I want to find the inverse Fourier transforms of:
$$u(\nu + 1) \ \exp(-\nu)$$
Attempt:
So the inverse Fourier transform is given by:
$$\int^\infty_{-\infty} u(\nu + 1) \ e^{-\nu} e^{j2 \pi t} \ d\nu = \int^\infty_{-\infty} u(\nu + 1) \ e^{\nu (j2 \pi t -1)} \ d\nu$$
I know that $FT \Big[ u(t) \Big]= \frac{1}{2} \left( \delta(\nu) – \frac{j}{\pi \nu} \right).$ So do we need to use this property or somehow proceed with direct integration? Also what can I do about the $+1$ in the argument of the Heaviside unit step function?
Any explanation on how to solve this is greatly appreciated.
Best Answer
We have most of the problem done through comments, just some minor tweaks:
It is easier to see the behavior of the result if we instead represent the integral as
$$ \int_{-\infty}^{\infty} u(\nu + 1) e^{- \nu} e^{2 \pi j \nu t} \mathrm{d}\nu = \int_{-\infty}^{\infty} u(\nu + 1) e^{- \nu (1 - 2 \pi j t)} \mathrm{d}\nu $$ As I mentioned before, the heaviside step function turns on to a value of $1$ after its argument is larger than or equal to zero, this means $\nu \geq -1$ otherwise the integral is zero thus: $$ \int_{-\infty}^{\infty} u(\nu + 1) e^{- \nu (1 - 2 \pi j t)} \mathrm{d}\nu = \int_{-1}^{\infty} e^{- \nu (1 - 2 \pi j t)} \mathrm{d}\nu = \left.\frac{-1}{1 - 2 \pi j t}e^{- \nu (1 - 2 \pi j t)} \right|^{\infty}_{-1} $$ We see that at infinity this exponential term goes to zero so we are left with: $$ -\left( \frac{-1}{1-2 \pi j t} e^{1 - 2 \pi j t} \right) \quad \text{ or } \quad \frac{1}{1-2 \pi j t} e^{1 - 2 \pi j t} $$