0
$\begingroup$

I'm studying for exams at the moment and I'm trying to reproduce a solution from my professor (I have the solutions). The following signal is given:

Signal

The excercise says:

Calculate the Fourier transforms for i = 13 to 15 and N = 64.

The signal in the time domain looks like this (this is where I'm still on par with the solution): $$ f(k) = \left( -\frac{1}{N}k + 1 \right)\left( \sigma(k) - \sigma(k-N) \right) $$

The problem starts when trying to transform the whole thing to the $z$-domain. This is my take: $$ f(k)=-\frac{1}{N}k\sigma(k) + \frac{1}{N}k\sigma(k-N) + \sigma(k) - \sigma(k-N) $$ Corrected for delay: $$ f(k)=-\frac{1}{N}k\sigma(k) + \frac{1}{N}(k-N)\sigma(k-N) + N\sigma(k-N) + \sigma(k) - \sigma(k-N) $$

Transformed: $$ F(z) = -\frac{z}{(z-1)^2}\frac{1}{N} + \frac{z}{(z-1)^2}\frac{1}{N}z^{-N} + \frac{z}{z-1}Nz^{-N} + \frac{z}{z-1} - \frac{z}{z-1}z^{-N} $$

But the solution only gives me three summands and I can't figure out how he is able to simplify the rest of the equation (Being well aware that the solution given by the professor might be wrong). He states:

\begin{align} F(z) &= - \frac{1}{N}\frac{z}{(z-1)^2} + \frac{1}{N}\frac{z}{(z-1)^2}z^{-N} + \frac{z}{z-1}\\ F(z) &= \frac{z}{(z-1)^2}\left[-\frac{1}{N} + \frac{1}{N}z^{-N} + {z-1}\right] \end{align}

  • What am I doing wrong / am I doing anything wrong?
  • Can you please sanity check me?
$\endgroup$
1
$\begingroup$

Your mistake is quite simple, originating from a subtle inaccuracy in the reformulation from here $$ f(k)=-\frac{1}{N}k\sigma(k) + \frac{1}{N}k\sigma(k-N) + \sigma(k) - \sigma(k-N) $$ to here: $$ f(k)=-\frac{1}{N}k\sigma(k) + \frac{1}{N}(k-N)\sigma(k-N) + N\sigma(k-N) + \sigma(k) - \sigma(k-N). $$

The 3rd term of the second equation is only $\sigma(k-N)$ because the $N$ is cancelled by the $1/N$ from the original term. Then, the (corrected) third term cancels with the last one. And then, transforming this to Z-domain gives exactly the result your professor has provided.

$\endgroup$
  • $\begingroup$ Oh snap! This happens if you actually focus on the z-transform itself and constantly forget the constants! $\endgroup$ – Jan Krüger Jan 21 '17 at 16:12

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.