Post on 27-Dec-2015
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Chapter 14Chapter 14
Frequency ResponseFrequency Response
電路學電路學 (( 二二 ))
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Frequency ResponseFrequency ResponseChapter 14Chapter 14
14.1 Introduction14.2 Transfer Function14.3 The Decible Scale14.4 Bode Plots14.5 Series Resonance14.6 Parallel Resonance14.7 Passive Filters14.8 Active Filters14.9 Scaling
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What is Frequency Response of a Circuit?What is Frequency Response of a Circuit?
It is the variation in a circuit’s
behavior with change in signal
frequency and may also be
considered as the variation of the gain
and phase with frequency.
14.1 Introduction (1)14.1 Introduction (1)
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14.2 Transfer Function (1)14.2 Transfer Function (1)• The transfer function H(ω) of a circuit is the
frequency-dependent ratio of a phasor output Y(ω) (an element voltage or current ) to a phasor input X(ω) (source voltage or current).
|)(|
)()(
)( HXY
H
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14.2 Transfer Function (2)14.2 Transfer Function (2)
• Four possible transfer functions:
)()(
gain Voltage )(i
o
VV
H
)()(
gain Current )(i
o
II
H
)()(
Impedance Transfer )(i
o
IV
H
)()(
Admittance Transfer )(i
o
VI
H
|)(|
)()(
)( HXY
H
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14.2 Transfer Function (3)14.2 Transfer Function (3)
)()(
)(
DN
H
The transfer function can be expressed in terms The transfer function can be expressed in terms of its numerator polynomial of its numerator polynomial NN(() and ) and denominator polynomial denominator polynomial DD(() )
zero zero 零點零點 : a root of the numerator : a root of the numerator polynomial.polynomial.
pole pole 極點極點 : a root of the denominator : a root of the denominator polynomial.polynomial.
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14.2 Transfer Function (3)14.2 Transfer Function (3)Example 1
For the RC circuit shown below, obtain the transfer function Vo/Vs and its frequency response.Let vs = Vmcosωt.
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14.2 Transfer Function (4)14.2 Transfer Function (4)
Solution:
The transfer function is
,
The magnitude is 2)/(1
1)(
o
H
The phase iso
1tan
1/RCo
jωω R/ jω j Rjωω
11
1
1
)(s
o
VV
H
Low Pass FilterLow Pass Filter
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14.2 Transfer Function (5)14.2 Transfer Function (5)Example 2
Obtain the transfer function Vo/Vs of the RL circuit shown below, assuming vs = Vmcosωt. Sketch its frequency response.
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14.2 Transfer Function (6)14.2 Transfer Function (6)
Solution:
The transfer function is
,
The magnitude is 2)(1
1)(
o
H
The phase iso
1tan90
R/Lo
jωω Rjωω R
Ljω
1
1)(
s
o
VV
H
High Pass FilterHigh Pass Filter
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14.3 The Decibel Scale (1)14.3 The Decibel Scale (1)
210
1
Number of bels logP
GP
2dB 10
1
10logP
GP
210
1
20logV
V
• log log PP11PP22 = log = log PP11 + log + log PP22
• log log PP11//PP22 = log = log PP11 - log - log PP22
• log log PPnn = n = n log log PP
• log 1 = 0log 1 = 0
belbel 為功率增益的單位為功率增益的單位
dB (decibel) dB (decibel) 為為 belbel 的的1/101/10
210
1
20logI
I
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14.4 Bode Plots14.4 Bode Plots (1)(1)
Bode plotsBode plots are are semilog plots of the semilog plots of the magnitude (in dB) and magnitude (in dB) and phase (in degree) of a phase (in degree) of a transfer function vs transfer function vs frequency.frequency.
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14.4 Bode Plots14.4 Bode Plots (2)(2)
jH He H
ln ln ln lnjH e H j H
1 21 1
21 2
( ) (1 / )[1 2 / ( / ) ]( )
(1 / )[1 2 / ( / ) ]k k
n n
K j j z j j
j p j j
H
dB 10
10 10 10 1
10 1
20log
20log | | 20 log | | 20 log |1 / |
20 log |1 / |
H H
K j j z
j p
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14.4 Bode Plots14.4 Bode Plots (2)(2)
1.1. A gain A gain KK
2.2. A pole (A pole (jj))-1-1 or zero ( or zero (jj) at origin) at origin
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14.4 Bode Plots14.4 Bode Plots (3)(3)
3.3. A simple pole 1/(1+A simple pole 1/(1+jj/p/p11) or zero (1+) or zero (1+jj/z/z11))
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3.3. A quadratic pole 1/[1+A quadratic pole 1/[1+jj2222//nn+(+(jj//nn))22] or ] or zero [1+zero [1+jj2211//kk+(+(jj//kk))22]]
14.4 Bode Plots14.4 Bode Plots (4)(4)
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14.4 Bode Plots14.4 Bode Plots (5)(5)
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14.4 Bode Plots14.4 Bode Plots (6)(6)
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14.4 Bode Plots14.4 Bode Plots (7)(7)Example 3Construct the Bode plots for the transfer function
200( )
( 2)( 10)
j
j j
H
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14.4 Bode Plots14.4 Bode Plots (8)(8)Example 4Draw the Bode plots for the transfer function
5( 2)( )
( 10)
j
j j
H
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14.4 Bode Plots14.4 Bode Plots (9)(9)Example 5Obtain the Bode plots for
2
10( )
( 5)
j
j j
H
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14.4 Bode Plots14.4 Bode Plots (10)(10)Example 6Draw the Bode plots for
2
1( )
60 100
s
s s
H
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14.3 Series Resonance (1)14.3 Series Resonance (1)
Resonance is a condition in an RLC circuit in which the capacitive and inductive reactance are equal in magnitude, thereby resulting in purely resistive impedance.
)C
1L ( jRZ
Resonance frequency:
HzLC2
1f
rad/sLC
1
o
oro
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14.3 Series Resonance (2)14.3 Series Resonance (2)
The features of series resonance:
The impedance is purely resistive, Z = R;• The supply voltage Vs and the current I are in phase, so
cos = 1;• The magnitude of the transfer function H(ω) = Z(ω) is
minimum;• The inductor voltage and capacitor voltage can be much
more than the source voltage.
)C
1L ( jRZ
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14.3 Series Resonance (3)14.3 Series Resonance (3)
Bandwidth B
The frequency response of the resonance circuit current is )
C
1L ( jRZ
22
m
)C /1L (R
V|I|I
The average power absorbed by the RLC circuit is
RI2
1)(P 2
The highest power dissipated The highest power dissipated occurs at resonance:occurs at resonance: R
V
2
1)(P
2mo
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14 3 Series Resonance (4)14 3 Series Resonance (4)
Half-power frequencies ω1 and ω2 are frequencies at which the dissipated power is half the maximum value:
The half-power frequencies can be obtained by setting Z
equal to √2 R.
4R
V
R
)2/(V
2
1)(P)(P
2m
2m
21
LC
1)
2L
R(
2L
R 21
LC
1)
2L
R(
2L
R 22 21 o
Bandwidth Bandwidth BBLR
ω ωB 12
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14.3 Series Resonance (5)14.3 Series Resonance (5)
Quality factor,RLω
RILI o
02
21
221
/resonance at period one in
circuit the bydissipated Energycircuit the in stored energy Peak
Q
• The quality factor is the ratio of its resonant frequency to its bandwidth.
• If the bandwidth is narrow, the quality factor of the resonant circuit must be high.
• If the band of frequencies is wide, the quality factor must be low.
The relationship between the B, Q and o:
CRωQω
LR
B oo 2
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14.3 Series Resonance (6)14.3 Series Resonance (6)
Example 7
A series-connected circuit has R = 2 Ω, L = 1 mH, and C = 0.4 F.
a. Find the resonant frequency and the half-power frequencies.
b. Calculate the quality factor and bandwidth.
c. Determine the amplitude of the current at 0
, 1 and 2.
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14.4 Parallel Resonance (1)14.4 Parallel Resonance (1)
Resonance frequency:
HzLCπ
fLC
ωo 2
11 o or rad/s
)L
1C ( j
R
1Y
It occurs when imaginary part of Y is zeroIt occurs when imaginary part of Y is zero
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Summary of series and parallel resonance circuits:Summary of series and parallel resonance circuits:
14.4 Parallel Resonance (2)14.4 Parallel Resonance (2)
LC
1
LC
1
RC
1or
R
L
o
o
ω
ωRCor
L
Ro
o
Qo
Qo
2Q )
2Q
1( 1 2 o
o
2Q
)2Q
1( 1 2 o
o
2
Bo
2
Bo
characteristic Series circuit Parallel circuit
ωo
Q
B
ω1, ω2
Q ≥ 10, ω1, ω2
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14.4 Parallel Resonance (3)14.4 Parallel Resonance (3)Example 8
In a parallel RLC circuit, R = 8 kΩ, L = 0.2 mH, and C = 8 F (a) Calculate 0, Q, and B. (b) Find 1 and 2 (c) Determine the power dissipated at 0, 1 and 2
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14.4 Parallel Resonance (4)14.4 Parallel Resonance (4)Example 9
Calculate the resonant frequency of the circuit in the figure shown below.
rad/s2.1792
19AnswerAnswer::
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14.5 Passive Filters (1)14.5 Passive Filters (1)• A filter is a circuit that
is designed to pass signals with desired frequencies and reject or attenuate others.
• Passive filter consists of only passive element R, L and C.
• There are four types of filters.
Low Pass
High Pass
Band Pass
Band Stop
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14.5 Passive Filters (2)14.5 Passive Filters (2)Example 10
For the circuit in the figure below, obtain the transfer function For the circuit in the figure below, obtain the transfer function
VVoo(()/V)/Vii((). Identify the type of filter the circuit represents and ). Identify the type of filter the circuit represents and
determine the corner frequency. Take determine the corner frequency. Take RR1 1 = 100 = 100 = = RR22
and and LL =2 mH. =2 mH.
krad/s25AnswerAnswer::