Schedule…
Date
27 Oct
Day
Mon
28 Oct
Tue
29 Oct
Wed
30 Oct
Thu
31 Oct
Fri
1 Nov
Sat
2 Nov
Sun
3 Nov
Mon
4 Nov
ECEN 301
Class
No.
Title
Chapters
16
Sinusoidal Frequency
Response
6.1
17
Operational
Amplifiers
Operational
Amplifiers
Lab
Due date
Exam
LAB 5
8.1 – 8.2
Recitation
18
HW
Due date
HW 7
8.3 – 8
Tue
LAB 6
Exam 1
Discussion #16 – Frequency Response
1
Easier to Maintain Than to Retake
Alma 59:9
9 And now as Moroni had supposed that there should
be men sent to the city of Nephihah, to the assistance
of the people to maintain that city, and knowing that it
was easier to keep the city from falling into the
hands of the Lamanites than to retake it from
them, he supposed that they would easily maintain
that city.
ECEN 301
Discussion #16 – Frequency Response
2
Lecture 16 – Frequency Response
Back to AC Circuits and Phasors
• Frequency Response
• Filters
ECEN 301
Discussion #16 – Frequency Response
3
Frequency Response
Frequency Response H(jω): a measure of how the
voltage/current/impedance of a load responds to the
voltage/current of a source
VL ( j )
H V ( j ) 
VS ( j )
I L ( j )
H I ( j ) 
I S ( j )
VL ( j )
H Z ( j ) 
I S ( j )
ECEN 301
Discussion #16 – Frequency Response
4
Frequency Response
VL(jω) is a phase-shifted and amplitude-scaled version of VS(jω)
VL ( j )
 H V ( j )
VS ( j )
VL ( j )  HV ( j )VS ( j )
ECEN 301
Discussion #16 – Frequency Response
5
Frequency Response
VL(jω) is a phase-shifted and amplitude-scaled version of VS(jω)
VL ( j  )
 H V ( j )
VS ( j )
NB : any complex number
A can be expressed as
VL ( j )  H V ( j )VS ( j )
VL e jL  H V e jHV VS e jVS
A  A e jA
A  A A
VL e jL  H V VS e j ( HV  VS )
ECEN 301
VL  H V VS
L  HV  VS
Amplitude scaled
Phase shifted
Discussion #16 – Frequency Response
6
Frequency Response
Example1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
R1
vs(t)
+
–
ECEN 301
C
+
RL
–
Discussion #16 – Frequency Response
7
Frequency Response
Example1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
R1
vs(t)
+
–
ECEN 301
1.
C
+
RL
–
Note frequencies of AC sources
Only one AC source so frequency
response HV(jω) will be the
function of a single frequency
Discussion #16 – Frequency Response
8
Frequency Response
Example1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
1.
2.
Note frequencies of AC sources
Convert to phasor domain
Z1 = R1
R1
vs(t)
+
–
ECEN 301
C
+
RL
–
ZC=1/jωC
Vs
+
~
–
Discussion #16 – Frequency Response
ZLD=RL
9
Frequency Response
Example1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
1.
2.
3.
Z1 = R1
Note frequencies of AC sources
Convert to phasor domain
Solve using network analysis
•
Thévenin equivalent
ZC=1/jωC
ZT  Z1 || Z C
ECEN 301
Discussion #16 – Frequency Response
10
Frequency Response
Example1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
1.
2.
3.
Z1 = R1
ZC=1/jωC
Vs
+
~
–
+
VT
–
Note frequencies of AC sources
Convert to phasor domain
Solve using network analysis
•
Thévenin equivalent
ZC
VT ( j )  VS ( j )
Z1  Z C
ZT  Z1 || Z C
ECEN 301
Discussion #16 – Frequency Response
11
Frequency Response
Example1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
1.
2.
3.
ZT
VT
+
~
–
+
VL
–
ZT  Z1 || Z C
ZC
VT ( j )  VS ( j )
Z1  Z C
ECEN 301
4.
Note frequencies of AC sources
Convert to phasor domain
Solve using network analysis
•
Thévenin equivalent
Find an expression for the load voltage
ZLD
Z LD
VL ( j )  VT ( j )
ZT  Z LD

ZC

 VS ( j )
Z1  Z C

Discussion #16 – Frequency Response

Z LD

 Z1 || Z C   Z LD
12
Frequency Response
Example1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
5.
ZT
VT
+
~
–
+
VL
–
ZLD
Find an expression for the frequency
response
H V ( j ) 
VL ( j )
VS ( j )
ZC
Z LD


Z1  Z C Z1 || Z C   Z LD
ECEN 301
Discussion #16 – Frequency Response
13
Frequency Response
Example1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
5.
ZT
Find an expression for the frequency
response
H V ( j ) 
VT
+
~
–
+
VL
–
ZLD

ZC
Z LD

Z1  Z C Z1 || Z C   Z LD
Z LD Z C
Z LD Z1  Z C   Z1Z C
10 4 /( j 10 5 )
 4 3
10 10  1 /( j 10 5 )  10 3 /( j 10 5 )
100

110  j


ECEN 301

  
  arctan

2
2
110


110  
100
Discussion #16 – Frequency Response
14
Frequency Response
Example1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
5.
ZT
VT
+
~
–
+
VL
–
ZLD
Find an expression for the frequency response
•
Look at response for low frequencies (ω = 10)
and high frequencies (ω = 10000)
H V ( j ) 
  
  arctan

2
2
 110 
110  
100
ω = 10000
ω = 10
HV ( j)  0.905   0.0907
ECEN 301
H V ( j )  0.010  
Discussion #16 – Frequency Response

2
15
Frequency Response
Example2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
L
is(t)
ECEN 301
R1
+
RL
–
Discussion #16 – Frequency Response
16
Frequency Response
Example2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
1.
L
is(t)
ECEN 301
R1
+
RL
–
Note frequencies of AC sources
Only one AC source so frequency
response HZ(jω) will be the
function of a single frequency
Discussion #16 – Frequency Response
17
Frequency Response
Example2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
1.
2.
Note frequencies of AC sources
Convert to phasor domain
ZL = jωL
L
is(t)
ECEN 301
R1
+
RL
–
Is(jω)
Discussion #16 – Frequency Response
Z1=R1
+
VL
–
ZLD = RL
18
Frequency Response
Example2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
1.
2.
3.
ZL = jωL
IL(jω)
Is(jω)
Z1=R1
+
VL
–
ZLD = RL
Note frequencies of AC sources
Convert to phasor domain
Solve using network analysis
•
Current divider & Ohm’s Law
Z1 || ( Z L  Z LD )
I L ( j )  I S ( j  )
( Z L  Z LD )
VL ( j )  I L ( j ) Z LD

Z1 || ( Z L  Z LD ) 
 Z LD
  I S ( j )
( Z L  Z LD ) 

ECEN 301
Discussion #16 – Frequency Response
19
Frequency Response
Example2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
4.
ZL = jωL
IL(jω)
Is(jω)
Z1=R1
ECEN 301
+
VL
–
ZLD = RL
Find an expression for the frequency
response
VL ( j )
H Z ( j ) 
I S ( j )

Z1 || ( Z L  Z LD )
Z LD
( Z L  Z LD )

RL
1  RL / R1  jL / R1

800
1  j 0.4 10 6
Discussion #16 – Frequency Response
20
Frequency Response
Example2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
4.
ZL = jωL
Find an expression for the frequency response
•
Look at response for low frequencies (ω = 10)
and high frequencies (ω = 10000)
IL(jω)
Is(jω)
Z1=R1
+
VL
–
ZLD = RL
800
H Z ( j ) 
1  j 0.4 10 6
ω = 10
ω = 10000
H Z ( j)  800  4.0 H Z ( j)  800  0.0040
ECEN 301
Discussion #16 – Frequency Response
21
st
1
and
nd
2
Order RLC Filters
Graphing in Frequency Domain
Filter Orders
Resonant Frequencies
Basic Filters
ECEN 301
Discussion #16 – Frequency Response
22
Frequency Domain
Graphing in the frequency domain: helpful in order
to understand filters
1.5
3.50
0.0
0.00
π
X(jw)
x(t)
π
2.00
4.00
0.00
-600.0
-1.5
-ω
0.0
ω
600.0
w
time
cos(ω0t)
Time domain
ECEN 301
π[δ(ω – ω0) + δ(ω – ω0)]
Frequency domain
Discussion #16 – Frequency Response
23
Frequency Domain
Graphing in the frequency domain: helpful in order
to understand filters
1
X(jw)
W/π
x(t)
π/W
-10.00
0.00
-0.5
0.0
10.00
-10.00
-W
0.00
Time domain
ECEN 301
10.00
w
t
sinc(ω0t)
W
1, |  | W
X ( j )  
0, |  | W
Frequency domain
Discussion #16 – Frequency Response
24
Electromagnetic (Frequency) Spectrum
ECEN 301
Discussion #16 – Frequency Response
25
Basic Filters
Electric circuit filter: attenuates (reduces) or
eliminates signals at unwanted frequencies
Low-pass
0.0
High-pass
0.00
.
ω
10 . 0 0
0.0
0.00
.
Band-pass
ω
10 . 0 0
Band-stop
0.0
0.00
.
ECEN 301
ω
10.00
0.0
0.00
Discussion #16 – Frequency Response
.
ω
10 . 0 0
26
Filter Orders
Higher filter orders provide a higher quality filter
50
30
10
Gain (dB)
-10
-30
1st Order
-50
2nd Order
3rd Order
-70
4th Order
-90
32nd Order
-110
-130
-150
0.1
1
10
100
Frequency (w)
ECEN 301
Discussion #16 – Frequency Response
27
Impedance
Impedance of resistors, inductors, and capacitors
I(jω)
Im
ωL
Phasor domain
Vs(jω) +~
–
ZL
π/2
-π/2
R
ZR
ZC
+
Z VZ(jω)
–
Re
+
R
–
+
L
–
+
C
–
-1/ωC
Z R ( j)  R
ECEN 301
Z L ( j)  jL
Discussion #13 – Phasors
Z C ( j ) 
1
jC
28
Impedance
Impedance of capacitors
Im
Re
+
C
–
+
C
–
+
C
–
-π/2
ZC
-1/ωC
ECEN 301
1
Z C ( j ) 
jC
as   0
1
Z C ( j ) 
0

Discussion #13 – Phasors
as   
1

0
Z C ( j ) 
29
Impedance
Impedance of inductors
Im
ωL
ZL
π/2
Re
+
L
–
+
L
–
Z L ( j)  jL
ECEN 301
as   0
Z C ( j )  0
Discussion #13 – Phasors
+
L
–
as   
Z C ( j )  
30
Resonant Frequency
Resonant Frequency (ωn): the frequency at which
capacitive impedance and inductive impedance are
equal and opposite (in 2nd order filters)
Z L ( j )   Z C ( j )
Impedances in series
1
jL  
jC
Impedances in parallel
L
+
vi(t)
–
ECEN 301
C
+
vo(t)
–
n 
1
LC
+
vi(t) L
–
Discussion #16 – Frequency Response
C
+
vo(t)
–
31
Resonant Frequency
Resonant Frequency (ωn): the frequency at which
capacitive impedance and inductive impedance are
equal and opposite (in 2nd order filters)
Z L  j L
Impedances in series
 1 
 j
L
 LC 
ZL=jωL
+
Vi(jω)
–
ECEN 301
ZC=1/jωC
+
Vo(jω)
–
 1   LC 
 j
L  
 LC   LC 
LC
LC
j LC

C
Z C  1 / jC
 1 
 1 / j
C
 LC 

LC
jC

j LC
C
 jL
Discussion #16 – Frequency Response
32
Resonant Frequency
Resonant Frequency (ωn): the frequency at which
capacitive impedance and inductive impedance are
equal and opposite (in 2nd order filters)
Impedances in series
+
Vi(jω)
–
ZEQ=0
+
Vo(jω)
–
j LC
ZL 
C
j LC
ZC  
C
Z EQ  Z L  Z C
0
Vo ( j )  0
ECEN 301
Discussion #16 – Frequency Response
33
Resonant Frequency
Resonant Frequency (ωn): the frequency at which
capacitive impedance and inductive impedance are
equal and opposite (in 2nd order filters)
Impedances in parallel
+
Vi(jω)
–
ZL=jωL ZC=1/jωC +
Vo(jω)
–
ECEN 301
Z L  j L
 1 
 j
L
 LC 
 1   LC 
 j
L  
 LC   LC 
LC
LC
j LC

C
Z C  1 / jC
 1 
 1 / j
C
 LC 

LC
jC

j LC
C
 jL
Discussion #16 – Frequency Response
34
Resonant Frequency
Resonant Frequency (ωn): the frequency at which
capacitive impedance and inductive impedance are
equal and opposite (in 2nd order filters)
Impedances in parallel
+
Vi(jω)
–
ZEQ=∞
+
Vo(jω)
–
j LC
ZL 
C
Z EQ  Z L || Z C
Vo ( j )  Vi ( j )
ECEN 301
j LC
ZC  
C
Discussion #16 – Frequency Response

Z L ZC
Z L  ZC
L/C

0

35
Low-Pass Filters
Low-pass Filters: only allow signals under the cutoff
frequency (ω0) to pass
Low-pass
+
vi(t)
–
0.0
0.00
ω1 ω0
ω2
.
 cos(3t )
ECEN 301
ω3
HL(jω)
+
vo(t)
–
C
10 . 0 0
R
vi (t )  cos(1t )
 cos(2t )
1st Order
R
vo (t )  cos(1t )
+
vi(t)
–
Discussion #16 – Frequency Response
2nd Order
L
C
+
vo(t)
–
36
Low-Pass Filters
1st Order Low-pass Filters:
ZR
Low-pass
ZC=1/jωC
+
Vi(jω)
–
0.0
0.00
ω1 ω0
ω1: ω → 0
.
ω3
ZR
+
Vi(jω)
–
10 . 0 0
ω3: ω → ∞
+
Vo(jω)
–
ZR
+
Vo(jω)
–
+
Vi(jω)
–
vo (t )  vi (t )
ECEN 301
+
Vo(jω)
–
vo (t )  0
Discussion #16 – Frequency Response
37
Low-Pass Filters
2nd Order Low-pass Filters:
Low-pass
ZR
ZL=jωL
+
Vi(jω)
–
0.0
0.00
ZR
ω1 ω0
ω2
.
ω3
ω1: ω → 0
+
Vi(jω)
–
vo (t )  vi (t )
ECEN 301
10 . 0 0
ZR
+
Vo(jω)
–
ZC=1/jωC +
Vo(jω)
–
ω2: ω = ωn
+
Vi(jω)
–
Resonant
frequency
+
Vo(jω)
–
ZR
ω3: ω → ∞
+
Vi(jω)
–
vo (t )  0
Discussion #16 – Frequency Response
+
Vo(jω)
–
vo (t )  0
38
High-Pass Filters
High-pass Filters: only allow signals above the
cutoff frequency (ω0) to pass
1st Order
High-pass
0.0
+
vi(t)
–
0.00
ω1
ω2
.
 cos(3t )
ECEN 301
10 . 0 0
2nd Order
R
vi (t )  cos(1t )
 cos(2t )
ω0 ω3
+
R vo(t)
–
C
HH(jω)
vo (t )  cos(3t )
+
vi(t)
–
Discussion #16 – Frequency Response
C
L
+
vo(t)
–
39
High-Pass Filters
1st Order High-pass Filters:
ZC=1/jωC
High-pass
+
Vi(jω)
–
0.0
0.00
ω1
.
ω0 ω3
ω1: ω → 0
+
Vi(jω)
–
10 . 0 0
ω3: ω → ∞
ZR
+
Vo(jω)
–
+
Vi(jω)
–
ZR
+
Vo(jω)
–
vo (t )  vi (t )
vo (t )  0
ECEN 301
ZR=R
+
Vo(jω)
–
Discussion #16 – Frequency Response
40
High-Pass Filters
2nd Order High-pass Filters:
ZC=1/jωC ZR
High-pass
+
Vi(jω)
–
0.0
0.00
ω1
ω2
.
ω0 ω3
10 . 0 0
ω1: ω → 0
+
Vi(jω)
–
ZR
vo (t )  0
ECEN 301
ZR
+
Vo(jω)
–
+
ZL=jωL Vo(jω)
–
ω2: ω = ωn
+
Vi(jω)
–
Resonant
frequency
+
Vo(jω)
–
+
Vi(jω)
–
vo (t )  0
Discussion #16 – Frequency Response
ω3: ω → ∞
ZR
+
Vo(jω)
–
vo (t )  vi (t )
41
Band-Pass Filters
Band-pass Filters: only allow signals between the
passband (ωa to ωb) to pass
2nd Order
Band-pass
0.0
0.00
ω1
ω3
ωa ω2 ωb
.
+
vi(t)
–
 cos(3t )
ECEN 301
L
+
R vo(t)
–
10.00
vi (t )  cos(1t )
 cos(2t )
C
HB(jω)
vo (t )  cos(2t )
Discussion #16 – Frequency Response
42
Band-Pass Filters
2nd Order Band-pass Filters:
Band-pass
0.0
0.00
ω1
ZC=1/jωC
ωa ω2 ωb
.
ω3
ω1: ω → 0
+
Vi(jω)
–
ZR
vo (t )  0
ECEN 301
10.00
+
Vi(jω)
–
ω2: ω = ωn
+
Vo(jω)
–
+
Vi(jω)
–
ZR
ZL=jωL
ω3: ω → ∞
Resonant
frequency
+
Vo(jω)
–
ZR
+
Vo(jω)
–
+
Vi(jω)
–
vo (t )  vi (t )
Discussion #16 – Frequency Response
ZR
+
Vo(jω)
–
vo (t )  0
43
Band-Stop Filters
Band-stop Filters: allow signals except those
between the passband (ωa to ωb) to pass
2nd Order
Band-stop
L
+
vi(t)
–
0.0
0.00
ω1
ω3
ωa ω2 ωb
.
10 . 0 0
vi (t )  cos(1t )
 cos(2t )
 cos(3t )
ECEN 301
C
+
R vo(t)
–
HN(jω)
vo (t )  cos(1t )
 cos(3t )
Discussion #16 – Frequency Response
44
Band-Stop Filters
2nd Order Band-stop Filters:
Band-stop
0.0
+
Vi(jω)
–
0.00
ω1
ωa ω2 ωb
.
ω3
ω1: ω → 0
+
Vi(jω)
–
ZR
vo (t )  vi (t )
ECEN 301
+
Vi(jω)
–
ZC=1/jωC
ZR
+
Vo(jω)
–
10 . 0 0
ω2: ω = ωn
+
Vo(jω)
–
ZL=jωL
ZR
ω3: ω → ∞
Resonant
frequency
+
Vo(jω)
–
+
Vi(jω)
–
vo (t )  0
Discussion #16 – Frequency Response
ZR
+
Vo(jω)
–
vo (t )  vi (t )
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