I know there are 3 properties of DTFT that help with my problem

  1. $$ a^{n}u[n]=\frac{1}{1-ae^{-jΩ}} $$

  2. $$ (n+1)a^{n}u[n]=\left(\frac{1}{1-ae^{-jΩ}}\right)^{2} $$

  3. $$ \frac{(n+r-1)!}{n!(r-1)!}a^{n}u[n]=\left(\frac{1}{1-ae^{-jΩ}}\right)^{r} $$

But I cannot find some use between them to calculate the DTFT of the following signal $$ x[n]=(n+5)\left(\frac{7}{45}\right)^{n}u[n] $$ where $u[n]$ is the unit step function. Can anyone help?



Note that you can express your signal as

$$x[n]=(n+1+4)\left(\frac{7}{45}\right)^{n}u[n] = (n+1)\left(\frac{7}{45}\right)^{n}u[n] + 4\left(\frac{7}{45}\right)^{n}u[n] $$

So now you have the sum of two signals:

$$x_1[n]=(n+1)\left(\frac{7}{45}\right)^{n}u[n]$$ $$x_2[n]=4\left(\frac{7}{45}\right)^{n}u[n]$$

Using your properties and the fact that the Fourier transform is linear, you should be able to solve it.

  • $\begingroup$ yep its pretty simple , but I didn't think of that . Thanks a lot $\endgroup$ – user33117 Jan 16 '18 at 19:06

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