# Convolution

• Oct 18th 2009, 09:51 AM
dsprice
Convolution
Given:

y'' +4y = sin3t
y'(0) = y(0) = 0

From this, I can pretty easily find Y(s) to be:

Y(s) = 3/[(s^2 + 3^2) (s^2 + 2^2)]

In the above, I can see sin3t and (1/2) sin2t, but I'm having trouble seeing how to get back to just plain y(t). Any help will be appreciated.
• Oct 18th 2009, 10:07 AM
Mush
Quote:

Originally Posted by dsprice
Given:

y'' +4y = sin3t
y'(0) = y(0) = 0

From this, I can pretty easily find Y(s) to be:

Y(s) = 3/[(s^2 + 3^2) (s^2 + 2^2)]

In the above, I can see sin3t and (1/2) sin2t, but I'm having trouble seeing how to get back to just plain y(t). Any help will be appreciated.

Use partial fractions!

$\frac{3}{(s^2 + 3^2)(s^2 + 2^2)} = \frac{As+B}{s^2 + 3^2} + \frac{Cs+D}{s^2+2^2}$

Multiply through by the denominator of the LHS:

$3 = (As+B)(s^2 + 4) + (Cs+D)(s^2 + 9)$

Expand it out:

$3 = As^3 + 4As + Bs^2 4B + Cs^3 + 9Cs + Ds^2 + 9D$

Not collect like terms in s:

$3 = (A+C) s^3 + (B+D)s^2 + (4A+9C)s + (4B+9D)$

Now, if you compare coefficients of both sides of the equation, you get the following:

$A+C = 0$
$B+D = 0$
$4A+9C = 0$
$4B + 9D = 3$

This is two groups of two equations in two unknowns. Easily solveable!