14.1
Singularity Function Response:
For a system described by the following differential equations:
 x 
a a   x  b b  v 
x     Ax  Bv  




 
 x 
a a   x  b b  v 
1
11
12
1
11
12
s1
21
22
2
21
22
s2
s
2
The following formulas can be used to determine the forced responses to sources
consisting of singularity functions:
g 
For a ramp source of the form vs  g t   t for t  0 , the solution can be written
g 
as:
m  n 
x   t   
m  n 
where
m 
m  n 
 A Bg    and A     
m 
m   n 
1
s
2
1
1
2
2
f
1
1
1
1
1
2
2
s
2
If the system has a single input, B becomes a vector b and g s becomes a scalar g s
14.2
g 
For a step input of the form vs  g s   1  for t  0 , the forced response is:
 g2 
 A Bg  x
Note this is the same as for DC sources examined earlier.
If the system has a single input, B becomes a vector b and g s becomes a scalar g s
1
s
f
g 
For impulse sources of the form vs  g  (t )    (t ) the forced solution is zero,
g 
since  (t )  0 for t > 0. Therefore, the complete solution is simply the natural
solution whose coefficients are evaluated based on the following initial conditions:
1
s
2
x(0 )  Bgs  x(0 )
If the system has a single input, B becomes a vector b and g s becomes a scalar g s
14.3
Example:
Derive the formula for each of singularity function responses through proper
substitution of a guessed a general response. For the impulse response formula you
must integrate both sides from 0- to 0+ after the substitution to get the proper result.
Example:
Independently find the impulse, step and ramp response for the system output
below:
 iL   5  1  iL  1 ;
v    1  3 v   0is
 C   
 C 
i 
v0  0 1 L  ;
vC 
show v0impulse (t )  t exp(4t )u(t )
1
1  exp( 4t )  4t exp( 4t )u (t )
16
1
1 1
v0 Ramp (t )  t   exp( 4t )  t exp( 4t )u (t )
16  2 2

v0 Step (t ) 
14.4
Multiple Source Example:
 v
v

is
c1
vc2
9
vs
1
F
2
vc1
c2
1
F
3
12 
vo
1 
 7
2 i

  18 6   v  2
 




9  
  1
 v 
1

 
    0 0  v 
 4
4
c1
s
c2
s
v 
i 
v  1  1     0 0  
v 
v 
c1
s
c2
s
o
Find the solution if:
v  9u(t ) and is (t )  5u(t )
s
Show: v0 (t )  27.83(exp(0.1038t )  exp(0.5351t ))u(t )
14.5
Use Matlab to help with computations, it is convenient to set up and perform
computation in matrix form. Matlab (short for Matrix Laboratory) is designed to
do computations this way.
% Define differential equation matrices
a = [-(7/18), (1/6); (1/4), -(1/4)];
b = [2, (2/9); 0, 0];
% Natural Solution - find roots of Char. Eqn.
% det|pI-a| = 0 = 72p^2 + 46p + 4
rts = roots([72 46 4])
rts =
-0.5351
-0.1038
% Roots are real and distinct so natural soln. is of
% the form
vc1(t) = C1*exp(rts(1)*t)+C2*exp(rts(2)*t)
%
vc2(t) = D1*exp(rts(1)*t)+D2*exp(rts(2)*t)
% Forced soln for vs
fc1 = -inv(a)*b*[5; 9]
fc1 =
54.0000
54.0000
% Initial conditions for x dot
14.6
x0p = [0; 0] % System at rest
x0p =
0
0
x0dotp = a*x0p+b*[5; 9]
x0dotp =
12
0
% Solution has form:
vc1(t) = [fc1(1) + C1*exp(rts(1)*t)+C2*exp(rts()*t)]
%
vc2(t) = [fc1(2) + D1*exp(rts(1)*t)+D2*exp(rts(2)*t)]
% Form of derivative equation: vc1dot(t) = [rts(1)*C1*exp(rts(1)*t)+rts(2)*C2*exp(rts(2)*t)]
%
vc2dot(t) = [rts(1)*D1*exp(rts(1)*t)+rts(2)*D2*exp(rts(2)*t)]
% Solve Resulting system of equation to find C1 and C2
s1 = [1 1; rts(1), rts(2)];
r1 = [x0p(1)-fc1(1); x0dotp(1)];
cc = inv(s1)*r1
cc =
-14.8254
-39.1746
% Solve Resulting system of equation to find D1 and D2
s2 = [1 1; rts(1), rts(2)];
r2 = [x0p(2)-fc1(2); x0dotp(2)];
dd = inv(s2)*r2
14.7
dd =
13.0021
-67.0021