network theory - ggn.dronacharya.info · property 1. l-c immittance function • 1. z lc (s) or y...
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NETWORK THEORYNETWORK THEORY
LECTURE 3SECTION-D:TYPES OF FILTERS AND THEIR CHARACTERISTICSD:TYPES OF FILTERS AND THEIR CHARACTERISTICS
Example - Passive Second Order Filter Function
T(dB)
0
01
0)(
22
2
sasjsasQ
sas
sQ
ssT o
oo
o
0 dBQ
3
* Low pass filter
Passive Second Order Filter Function
20
02
12
2
)()()(
sQ
s
asasasVsVsT o
i
o
General form of transfer function
20012
22
2
22
0a,0a
thfilter wiorder second a of form thehas This
1
11
1
1
1
11
1
1
1)(
111
111
)()()(
a
LCRRCQand
LCwhere
sQ
sLCRC
ss
LC
LCsRLs
sRCRsL
RsRC
sRCRsL
sRCR
ZRZRZ
sT
sosRCR
RsC
RZ
RZwhere
ZRZRZ
sVsVsT
oo
oo
o
LC
C
C
C
LC
C
i
o
Example - Passive Second Order Filter Function
00
1
0
)(
2
2
22
0
sassjsat
sasss
sQ
s
Qs
sT
o
o
oo
T(dB)
0
0 dB
-3 dB
Qo
4
Passive Second Order Filter Function* Bandpass filter
01a,0a
thfilter wiorder second a of form thehas This
1
1
11
1
11
1)(
111
111
)()()(
012
222
2
2
2
2
aQRC
LCRRCQand
LCwhere
sQ
s
sRC
LCRCss
RCs
LCsRsLsL
LCssLR
LCssL
RZZZZ
sT
soLCs
sL
sLsC
ZZ
ZZwhere
RZZZZ
sVsVsT
o
oo
oo
LC
LC
LC
LC
LC
LC
i
o
20
02
12
2
)()()(
sQ
s
asasasVsVsT o
i
o
General form of transfer function
Low Pass Butterworth Filter Design
CRRCss
CRVVT(s)
i
o
12
1
222
22
o CRso
RC 11
* Given the filter specification (0), we can determine the R and C.* One specification, two parameters – R and C * Pick a convenient value, say C = 5 nF.* Calculate R from C and ωo.
Hurwitz Polynomials
5
Low Pass Butterworth Filter Design
sasjsasj
sas
o
02/
01
T(dB)
0
0 dBQ(dB)
o1
), we can determine the R and C.R and C
NOTE 40 dB/dec
16sec)/1026.1)(105(
11 ,5
sec/1026.122Given :
79
70
radxFxCR
thennFCChooseradxMHzExample
o
Hurwitz Polynomials
PROPERTY 1. L-C IMMITTANCE FUNCTION
• 1. ZLC (s) or YLC (s) is the ratio of odd to even or even to odd
polynomials.
• Consider the impedance Z(s) of passive one
As we know, when the input current is dissipated by one-port network is zero:
Average Power= =0
1 1
2 2
( ) ( )( )( ) ( )
M s N sZ sM s N s
1 Re[ ( )]2
C IMMITTANCE FUNCTION
(s) is the ratio of odd to even or even to odd
Consider the impedance Z(s) of passive one-port network.
(M is even N is odd)
As we know, when the input current is I, the average power port network is zero:
= =0
( ) ( )( ) ( )
M s N sM s N s
21 Re[ ( )]2
Z j I
1 2 1 2( ) ( ) ( ) ( ) 0M j M j N j N j
1 2 1 22 2
2 2
( ) ( ) ( ) ( )( ) 0( ) ( )
M s M s N s N sEvZ sM s N s
1 20M N OR
1 1
2 2
( ) , ( )M NZ s Z sN M
Z(s) or Y(s) is the ratio of even to odd or odd to even!!
1 2 1 2( ) ( ) ( ) ( ) 0M j M j N j N j
1 2 1 22 2
2 2
( ) ( ) ( ) ( )( ) 0( ) ( )
M s M s N s N sM s N s
2 10M N
1 1
2 2
( ) , ( )M NZ s Z sN M
Z(s) or Y(s) is the ratio of even to odd or odd to even!!
PROPERTY 2. L-C IMMITTANCE FUNCTION
• 2.The poles and zeros are simple and lie on the axis.
• Since both M and N are Hurwitz, they have only imaginary roots, and it follows that the poles and zeros of Z(s) or Y(s) are on the imaginary axis.
• Consider the example
1 1
2 2
( ) , ( )M NZ s Z sN M
C IMMITTANCE FUNCTION
2.The poles and zeros are simple and lie on the
Since both M and N are Hurwitz, they have only imaginary roots, and it follows that the poles and zeros of Z(s) or Y(s) are on the
j
1 1
2 2
( ) , ( )M NZ s Z sN M
4 24 2 0
5 35 3 1
( ) a s a s aZ sb s b s b s
Ex) highest order of the numerator : 2n denominator can either be 2n-1 (simple pole at s= ) or
the order can be 2n+1 (simple zero at s= ).
Impedance function cannot have multiple poles or zeros
axis.
In order for the impedance to be positive real positive.
4 24 2 0
5 35 3 1
( ) a s a s aZ sb s b s b s
The highest powers of the numerator and the denominator polynomials can differ by, at most, unity.
Ex) highest order of the numerator : 2n -> highest order of the 1 (simple pole at s= ) or
the order can be 2n+1 (simple zero at s= ).
cannot have multiple poles or zeros on the
n order for the impedance to be positive the coefficients must be real and
positive.
j
The highest powers of the numerator and the denominator polynomials can differ by, at most, unity.