chapter 6 frequency response & systems concepts
DESCRIPTION
Chapter 6 Frequency Response & Systems Concepts. AC circuit analysis methods to study the frequency response of electrical circuits Understanding of frequency response aided by the concepts of phasors and impedance. Filtering – a new concept will be explored. Objectives. - PowerPoint PPT PresentationTRANSCRIPT
Këpuska 2005 1
Chapter 6
Frequency Response & Systems Concepts
•AC circuit analysis methods to study the frequency response of electrical circuits
•Understanding of frequency response aided by the concepts of phasors and impedance.
•Filtering – a new concept will be explored
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Objectives
• Understand significance of frequency domain analysis
• Introduction of Fourier series as a tool for computation of Fourier spectrum.
• Analyze first and second-order electrical filters by determining their filtering properties.
• Computation of frequency response and its graphical representation as Bode plot.
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Sinusoidal Frequency Response
• Provides a circuit response to a sinusoidal input of arbitrary frequency.
• The frequency response of a circuit is a measure of voltage or current (magnitude and phase) as a function of the frequency of excitation (source) signal.
)(
)()(
jV
jVjH
S
Lv
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Methods to Compute Frequency Response
• Thevenin equivalent source circuit:
~
Zs Z1
Z2VTVL
+
-
~
ZT
VT
21
21
ZZZ
ZZZZ
s
sT
21
2
ZZZ
ZVV
sST
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Load Voltage VT
~
ZT
VTZL
SSSL
L
SS
S
SL
L
TTL
LL
VZZZZZZZ
ZZ
VZZZ
Z
ZZZ
ZZZZ
Z
VZZ
ZV
2121
2
21
2
21
21
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Frequency Response
• From definition:
• VL(j) is a phase-shifted and amplitude-scaled version of VS(j) ⇨
2121
2
ZZZZZZZ
ZZ
jV
jVjH
SSL
L
S
LV
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Frequency Response (cont)
• Phasor form of the load voltage:
SVL
SVL
jHjSV
jL
jS
jHjV
jL
SVL
jH
jVjHjV
ejVjHejV
ejVejHejV
jVjHjV
SVL
SVL
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Example 6.1
• Compute the frequency response Hv(j) of the circuit for R1= 1k, C=10F; and RL= 10k.
VS C
R1
RL VL
110arctan
110
10022
2121
2
2121
2
jH
ZZZZZ
ZZjH
VZZZZZ
ZZ
VZZ
ZV
V
L
LV
SL
L
TTL
LL
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Magnitude & Phaze
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Fourier Analysis
• Let x(t) be a periodic signal with period T.– x(t) = x(t+nT) for n=1,2,3,…
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Fourier Series• A signal x(t) can be expressed as an infinite summation
of sinusoidal components know as Fourier Series:• Sine-cosine (quadrature) representation
• Magnitude and Phase form:
• Fundamental Frequency and Period T:
110
2sin
2cos
nn
nn t
Tnbt
Tnaatx
10
10
2cos
2sin
nnn
nnn
tT
ncctx
tT
ncctx
Tf
22 00
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Fourier Series
• It can be shown
• Or similarly
n
n
n
n
nnn
n
n
nnn
a
bcba
a
bcba
tan
cot
22
22
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Fourier Series Aproximation
• Infinite summation practically not possible
• Replaced by finite summation that leads to approximation.– Higher order coefficients; n, are associated
with higher frequencies; (2/T)n. ⇒– Better approximations require larger
bandwidths.
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Fourier Series
• Odd and Even Functions
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Frequency Spectrum
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Computation of Fourier Series Coefficients
dttT
ntxT
dttT
ntxT
b
dttT
ntxT
dttT
ntxT
a
dttxT
dttxT
a
T
T
T
n
T
T
T
n
T
T
T
2sin
22sin
2
2cos
22cos
2
x(t)of valueaverage11
2
20
2
20
2
20
0
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Example of Fourier Series Approximation
• Square wave and its representation by a Fourier series. (a) Square wave (even function); (b) first three terms; (c) sum of first three terms
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Example 6.3 Computation of Fourier Series Coefficients
• Problem: Compute the complete Fourier Spectrum of the sawtooth function shown in the Figure below for T=1 and A=1:
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Solution
• x(t) is an odd function.
• Evaluate the integral in equation
TtT
tAtx
0 ,
21)(
1,2,3,...n 2
2cos2
4
2cos
/2
2sin
/2
140
2sin
42cos
2
2
2sin
222sin
2
2sin
21
2
2
2
0
222
02
0
00
0
πn
Aπn
πn
T
T
A
tT
πn
Tπn
tt
T
πn
TπnT
A
dttT
πn-t
T
At
T
πn
πn
T-
T
A
dttT
πn
T
t-
T
A dtt
T
πn
T
A
dttT
πn
T
t-A
Tb
T
TT
TT
T
n
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Solution (cont)• Spectrum computation:
00
cotcot 11
22
n
n
n
nnn
b
a
b
bbac
n
n
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Matlab Simulation• Components of the sawtooth wave
function:
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Matlab Simulation• Fourier Series approximation of sawtooth
wave function
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Example 6.4
• Problem: Compute the complete Fourier series expansion of the pulse waveform shown in the Figure for /T=0.2
• Plot the spectrum of the signal
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Solution
• Expression for x(t)
• Evaluate Integral Equations:
Tt
tAtx
0
0)( {
52.0|
10
0
0
AA
T
At
T
AAdt
Ta
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Solution (cont)
5
2cos1
2cos
2
2
2sin
2
5
2sin
2sin
2
2
2cos
2
0
0
0
0
πn
πn
At
T
πn
πn
AT
T
dttT
πn A
Tb
πn
πn
At
T
πn
πn
AT
T
dttT
πn A
Ta
n
n
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Spectrum Computation
• Magnitude:
• Phase:
22
22
5
2cos1
5
2sin
n
n
An
n
Abac
nnn
52sin
52cos1cotcot 11
nnA
nnA
a
b
n
n
n
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Graphical Representation
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Matlab Simulation
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Matlab Simulation
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Linear Systems Response to Periodic Inputs
• Any periodic signal x(t) can be represented as a sum of finite number of pure periodic terms:
10
2sin
nnn t
Tncctx
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General Input-Output Representation of a System
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Linear Systems• For Linear Systems - by definition
Principle of superposition applies:T{ax1(t) + bx2(t)} = aT{x1(t)} + bT{x2(t)}
x1
x2
a
b
T{} y= T{ax1(t) + bx2(t)} ax1[n] + bx2[n]
x1
x2
T{}
T{}
a
bbT{x2[n]}
y= aT{x1(n)}+bT{x2(n)}
aT{x1[n]}
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Linear System View of a Circuit
• Output of a circuit y(t) as a function of the input x(t):
1
sinn
nnnnn jHtcjHty
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Example 6.6 Response of Linear System to Periodic Input
• Problem: – Linear system:
– Input: sawtooth waveform approximated with only first two Fourier components of the input waveform.
2.01
2
jjH
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Solution• Approximation of the sawtooth function
with first two terms of Fourier Series:
• Spectrum Computation:
tttA
tA
tx
16sin2
8sin4
25.0
4sin
25.0
2sin
2)(
1622
814
01222
22
0122
1
2
111
bbac
bbac
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Frequency Response• Magnitude and Phase
• Computation of Frequency Response for two frequency values of 1 = 8 and 2 = 16:
5arctan
2.01
2
2.01
22
jjH
022
222
2
011
221
1
32.8447.116*2.0arctan2.0arctan
1980.016*2.01
2
2.01
2
75.7837.18*2.0arctan2.0arctan
3902.08*2.01
2
2.01
2
radj
jH
radj
jH
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Frequency Response (cont)
• Computation of steady-state periodic output of the system:
47.116sin2
1980.037.18sin4
3902.0
sin2
1
tt
jHtcjHtyn
nnnnn
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Matlab Simulation
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Matlab Simulation
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Filters
• Low-Pass Filters
C
R
VoVi
+ +
--
Simple RC Filter
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Low-Pass Filter
RCjjV
jV
jVRCj
CjR
CjjVjV
jV
jVjH
i
o
iio
i
o
1
1
1
11
1
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Low-Pass Filter
=0– H(j)=1 ⇨ Vo(j)=Vi(j)
>0
RC
RC
j
i
o
eRC
e
e
RC
RCjjV
jVjH
arctan
2
1arctan
0
2
1
1
1
1
1
1
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Low-Pass Filter
RC
RCjH
RCjH
ejHjH
o
o
o
jHj
1
arctanarctan
1
1
1
122
Cutoff Frequency
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Example 6.7
• Compute the response of the RC filter to sinusoidal inputs at the frequencies of 60 and 10,000 Hz.
• R=1k, C=0.47F, vi(t)=5cos(t) V
0=1/RC= 2,128 rad/sec = 120 rad/sec ⇒ /0 = 0.177
= 20,000 rad/sec ⇒ /0 = 29.5
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Solution
537.1000,102cos923.4
175.0602cos923.4
537.150345.0000,102
128,2000,102
1
1000,102
175.05985.0602
128,2602
1
1602
1
1
0
tv
tv
VjVj
jV
VjVj
jV
jVj
jV
o
o
io
io
io
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High-Pass Filters
VoVi
+ +
--
C
R
RCj
RCj
jV
jV
jVRCj
RCj
CjR
RjVjV
jV
jVjH
i
o
iio
i
o
1
11
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High-Pass Filter• The expression in previous slide can be
written in magnitude-and-phase form:
RCjH
RC
RCjH
eRC
RC
eRC
RCe
RCj
RCjjH
RCj
RCj
j
arctan90
1
1
11
0
2
arctan2
2
1arctan
2
2
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High-Pass Filter Response
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Band-Pass Filters
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Frequency Response of Band-Pass Filter
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Analysis of the Second Order Circuit
21
2
111
1
jj
jA
LCjRCj
RCj
LjCj
R
R
jV
jVjH
i
o
1, 2 and are the two frequencies that determine the pass-band (bandwidth) of the filter.
•A – gain of the filter.
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Magnitude and Phase form
21
21
arctanarctan2
2
2
2
1
arctanarctan
2
2
2
2
1
21
11
11
11
j
jj
j
eA
ee
eA
jj
jAjH
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Magnitude and Phase form
21
2
2
2
1
arctanarctan2
11
jH
AjH
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Frequency Response and Bandwidth
2
2
2
11
1
21
2
1
nn
n
nn
n
i
o
jQ
j
Qj
jj
j
LCjRCj
RCj
jV
jVjH
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Frequency Response and Bandwidth
Ratio Damping; 22
1
FactorQuality ; R
11
2
1Q
FrequencyResonant or Natural; 1
L
CR
Q
C
L
RC
LC
n
n
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Normalized Frequency Response & Bandwidth
Bandwidth ; Q
B n
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Frequency Response of the Filter with R=1 k; C=10 F; L=5 mH
rad/s 10,000Q
B n
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Frequency Response of the Filter with R=10 ; C=10 F; L=5 mH
rad/s 100Q
B n
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Bode Plots
• Logarithmic Plots of System’s Frequency Response:
• Both plots are function of frequency also represented in log scale.
Re
Imarctan
(dB) decibelsin log20
jH
A
A
A
AjH
i
o
i
o
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Bode Plot
• Example of Low-Pass Filter:
02
0
0
arctan
1
1
1
1
jV
jV
jjV
jVjH
i
o
i
o
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Bode Plot of Low-Pass Filter
0103.3log202log10log20 1
log20log20log20 1
log20 1
1log10log20
1
log20
0
00
0
2
02
0
KKjV
jV
KjV
jV
KjV
jV
KK
jV
jVjH
dBi
o
dBi
o
dBi
o
dBi
o
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Bode Plot of Low-Pass Filter
n⇨ Cut-off Frequency
-20 dB Slope
3dB
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Bode Plot of Low-Pass Filter
0
0
0
0
2
4
0
arctan
when
when
when
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Bode Plot of Low-Pass Filter
-450 dB Slope
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Correction Factors
/0Magnitude
Response Error in dB
Phase Response Error in dB
0.1 0 -5.7
0.5 -1 4.9
1 -3 0
2 -1 -4.9
10 0 +5/7
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Bode Plots of Higher-Order Filters
• Bode Plots Higher-Order Filters may be obtained by combining Bode-Plots of lower-order functions:– H(j) = H1(j) H2(j) H3(j) … Hn(j)
– |H(j)|dB = |H1(j)|dB + |H2(j)|dB + |H3(j)|dB + … + |Hn(j)|dB
H(j) = H1(j) + H2(j) + H3(j) + … + Hn(j)
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Example of High-Order Filter
1001
101
51005.0
1001log20
101log20
51log20005.0log20
1001
101
51005.0
1001100
10110
515
10010
5
jjjjH
jjjjH
jj
j
jj
jjH
jj
jjH
dB
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Magnitude and Phase in Bode Plots
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Composite Bode Plot
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Bode Plot Approximation Example
50001
101
200120
50001log20
101log20
2001log2020log20
50001
101
200120
1002.0102
1.020235
jjjjH
jjjjH
jjj
jjH
jjj
jjH
dB
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Straight Line Aproximation of Bode Plots
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Actual Magnitude and Phase
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Bode Plot Approximation Example
30001
1001
30111.0
30001log20
1001log20
301log201log201.0log20
30001
301
10011.0
1000,90
030,31091
1.0102
4
23
jjjjH
jjjjjH
jj
jjjH
jj
jjjH
dB
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Bode Plots
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Bode Plots