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Net work analysis
Dr. Sumrit Hungsasutra
Text : Basic Circuit Theory, Charles A. Desoer & Kuh, McGrawHill
Linear Time-invariant Circuits
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Definition and properties
Node and mesh analysis
Input-output representation
Responses to an arbitrary input
Computation of convolution integrals
Definition and properties
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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
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
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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
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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
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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
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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
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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.
iL(0)= Io
Simple circuit
Fig 1
is
L
R1
C
+
vC(0)=V0 R2
-
+
v2
-
Node and mesh analysis


Pick a reference node as a datum (ground)
Apply KCL at each node
L
Redrawn circuit of Fig1
+
C
R1
Fig 2
2
1
is
v1
-
+
v2
-
3
R2
Node and mesh analysis
KCL at node 1
t
dv1 v1
1
C

 I0 
(v1  v2 )dt '  is (t )
dt R1
L

0
KCL at node 2
t'
v2
1
I0 
(v2  v1 )dt '
0
L
R2

0
Initial condition
v1 (0)  V0
Node and mesh analysis
Adding the two equations
dv1 v1 v2
C
 
 is (t )
dt R1 R 2
Diff KCL from node 2
1
1
1 dv2
v2  v1 
0
L
L
R2 dt
or
L dv2
v1  v2 
R2 dt
dv1 dv2 L d 2 v2


dt
dt R2 dt 2
Node and mesh analysis
substitute
dv1
dt
Differential equation for voltage at node 2
LC
d 2 v2
dt 2
R2
L dv2
 ( R2C  )
 (1  )v2  R2is (t )
R1 dt
R1
v2 (0)  R2 I 0
dv2
R2
R2
(0) 
[v1 (0)  v2 (0)] 
[V0  R2 I 0 ]
dt
L
L
Mesh and mesh analysis
Redrawn Fig 1 using Thevenin equivalent
i1
R1
Vs=R1is
Fig 3
i1
L
1
i1  i2
C
i2
i2
R2
Mesh and mesh analysis
Mesh 1 KVL
t

1
R1i1  V0 
(i2  i1 )dt '  vs (t )
C
0
Mesh 2 KVL
t
di2
1
L
 R2i2  V0 
(i2  i1 )dt '  0
dt
C

0
Initial condition
i2 (0)  I 0
Mesh and mesh analysis
Adding the two equations
di2
R1i1  L
 R2i2  vs
dt
Or
vs
L di2 R2
i1  

i2 
R1 dt R 1
R1
Diff both sides
2
L
d i2
dt
2
di 2 i2 i1
 R2
  0
dt C C
Mesh and mesh analysis
Substitute for i1
d 2i2
LC
dt 2
vs
R2
L di 2
 ( R2C  )
 (1  )i2 
R1 dt
R1
R1
Initial conditions
i2 (0)  I 0
Substitute
di2
1
(0)  (V0  R2 I 0 )
dt
L
v2  R2i2 , vs  R1is
LC
d 2 v2
dt 2
R2
L dv 2
 ( R2C  )
 (1  )v2  R2is
R1 dt
R1
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.”
iL(0)= Io
input
L
is
LC
R1
d 2 v2
dt 2
C
+
vC(0)=V0 R2
-
+
v2
-
R2
L dv2
 ( R2C  )
 (1  )v2  R2is (t )
R1 dt
R1
output
Input-output Representation
General equation for a single-input single-output
dn y
dt
n
 a1
d n 1 y
dt
n 1
 ..  an y  b0
d mw
dt
y
Is the output from the circuit
w
Is the input to the circuit
Constants
a1 , a2 .., b1 , b2
m
 b1
d m 1w
dt
m 1
 ..  bm w
depend on circuit element values
and network topology. The initial conditions are
dy
d n 1 y
y(0), (0),.., n 1 (0)
dt
dt
Input-output Representation
Zero-input response
The general equation becomes homogeneous and the nth-degree
characteristic Polynomial is
s n  a1s n1  ..  an1s  an
si , i  1,2.... n
are the natural frequency of the network variable y
The zeroes of this polynomial
n
The solution of the homogeneous equation becomes
y(t ) 

i 1
If s1 is a repeated root
y(t )  k1e  k2te  k3t e  ...
s1t
s1t
2 s1t
ki e si t
Input-output Representation
Zero-state response
The general zero-state response of the circuit is
n
y(t ) 
k e
i
si t
 y p (t )
i 1
Where
y p (t ) is any particular solution of the circuit due to the input w
Input-output Representation
Example 1
Figure 4 shows a simple RC circuit with zero initial condition. The input is
v(s)  Vm cos t , t  0
v(s)  u(t ) Vm cos t
i
C
+
R
vs
-
Fig 4
KVL:
t

1
Ri (t ) 
idt ' u (t )  Vm cos tdt
C
or
di 1 0 dvs
R  i
dt C
dt
dvs
du
d
 Vm
cos t  Vm u (t ) cos t
dt
dt
dt
 Vm (t )  Vm u (t ) sin t
Input-output Representation
Initial condition:
vc (0 )  vc (0 )  0
From KVL
vR (0 )  vs (0 )  vc (0 )  Vm

v
(0
) Vm

R
iR (0 ) 

R
R
T 20.00
Voltage (V)
10.00
0.00
Tina simulation
-10.00
0.00
2.50m
5.00m
Time (s)
7.50m
10.00m
Input-output Representation
Impulse Response
From the general equation,
y(n)  a1 y(n1)  ..  an y  b0 w(m)  b1w(m1)  ..  bm w
With initial conditions


y(0 )  y (0 )  y
If
(1)
(2)

(0 )  ..  y
( n1)

(0 )  0
w   (t ) the RHS are impulse and its derivatives and the response is

h(t )  



ki e  u (t )

i 1

n

si t
The impulse function and derivatives
du

dt
d
  (1)
dt
d  (1)
  (2)
dt
d ( n)
  ( n 1)
dt

and


and
t

and
and
t
 (1) (t )dt    (t )

t
t
 ( n 1) (t )dt    ( n) (t )


 (t )dt   u(t )

 (2) (t )dt    (1) (t )
Input-output Representation
Example 2
Suppose that the differential equation relating the output
of a circuit is
d2y
dy
dw
 4  3y 
 2w
2
dt
dt
dt
Find the impulse response of the circuit.
The characteristic polynomial is
and the roots are
The response

s 2  4s  3  0
s1  1, s2  3

h(t )  k1et  k2e3t u (t )
y and the input w
Input-output Representation
Solve for constants
from

t
h(t )  k1e  k2e

3t
 u(t )
 

  (t )   k e  3k e  u (t )
h(1) (t )  k1e t  k2 e 3t  (t )  k1et  3k2 e 3t u (t )
  k1  k2
t
1
3t
2
h(2) (t )   k1  k2   (1) (t )   k1  3k2   (t ) 


k1et  9k2 e3t u (t )
Input-output Representation
Substitutew   (t )
have
and
y  h(t )
we
h(2) (t )  4h(1) (t )  3h(t )   (1) (t )  2 (t)
(k1  k2 ) (1) (t )  (3k1  k2 ) (t )   (1) (t )  2 (t )
(k1  k2 )  1
k1 
1
(3k1  k2 )  2
k2 
and
2
 h(t ) 
and
1
2
e
t
e
3t
1
 u(t )
2
Response to an arbitrary input
An arbitrary input signal can be divided into many impulse functions.
p (t )
t0
 0
1

p (t )  

 0
1

0t 
t
0
is (t0 ) p (t ' t0 )

is

t’
t 0 t1
isa

Time t’
t0
t1
t2
tk
tk 1
tn 1
is (t1 ) p (t ' t1 )
t
t1
t2
t’
Response to an arbitrary input
is (t ' )  is (t0 ) p (t 't0 )  is (t1 ) p (t 't1 )  ..  is (t n1 ) p (t 't n1 )
is (t ')  is (t0 )h (t ' t0 )  is (t1 )h (t ' t1 )  .. is (tn1 )h (t ' tn1 )
As n
    0 the output response is the sum of all impulse responses
t

v(t )  is (t ' )h(t  t ' )dt ' , t  t 0
t0
Conclusion
1 Determine the impulse response h(t )
2 Calculate the integral
3 This type of integral is called convolution integral
Response to an arbitrary input
The complete response
y(t )  z (t )  v(t )
Where z (t ) is the zero-input response
t

y (t )  z (t )  is (t ')h(t  t ')dt ',
t  t0
t0
Note that the complete response is a linear function of input only if the
Zero-input response is identically zero.
Computation of convolution
integrals
From
v(t )   h(t  t ')i (t ')dt ', t  t
t
s
0
t0
For the unit impulse at
t1  t 0
is (t )   (t  t1 )
t

v(t )  h(t  t ') (t ' t1 )dt ',
t  t0
t0
t1

 h(t  t ') (t ' t )dt '
1
t1
t1

 h(t  t1 )  (t ' t1 )dt '  h(t  t1 )
t1
Computation of convolution
integrals
Let
t  t'  
t '  t   , dt'  d
t t0
Then
v(t ) 
 h( )i (t   )d
s
0
t t0
v(t ) 
 h(t ')i (t  t ')dt '
s
0
Thus
t
t
 h(t  t ')i (t ')dt '   h(t ')i (t  t ')dt ', t  0
s
0
s
0
Convolution integral is symmetric role
Computation of convolution
i (t )
h(t )
integrals
s
2
Example 3
1
Let the input be a step function and
0
the impulse response be a triangular
waveform
h(t ')
2
t
t
0
is (t ')
2
1
0
2
t
h(t ')
t
0
is (t ')
2
1
-2
0
t
0
t
Computation of convolution
integrals
h(t  t ')
is (t  t ')
2
1
-1
0
is (t ')h(t  t ')
t 1
2
t
0
t 1
t
t 1
t
is (t  t ')h(t ')
1
-1
0
t 1
t
0
is (t  t ')h(t ')
is (0  t ')
2
h(t ')
1
0
t=0
Area = 0
t'
2
1
2
is (1  t ')
h(t ')
0
t'
2
2
t=1
Area = 3/2
0
2
0
t=2
t'
2
2
Area = 2
0
2
t'
2
h(t ')
1
0
t'
2
h(t ')
1
is (3  t ')
t'
2
1
is (2  t ')
2
t=3
t'
2
Area = 2
0
2
v(t )
2
3/ 2
t=3
Area = 2
1
2
t'
t'
Computation of convolution
integrals
Example
4
Find the zero-state response for the input and impulse response shown
is (t )
h(t )
0
e t
1
1
t
1
is (t )  u (t )  u (t  1)
t
0
h(t )  et u(t )
Computation of convolution
t
integrals

v(t )  h(t  t ')is (t ')dt '
0
For 0  t  1
h(t   )
1
0
is (t )  1

v(t )  e
For
t 1
0
dt '  1  e
t
is (t )  0
1

v(t )  e  (t t ') dt '  (e  1)e t
0
t

1
is ( )
t
 ( t t ')
is ( )
1
h(t   )
0
1  e 1
0
1
t

v(t )
t
1
Computation of convolution
integrals
Example 5
Find the zero-state response for the input and impulse response shown
is (t )
h(t )
sin  t
0
1
1
2
is (t )  u(t )sin  t
t
0
1
t
t
h(t )  u(t )  u (t  1)
Computation of convolution
integrals
t  0 v(t )  0
For
For
0  t 1
t
1 t
For
t

v(t )  sin  (t   )d 
0

v(t )  sin  (t   ) d 
0
1

1

(1  cos  t )
(cos  (t  1)  cos  t )  
2

cos  t
Computation of convolution
integrals
sin  (0   )
h( )
t0
1
0

1
h( )
0  t 1
sin  (t   )1
0

1
h( )
1 t
sin  (t   )
1
0
1
