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 1


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.

  • \$\begingroup\$ You're completely right. I got mixed up after separating the functions. Thank you! \$\endgroup\$ Commented Nov 25, 2013 at 2:12
  • \$\begingroup\$ You are welcome. Good luck \$\endgroup\$ Commented Nov 25, 2013 at 3:26

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