Discrete Time Fourier Transform of the signal represented by $x[n] = n^2 a^n u[n]$

Question

I have a homework problem that I am just not sure where to start with. I have to take the Discrete Time Fourier Transform of a signal represented by:

$$x[n] = n^2 a^n u[n]$$

given that $|a| < 1$, $\Omega_0 < \pi$, and u[n] being the unit step function.

There is a hint saying that "Calculus and derivatives will help!" but that actually confuses me more than it helps. However, regardless, I'm just not sure how to even get started evaluating that. I know how to take the DTFT of a signal, but just can't figure out how to get it into a usable form. I also am allowed to use the following conversion:

$$ x[n] = a^nu[n] \iff X(\Omega) = \frac{e^{j\Omega}}{e^{j\Omega} - a}$$

I am not looking for the answer, but rather just some pointers as to how to start manipulating the original signal.

Answer 1

To include this as the answer that most helped me, since it was just as a comment to my original post. The most straightforward solution seemed to be by using the property of the Discrete Time Fourier Transform for when you are multiplying by n in the time domain, that corresponds to a derivation in the Fourier/frequency domain.

Therefore, the solution would just be to take the Fourier transform from the provided conversion, and then taking the second derivative of that. This is the solution that I got after doing so:

$$X(\Omega) = \frac{a^3e^{j\Omega} - ae^{j3\Omega}}{(e^{j\Omega} - a)^4}$$

Answer 2

You haven't given a definition for the discrete Fourier transform but I presume it is something like $$X(\omega) = \sum_{n=-\infty}^{\infty}x(n) e^{-i\omega n}$$ which in this case would be $$X(\omega) = \sum_{n=0}^{\infty}n^2 a^n e^{-i\omega n} = \sum_{n=0}^{\infty}n^2 z^n$$ where $z = ae^{-i\omega}$. Now suppose we set $$f(z) = \sum_{n=0}^{\infty} z^n$$ This series converges provided that $|z| < 1$. Fortunately, we are given that $|z| = |a| < 1$. We can differentiate power series term by term in the interior of their disc of convergence, so $$f'(z) = \sum_{n=1}^{\infty} nz^{n-1}$$ and $$f''(z) = \sum_{n=2}^{\infty}n(n-1)z^{n-2} = z^{-2}\sum_{n=2}^{\infty}n(n-1)z^{n}$$ Try rearranging this to see if you can get an expression for $\sum_{n=2}^{\infty}n^2 z^n$ in terms of $f'(z)$ and $f''(z)$. Then you can manually add the $n=0$ and $n=1$ terms.

You will also want to use the fact that $$f(z) = \frac{1}{1-z}$$ for $|z| < 1$.