1
The Laplace
Transform in Circuit Analysis
Circuit Elements in the s Domain
The Transfer Function and Natural Response
The Transfer Function and the Convolution Integral
The Transfer Function and the Steady-
State Sinusoidal Response
The Impulse Function in Circuit Analysis
4.14.2-3 Circuit Analysis in the s Domain4.4-54.6
4.7
4.8
Chapter 4
2
Key points
How to represent the initial energy of L, C in the s-domain?
Why the functional forms of natural and steady- state responses are determined by the poles of transfer function H(s) and excitation source X(s), respectively?
Why the output of an LTI circuit is the convolution of the input and impulse response? How to interpret the memory of a circuit by convolution?
4
A resistor in the s domain
iv-relation in the time domain:
).()( tiRtv
By operational Laplace transform:
).()(
,)()()(sIRsV
tiLRtiRLtvL
Physical units: V(s) in volt-seconds, I(s) in ampere-seconds.
5
An inductor in the s domain
).()( tidtdLtv
.)()()(
,)()()(
00 LIsIsLIssILsVtiLLtiLLtvL
initial current
iv-relation in the time domain:
By operational Laplace transform:
7
A capacitor in the s domain
).()( tvdtdCti
.)()()(
,)()()(
00 CVsVsCVssVCsItvLCtvCLtiL
initial voltage
iv-relation in the time domain:
By operational Laplace transform:
9
Circuit Analysis in the s Domain
1. Procedures2. Nature response of RC circuit3. Step response of RLC circuit4. Sinusoidal source5. MCM6. Superposition
Section 4.2, 4.3
10
How to analyze a circuit in the s-domain?
1. Replacing each circuit element with its s-domain equivalent. The initial energy in L or C is taken into account by adding independent source in series or parallel with the element impedance.
2. Writing & solving algebraic equations by the same circuit analysis techniques developed for resistive networks.
3. Obtaining the t-domain solutions by inverse Laplace transform.
11
Why to operate in the s-domain?
It is convenient in solving transient responses of linear, lumped parameter circuits, for the initial conditions have been incorporated into the equivalent circuit.
It is also useful for circuits with multiple essential nodes and meshes, for the simultaneous ODEs have been reduced to simultaneous algebraic equations.
It can correctly predict the impulsive response, which is more difficult in the t-domain (Sec. 4.8).
12
Nature response of an RC circuit (1)
Q: i(t), v(t)=?
.)(1
)( , 1000
RCsRV
RCsCVsIIR
sCI
sV
Replacing the charged capacitor by a Thévenin equivalent circuit in the s-domain.
KVL, algebraic equation & solution of I(s):
Nature response of an RC circuit (2)
The t-domain solution is obtained by inverse Laplace transform:
).(
1)(
)(
)(0
1)(01
01
tueRV
sLe
RV
RCsRVLti
RCt
RCt
i(0+) = V0 /R, which is true for vC (0+) = vC (0-) = V0 .
i() = 0, which is true for capacitor becomes open (no loop current) in steady state.
4
14
Nature response of an RC circuit (3)
To directly solve v(t), replacing the charged capacitor by a Norton equivalent in the s-domain.
.)(
)( , 10
0
RCsVsV
RVsCVCV
Solve V(s), perform inverse Laplace transform:
).()()( )( )(0
10
1 tRitueVRCsVLtv RCt
16
Step response of a parallel RLC (2)
KCL, algebraic equation & solution of V(s):
.)()(
)( , 112
LCsRCsCIsV
sLV
RVsCV
sI dcdc
Solve IL (s):
.)106.1()104.6(
1084.3
)()()()()(
942
7
112
1
sss
LCsRCssLCI
sLsVsI dc
L
17
Step response of a parallel RLC (3)
Perform partial fraction expansion and inverse Laplace transform:
.s)(mA )k24k32(
12720)k24k32(
1272024)(
jsjss
sIL
.(mA) )( )k2432sin()k2424cos(24
(mA) )( 127k)24(cos4024
..)(20)(24)(
k)32(
k)32(
)k24(k)32(127
tutte
tute
cctueeetuti
t
t
tjtjL
18
Transient response due to a sinusoidal source (1)
For a parallel RLC circuit, replace the current source by a sinusoidal one: The algebraic equation changes:
.)()(
)()(
,)()(
)(
,
11222
1
11222
2
22
LCsRCsssLCI
sLVsI
LCsRCsssCIsV
ssII
sLV
RVsCV
mL
m
mg
).(cos)( tutIti mg
19
.)()(
)(*22
*11
jsK
jsK
jsK
jsKsIL
Driving frequency
Neper frequency
Damped frequency
).( cos2cos2)( 2211 tuKteKKtKti tL
Steady-state response (source)
Natural response (RLC parameters)
Transient response due to a sinusoidal source (2)
Perform partial fraction expansion and inverse Laplace transform:
21
Step response of a 2-mesh circuit (2)
MCM, 2 algebraic equations & solutions:
)2(0)4810()(42
)1(336)(424.8
212
211
IsIIs
IIsI
.0
336109042
424.842
2
1
s
II
ss
.
124.1
24.87
121
21415
0336
109042424.842 1
2
1
sss
sssss
sII
22
Step response of a 2-mesh circuit (3)
Perform inverse Laplace
transform:
.A 15)48//42(
336)(1415)( 1221 tueeti tt
.A 74842
4215)(4.14.87)( 1222
tueeti tt
23
Use of superposition (1)
Given 2 independent sources vg , ig and initially charged C, L, v2 (t)=?
24
Use of superposition: Vg acts alone (2)
1 2
.01
,11
.0)(
,0)(
22
1
121
1
2
2112
1211
1
1
VsCR
VsC
RV
VsCVsCsLR
RV
sCVV
sCVV
sLV
RVV gg
25
Use of superposition (3)
.0
1
11
1
2
1
2212
1211
2
1
2
1
RVVV
YYYY
VV
sCR
sC
sCsCsLR
g
.2122211
1122 gV
YYYRYV
For convenience, define admittance matrix:
26
Use of superposition: Ig acts alone (4)
1 2
. ,0
2122211
112
2
1
2212
1211g
gI
YYYYV
IVV
YYYY
Same matrix Same denominator
27
Use of superposition: Energized L acts alone (5)
1 2
. ,0 2
122211
122
2
1
2212
1211
YYYsYV
sVV
YYYY
Same matrix Same denominator
28
Use of superposition: Energized C acts alone (6)
12
.)(
,
2122211
1211""2
""2
""1
2212
1211
YYYCYYV
CC
VV
YYYY
The total voltage is: .""22222 VVVVV
30
What is the transfer function of a circuit?
The ratio of a circuit’s output to its input in the s-domain:
)()()(
sXsYsH
A single circuit may have many transfer functions, each corresponds to some specific choices of input and output.
31
Poles and zeros of transfer function
For linear and lumped-parameter circuits, H(s) is always a rational function of s.
Poles and zeros always appear in complex conjugate pairs.
The poles must lie in the left half of the s-plane if bounded input leads to bounded output.
Re
Im
32
Example: Series RLC circuit
If the output is the loop current I:
.1)(
1)( 21
sRCLCs
sCsCsLRV
IsHg
If the output is the capacitor voltage V:
.1
1)(
)()( 21
1
sRCLCssCsLRsC
VVsH
g
input
33
How do poles, zeros influence the solution?
Since Y(s) =H(s) X(s), the partial fraction expansion of the output Y(s) yields a term K/(s-a) for each pole s =a of H(s) or X(s).
The functional forms of the transient (natural) and steady-state responses ytr (t) and yss (t) are determined by the poles of H(s) and X(s), respectively.
The partial fraction coefficients of Ytr (s) and Yss (s) are determined by both H(s) and X(s).
35
Only one essential node, use NVM:
,01005.02501000 6
s
Vs
VVV oogo
.105.26000)5000(1000)(
72
sss
VVsH
g
o
H(s) has 2 complex conjugate poles:
Vg (s) = 50/s2 has 1 repeated real pole: s = 0(2).
.40003000 js
Example 4.2 (2)
36
The total response in the s-domain is:
).(10410
)()80000,4cos(105)(4
000,33
tut
tuteyytv tsstro
sstrgo YYsssssVsHsV
722
4
105.26000)5000(105)()()(
The total response in the t-domain:
poles of H(s): -3k
j4k pole of Vg (s): 0(2)
expansion coefficients depend on H(s) & Vg (s)
.1041040003000
80105540003000801055 4
2
44
ssjsjs
Example 4.2 (3)
38
The Transfer Function and the Convolution Integral
1. Impulse response2. Time invariant3. Convolution integral4. Memory of circuit
Section 4.6
39
Impulse response
If the input to a linear, lumped-parameter circuit is an impulse (t), the output function h(t) is called impulse response, which happens to be the natural response of the circuit:
).()()()(
),(1)()( ,1)()(11 thsHLsYLty
sHsHsYtLsX
The application of an impulse source is equivalent to suddenly storing energy in the circuit. The subsequent release of this energy gives rise to the natural response.
40
Time invariant
For a linear, lumped-parameter circuit, delaying the input x(t) by simply delays the response y(t) by as well (time invariant):
).()(
)(),(),(
),()()(),()(),(),()()(),(
11
tuty
sYLsYLty
sYesXsHesXsHsYsXetutxLsX
tt
ss
s
41
Motivation of working in the time domain
The properties of impulse response and time- invariance allow one to calculate the output function y(t) of a “linear and time invariant (LTI)” circuit in the t-domain only.
This is beneficial when x(t), h(t) are known only through experimental data.
42
Decompose the input source x(t)
We can approximate x(t) by a series of rectangular pulses rec
(t-i ) of uniform width :
By having , rec
(t-i )/ (t-i ), x(t) converges to a train of impulses:
)(sq it
)(sq t
.)()(
)(reclim)()(
0
0 0
iii
iii
tx
txtx
43
Synthesize the output y(t) (1)
Since the circuit is LTI:
.)()(
)()(
0
0
iii
iii
thx
tx
;)()(),()(
),()(
tyatxatht
tht
iiii
ii
44
As , summation integration:
.)()()()()(0
dthxdthxty
if x(t) extends (-, )
By change of variable u= t-,
.)()()(
duuhutxty
The output of an LTI circuit is the convolution of input and the impulse response of the circuit:
.)()()()(
)()()(
dhtxdthx
thtxty
Synthesize the output y(t) (2)
45
Convolution of a causal circuit
For physically realizable circuit, no response can occur prior to the input excitation (causal), {h(t) =0 for t <0}.
Excitation is turned on at t =0, {x(t)=0 for t <0}.
.)()(
)()()(
0
t
dhtx
thtxty
46
Effect of x(t) is weighted by h(t)
The convolution integral
If h(t) is monotonically decreasing, the highest weight is given to the present x(t).
t0
t
dhtxty0
)()()(
shows that the value of y(t) is the weighted average of x(t) from t =0 to t = t [from
= t to =0 for x(t-)].
47
Memory of the circuit
.)()()(0 t
dhtxty
implies that the circuit has a memory over a finite interval t =[t-T,t].
T
If h(t) only lasts from t =0 to t =T, the convolution integral
If h(t) =(t), no memory, output at t only depends on x(t), y(t) =x(t)*(t) =x(t), no distortion.
48
.1
1)( ,1
1
sV
VsHVs
Vi
oio
).(1
1)( 1 tues
Lth t
Q: vo (t)=?
Example 4.3: RL driven by a trapezoidal source (1)
50
Since the circuit has certain memory, vo (t) has some distortion with respect to vi (t).
Example 4.3 (3)
52
How to get sinusoidal steady-state response by H(s)?
In Chapters 9-11, we used phasor analysis to get steady-state response yss (t) due to a sinusoidal input
If we know H(s), yss (t) must be:
.)()()( where
,)(cos)()()(
j
js
ss
ejHsHjH
tAjHty
The changes of amplitude and phase depend on the sampling of H(s) along the imaginary axis.
.cos)( tAtx
53
Proof
.sinsincoscoscos)( tAtAtAtx
.sincossincos)( 222222
s
sAs
As
sAsX
),()(sincos)()()()( 22 sYsYs
sAsHsXsHsY sstr
.)(cos)(..)(2
)()(
)(1
tjHAcc
jsAeejH
Ltyjj
ss
.2)(
2sincos)(sincos)(
))(( ,)( where 1
*11
j
js
jsss
AejHj
jAjHjs
sAsH
jssYKjs
Kjs
KsY
54
Obtain H(s) from H( j)
We can reverse the process: determine H( j) experimentally, then construct H(s) from the data (not always possible).
Once we know H(s), we can find the response to other excitation sources.
56
E.g. Impulsive inductor voltage (1)
The opening of the switch forces the two inductor currents i1 , i2 change immediately by inducing an impulsive inductor voltage [v=Li'(t)].
i1 (0-)=10 A i2 (0-)=0
Q: vo (t)=?
57
E.g. Equivalent circuit & solution in the s-domain (2)
.5
106012)5(
)150656(2)(2
0
ssss
sssVimproper rational
initial current
,0215310
)30100( 00
s
VssV
58
E.g. Solutions in the t-domain (3)
).()1060()(125
106012)( 510 tuet
ssLtv t
To verify whether this solution vo (t) is correct, we need to solve i(t) as well.
).()24()( ,5
24215310
30100)( 5 tuetissss
ssI t
jump
jump
59
Impulsive inductor voltage (4)
The jump of i2 (t) from 0 to 6 A causes , contributing to a voltage impulse
After t > 0+,
consistent with that solved by Laplace transform.
)(6)(2 tti
,1060)10(2)24(15
)()H2()()15()(555
22ttt
o
eee
tititv
).(12)(22 ttiL
60
Key points
How to represent the initial energy of L, C in the s-domain?
Why the functional forms of natural and steady- state responses are determined by the poles of transfer function H(s) and excitation source X(s), respectively?
Why the output of an LTI circuit is the convolution of the input and impulse response? How to interpret the memory of a circuit by convolution?
62
Why can a voltage surge occur?
Q: Why a voltage surge is created when a load is switched off?
Model: A sinusoidal voltage source drives three loads, where Rb is switched off at t =0.
Since i2 (t) cannot change abruptly, i1 (t) will jump by the amount of i3 (0-), voltage surge occurs.
63
Example
Let Vo =1200
(rms), f =60 Hz, Ra =12 , Rb =8 , Xa =41.1
(i.e. La =Xa /=109 mH), Xl =1
(i.e.
Ll =2.65 mH). Solve vo (t) for t >0-.
To draw the s-domain circuit, we need to calculate the initial inductor currents i2 (0-), i0 (0-).
64
Steady-state before the switching
The three branch currents (rms phasors) are:
I1 =Vo /Ra =(1200)/(12 )=100
A,
I2 =Vo /(jXa ) = (1200)/(j41.1 )=2.92-90
A,
I3 =Vo /Rb =(1200)/(8 )=150
A,
The line current is: I0 = I1 + I2 + I3 =25.2-6.65
A.
Source voltage: Vg =Vo + I0 (jXl )=125-11.5
V.
The two initial inductor currents at t =0- are:
i2 (t)=2.92(2)cos(120t-90), i2 (0-)=0;
i0 (t)=25.2(2)cos(120t-6.65), i0 (0-) =35.4 A.
65
S-domain analysis
The s-domain circuit is:
By NVM:
(Ll = 2.65 mH)
(I0 = 35.4 A)
(12 ) (La = 109 mH)
,00
a
o
a
o
l
glo
sLV
RV
sLVILV
120
85.686120
85.6861475253
)()( 0
jsjssLLLLRsRIsVLR
Vlalaa
aglao
Vg = 125- 11.5
V (rms)
66
Inverse Laplace transform
Given ,120
85.686120
85.6861475253)(
jsjsssVo
).()85.6120cos(173 253)( 1475 tutetv to
1
Two Post Network:
Two Post Network
+
-
+
-
Post-IInput port
Port-IIOutput port
Fig.1
A network which has two terminals (one port) on the one side and another two terminals on
the opposite side forms a two port network. One port functions as input and the other as
output to the network.
The networks may composed of active or passive elements, but we are not concerned with its
internal functioning and simple we assumed as a black box
From above fig.1 it is noted that there are four variable V1,I1,V2,I2 of the four , any two can be
dependent and other two independent, which gives rise to different parameters-
(a) Impedance parameters or Z parameters.
(b)Admittance parameters or Y parameters.
(C) Hybrid parameters or h parameters
(d)Transmission parameters.
(2) Reason why to study two port – network:
(a) Such networks are useful in communication, control system, power systems
and electronics.
(b) Knowing the parameters of a two – port network enables us to treat it as a
“black box” when embedded within a larger network.
(a) Impedance parameters or Z parameters.
Linear Network
+
-
+
-
Fig.2
2
Impedance parameters are commonly used in the synthesis of filters and also useful in the
design and analysis of impedance matching networks and power distribution networks.
V1, V2 are dependent variables.
II , I2 are independent variables.
The defining equations are-
...........(1)1 11 1 12 2............(2)2 21 1 22 2
V z I z I
V z I z I
= +
= +
In matrix form as:
11 121 1
21 222 2
z zV I
z zV I
= ………….. ……..(3)
To obtain Z parameters, we alternatively open circuit the output and input ports.
Thus-
111
1 02
221
1 02
Vz
II
Vz
II
==
==
11 2
2 01
22 2
2 01
Vz
II
Vz
II
==
==
These parameters are as follows:
z11 Open circuit input impedance
z12 Open circuit transfer impedance from port 1 to port 2
z21 Open circuit transfer impedance from port 2 to port 1
z22 Open circuit output impedance
When z11=z22, the network is said to be symmetrical.
• When the network is linear and has no dependent sources, the transfer impedances
are equal (z12=z21), the network is said to be reciprocal.
3
• This means that if the input and output are switched, the transfer impedances remain
the same.
Any two-port network that is composed entirely of resistors, capacitors, and inductors
must be reciprocal
Since, the Z parameters are obtained by opening the input or output port, they are also called
the open circuit impedances.
Z11 and Z22 are called driving point impedances and Z12 and Z21 are called transfer
impedances.
Z- Parameters for a T-Network-
-
+
-
Fig.3
Z11 = ZA +ZC Z12 = Z21 =Zc Z22 =ZB +ZC
Network in which, Z12 = Z21 as in case in above T network are called reciprocal networks.
If ZA =ZB, the network is symmetrical.
Equivalent circuit for the Z parameters is here under-
4
+ +
1V2V
11Z 22Z
12 2Z I 21 1Z I
1I2I
Fig.4
(b)Admittance parameters or Y parameters.
Linear Network
+
-
+
-
Fig.2
Fig.5
Y – Parameter also called admittance parameter and the units is Siemens (S).
V1, V2 are independent variables.
II , I2 are dependent variables.
The defining equations are-
11 121 1
21 222 2
y yI V
y yI V
=
It is represented as a circuit as follows-
5
1V2V
11Y
22Y12 2VY
21 1VY
1I2I
Fig.6
This is a voltage controlled (dependent) current source (VCCS)
In the matrix form-
11 121 1
21 222 2
y yI V
y yI V
=
To obtain the Y parameters, we alternatively short circuited the input and output ports. Thus-
111
1 02
221
1 02
IY
VV
IY
VV
==
==
112
2 01
222
2 01
IY
VV
IY
VV
==
==
These parameters are as follows:
y11 Short circuit input admittance
y12 Short circuit transfer admittance from port 1 to port 2
y21 Short circuit transfer admittance from port 2 to port 1
y22 Short circuit output admittance
The impedance and admittance parameters are collectively called the immitance
parameters.
6
Since, the Y parameters are obtained by short circuiting the input or output, therefore they
are also called the short circuit admittance parameters.
Y parameters for π network-
1V 2V
CY
AYBY
1I 2I
Fig.7
Y parameters can also be obtained from the Z parameters-
[V] = [Z][I] or, [I] = [Z]-1[V]
Therefore-
111 12 22 12
21 22 21 11
Y Y Z Z
ZY Y Z Z
−=
∆ −
Where Z∆ = determinant of Z
Reciprocal network
Symmetrical Network condition
(C) Hybrid parameters or h parameters
7
Linear Network
+
-
+
-
Fig.2
Fig.8
I1, V2 are independent variables.
I2 ,V1 are dependent variables the, defining equations are-
1 11 1 12 2
2 21 1 22 2
V h I h V
I h I h V
= +
= +
In the matrix form-
11 12
21 222
11
2
Ih
Vh
V h
hI
=
The h terms are known as the hybrid parameters, or simply h-parameters. The name comes
from the fact that they are a hybrid combination of ratios. These parameters tend to be much
easier to measure than the z or y parameters. They are particularly useful for characterizing
transistors.Transformers too can be characterized by the h parameters.
The h parameters are obtained from above equations by setting V2=0 i.e output port short
circuited and I1=0 i.e input open circuited.
1 111 12
1 20 02 1
2 221 22
1 20 02 1
V Vh h
I VV I
I Ih h
I VV I
= == =
= == =
The parameters h11, h12, h21, and h22represent an impedance, a voltage gain, a current gain,
and an admittance respectively.
8
The h-parameters correspond to:
h11 Short circuit input impedance
h12 Open circuit reverse voltage gain
h21 Short circuit forward current gain
h22 Open circuit output admittance
In a reciprocal network, h12=-h21.
The equivalent network is shown below:
The inverse of hybrid parameters are called g parameters, inverse hybrid parameters-
1 11 1 12 2
2 21 1 22 2
I g V g I
V g V g I
= +
= +
The values of the g parameters are determined as:
1 111 12
1 20 02 1
2 221 22
1 20 02 1
I Ig g
V II V
V Vg g
V II V
= == =
= == =
The equivalent model is shown below:
+
1V2V
12 2h V 21 1h I
11h
22h
Fig.9
The g parameters correspond to:
9
g11 Open circuit input admittance
g12 Short circuit reverse current gain
g21 Open circuit forward voltage gain
g22 Short circuit output impedance
(d)Transmission parameters or ABCD parameters:
Linear Network
+
-
+
-
Fig.2
Fig.10
T – parameter or ABCD – parameter is a set of parameters relates the variables at the
input port to those at the output port.
T – parameter also called transmission parameters because this parameter are useful in
the analysis of transmission lines because they express sending – end variables (V1 and I1)
in terms of the receiving – end variables (V2 and -I2).
The equation is:
.......(1)1 2 2
.......(2)1 2 2
V AV BI
I CV DI
= −
= −
In matrix form is:
1 2
1 2
V VA B
I IC D
=−
The T – parameter that we want determine are A, B, C and D where A and D are
dimensionless, B is in ohm (Ω) and C is in Siemens (S).
The values can be evaluated by setting
10
I2 = 0 (input port open – circuit)
V2 = 0 (output port short circuit)
Thus;
1
2 02
1
2 02
VA
VI
IC
VI
==
==
1
2 02
1
2 02
VB
IV
ID
IV
==
==
In term of the transmission parameter, a network is reciprocal if;
AD-BC=1 and for symmetrical A=D.
Condition for reciprocity and symmetry:
Parameters Condition of Reciprocity Condition of symmetry
[Z] Z12=Z21 z11=z22
[Y] Y12=Y21 Y11=Y22
[h] h12=-h21 h11h22-h12h21=1
[g] g12=-g21 g11g22-g12g21=1
[ABCD] AD-BC=1 A=D
Interconnections of Networks
Often it is worthwhile to break up a complex network into smaller parts.
The sub-network may be modeled as interconnected two port networks.
From this perspective, two port networks can be seen as building blocks for constructing
a more complex network.
These connections may be in series, parallel, or cascaded.
(1)Series Connection
Two port networks with ABCD parameters are A,B,C,D and A’,B’,C’,D’ are connected in
series as shown in fig below-
11
'A 'B
'D'C
2V1V '2V'
1V
1I 2I '1I
'2I
Fig.11
1 2
1 2
V VA B
I IC D
=−
and
' '' '
' ' ' '
1 2
1 2
V VA B
I C D I
=−
Here I2= '
1I and V2= '1V
'' '
' ' '
1 2
1 2
V VA B A B
I C D C D I
=−
Thus the overall transmission matrix for two port networks in tandem is the matrix
product of individual transmission network matrix.
When two port networks are connected in series, the Z parameter of the combination is
equal to the sum of the individual Z parameters.
11 12 11 1211 12
21 22 21 22 21 22
Z ZZ ZZ Z
Z Z Z Z Z Z
a a b b
a a b b
= +
(2)Parallel Connection
Two port networks are in parallel when their port voltages are equal and the port
currents of the larger network are the sums of the individual port currents. Consider the
network shown-
12
1V
+
--
2VNetwork
a
Network
b
1I 2I2aI1aI
2aI2bI
2bV1bV
2aV1aV
Fig.12
Va=Vb=V, I1=Ia1+Ib1, I2=Ia2+Ib2
[I]=[IYaI+IYbI][V]. the Y parameter of the parallel network is equal to the sum of the
individual network’s Y parameter.
[Y]=[Ya]+[Yb]
(3)Series-Parallel Connection:
In this connection the input ports are connected in series while the output ports are
connected in parallel. The connection requires that-
V1=V1a+V1b, I1=I1a=I1b and V2=V2a=V2b, I2=I1a+I2b
1V 2V
1I 2I2aI1aI
1bI2bI
2bV1bV
2aV1aV
Fig.13
The overall h-parameters of the combined series-parallel connection can be obtained by
adding their individual h-parameters.
13
So, overall we have- 11 12 11 1211 12
21 22 21 22 21 22
h hh hh h a a b bh h h h h ha a b b
= +
(4) Parallel-Series connection.
In this connection the input ports are connected in parallel while the output ports are
connected in series. The connection requires that-
V1=V1a=V1b, I1=I1a+I1b and V2=V2a+V2b, I2=I1a+I2b
1V
+
-
-
2V
Network
a
Network
b
1I 2I2aI1aI
1bI2bI
2bV1bV
2aV1aV
Fig.14
The overall inverse h-parameters or g-parameters of the combined parallel series
connection can be obtained by adding their individual g-parameters.
So, overall we have-
11 12 11 1211 12
21 22 21 22 21 22
g gg gg g
g g g g g g
a a b b
a a b b
= +
Image Impedance:
14
In a two port network, if two impedances Zi1 and Zi2 are such that Zi1 is the driving point
impedance at port 1 with impedance Zi2 connected across port 2 and Zi2 is the driving
point impedance at port2 with impedance Zi1 connected across port1, then impedances Zi1
and zi2 are called the image impedances of the network.
For symmetrical network image impedances are equal to each other i.e Zi1=Zi2 and is
called the characteristic or iterative impedance Zc.
Two port
Network1iZ 2V
1V2iZ
2I1I
Fig.a
1iZ 2V1V 2iZ
2I1I
In fig.a 11
1
VZi I
= =driving point impedance at port 1.
In fig.b 2
22
VZi I= = driving point impedance at port 2.
T to π Transformation:
cY
bYaY
1Z 2Z
3Z
15
Impedances Z1, Z2, and Z3 of the above fig.a T network are known and it is require to
calculate equivalent admittances Ya , Yb and Yc of the π network in the fig.b
2
1 2 2 3 3 1
ZYa Z Z Z Z Z Z
=+ +
, 1
1 2 2 3 3 1b
ZYZ Z Z Z Z Z
=+ +
and 3
1 2 2 3 3 1c
ZY
Z Z Z Z Z Z=
+ +
π to T Transformation:
cY
bYaY
1Z 2Z
3Z
In this case, admittances Ya,Yb ,and Yc of the above fig.a π network are known and it is
require to calculate equivalent impedances Z1 ,Z2 and Z3 of the T network in the fig.b
1
YbZY Y Y Y Y Ya c c ab b
=+ +
, 2aY
ZY Y Y Y Y Ya c c ab b
=+ +
and 3cY
ZY Y Y Y Y Ya c c ab b
=+ +