Chapter 4 Linear Wire Antennas 1 ECE 5318/6352 Antenna Engineering Dr. Stuart Long.
ECE 550 LINEAR SYSTEM THEORY LECTURE NOTES
Transcript of ECE 550 LINEAR SYSTEM THEORY LECTURE NOTES
115
Passive Filters
Up to now, we have driven a circuit with a dc or a fixed frequency ac signal and looked at the effects of changing the magnitudes and phase angles of thevoltages, currents and impedances. In some systems, you really want toknow the response of the system when the frequency is varied, i.e., thefrequency response. Often, filters weaken (attenuate) the input signal outside a particular frequency band. For example, you might have a graphic equalizer an your stereo which is a collection of filter circuits designed to allow you to amplify the sound (or audible frequency kHzHz 3~300 ) in a particular frequency range and attenuate frequencies outside that band.
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Transfer function:
)( ωjV
)( ωjI
System
)()()(
ωωω
jIjVjZ = transfer impedance
)()(
)( 0
ωω
ωjVjV
jGi
V = voltage gain
)()()(
ωωωjVjIjY = transfer admittance
)()(
)( 0
ωω
ωjIjI
jGi
I = current gain
)( ωjZ , )( ωjY , )( ωjGV , and )( ωjGI are called the ‘transfer functions’
which describe the system characteristics. We generally represent it as )( ωjH
System
)( ωjI i
)( ωjVi
)(0 ωjI
)(0 ωjV
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Consider a linear system (current proportional to the input voltage, output response proportional to the input)
)( ωjX )( ωjY)( ωjH
Linear System
The Linear System can be characterized by a Transfer Fct.: )()()(
ωωωjXjYjH =
Let us consider an example:
RI
+_ Ljω
+
_ 0ViV
The transfer function,
LjRLj
jVjV
jGjHi
V ωω
ωω
ωω+
===)()(
)()( 0
By Voltage Divider
LjRLjVV i ω
ω+
=0
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We show voltages and currents as functions of “ ωj ”indicating that we are interested in the frequency characteristics or response of a circuit /system. In ENGR 203 you will see this developed into Fourier Transform Analysis.
We can write )( ωjH as: (multiply by L
L1
1 to isolate jω)
LRj
jjH+
=ω
ωω)( and: (recall, for a complex number c=a+jb, |c|=sqrt(a2+b2)
( )( ωjH is a complex function, so it can be represented in polar form, i.e. using mag. and phase, )(|)(|)( ωθωω jjHjH ∠= , and you can give the “phase angle plot” of )( ωθ j vs. freq. ω)
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2
222 11
1)(
ωω
ωω
⎟⎠⎞
⎜⎝⎛+
=
⎟⎠⎞
⎜⎝⎛+
=
LR
LR
jH
0)(0=
=ωωjH ; 1)( =
∞=ωωjH
Magnitude Plot ( ωω ..)( freqvsjH )
ω
)( ωjH
707.02
1=
cω
1.0
This network behaves like a “High-Pass” filter. As you can see for lowfrequencies )( ωjH is smaller. “ωc” gives an indication of the frequencies that can be passed by the filter. Itis called the “cut-off frequency” or the “half-power frequency” at which theaverage power delivered to the load is half of the maximum power.
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For this example, when
LR
=ω
21
2)( =
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
LR
LR
jH ω LR
c =⇒ω
Consider another example:
R
I
+_
Ljω
+
_iV
0V
By simple inspection, V0=Vi at dc since jωL=0 V0=0 at ∞=ω , since ∞=Ljω (open circuit) This implies that this is a ‘low-pass’ filter!
2)( maxH
jH c =ω
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LRj
LR
jH+
=ω
ω)( ⇒ ( )22
)(
LR
LR
jH+
=ω
ω
ω
)( ωjH
707.02
1=
cω
1.0
Low pass behavior
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Some other filter examples:
RI
+_
+
_iV
0VC
R
I
+_
+
_iV
0V
C
0)(1)0(=∞=
HH } Low-pass
RCc1
=ω
RCc1
=ω
RCc1
=ω
RCc1
=ω
RCjRCjH
1
1)(
+=
ωω
} High-pass1)(0)0(=∞=
HH
RCjjjH
1)(
+=
ωωω
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Parallel Resonant Circuits: Consider:
+
_
outVinI C R L
Let us first analyze the circuit by inspection:
0=outV @ dc since ⇒= 0Ljω inductor shorts the output.
0=outV @ ∞=ω since ⇒= 01Cjω
capacitor shorts the output.
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The transfer function of the circuit can be given as (Transfer Impedance)
⎥⎦⎤
⎢⎣⎡ −+
==
LCj
RjIjV
jHi
out
ωωω
ωω
111
)()(
)( =Zeq = ...)111(
1
321 ZZZ++
or 22 11
1)(
⎥⎦⎤
⎢⎣⎡ −+⎟
⎠⎞
⎜⎝⎛
=
LC
R
jH
ωω
ω ,
when the capacitor and inductor are in resonance (exchange energy) 01
=−L
Cω
ω , )1(L
Cω
ω = , ω2= LC1 ,
LC1
=ω , then ω = ωo (resonant frequency)
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RHjH == max)( ω , when 0ωω = , such that
01
00 =−
LC
ωω
or: LC1
0 =ω , again is called the “resonant frequency”
or the center frequency of the “bandpass” filter
ω1 and ω2 are called the “half-power frequencies” where 2
)( maxHjH =ω
(or lower and upper cutoff frequencies) We can calculate ω1 and ω2 to be
LCRCRC1
21
21 2
2,1 +⎟⎠⎞
⎜⎝⎛++=ω
We can also show that 210 ωωω = 12 ωω −=B is called the bandwidth of the filter. (defines the pass band
– the range of frequencies passed with appreciable output)
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Quality Factor Quality Factor (Q Factor), or “selectivity factor”, indicates the quality of the filter circuit
Q = 12
00
ωωω
βω
−= (dimensionless)
Where: High Q – good selectivity (narrow bandwidth passed) Low Q – poor selectivity (broad bandwidth passed)
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Series Resonance:
R
+_
+
_
iV 0V
C
L
By inspection: ( 0=ω ) V0=Vi @ dc since ⇒∞==
CjZc ω
1 capacitor is dominant, like an open circuit
iVV
H 01)0( ==⇒
AND V0=Vi @ ∞=ω , since ⇒∞== LjZ ω inductor is dominant, like an open circuit
1)( =∞⇒ H
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At 0ωω = whenCjLj0
0 ωω = ; 0)(0 00 =⇒= ωHV (Zeq = 0
00 =
−+
CjLj
ωω )
This implies that the circuit exhibits “band-stop” characteristic.
⎟⎠⎞
⎜⎝⎛ −+
⎟⎠⎞
⎜⎝⎛ −
==
CLjR
CLj
VV
jHi
ωω
ωω
ω1
1
)( 0
22 1
1
)(
⎟⎠⎞
⎜⎝⎛ −+
⎟⎠⎞
⎜⎝⎛ −
=
CLR
CL
jH
ωω
ωω
ω
0)( =ωjH , when C
L0
01
ωω = , ⇒
LC1
0 =ω resonant frequency
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Bandstop Characteristic.
ω1 and ω2 can be determined by equating 2
)( maxHjH =ω
Where,
LCLR
LR 1
22
2
2,1 +⎟⎠⎞
⎜⎝⎛++=ω
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Multi-Section Filter Design References:
• ‘Microwave Filters, Impedance-Matching Networks and Coupling Structures’ by Matthaei, Young and Jones
• “Micro Engineering’ by David M. Pozar.
maxH
)( ωjH
ωcωLow-pass
maxH
)( ωjH
ωcωHigh-pass
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maxH
)( ωjH
ω0ω
Band-pass
maxH
)( ωjH
ωBandstop
0ω
N-SectionFilter
R(load)
Rs
Vs Pinc
Zin
Pload
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In this filter design procedure, we characterize the performance of a filter using a transfer function called Power Transfer Ratio or Power Loss Ratio, PLR defined as
load
incLR P
PloadtodeliveredPowersourcefromavailablePower
P ==
It can be expressed in terms of the input impedance inZ as Eqn. A
)(21
*
2
inin
inLR ZZ
ZP
+
+=
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Design Procedure:
Filter Specs.
Low-passprototype
design
Scaling andconversion
Final Circuit
Type of filter No. of sections
csR ω, for Low Pass / High Pass
BRs ,0, ω etc., Bandpass / Bandstop.
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Lowpass Prototype Design:
R0=g0=1
+_ C1=g1
Zin
L2=g2
C3=g3
L4=g4
C5=g5 gn+1
Ladder Circuit (a)
g0=1
+_
Zin
L1=g1
C2=g2
L3=g3
C4=g4 gn+1
L5=g5
Ladder Circuit (b)
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In order to have a maximally flat response in the pass band, with a cut-off frequency of 1=cω and a source resistance of Ω= 10R , The power loss ratio, PLR for the networks shown above should be Eqn. B Where N is the No. of sections (counting only L’s and C’s). We can find out g1, g2 … gn+1 for the networks (a) and (b) by equating the two equations A and B and comparing the coefficients of Kω
N
LRP 21 ω+=
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For example: Consider the network with N=1 (network (b))
Zin
Ω1 L1=g1
g2=R
According to A, )(2
1)(2
1 2
*
2
LjRLjRLjR
ZZZ
Pinin
inLR ωω
ω−++
++=
+
+=
( )R
LRR
LR4)1(
4)()1( 222
222 ωω ++
=++
=
LjRZin ω+=
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And this should be equal to 22 11 ωω +=+ N
Equating 0ω th coefficient, 0)1(14
)1( 22
=−⇒=+ RR
R or R=1
14
2
=R
L or L=2
21 =∴ g and 12 =g
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Design table for Low-pass filter prototype
Impedance Scaling: If source resistance is Ω0R instead of Ω1 , the modified filter components are given by:
0' RRS = ,
' '0
0
, CL R L CR
= = LL RRR 0' =
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For example: (For a source resistance of 100Ω)
Ω1 L=2
R=1
Ω100 200H
Ω100R=
Ω1
R=1
Ω100
Ω100R=2 0.02F
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Frequency Scaling: To change the cutoff frequency from 1=cω to “ cω ”, New L & C’s are given by: No change in R’s.
c
kk
LL
ω='
c
kk
CC
ω='
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Impedance and frequency scaling: (will show how we use this in an example) New filter elements are given by Filter to Filter Transformations: The main advantage of using this method of low-pass filter prototype is to be able to come up with other types of filter designs such as high-pass, bandpass and bandstop from the original low-pass prototype itself.
0' RRS = , LL RRR 0
' =
c
kok
LRL
ω=' ,
oc
kk R
CC
ω='
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Low-pass to High-pass Transformation:
g0=1 L1=g1
C2=g2
L3=g3
Low pass
R0
High pass
C1' C3'
L2' R0
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Transformation: Eqn. C Examples: Design a 5-section maximally flat low-pass filter with a cut-off frequency of 100kHz. The source resistance is 1kΩ. From the design table, for N=5, g1=0.618; g2=1.618; g3=2; g4=1.618; g5=0.618; g6=1 R0=1000; sec/10283.6101002 53 radc ×=××= πω
kck LR
Cω0
' 1=
kck C
RL
ω0' =
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Considering the network (a) for the low-pass prototype.
nFR
gCc
9836.010283.61000
618.05
0
1'1 =
××==
ω
mHgR
Lc
6.220'2 ==
ω nF
Rg
Cc
183.30
3'3 ==
ω
mHgR
Lc
6.240'4 ==
ω nF
Rg
Cc
9836.00
5'5 ==
ω
The final network will be
1000
0.984nF
2.6mH
3.183nF
2.6mH
0.984nF 1000
Ω
Ω
Note: You can use the network (b) (starting with an inductor) and do the design. Both designs should give you identical response.
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Example 2: Design a 5-section maximally flat high-pass filter with a cut-off frequency of 100kHz. Source resistance is 1kΩ. Again from the design table: g1=0.618; g2=1.618; g3=2; g4=1.618 and g5=0.618. R0=1000Ω; sec/10283.6 5 radc ×=ω Considering network (a) (capacitance first in the low-pass prototype), applying equations given in C,
mHg
RL
c
6.21
01 ==
ω nF
gRC
c
984.01
202 ==
ω
mHg
RL
c
796.03
03 ==
ω nF
gRC
c
984.01
404 ==
ω
mHg
RL
c
6.25
05 ==
ω
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The high-pass filter is
1000Ω
1000Ω
0.984nF 0.984nF
2.6mH 0.796mH 2.6mH
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Low-pass to Band-pass Transformation:
Transforms toL L' 'C
In a low-pass prototype where Eqn. D1
Δ=
0
0'
ωLR
L
oLRC
0
'
ωΔ
=
ω0 center frequency of the bandpass filter
0
12
ωωω −
=Δ is called the fractional
bandwidth.
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Transforms toC L" "C
In a low-pass prototype Eqn. D2
CR
L0
0"
ωΔ
=
oRCCΔ
=0
"
ω
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Low-pass to Band-stop Transformation:
LTransforms to
L
'
'
CIn a low-pass prototype Where Eqn. E1
0
0'
ωLR
LΔ
=
oLRC
Δ=
0
' 1ω
ω0 center frequency
0
12
ωωω −
=Δ fractional bandwidth.
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Transforms toL"
"C
C
Where Eqn. E2
CR
LΔ
=0
0"'
ω
oRCC
0
"
ωΔ
=
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Example for Bandpass filter design: Design a maximally flat bandpass filter with N=2; center frequency is100kHz; bandwidth is 10%; Source resistance is 50Ω. From the design table, g1=1.4142, g2=1.4142 Let us use the low-pass prototype network given in (b) (First element: inductor)
sec/10283.6101002 530 rad×=××= πω , %)10(1.0=Δ
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From Eqn. D1, mHRg
L 1.11.010283.6
504142.15
0
011 =
×××
=Δ
=ω
nFRg
C 25.2504142.110283.6
1.05
0101 =
×××=
Δ=ω
From Eqn. D2, uHgR
L 63.520
02 =
Δ=ω
nFR
gC 15.450
501.010283.64142.1
500
22 =
×××=
Δ=ω
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The bandpass circuit will be,
50 L1=1.1mH C1=2.25nF
L2=5.63uH C2=450.15nF 50
Ω
Ω
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Example for Bandstop filter design: Design a maximally flat bandstop filter with N=2; center frequency is100kHz; bandwidth is 10%; Source resistance is 50Ω. Again, start with low-pass prototype table: g1=1.4142, g2=1.4142 Let us start with the low-pass prototype given in (a) (First element: capacitor)
From Eqn. E2, mHg
RL 563.0
4142.11.010283.650
510
01 =
×××=
Δ=ω
nFRg
C 5.45010283.6
4142.11.05
00
11 =
×××
=Δ
=ω
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From Eqn. E1, uHRg
L 25.1110283.6
504142.11.05
0
022 =
×××
=Δ
=ω
nF
RgC 1.225
504142.11.010283.611
5020
2 =××××
=Δ
=ω
The bandstop circuit will be,
Ω
Ω