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ECEN 301 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
)()(
)(
jVjV
jHS
LV
)()(
)(
jIjI
jHS
LI
)()(
)(
jIjV
jHS
LZ
ECEN 301 Discussion #16 – Frequency Response 4
Frequency Response
VL(jω) is a phase-shifted and amplitude-scaled version of VS(jω)
)()()(
)()()(
jVjHjV
jHjVjV
SVL
VS
L
ECEN 301 Discussion #16 – Frequency Response 5
Frequency Response
VL(jω) is a phase-shifted and amplitude-scaled version of VS(jω)
)(
)()()(
)()()(
SVL
SVL
VHjSV
jL
VjS
HjV
jL
SVL
VS
L
eVHeV
eVeHeV
jVjHjV
jHjVjV
SVL VHV SVL VH
AAA
eAA Aj
as expressed becan
numbercomplex any :
A
NB
Amplitude scaled Phase shifted
ECEN 301 Discussion #16 – Frequency Response 6
Frequency Response
Example 1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
R1
Cvs(t)+–
+RL
–
ECEN 301 Discussion #16 – Frequency Response 7
Frequency Response
Example 1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
R1
Cvs(t)+–
+RL
–
1. Note frequencies of AC sources
Only one AC source so frequency response HV(jω) will be the function of a single frequency
ECEN 301 Discussion #16 – Frequency Response 8
Frequency Response
Example 1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
R1
Cvs(t)+–
+RL
–
1. Note frequencies of AC sources2. Convert to phasor domain
Z1 = R1
ZLD=RL
ZC=1/jωC
Vs+–~
ECEN 301 Discussion #16 – Frequency Response 9
Frequency Response
Example 1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF1. Note frequencies of AC sources2. Convert to phasor domain3. Solve using network analysis
• Thévenin equivalent
Z1 = R1
ZC=1/jωC
CT ZZZ ||1
ECEN 301 Discussion #16 – Frequency Response 10
Frequency Response
Example 1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
C
CST ZZ
ZjVjV
1
)()(
1. Note frequencies of AC sources2. Convert to phasor domain3. Solve using network analysis
• Thévenin equivalent
Z1 = R1
+VT
–
ZC=1/jωC
Vs+–~
CT ZZZ ||1
ECEN 301 Discussion #16 – Frequency Response 11
Frequency Response
Example 1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
C
CST ZZ
ZjVjV
1
)()(
1. Note frequencies of AC sources2. Convert to phasor domain3. Solve using network analysis
• Thévenin equivalent4. Find an expression for the load voltage
CT ZZZ ||1
ZT
ZLDVT
+–~
+VL
–
LDC
LD
C
CS
LDT
LDTL
ZZZ
Z
ZZ
ZjV
ZZ
ZjVjV
||)(
)()(
11
ECEN 301 Discussion #16 – Frequency Response 12
Frequency Response
Example 1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF5. Find an expression for the frequency
response ZT
ZLDVT
+–~
+VL
–
LDC
LD
C
C
S
LV
ZZZ
Z
ZZ
Z
jV
jVjH
||
)(
)()(
11
ECEN 301 Discussion #16 – Frequency Response 13
Frequency Response
Example 1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF5. Find an expression for the frequency
response ZT
ZLDVT
+–~
+VL
–
110arctan
110
100
110
100
)10/(10)10/(11010
)10/(10
||)(
22
53534
54
11
11
j
jj
j
ZZZZZ
ZZ
ZZZ
Z
ZZ
ZjH
CCLD
CLD
LDC
LD
C
CV
ECEN 301 Discussion #16 – Frequency Response 14
Frequency Response
Example 1: compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
2010.0)(
jHV
5. Find an expression for the frequency response• Look at response for low frequencies (ω = 10)
and high frequencies (ω = 10000)
ZT
ZLDVT
+–~
+VL
–
110arctan
110
100)(
22
jHV
0907.0905.0)( jHV
ω = 10ω = 10000
Lowpass filter
Frequency Response
Lowpass filter
ECEN 301 Discussion #16 – Frequency Response 16
Frequency Response
Example 2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
R1is(t)
+RL
–
L
ECEN 301 Discussion #16 – Frequency Response 17
Frequency Response
Example 2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
R1is(t)
+RL
–
L 1. Note frequencies of AC sources
Only one AC source so frequency response HZ(jω) will be the function of a single frequency
ECEN 301 Discussion #16 – Frequency Response 18
Frequency Response
Example 2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
R1is(t)
+RL
–
L
1. Note frequencies of AC sources2. Convert to phasor domain
Is(jω)
ZLD = RL
ZL = jωL
Z1=R1
+ VL
–
ECEN 301 Discussion #16 – Frequency Response 19
Frequency Response
Example 2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
LDLDL
LDLS
LDLL
ZZZ
ZZZjI
ZjIjV
)(
)(||)(
)()(
1
1. Note frequencies of AC sources2. Convert to phasor domain3. Solve using network analysis
• Current divider & Ohm’s Law
Is(jω)
ZLD = RL
ZL = jωL
Z1=R1
+ VL
–
IL(jω)
)(
)(||)()( 1
LDL
LDLSL ZZ
ZZZjIjI
ECEN 301 Discussion #16 – Frequency Response 20
Frequency Response
Example 2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
6
11
1
104.01
800
//1
)(
)(||
)(
)()(
j
RLjRR
R
ZZZ
ZZZ
jI
jVjH
L
L
LDLDL
LDL
S
LZ
Is(jω)
ZLD = RL
ZL = jωL
Z1=R1
+ VL
–
IL(jω)
4. Find an expression for the frequency response
ECEN 301 Discussion #16 – Frequency Response 21
Frequency Response
Example 2: compute the frequency response HZ(jω)
R1 = 1kΩ, RL = 4kΩ, L = 2mH
262
6
6
)104.0(1
)104.01(800
104.01
800)(
j
jjH ZIs(jω)
ZLD = RL
ZL = jωL
Z1=R1
+ VL
–
IL(jω)
4. Find an expression for the frequency response• Look at response for low frequencies (ω = 10)
and high frequencies (ω = 10000)
0.4800)( jH Z 0040.0194)( jH Z
ω = 10 ω = 10000
Lowpass filter
ECEN 301 Discussion #16 – Frequency Response 22
1st and 2nd Order RLC Filters
Graphing in Frequency DomainFilter Orders
Resonant FrequenciesBasic Filters
ECEN 301 Discussion #16 – Frequency Response 23
Frequency Domain
Graphing in the frequency domain: helpful in order to understand filters
-1.5
0.0
1.5
0.00 2.00 4.00
time
x(t
)
0.00
3.50
-600.0 0.0 600.0
w
X(j
w)
-ω ω
π π
cos(ω0t) π[δ(ω – ω0) + δ(ω – ω0)]
Time domain Frequency domainFourier Transform
ECEN 301 Discussion #16 – Frequency Response 24
Frequency Domain
Graphing in the frequency domain: helpful in order to understand filters
0.0
-10.00 0.00 10.00
wX
(jw
)-W W
1
sinc(ω0t)
Time domain Frequency domain
-0.5-10.00 0.00 10.00
t
x(t
)
W/π
π/W
W
WjX
||,0
||,1)(
Fourier Transform
ECEN 301 Discussion #16 – Frequency Response 25
Basic Filters
Electric circuit filter: attenuates (reduces) or eliminates signals at unwanted frequencies
0.00. 00 10. 00
.
Low-pass
0.00. 00 10. 00
.
High-pass
0.00.00 10.00
.
Band-pass
0.00. 00 10. 00
.
Band-stop
ω ω
ω ω
ECEN 301 Discussion #16 – Frequency Response 26
Filter Orders
Higher filter orders provide a higher quality filter
-150
-130
-110
-90
-70
-50
-30
-10
10
30
50
0.1 1 10 100
Frequency (w)
Gai
n (
dB
) 1st Order
2nd Order
3rd Order
4th Order
32nd Order
ECEN 301 Discussion #13 – Phasors 27
Impedance
LjjZL )(RjZR )(
+L–
+C–
Re
Im
-π/2
π/2R
-1/ωC
ωL
ZR
ZC
ZL
Phasor domain
+R–
CjjZC 1
)(
Vs(jω) +–~
+VZ(jω)
–
I(jω)
Z
Impedance of resistors, inductors, and capacitors
ECEN 301 Discussion #13 – Phasors 28
Impedance
+C–
Re
Im
-π/2
-1/ωC
ZC
CjjZC 1
)(
Impedance of capacitors
0
1)(
0 as
jZC
+C–
0
1)(
as
jZC
+C–
ECEN 301 Discussion #13 – Phasors 29
Impedance
LjjZL )(
+L–Re
Im
π/2
ωL
ZL
Impedance of inductors
0)(
0 as
jZC
+L–
+L–
)(
as
jZC
ECEN 301 Discussion #16 – Frequency Response 30
Resonant Frequency
Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters)
LCn
1
C+
vi(t)–
+vo(t)
–
L
C+
vi(t)–
+vo(t)
–L
Impedances in series Impedances in parallelCjLj
jZjZ CL
1
)()(
ECEN 301 Discussion #16 – Frequency Response 31
Resonant Frequency
Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters)
C
LCj
LC
LCjL
LC
LCL
LCj
LLC
j
LjZL
1
1
ZL=jωL
+Vi(jω)
–
+Vo(jω)
–
ZC=1/jωC
C
LCj
jC
LC
CLC
j
CjZC
1/1
/1 Impedances in series
ECEN 301 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)
0)( jVo
C
LCjZL
+Vi(jω)
–
+Vo(jω)
–
ZEQ=0C
LCjZC
0
CLEQ ZZZ
Impedances in series
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)
C
LCj
LC
LCjL
LC
LCL
LCj
LLC
j
LjZL
1
1
C
LCj
jC
LC
CLC
j
CjZC
1/1
/1 Impedances in parallel
ZL=jωL+Vi(jω)
–
+Vo(jω)
–
ZC=1/jωC
ECEN 301 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)
)()( jVjV io
C
LCjZL
+Vi(jω)
–
+Vo(jω)
–
ZEQ=∞
C
LCjZC
0
/
||
CL
ZZ
ZZ
ZZZ
CL
CL
CLEQ
Impedances in parallel
ECEN 301 Discussion #16 – Frequency Response 35
Low-Pass FiltersLow-pass Filters: only allow signals under the cutoff
frequency (ω0) to pass
)cos()( 1ttvo
0.00. 00 10. 00
.
Low-pass
ω1 ω0 ω2 ω3
)cos(
)cos(
)cos()(
3
2
1
t
t
ttvi
HL(jω)
R
C+
vi(t)–
+vo(t)
–
R
C+
vi(t)–
+vo(t)
–
L
1st Order
2nd Order
ECEN 301 Discussion #16 – Frequency Response 36
Low-Pass Filters
1st Order Low-pass Filters:
0)( tvo
ZR
+Vi(jω)
–
+Vo(jω)
–
ZC=1/jωC
ZR
+Vi(jω)
–
+Vo(jω)
–
ω1: ω → 0 ω3: ω → ∞
)()( tvtv io
0.00. 00 10. 00
.
Low-pass
ω1 ω0 ω3
ZR
+Vi(jω)
–
+Vo(jω)
–
ECEN 301 Discussion #16 – Frequency Response 37
Low-Pass Filters
2nd Order Low-pass Filters:
ZR
+Vi(jω)
–
+Vo(jω)
–
ZC=1/jωC
ZL=jωL
ω1: ω → 0
)()( tvtv io
ZR
+Vi(jω)
–
+Vo(jω)
–
ω2: ω = ωn
0)( tvo
ω3: ω → ∞
0)( tvo
ZR
+Vi(jω)
–
+Vo(jω)
–
0.00. 00 10. 00
.
Low-pass
ω1 ω0 ω2 ω3
ZR
+Vi(jω)
–
+Vo(jω)
–
Resonantfrequency
ECEN 301 Discussion #16 – Frequency Response 38
High-Pass FiltersHigh-pass Filters: only allow signals above the cutoff
frequency (ω0) to pass
)cos()( 3ttvo )cos(
)cos(
)cos()(
3
2
1
t
t
ttvi
HH(jω)
0.00. 00 10. 00
.
High-pass
ω1 ω0ω2 ω3
+vi(t)
–
+vo(t)
–
1st Order
RC
R
+vi(t)
–
+vo(t)
–
2nd Order
C L
ECEN 301 Discussion #16 – Frequency Response 39
High-Pass Filters
1st Order High-pass Filters:
)()( tvtv io
ZC=1/jωC
+Vi(jω)
–
+Vo(jω)
–
ZR=R
ZR
+Vi(jω)
–
+Vo(jω)
–
ω1: ω → 0 ω3: ω → ∞
0)( tvo
0.00. 00 10. 00
.
High-pass
ω1 ω0 ω3
ZR
+Vi(jω)
–
+Vo(jω)
–
ECEN 301 Discussion #16 – Frequency Response 40
High-Pass Filters
2nd Order High-pass Filters:
ZR
+Vi(jω)
–
+Vo(jω)
–
ZC=1/jωC
ZL=jωL
ω1: ω → 0
0)( tvo
ω2: ω = ωn
0)( tvo
ω3: ω → ∞
)()( tvtv io
ZR
+Vi(jω)
–
+Vo(jω)
–
0.00. 00 10. 00
.
High-pass
ω1 ω0ω2 ω3
ZR+
Vi(jω)–
+Vo(jω)
–
ZR+
Vi(jω)–
+Vo(jω)
–
Resonantfrequency
ECEN 301 Discussion #16 – Frequency Response 41
Band-Pass FiltersBand-pass Filters: only allow signals between the
passband (ωa to ωb) to pass
)cos()( 2ttvo )cos(
)cos(
)cos()(
3
2
1
t
t
ttvi
HB(jω)
0.00.00 10.00
.
Band-pass
ω1 ωa ω2 ω3ωb
+vi(t)
–
+vo(t)
–
2nd Order
RC
L
ECEN 301 Discussion #16 – Frequency Response 42
Band-Pass Filters
2nd Order Band-pass Filters:
ZR
+Vi(jω)
–
+Vo(jω)
–
ZC=1/jωC
ZL=jωL
ω1: ω → 0
0)( tvo
ω2: ω = ωn
)()( tvtv io
ω3: ω → ∞
0)( tvo
0.00.00 10.00
.
Band-pass
ω1 ωa ω2 ω3ωb
ZR
+Vi(jω)
–
+Vo(jω)
–
+Vi(jω)
– ZR
+Vo(jω)
–
ZR
+Vi(jω)
–
+Vo(jω)
–
Resonantfrequency
ECEN 301 Discussion #16 – Frequency Response 43
Band-Stop FiltersBand-stop Filters: allow signals except those between
the passband (ωa to ωb) to pass
)cos(
)cos()(
3
1
t
ttvo
)cos(
)cos(
)cos()(
3
2
1
t
t
ttvi
HN(jω)
+vi(t)
–
+vo(t)
–
2nd Order
RC
L
0.00. 00 10. 00
.
Band-stop
ω1 ωa ω2 ω3ωb
ECEN 301 Discussion #16 – Frequency Response 44
Band-Stop Filters
2nd Order Band-stop Filters:
ω1: ω → 0
)()( tvtv io
ω2: ω = ωn
0)( tvo
ω3: ω → ∞
)()( tvtv io
ZR
+Vi(jω)
–
+Vo(jω)
– ZC=1/jωC
ZL=jωL
0.00. 00 10. 00
.
Band-stop
ω1 ωa ω2 ω3ωb
ZR
+Vi(jω)
–
+Vo(jω)
– ZR
+Vi(jω)
–
+Vo(jω)
– ZR
+Vi(jω)
–
+Vo(jω)
–
Resonantfrequency