Thread: using Fourier transform to solve ODE

1. using Fourier transform to solve ODE

For part i) I got the answer 1/((jw)^2 + 5jw +6)

For part ii)
I first consider input to be a unit impulse

Thus, Y(w)=H(w)F(w) and F(w)=1
yI(t)=-1/2pi integrate from -infinity to infinity (e^jwt)/(w^2 - 5jw - 6) dw
where yI(t) is the output when f(t) is a unit impulse
using complex contour integration,
I got yI(t) = 0 for t<0
and yI(t) = 2je^(-2t) - 3je^(-3t) for t>0
Then using y(t)=yI(t)*f(t)
I got y(t) = integrate from -infinity to infinity ( 2je^(-2(t-k)) - 3je^(-3(t-k)) )f(k)dk

For a similar example in my notes, it just stops at this step.

However, for this question I am not sure whether should I stop here as I noticed that I havent use the boundary conditions yet.
Please gives me some idea on this.

2. Originally Posted by progrocklover

For part i) I got the answer 1/((jw)^2 + 5jw +6)

For part ii)
I first consider input to be a unit impulse

Thus, Y(w)=H(w)F(w) and F(w)=1
yI(t)=-1/2pi integrate from -infinity to infinity (e^jwt)/(w^2 - 5jw - 6) dw
where yI(t) is the output when f(t) is a unit impulse
using complex contour integration,
I got yI(t) = 0 for t<0
and yI(t) = 2je^(-2t) - 3je^(-3t) for t>0
Then using y(t)=yI(t)*f(t)
I got y(t) = integrate from -infinity to infinity ( 2je^(-2(t-k)) - 3je^(-3(t-k)) )f(k)dk

For a similar example in my notes, it just stops at this step.

However, for this question I am not sure whether should I stop here as I noticed that I havent use the boundary conditions yet.
Please gives me some idea on this.
I agree with your Fourier transform this gives that

$\displaystyle [(jw)^2+5(jw)+6]\hat{y}=\hat{f}(w) \iff (jw+2)(jw+3)\hat{y}=\hat{f}(w)$

Now solving for $\hat{y}$ and using partial fractions I get

$\displaystyle \hat{y}=\frac{\hat{f}(w)}{jw+2}-\frac{\hat{f}(w)}{jw+3}$

Since we don't know what $f$ is we will have to make some assumptions later.

Since we have the product of Fourier transforms we will use the convolution theorem

Note that

$\displaystyle \mathcal{F}^{-1}\left( \frac{1}{a+jw}\right)=e^{-ax}u(x)$

Where $u(x)=\begin{cases}1, \text{ if } x> 0 \\ 0, \text{ if } x< 0\end{cases}$

The Heaviside step function.

Now using the convolution theorem we get that

$\displaystyle y=\int_{-\infty}^{\infty}f(\tau)e^{-2(x-\tau)}u(x-\tau)d\tau-\int_{-\infty}^{\infty}f(\tau)e^{-3(x-\tau)}u(x-\tau)d\tau$

So we will need to make assumptions on $f$ so the above integrals converge.