# DTFT fourier transform (modified property)

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?

## 1 Answer

Hint:

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.

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