I have the following information:

H(jw) = (3+jw)/(e^(2jw))
f(t) = e^(-3t)u(t)

Therefore,
F(w) = 1/(3+jw)
Y(w) = H(w)*F(w) = 1/(e^(2jw))

I have to find y(t), but I am having troubling finding.

To find y(t), we need to take the inverse Fourier transform of Y(w). In this case, Y(w) is given as 1/(e^(2jw)).

To simplify the expression, we can rewrite the denominator using the Euler's formula, which states that e^(ix) = cos x + jsin x.

So, e^(2jw) = cos(2w) + jsin(2w).

Now, we can substitute this value back into the expression for Y(w):

Y(w) = 1/(cos(2w) + jsin(2w)).

To find y(t), we need to take the inverse Fourier transform of Y(w), which can be obtained using the formula:

y(t) = (1/2π) ∫[Y(w) * e^(jwt)] dw.

Let's substitute the value of Y(w) into the formula:

y(t) = (1/2π) ∫ [ 1 / (cos(2w) + jsin(2w)) ] * e^(jwt) dw.

Now, we need to solve the integral to obtain y(t).

Integrating this expression can be quite complex, involving techniques like residue calculus. However, it is worth noting that the given expression for Y(w) can be rewritten as:

Y(w) = 1 / [e^(2jw)],

which can also be expressed as:

Y(w) = 1 / [e^(-j2w)].

Recognizing this form, we can use the inverse Fourier transform of a unit impulse function, which is given by:

1 / [2π] ∫[delta(w - a) * e^(jwt)] dw = e^(jwt) / [2π],

where delta(w) is the Dirac delta function.

Therefore, we can rewrite Y(w) as:

Y(w) = 1 / [e^(-j2w)] = 1 / [2π] * e^(j2w).

Now, using the above inverse Fourier transform formula, we can directly obtain y(t) as:

y(t) = 1 / [2π] * e^(j2t).

So, y(t) = e^(j2t) / [2π].