What is the frequency transfer function when $x(t) = e^{-3t}u(t)$ and $y(t) = 2u(t)[e^{-t}-e^{-4t}]$

Question

What I did was take the fourier transform of both $x(t)$ and $y(t)$ and then divided $Y(j\omega)/X(j\omega)$. So $$Y(j\omega) = \frac{2}{1+j\omega}-\frac{2}{4+j\omega}\quad\text{and}\quad X(j\omega) = \frac{1}{3+j\omega}$$

However, the answer is of the form: $\displaystyle \frac{A_1}{1+j\omega}+\frac{A_2}{4+j\omega}$ where the $A$'s are constant. I am not getting the solution of this form. I got $Y(j\omega)$ in this form, but not $H(j\omega)$.

Am I doing something wrong?

Answer 1

Your answer is correct; it's just in a different form than the given answer. In order to get your answer in the given form, do the following:

1. rewrite $Y(j\omega)$ by combining the two terms: $Y(j\omega)=\displaystyle\frac{N(j\omega)}{(j\omega+1)(j\omega+4)}$, where $N(j\omega)$ is a (very simple) polynomial. I'm sure you know how to obtain $N(j\omega)$.
2. write the frequency response as $\displaystyle H(j\omega)=\frac{N(j\omega)(j\omega+3)}{(j\omega+1)(j\omega+4)}$ and use partial fraction expansion to obtain the answer in the given form.

Alternatively, you could also determine $h(t)$ via inverse Fourier transform of $H(j\omega)$ as you obtained it (and noting that multiplication by $j\omega$ corresponds to differentiation in the time domain). You will see that $h(t)$ is in the form $A_1e^{-t}u(t)+A_2e^{-4t}u(t)$, which directly leads to $H(j\omega)$ in the given form.