0
$\begingroup$

Hi everybody i'm a student. Yesterday i had a test about my Engineering subject about signal processing and there was this problem:

You have the sequence $x(n) = N+1 - |n|$. With $|n|\leq N.$ Determine the $\mathcal{Z}$-transformation of the signal and the fourier transformation.

I'm going mad, me and my colleagues feel to miss something, each way we try gets us drowning in calculations...

$\endgroup$
  • $\begingroup$ Consider adding the "self-study" tag. $\endgroup$ – Gilles Jan 20 '16 at 18:17
  • $\begingroup$ Done. Even though it's not self for me, someone who is self studying could find it useful! $\endgroup$ – magicleon94 Jan 20 '16 at 18:29
1
$\begingroup$

You're a student, so I'll try to help you figure out the solution by yourself. Make sure you understand the following points:

  1. The signal is a triangle of length $2N+1$.
  2. A triangular sequence can be written as the convolution of two even more basic sequences, each of length $N+1$.
  3. Convolution in the time domain corresponds to multiplication of the $\mathcal{Z}$-transforms.
  4. Define the basic signal determined in point 2. in the interval $0\le n\le N$ and compute its $\mathcal{Z}$-transform.
  5. Take the square of that $\mathcal{Z}$-transform, which gives you the $\mathcal{Z}$-transform of a triangle in the range $0\le n\le 2N$.
  6. In order to get the transform of $x[n]$, you need to shift the triangle to the left by $N$, which corresponds to a multiplication by $z^N$.

If you did everything right, the resulting $\mathcal{Z}$-transform should be

$$X(z)=\left(\frac{1-z^{N+1}}{1-z^{-1}}\right)^2z^N\tag{1}$$

You obtain the Fourier transform by substituting $z=e^{j\omega}$ in $(1)$. Note that the result must be real-valued due to the symmetry of $x[n]$. I leave the intermediate steps up to you, but your solution should be

$$X(e^{j\omega})=\left(\frac{\sin\left(\frac{(N+1)\omega}{2}\right)}{\sin\left(\frac{\omega}{2}\right)}\right)^2\tag{2}$$

$\endgroup$
  • $\begingroup$ Yeah this is good, i didn't think of it as the convolution of two rectangular signals. We wrote it in terms of step functions and transformed them but it was a mess. $\endgroup$ – magicleon94 Jan 20 '16 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.