For a System Described by the Following Differential Equations

Singularity Function Response: For a system described by the following differential equations:

x˙ 1  a11 a12 x1  b11 b12 vs1  x˙   Ax  Bv     s       x˙ 2  a21 a22 x2  b21 b22 vs2 

The following formulas can be used to determine the forced responses to sources consisting of singularity functions:

g1   g t  t t  0 s g  For a ramp source of the form vs  2  for , the solution can be written as:

m1  n1  x  t  f     m2  n2  where

m1  m1  n1   A 1Bg  A 1  s   and     m2  m2  n2 

If the system has a single input, B becomes a vector b and gs becomes a scalar gs 14.2

 g1   gs    t  0 g2 For a step input of the form vs   for , the forced response is: 1  A Bgs  x f Note this is the same as for DC sources examined earlier. If the system has a single input, B becomes a vector b and gs becomes a scalar gs

g1   g  (t)   (t) s g  For impulse sources of the form vs  2  the forced solution is zero, 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:   x(0 )  Bgs  x(0 ) If the system has a single input, B becomes a vector b and gs becomes a scalar gs 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:

˙ iL   5 1iL  1 iL          is ; v0  0 1 ; v˙C   1  3vC  0 vC  showv0impulse (t)  t exp(4t)u(t) 1 v (t)  1 exp(4t)  4t exp(4t)u(t) 0Step 16 1  1 1  v (t)  t   exp(4t)  t exp(4t) u(t) 0Ramp 16  2 2  14.4

Multiple Source Example:

 7 1    2 v˙ c1  vc1  is  is  18 6  2         9  v v˙ 1 1 v v 9  c2  c2     c 2  0 0 s   4 4 1 1 F F vc1 vs 3 2 vo vc1  is  v  1  1  0 0 12  o       vc2  vs 

Find the solution if:

vs  9u(t) and is (t)  5u(t)

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