eq  Z1  Z 2  Z 3  ...
Complex Impedance
Resistance, Reactance
Example
=5000 rad/sec
Apply ZL=jL, ZC=1/j C
Zab=90+j(160-40)=90+j120=sqrt(902+1202)exp{jtan-1(120/90)}
=150  53.13 degree
o
I=750  30 deg / 150  53.13 deg = 5  -23.13 deg=5exp(-j23.13 )
Impedance in Parallel
Yab  Y1  Y2  Y3  ...
Y
1
1
1
1



 ...
Z ab Z1 Z 2 Z3
1
 G  jB
Z
Complex Admittance
Conductance, Susceptance
Example
=200000 rad/sec
Apply ZL=jL, ZC=1/j C
Series: Use Z; Parallel: Use Y
Y=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
Z1 
ZbZc
Za  Zb  Zc
Za 
Z1 Z 2  Z 2 Z 3  Z 3 Z1
Z1
Z2 
ZcZa
Za  Zb  Zc
Zb 
Z1 Z 2  Z 2 Z 3  Z 3 Z1
Z2
Z3 
Za Zb
Za  Zb  Zc
Zc 
Z1 Z 2  Z 2 Z 3  Z 3 Z1
Z3
Example
Delta-T Transformation
Series, Parallel, Series
Another Delta-T Transformation
Thevenin and Norton
Transformation
Thevenin Equivalent Circuit
Norton Equivalent Circuits
Voltage divider
Vo=36.12-j18.84 (V)
Find VTh
Vx=100-I*10, Vx=I*(120-j40)-10*Vx; solve Vx and I
VTH=10Vx+I*120=784-j288 (V)
Find ZTh
Calculate Ia
Determine Vx
Calculate Ib
ZTh=VT/IT=91.2-j38.4 (Ohm)
Transformer
Vs  ( Z s  R1  jL1 ) I1  jMI 2
0   jMI1  ( R2  jL2  Z L ) I 2
Z ab  R1  jL1  Z r
2M 2
Zr 
R2  jL2  Z L
Time differentiation replaced by  j
AC Sine Wave, Ideal Transformer
Voltage and Current
v2 N 2

v1 N1
i1 N 2

i2 N1
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 Filters
High Pass
Low Pass
Frequency Response
• High Pass
Vo
R
1
j



e
2
Vin R  1 j C 1   RC
/ 2
• Low Pass
Vo
1 j C
1
j



e
2
Vin R  1 j C 1   2 / RC
H
L
Bode Plot
• Log10 (f)
– Compress many orders of magnitude
• Vertical scale
– Linear
– log: 10log10(Vo/Vin)
Bode Plot
Appendix D, Appendix E
Summary
• Sinusoidal, Steady-State Analysis
• Complex Impedance
Z=jL
Z=1/jC
• All Circuit Analysis Methods Apply
• Analysis Power Systems
• Frequency Response of Circuits
Homework
• Problem7.pdf
• Solution7.pdf
Frequency Response & Passive Filters
1.
2.
3.
4.
5.
6.
4
Filters
Low Pass Filter
High Bass Filter
Band Pass Filter
Band Stop Filter
Series RLC Resonance
Parallel RLC Resonance
Tuning a Radio
• Consider tuning in an FM radio station.
• What allows your radio to isolate one station from all of
the adjacent stations?
90.3
WHID
Salisbury
90.7
91.1
91.5
WSDL
WHFC
WBJC
90.5 Ocean City 90.9
Bel Air 91.3 Baltimore
WKHS
WETA
WMLU
Worton
Washington
Farmville
Frequency
(MHz)
Filters
• A filter is a frequency-selective circuit.
• Filters are designed to pass some frequencies and reject
others.
90.9
WETA
Washington
Frequency
(MHz)
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.
Z L  jL
 1 
ZC   j 


C


f
100 Hz
1000 Hz
10,000 Hz
R
100 W
100 W
100 W
ZL
j10 W
j100 W
j1000 W
ZC
-j1000 W
-j100 W
-j10 W
RC Low-Pass Filter
A simple low pass filter can be constructed
using a resistor and capacitor in series.
Transfer Function H()
A Transfer function H(ω) is the ratio of the output to the input
Output( )
H( ) 
Input( )
Hv(ω) = Transfer function for Voltage
Vo ( )
H V ( ) 
VS ( )
Hv(ω) describes what the phase shift and amplitude scaling are.
Vo ( )  H V ( )VS ( )
HV  Amplitude of transfer function
HV  Phase shift of transfer function
VC e jC  HV e jHV VS e jS
VC  HV VS
C  HV  S
Different Kinds of Filters
Ideal frequency response of four types of filters:
a) lowpass
c) bandpass
b) highpass
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 power
Pout
AP 
Pin
Vout
Voltage gain is ratio of output voltage to input voltage Av 
Vin
The Decibel
Bel is a logarithmic unit that represents a tenfold increase or
decrease in power
AP (bels)
Pout
 log10
Pin
Because the bel is such a large unit, the decibel (dB) is
often used
AP (dB)
For voltage
Pout
 10 log 10
Pin
 Vo 
AV (dB)  10 log10  
 Vs 
2
66
RC Low-Pass Filter
For this circuit, we want to compare the output (Vo)
to the input (Vs):
Vo 
ZC
Vs
R  ZC
1
V
1
jC
H v ( )  o 

Vs R  1
1  j RC
jC
H v ( ) 
Vo
1

2
Vs
1   RC 
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
V 
2
AV (dB)  10 log10  o   10 log10 .707   3dB
 Vs 
So for our RC circuit:
passband
V
1
1
H( )  o 

2
Vs
2
1   RC 
which implies: co 
or
f co 

2 RC

RC
reject-band
Vo
Vs
0 dB
–3 dB
V 
AV (dB)  10 log10  o 
 Vs 
“ideal” filter output
actual filter output
2
fco
f (Hz)
[Hz]
cutoff frequency
Example
What is the cutoff frequency for this filter?
Given:
R  8.2k W
C  0.0033F
co 
or

RC

f co 
2 RC
fco  5.88 kHz
[Hz]
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):
R
Vs
Vo 
R  ZL
1
R
Vo


Vs R  j L 1  j L
R
Vo

Vs
1
 L 
1 

R


2
RL Low-Pass Filter
The cutoff frequency for this circuit design is given by:
So for our RL circuit:
Vo

Vs
1
 L 
1 

 R 
2

1
2
R
which implies: co 
L
R
or f co 
[Hz]
2 L
passband
reject-band
Vo
Vs
0 dB
–3 dB
“ideal” filter output
V 
AV (dB)  10 log10  o 
 Vs 
2
actual filter output
fco
f (Hz)
cutoff frequency
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 = 2fco. Therefore, let us choose L=100 mH. Then,
R  co L  (2 )(10)(100 103 )  6.28W
Vo ( ) 
Vo

Vs
R
L
 
2
1
 L 
1 

 R 
2

R 2
L
Vs 
which implies: co 
20
  400
2
R
or
L
f co 
R
2 L
2
Vs
[Hz]
f(Hz)
|Vs|
|Vo|
1
1.0
0.995
10
1.0
0.707
60
1.0
0.164
Filters
Notice the placement of the elements in RC and
RL low-pass filters.
RC low-pass filter
RL low-pass filter
What would result if the position of the elements
were switched in each circuit?
RC and RL High-Pass Filter Circuits
Switching elements results in a High-Pass Filter.
fco 
1
R
or fco 
[Hz]
2 RC
2 L
reject-band
Vo
Vs
passband
“ideal”
0 dB
–3 dB
fco
f (Hz)
actual
cutoff frequency
Example
What 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?
This is a High-Pass Filter
1
f co 
[Hz]
2 RC
1
 R=
2 C f co
R  1004 W
Bandpass Filter
A bandpass filter is designed to pass all frequencies
within a band of frequencies, ω1 < ω0 < ω2
B
Bandwidth B of a Filter
B   2  1 
R 0

L Q
Bandpass Filters
Bandwidth B of a Filter
B   2  1 
R 0

L Q
B
Transfer function:
Center frequency
H() 
V0
R

Vi R  jL  1 / C 
1
0 
LC
0  12
Maximum occurs when
 L  1/ C
H ()  1
Example – RLC Bandpass Filters
Design 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.
f1=1kHz  1 = 2f1 = 6280 rad/s
f2 = 10 kHz  2 = 2f2 = 62,800 rad/s
0  12 0  12  (6280)(62800)  19,867rad/s
fo 
0 
Q
R
1
LC
0
2  1
L0
L
1 L
L



Q
Q 2C
Q LC Q C
L
Q
0
 3162.28Hz
2
1
1

 2.533 mH
2
2
6
o C  2 (3162.28)  (10 )
0
2  1

19,867rad/s
 0.3514
(2 *10000  2 *1000)rad/s
L
2.533(103 )
R

 143.24W
2
6
2
CQ
(10 )(0.3514)
Bandstop Filter
A bandstop filter is designed to stop or eliminate all
frequencies within a band of frequencies, ω1 < ω0 < ω2
Bandwidth B of a Filter
B   2  1 
B
R 0

L Q
Bandstop Filters
Bandwidth B of a Filter
B   2  1 
R 0

L Q
B
Transfer function:
V
jL  1 / C 
H()  0 
Vi R  jL  1 / C 
Minimum occurs when
 L  1/ C
Center frequency
0 
H ( )  0
1
LC
Formulas for Band Pass and Band Stop Filters
0 
1
LC
0  12
Q
0
2  1
B  2  1
B
82
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.
Input impedance:
The series resonant circuit
Z  H() 
Vs
1
 R  jL 
I
jC
1 

Z  R  j L 


C


Resonant/center frequency:
Resonance occurs when
imaginary part is 0
1 

Z  R  j L 

C 

Resonance occurs when
imaginary part is 0
L 
1
C
1
0 
rad / s
LC
Resonant/center frequency:
Resonance occurs when
imaginary part is 0
1 

Z  R  j L 


C


L 
1
C
1
0 
rad / s
LC
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
The current amplitude vs. frequency for the series resonant circuit
Maximum power:
1 Vm2
P(0 ) 
2 R
Half of this power is
obtained at 1 and 2
Vm2
P(1 )  P(2 ) 
4R
0  12
0 
1
rad/s
LC
Half power frequencies:
2
R
1
R 
1  
   
2L
 2L  LC
2
R
1
R 
2 
   
2L
 2L  LC
Quality Factor
The “sharpness” of the resonance in a resonant circuit is
measured quantitatively by the quality factor Q
Q
0
2  1
0 L
0
1
Q


R
0 CR B
The quality factor of a resonant circuits is the ratio of its
resonant frequency to its bandwidth
Series Resonance
Relation between Q and bandwidth B:
R 0
B   2  1  
L Q
The higher the circuit Q, the smaller the bandwidth
Series Resonance
High Q circuit if,
Q  10
and half power frequency can be approximated as:
Q  10
B
1  0 
2
B
 2  0 
2
Example - Series Resonance
• Find the resonant frequency in
the following circuit in Hz.
• The problem requires the
formula for the frequency f.
f0 
100 W
10 V
250 mH
0.1 F
1
2 LC
• Only the inductance and
capacitance matter.
– 1/2 (0.25 H 10-7 F)1/2 = 1 kHz
Series Resonant Circuit
Parallel Resonance
The parallel-resonant circuit
Parallel Resonance
Input admittance:
I 1
1
Y  H()    jC 
V R
jL
1 
1 
Y   j C 

R 
L 
Resonance occurs when
imaginary part is 0
1 
1 
Y   j C 

R 
L 
Resonance
occurs when
imaginary part
is 0
Resonant frequency:
1
C 
L
1
0 
rad / s
LC
Half power frequency:
2
1
1
 1 
1  
 
 
2RC
 2RC  LC
2
1
1
 1 
2 
 
 
2RC
 2RC  LC
Bandwidth B:
1
B  2  1 
RC
0
R
Q
 0 RC 
B
0 L
B
1  0 
2
half power frequencies can be approximated as:
B
 2  0 
2
High Q circuit if, Q  10
Homework Assignment
Chapter 11 of 8th Edition of Textbook
Problems 31, 36, 41
Filter Circuit Problems 1, 2
Filter Analysis
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.
1.8 kW
VIN
VIN = 10 V0
0V
VOUT
100 nF
Solution:
i)
calculate the reactance of the capacitor at 10Hz, 100Hz, 1kHz, 10kHz and 100kHz.
At 10 Hz:
XC 
1
1

 159,154W
2fC 2   10100109
At 100 Hz:
XC 
1
1

 15,915W
2fC 2   100100109
At 1 kHz:
XC 
1
1

 1,591W
2fC 2   1000100109
At 10 kHz:
XC 
1
1

 159W
2fC 2   10103 100109
At 100 kHz:
XC 
1
1

 15.9W
2fC 2   100103 100109
ii)
calculate the output voltage at each of these frequencies.
At 10 Hz:
VOUT 
XC
R X
2
2
C
VIN 
159154
18002  1591542
10  9.999V
At 100 Hz:
VOUT 
XC
R X
2
2
C
VIN 
15915
18002  159152
10  9.936V
At 1 kHz:
VOUT 
XC
R 2  X C2
VIN 
1591
18002  15912
10  6.622V
At 10 kHz:
VOUT 
XC
R 2  X C2
VIN 
159
18002  1592
10  0.879V
At 100 kHz:
VOUT 
XC
R X
2
iii)
fb 

•
2
C
VIN 
15.9
18002  15.92
10  0.088V
calculate the break frequency of this circuit.
1
2RC
1
 884.19Hz
2    1800 100 109
calculate VOUT at the break frequency.
1
 VIN
2
 0.707 10
 7.07V
VOUT 
plot a graph of output voltage against frequency on log graph paper below.
Frequency Response of Low Pass Filter
Theoretical Break
Frequency.
12
10
Vout (V)
8
6
4
2
0
1
10
100
1000
Frequency (Hz)
10000
100000
Filter Analysis
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.
47 nF
VIN
VIN = 10 V0
0V
VOUT
3.3 kW
Solution :
i)
calculate the reactance of the capacitor at 10Hz, 100Hz, 1kHz, 10kHz and 100kHz.
At 10 Hz:
XC 
1
1

 338,627W
2fC 2   10 47109
At 100 Hz:
XC 
1
1

 33,863W
2fC 2   100 47 109
At 1 kHz:
XC 
1
1

 3,386W
2fC 2   1000 47109
At 10 kHz:
XC 
1
1

 339W
2fC 2   10103  47 109
At 100 kHz:
XC 
1
1

 33.9W
2fC 2   100103  47 109
ii)
calculate the output voltage at each of these frequencies.
At 10 Hz:
VOUT 
R
R X
2
2
C
VIN 
3300
3300  3386272
2
10  0.097V
At 100 Hz:
VOUT 
R
R X
2
2
C
VIN 
3300
33002  338632
10  0.969V
At 1 kHz:
VOUT 
R
R X
2
2
C
VIN 
3300
33002  33862
10  6.979V
At 10 kHz:
VOUT 
R
R 2  X C2
VIN 
3300
33002  3392
10  9.947V
At 100 kHz:
VOUT 
R
R 2  X C2
iii)
fb 

iv)
VIN 
3300
33002  33.92
10  9.999V
calculate the break frequency of this circuit.
1
2RC
1
 1026.14Hz
2    3300 47 109
calculate VOUT at the break frequency.
1
 VIN
2
 0.707 10
 7.07V
VOUT 
iv)
plot a graph of output voltage against frequency on log graph paper below.
Frequency Response of High Pass Filter
Theoretical
12
Break
Frequency.
10
Vout (V)
8
6
4
2
0
1
10
100
1000
Frequency (Hz)
10000
100000
Example
What is the cutoff frequency for this filter?
Given:
R  8.2k W
C  0.0033F
co 
or

RC

f co 
2 RC
[Hz]
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,
R  c L  (2 )(10)(100103 )  6.28W
Vo ( ) 
R
L
 2   RL 
2
Vi 
20
 2  400 2
Vi
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).
Vs
Vo
Vs
Vo
Vs
Vo
Vs
Vo
Answers: (b) and (c) are high-pass;
(a) and (d) are low-pass
(2) If the cut-off frequencies for each of the circuits is
1kHz and the resistance in each circuit is R=1000W, 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).