net work analysis dr. sumrit hungsasutra text : basic circuit theory, charles a. desoer & kuh,...

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