Convolution property of Fourier transform

Question

I have a signal $$ x(t) = \frac{1}{T} e^{- \frac{t}{T} } u(t) $$ and I know his Fourier transform $$ X(f) = \frac{1}{1+ i 2\pi f T } $$ and I have to find $$ z(t) = x(t) \circledast x(t) $$ using convolution property or in time domain and after I have to calculate $$ Z(f) $$. I found easily that $$ Z(f) = [ \frac{1}{1+ i 2 \pi T } ] ^{2} $$ that’s because I studied that the Fourier transform of $$ z(t) = x(t) \circledast x(t) $$ is $$ Z(f) = X(f) \cdot X(f) $$. Now I don’t know how to find z(t). I tried to anti transform but I didn’t obtain a result similar of my book :/ the result should be $$ \frac{i}{2 \pi T } (- i 2 \pi T ) \frac{1}{T} e^{\frac{-t}{T}} u(t) $$ and it appears to have been applied derivative property but I don’t know why. Thank you so much !

Answer 1

You need to work with the frequency derivative property $$-itx(t) \longleftrightarrow \frac{dX(\omega)}{d\omega}=\frac{1}{2\pi}\frac{dX(f)}{df},$$ with $\omega=2\pi f$ and '$\longleftrightarrow$' tying the Fourier transform pair. Work on the derivative: $$-itx(t) \longleftrightarrow \frac{1}{2\pi}\frac{d}{df}\left(\frac{1}{1+i2\pi fT}\right)$$ $$-itx(t) \longleftrightarrow -\frac{iT}{(1+i2\pi fT)^2}.$$ Now modify the pair using the linearity property until you get $Z(f)$ on the right-hand side: $$-itx(t) \longleftrightarrow -\frac{i T}{(1+i2\pi fT)^2}$$ $$\frac{t}{T}x(t) \longleftrightarrow \frac{1}{(1+i2\pi fT)^2}$$ and therefore $$z(t)=\frac{t}{T}x(t)=\frac{1}{T^2}te^{-\frac{t}{T}}u(t).$$ This should be the right answer. I believe you have some typos in your proposed book solution, since the self-convolution of a causal exponential $e^{-at}u(t)$, $a>0$, is equal to $te^{-at}u(t)$, as you can now easily verify using similar procedure :)