net work analysis dr. sumrit hungsasutra text : basic circuit theory, charles a. desoer & kuh,...
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Net work analysis
Dr. Sumrit Hungsasutra
Text : Basic Circuit Theory, Charles A. Desoer & Kuh, McGrawHill
Linear Time-invariant Circuits Definition and properties Node and mesh analysis Input-output representation Responses to an arbitrary input Computation of convolution integrals
Definition and properties
Linear circuit contains linear elements. Linear time invariant circuit contains linear
time invariant elements and independent sources.
Circuits with nonlinear elements are nonlinear circuits.
Circuits with time varying elements are time varying circuits.
Definition and properties
Voltage sources and current sources play important roles in circuit analysis.
Dependent sources are non-linear and time varying.
All sources are inputs to the circuit. The input is a waveform of either
independent voltage or current source. The wave form can be of a constant, step
and a function of time.
Definition and properties
The output response is a branch voltage or branch current at the desired point or charge in a capacitor or flux in an inductor.
Differential equations can be written on all lumped circuit from which branch currents or branch voltages are solved.
The unique solution (circuit response) requires the input information and its initial solution.
Definition and properties
Common initial conditions are capacitor’s voltage and inductor’s current.
State of a circuit at time t0 to any set of initial conditions together with the inputs uniquely determine all the network variables of the circuit at time t>t0.
If all initial conditions are zero, it is called zero state.
Definition and properties
In linear circuit with zero state and no inputs all network variables remain equal to zero forever after.
When inputs are applied to the circuit, initial states (can also be zero state) are required to solve for all circuit variables.
Zero-state response is the solution of the circuit with inputs and zero state.
Definition and properties Zero-input response is the response with no
inputs. Complete responses are the responses of the
circuit to both inputs and initial states (zero states).
For linear time-invariant or time-varying circuits: Complete response is the sum of zero-input
response and zero-state response. Zero-state response is a linear function of input. Zero-input response is a linear function of initial
states
Node and mesh analyses
Simple topology circuits can be analyzed more easy (simple loop simple node circuits) using KCL and KVL.
More complex circuits requires a more advanced techniques.
Simple circuit
is R1 CL
iL(0)= Io
R2+ +
--v2vC(0)=V0Fig 1
Node and mesh analysis
Pick a reference node as a datum (ground) Apply KCL at each node
Redrawn circuit of Fig1 12
3
is
R1
R2
C
L
1v
2v+
+
-
-
Fig 2
Node and mesh analysis
1 10 1 2
1 0
1( ) ' ( )t
sdv v
C I v v dt i tdt R L
'2
0 2 120
1( ) ' 0
tv
I v v dtL R
KCL at node 1
KCL at node 2
Initial condition
01 )0( Vv
Node and mesh analysis
1 1 2
1
( )2 s
dv v vC i tdt R R
22 1
2
1 1 10
dvv v
L L R dt
21 2
2
dvLv v
R dt
Adding the two equations
Diff KCL from node 2
or2
1 2 22
2
dv dv d vL
dt dt R dt
Node and mesh analysis
dt
dv1
22 2 2
2 2 221 1
( ) (1 ) ( )sd v dv RL
LC R C v R i tR dt Rdt
022 )0( IRv
substitute
Differential equation for voltage at node 2
][)]0()0([)0( 0202
2122 IRV
L
Rvv
L
R
dt
dv
Mesh and mesh analysisRedrawn Fig 1 using Thevenin equivalent
Vs=R1is
R1
R2
C
L1i
2i1 2i i
1
1i 2i
Fig 3
Mesh and mesh analysis
1 1 0 2 1
0
1( ) ' ( )t
sR i V i i dt v tC
22 2 0 2 1
0
1( ) ' 0t
diL R i V i i dtdt C
Mesh 1 KVL
Mesh 2 KVL
Initial condition
02 )0( Ii
Mesh and mesh analysis
Adding the two equations
21 1 2 2 s
diR i L R i v
dt
Or
2 21 2
1 1 1
svdi RLi i
R dt R R
Diff both sides2
2 2 2 122
0d i di i i
L Rdt C Cdt
Mesh and mesh analysis
22 2 2
2 221 1 1
( ) (1 ) svd i di RLLC R C i
R dt R Rdt
02 )0( Ii
2 2 2 1, s sv R i v R i
Substitute for i1
Initial conditions
Substitute
)(1
)0( 0202 IRV
Ldt
di
22 2 2
2 2 221 1
( ) (1 ) sd v dv RL
LC R C v R iR dt Rdt
From this simple example we can see the general fact that
“Given any single-input single-output linear time-invariant circuit, it is always possible to write a single differential equation relating
the output to the input.”
is R1 CL
iL(0)= Io
R2+ +
--v2vC(0)=V0
22 2 2
2 2 221 1
( ) (1 ) ( )sd v dv RL
LC R C v R i tR dt Rdt
input output
Input-output Representation
1 1
1 0 11 1.. ..
n n m m
n mn n m m
d y d y d w d wa a y b b b w
dt dt dt dt
2121 ,..,, bbaa
General equation for a single-input single-output
Constants depend on circuit element values
and network topology. The initial conditions are
y
w
Is the output from the circuit
Is the input to the circuit
1
1(0), (0),.., (0)
n
n
dy d yy
dt dt
Input-output Representation
nnnn asasas
11
1 ..
nisi ....2,1, y
Zero-input responseThe general equation becomes homogeneous and the nth-degreecharacteristic Polynomial is
The zeroes of this polynomial
are the natural frequency of the network variable
The solution of the homogeneous equation becomes
1
( ) i
ns t
ii
y t k e
If s1 is a repeated root
1 1 121 2 3( ) ...s t s t s ty t k e k te k t e
Input-output Representation
Zero-state response
1
( ) ( )i
ns t
i pi
y t k e y t
The general zero-state response of the circuit is
Where is any particular solution of the circuit due to the input w ( )py t
Input-output RepresentationExample 1
0,cos)( ttVsv m
Fig 4
Figure 4 shows a simple RC circuit with zero initial condition. The input is
+
-sv
C
R
i ( ) ( ) cosmv s u t V t KVL:
0
1( ) ' ( ) cos
t
mRi t idt u t V tdtC
1 sdvdi
R idt C dt
or
cos ( ) cos
( ) ( )sin
sm m
m m
dv du dV t V u t t
dt dt dtV t V u t t
Input-output Representation
Initial condition:
0)0()0( cc vv
(0 ) (0 ) (0 )
(0 )(0 )
R s c m
mRR
v v v V
Vvi
R R
From KVL
Tina simulation
T
Time (s)
0.00 2.50m 5.00m 7.50m 10.00m
Vol
tage
(V
)
-10.00
0.00
10.00
20.00
Input-output Representation
Impulse Response
( ) ( 1) ( ) ( 1)1 0 1.. ..n n m m
n my a y a y b w b w b w
From the general equation,
With initial conditions
(1) (2) ( 1)(0 ) (0 ) (0 ) .. (0 ) 0ny y y y If ( )w t the RHS are impulse and its derivatives and the response is
1
( ) ( )i
ns t
ii
h t k e u t
The impulse function and derivativesdu
dt and ( ) ( )
tt dt u t
(1)d
dt
and(1) ( ) ( )
tt dt t
(1)(2)d
dt
and(2) (1)( ) ( )
tt dt t
( )( 1)
nnd
dt
and( 1) ( )( ) ( )
tn nt dt t
Input-output RepresentationExample 2
2
24 3 2
d y dy dwy w
dt dtdt
Suppose that the differential equation relating the output y and the input
wof a circuit is
Find the impulse response of the circuit.
The characteristic polynomial is 0342 ss
and the roots are 1 21, 3s s
31 2( ) ( )t th t k e k e u t The response
Input-output Representation
Solve for constants
31 2( ) ( )t th t k e k e u t
(1) 3 31 2 1 2
31 2 1 2
( ) ( ) 3 ( )
( ) 3 ( )
t t t t
t t
h t k e k e t k e k e u t
k k t k e k e u t
from
(2) (1)1 2 1 2
31 2
( ) ( ) 3 ( )
9 ( )t t
h t k k t k k t
k e k e u t
Input-output Representation
Substitute and we have
)(tw )(thy
(2) (1) (1)( ) 4 ( ) 3 ( ) ( ) 2 ( )h t h t h t t t
(1) (1)1 2 1 2( ) ( ) (3 ) ( ) ( ) 2 ( )k k t k k t t t
1 2( ) 1k k and 1 2(3 ) 2k k
11 2k and 1
2 2k
312
( ) ( )t th t e e u t
Response to an arbitrary input
0 0
1( ) 0
0
t
p t t
t
Time t’
si
0t 1t 2t kt 1kt 1nt t
sai
t’0t 1t
2t
0 0( ) ( ' )si t p t t
t’1t
1 1( ) ( ' )si t p t t
An arbitrary input signal can be divided into many impulse functions.
( )p t
1
0
Response to an arbitrary input )'()(..)'()()'()()'( 111100 nnssss ttptittptittptiti
0 0 1 1 1 1( ') ( ) ( ' ) ( ) ( ' ) .. ( ) ( ' )s s s s n ni t i t h t t i t h t t i t h t t
nAs
0
t
t
s ttdttthtitv
0
0,')'()'()(
the output response is the sum of all impulse responses
Conclusion
1 Determine the impulse response 2 Calculate the integral 3 This type of integral is called convolution integral
( )h t
Response to an arbitrary inputThe complete response
)()()( tvtzty Where is the zero-input response( )z t
0
0( ) ( ) ( ') ( ') ',t
s
t
y t z t i t h t t dt t t Note that the complete response is a linear function of input only if theZero-input response is identically zero.
Computation of convolution integrals
0
0( ) ( ') ( ') ',t
s
t
v t h t t i t dt t t From
For the unit impulse at 01 tt )()( 1tttis
0
1
1
1
1
1 0
1
1 1 1
( ) ( ') ( ' ) ',
( ') ( ' ) '
( ) ( ' ) ' ( )
t
t
t
t
t
t
v t h t t t t dt t t
h t t t t dt
h t t t t dt h t t
Computation of convolution integrals 'tt ddttt ','
0
0
( ) ( ) ( )
t t
sv t h i t d
Let
Then
0
0
( ) ( ') ( ') '
t t
sv t h t i t t dt
Thus
0 0
( ') ( ') ' ( ') ( ') ', 0t t
s sh t t i t dt h t i t t dt t Convolution integral is symmetric role
Computation of convolution integralsExample 3
Let the input be a step function and the impulse response be a triangularwaveform
t
( )h t
2
2
0 t
( )si t
1
0
t
( ')h t
2
2
0 t
( ')si t
1
0
t
( ')h t
-2
2
0 t
( ')si t
1
0
Computation of convolution integrals
t
( ')h t t
-1
2
0t
( ')si t t
1
01t 1t
t
( ') ( ')si t h t t
-1
2
0
( ') ( ')si t t h t
1
01t 1t t
(0 ')si t1
0
1
0
't
't
2
2( ')h t
't
( ') ( ')si t t h t
t = 0
Area = 0
21
2
(1 ')si t
0't
2
2
t = 1
Area = 3/2
1
0 't2
(2 ')si t
0't
2
2
t = 2
Area = 2
( ')h t
( ')h t
2
2
1
0 't2
(3 ')si t
0't
2
2
t = 3
Area = 2
( ')h t2
't2
t = 3
Area = 2
2
3/ 2
1
( )v t
Computation of convolution integralsExample 4
t
( )si t
1
10
t
( )h t
1
0
te
Find the zero-state response for the input and impulse response shown
)1()()( tututis ( ) ( )th t e u t
Computation of convolution integrals
0
( ) ( ') ( ') 't
sv t h t t i t dt
( )si 1
10 t
( )h t
1
10 t
( )h t
( )si
t1
0
( )v t11 e
For 10 t 1)( tis( ')
0
( ) ' 1t
t t tv t e dt e For 1t ( ) 0si t
1( ')
0
( ) ' ( 1)t t tv t e dt e e
Computation of convolution integralsExample 5Find the zero-state response for the input and impulse response shown
( )si t
10
( )h t
1
10 t tt2
sin t
( ) ( )sinsi t u t t ( ) ( ) ( 1)h t u t u t
Computation of convolution integrals
( ) 0v t For 0t
0
1( ) sin ( ) (1 cos )
t
v t t d t
For 0 1t
0
1 2( ) sin ( ) (cos ( 1) cos ) cos
t
v t t d t t t
For 1 t
Computation of convolution integrals ( )h
1
10
0t
sin (0 )
( )h 1
10
0 1t
sin ( )t
( )h 1
10
1 t
sin ( )t