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CT-LTI System Analysis: Approach #I

The document summarizes two approaches for analyzing the response of a generalized input-output differential equation (DE). Approach I involves deriving the input-output DE, solving the ordinary differential equation (ODE) using standard techniques, and separating the natural and forced responses. Approach II determines the zero-input response (ZIR) from the homogeneous solution and impulse response, then finds the zero-state response (ZSR) by convolving the impulse response with the input to get the total response. The techniques are illustrated using a series RLC circuit example, and the nature of overdamped, critically damped, underdamped, and undamped system responses are discussed.

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BAYAT MOBIN BAMIM
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110 views18 pages

CT-LTI System Analysis: Approach #I

The document summarizes two approaches for analyzing the response of a generalized input-output differential equation (DE). Approach I involves deriving the input-output DE, solving the ordinary differential equation (ODE) using standard techniques, and separating the natural and forced responses. Approach II determines the zero-input response (ZIR) from the homogeneous solution and impulse response, then finds the zero-state response (ZSR) by convolving the impulse response with the input to get the total response. The techniques are illustrated using a series RLC circuit example, and the nature of overdamped, critically damped, underdamped, and undamped system responses are discussed.

Uploaded by

BAYAT MOBIN BAMIM
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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CT-LTI System Analysis

Generalized I/O DE is d2y dy


2
 a1  a0 y  b0 x
dt dt

Approach #I Steps:
(i) Derive the I/O DE
(ii) Solve the ODE using standard technique

y (t )  yn (t ) natural response/homogeneous solution of the ODE 


y (t ) forced response/particular solution of the ODE
Approach #II
Steps:
(i) Derive the I/O DE
(ii) Find ZIR from the homogeneous solution of the ODE using ICs
(iii)Find the impulse response(h(t)) of the system from the
homogeneous solution of the ODE using auxiliary ICs
(iii) Find ZSR by convolving h(t) with input x(t)
n
y (t )   c j e  x(t )  h(t )
 jt

j 1

ZSR
ZIR
The technique discussed above is illustrated using
R= 3 W, L= 1 H, C=0.5 F and vs(t)=x(t)=10e-3tu(t).
The generalized ODE can be written in the operator notation as
( D 2  3D  2) y  2 x
The characteristic equation is ( 2  3  2)  0
The characteristic roots are   1 and -2
Natural response/
Homogeneous solution yn (t )  C1e 1t  C2 e 2 t
of the ODE is
Forced response/Particular solution of the ODE

Use method of undermined coefficient (page 140


Art. 2.5-1 BP Lathi
Here input x(t)=10e-3tu(t), the response will have form
Ae-3tu(t). Substituting the solution to the DE gives
9 Ae 3t  9 Ae 3t  2 Ae 3t  2  10e 3t
since e 3t  0; A  10.
y (t )  10e u (t )
3 t

y (t )  C1e  t  C2 e 2 t  10e 3t u (t )


ICs : vc (0 )  vc (0 )  5V  y (0)
dvc dy
i (0 )  i (0 )  C t 0
0 t 0
0
dt dt
y (0)  C1  C2  10  5
dy
t 0
 C1  2C2  30  0
dt
Solving : C1  20, C2  25
y (t )  (20e  25e
t 2 t
 10e )u (t )
3 t

Natural Response Forced Response


Approach #II
Determination of ZIR
The characteristic roots are   1 and -2
Natural response/
Homogeneous solution yn (t )  C1e 1t  C2 e 2 t
of the ODE is

y (t )  C1e  t  C2 e 2 t
ICs : vc (0 )  vc (0 )  5V  y (0)
dvc dy
i (0 )  i (0 )  C t 0
0 t 0
0
dt dt
y (0)  C1  C2  5
dy
t 0
 C1  2C2  0
dt
Solving : C1  10, C2  5
yZSR (t )  (10e  5e )u (t )
t 2 t

Determination of Impulse Response


h(t )  bn (t )   P( D) yn (t )  u (t )
yn (t ) is the linear combination of the characteristics modes
of the system. Here n is the order of the system.
Here n=2, b n =b 2 =0, P(D)=2
y n (t)=C1e-t +C2 e-2t
Using auxiliary ICs( y'n (0)  1, yn (0)  0)
C1  2C2  1; C1  C2  0
Solving: C1 =1, C2 =-1
y n (t)=e-t -e-2t
h(t )  0   (t )  2(e -t -e-2t )u (t )  2(e -t -e -2t )u (t )
yZSR (t )  h(t )  x(t )  2(e-t -e-2t )u (t ) 10e 3t u (t )
 o 2(e  e )10e d  10e  20e  10e
t  2 3( t  ) t 2 t 3 t

y (t )  yZIR (t )  yZSR (t )  (20e  t  25e 2 t  10e 3t )u (t )


Nature of Response of a SO System
The nature of the response of a SO system depends on the
nature of the characteristic roots of the system.

( 2  a1  a0 )  0
a1 1
1 / 2    (a12  4a02    ( 2  n2
2 2
a
Where   is the damping factor (neper/s) and
2
n = a0 is the undamped natural frequency(rad/s).

The ratio   is called the damping ratio of the system.
n
Using these notations, the characteristics equation of SOS
can be written as
 2  2n   n2  0
Case I: Overdamped response(when  >n )
1 and 1 are real and negative. Natural response
y n (t )  A1e t  A2 e t
1 2
Case I: Critically damped response(when   nI )
1 and 1 are real, equal and negative. Natural response
y n (t )  ( A1  A2t )e  t
Case III: underdamped response(when   n )
1 and 1 are complex conjugate with negative real part.
1 /2 =-  j n2   2 =-  jd
Where d = n2   2 is the damped natural frequency of the SOS
Natural response
y n (t )  A1e(-  j ) t  A2 e(-  j ) t  e  t ( B1 cos d t  B2 sin d t )
d d
Case IV: undamped response(when   0)
1 and 1 are complex conjugate with negative real part.
1 /2 =  j n2   2
Natural response
y n (t )  A1e(j ) t  A2 e( j ) t  ( B1 cos d t  B2 sin d t )
d d
Symbolic Solution of ODE using Matlab

syms y(t)
Dy=diff(y)
vc=dsolve(diff(y,2)+3*diff(y)+2*y==20*exp(-3*t),y(0)==5,Dy(0)==0)
i=0.5*diff(vc)

vc =20*exp(-t) - 25*exp(-2*t) + 10*exp(-3*t)


i =25*exp(-2*t) - 10*exp(-t) - 15*exp(-3*t)
ezplot(vc, [0 10])
VC

i
Undamped case

VC

i
Finding Initial Values

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