# How to apply fourier transform to $0.5^n u(n)$

I'm working in a signals class for continuous signals, and we have this problem shown above. I have tried using this function $f_1*f_2 = F_1 * F_2$, where I'm assuming this means multiplication of two functions is equal to the convolution of their fourier transforms. I'm using $f_1 = 0.5^n$ and $f_2 = u(n)$.

So I can calculate the fourier transorm of $u(n)$ fine. It is $\pi\delta(\omega) + 1/(j\omega))$. However, I cannot for the life of me figure out $0.5^n$. I tried to put it into the fourier transform integral integral of$(0.5^t)/(e^{j \omega t})dt$ from negative infinity to infinity, but I end up with $0.5t/e^{jw}$, and when evaluated from negative infinity to infinity, I end up with $\infty$ as my answer, unless of course the integration is wrong.

Therefore, either the answer is $\infty * \pi\delta(w) + 1/(j\omega)$, which when convoluted would equal just the second function..? OR am I going about this problem completely wrong?

## 1 Answer

I think you'd better not split the functions. If you want to calculate the fourier transform of $0.5^t u(t)$, then you might put all that into the integral. Therefore, your integral limits will become $0$ to $+\infty$ because of $u(t)$. And your integral result should not be infinity because your function $0.5^t$ goes to zero.

Solving the integral:

$\large\int_0^\infty e^{(-jwt)} . 0.5^t dt$

the inside part could be rewrite:

$\huge\frac{e^{(-jwt)}}{2^t} => (\frac{e^{(-jw)}}{2})^t$

You could name $\large\frac{e^{(-jw)}}{2} = u$ so you get integral of $\large u^t dt = \frac{u^t}{ln(u)} + C$

Then you get (replacing u):

$\huge\frac{(\frac{e^{-jw}}{2})^t}{ ln(e^{-jw}/2)}$

When t goes to $+\infty$ you get zero. When it goes to zero you get:

$\large\frac{1}{ln(e^{-jw}/2)}$

which left us:

$\large\frac{1}{-jw - ln(2)}$

I'm not sure I did not get lost in calculations but I believe its correct.

• You're completely right. I got mixed up after separating the functions. Thank you! – nathpilland Nov 25 '13 at 2:12
• You are welcome. Good luck – Felipe_Ribas Nov 25 '13 at 3:26