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1. Circuit Analysis – II (EE-201)
Variable-frequency Networks
Part 1

2. Variable-Frequency Response Analysis
Network performance as function of frequency.
Transfer function
VARIABLE-FREQUENCY NETWORK
LEARNING GOAL

3.  0RRZRResistor
VARIABLE FREQUENCY-RESPONSE ANALYSIS
In AC steady state analysis the frequency is assumed constant (e.g., 60Hz).
Here we consider the frequency as a variable and examine how the performance
varies with the frequency.
Variation in impedance of basic components

4.  90LLjZL Inductor

5. Capacitor  90
11
CCj
Zc


6. Frequency dependent behavior of series RLC network
Cj
RCjLCj
Cj
LjRZeq




1)(1 2


C
LCjRC
j
j

 )1( 2





C
LCRC
Zeq

 222
)1()(
||

 




 
 
RC
LC
Zeq

 1
tan
2
1
sC
sRCLCs
sZ
sj
eq
1
)(
2


notation"intionSimplifica"

7. For all cases seen, and all cases to be studied, the impedance is of the form
01
1
1
01
1
1
...
...
)(
bsbsbsb
asasasa
sZ n
n
n
n
m
m
m
m


 



sC
ZsLsZRsZ CLR
1
,)(,)( 
Simplified notation for basic components
Moreover, if the circuit elements (L,R,C, dependent sources) are real then the
expression for any voltage or current will also be a rational function in s
LEARNING EXAMPLE
sL
sC
1
R
So V
sCsLR
R
sV
/1
)(

 SV
sRCLCs
sRC
12


So V
RCjLCj
RCj
V
js
1)( 2









 

010
1)1053.215()1053.21.0()(
)1053.215(
332
3


jj
j
Vo
MATLAB can be effectively used to compute frequency response characteristics

8. For all cases seen, and all cases to be studied, the impedance is of the form
01
1
1
01
1
1
...
...
)(
bsbsbsb
asasasa
sZ n
n
n
n
m
m
m
m


 



sC
ZsLsZRsZ CLR
1
,)(,)( 
Simplified notation for basic components
Moreover, if the circuit elements (L,R,C, dependent sources) are real then the
expression for any voltage or current will also be a rational function in s
LEARNING EXAMPLE
sL
sC
1
R
So V
sCsLR
R
sV
/1
)(

 SV
sRCLCs
sRC
12


So V
RCjLCj
RCj
V
js
1)( 2









 

010
1)1053.215()1053.21.0()(
)1053.215(
332
3


jj
j
Vo
MATLAB can be effectively used to compute frequency response characteristics

9. LEARNING EXAMPLE A possible stereo amplifier
Desired frequency characteristic
(flat between 50Hz and 15KHz)
Postulated amplifier
Log frequency scale

10. Frequency domain equivalent circuit
Frequency Analysis of Amplifier
)(
)(
)(
)(
sV
sV
sV
sV
in
o
S
in

)(
/1
)( sV
sCR
R
sV S
inin
in
in

 ]1000[
/1
/1
)( in
oo
o
o V
RsC
sC
sV


)(
)(
)(
sV
sV
sG
S
o

Voltage Gain

11. 













ooinin
inin
RsCRsC
RsC
sG
1
1
]1000[
1
)(














 000,40
000,40
]1000[
100 ss
s
required
actual



000,40
000,40
]1000[)(000,40||100
s
s
sGs 
Frequency dependent behavior is
caused by reactive elements

12. NETWORK FUNCTIONS
INPUT OUTPUT TRANSFER FUNCTION SYMBOL
Voltage Voltage Voltage Gain Gv(s)
Current Voltage Transimpedance Z(s)
Current Current Current Gain Gi(s)
Voltage Current Transadmittance Y(s)
When voltages and currents are defined at different terminal pairs we
define the ratios as Transfer Functions
If voltage and current are defined at the same terminals we define
Driving Point Impedance/Admittance
EXAMPLE




admittanceTransfer
tanceTransadmit
)(
)(
)(
1
2
sV
sI
sYT
gainVoltage
)(
)(
)(
1
2
sV
sV
sGv 
To compute the transfer functions one must solve
the circuit. Any valid technique is acceptable

13. LEARNING EXAMPLE

14. LEARNING EXAMPLE

15. POLES AND ZEROS (More nomenclature)
01
1
1
01
1
1
...
...
)(
bsbsbsb
asasasa
sH n
n
n
n
m
m
m
m


 


 Arbitrary network function
Using the roots, every (monic) polynomial can be expressed as a
product of first order terms
))...()((
))...()((
)(
21
21
0
n
m
pspsps
zszszs
KsH



functionnetworktheofpoles
functionnetworktheofzeros


n
m
ppp
zzz
,...,,
,...,,
21
21
The network function is uniquely determined by its poles and zeros
and its value at some other value of s (to compute the gain)
EXAMPLE
1)0(
22,22
,1
21
1



H
jpjp
z
:poles
:zeros




)22)(22(
)1(
)( 0
jsjs
s
KsH
84
1
20


ss
s
K
 1
8
1
)0( 0KH
84
1
8)( 2



ss
s
sH

16. The network function is uniquely determined by its poles and zeros
and its value at some other value of s (to compute the gain)
EXAMPLE
1)0(
22,22
,1
21
1



H
jpjp
z
:poles
:zeros




)22)(22(
)1(
)( 0
jsjs
s
KsH
84
1
20


ss
s
K
 1
8
1
)0( 0KH
84
1
8)( 2



ss
s
sH

17. LEARNING EXTENSION Find the driving point impedance at )(sVS
)(
)(
)(
sI
sV
sZ S

)(sI
)(
1
)()(: sI
sC
sIRsV
in
inS KVL

in
in
sC
RsZ
1
)(
Replace numerical values




 M
s
100
1

18. LEARNING EXTENSION
)104(
000,20,50
0
7
0
21
1



K
HzpHzp
z
:poles
:zero
)(
)(
)(
sV
sV
sG
K
S
o
o
gainvoltagethefor
ofvaluetheandlocationszeroandpoletheFind














ooinin
inin
RsCRsC
RsC
sG
1
1
]1000[
1
)( 













 000,40
000,40
]1000[
100 ss
s
For this case the gain was shown to be
))...()((
))...()((
)(
21
21
0
n
m
pspsps
zszszs
KsH



Zeros = roots of numerator
Poles = roots of denominator

19. Book and Notes available at www.eedmd.weebly.com/ca2