If we have a signal $y[n]$ and its unit delayed version $y[n-1]$, can we write $y[n-1]$ in terms of $y[n]$ times some exponential?

The reason I want to do this is to then take $y[n]$ common and simplify my analysis that I am doing for my research. The above is a generic form.

  • $\begingroup$ Do you know z-transform or Laplace-transform? $\endgroup$ – Memming Oct 31 '17 at 12:18
  • $\begingroup$ @Memming: Yes I know z transform. And I know that unit delay is represented by $z^{-1}$ but I can't really put this in the form of y[n] times exp(something). $\endgroup$ – S. Khan Oct 31 '17 at 12:54
  • $\begingroup$ It won't happen in the time domain, unless your signal is geometric (or exponential). $\endgroup$ – Memming Oct 31 '17 at 13:32
  • $\begingroup$ @ Memming: The signal is sampled CPM, and I want to average two samples that are one time delay apart. $ e^{j2 \pi h\sum_{i=0}^m \alpha_i q_{n-iN_{s}}} + e^{j2 \pi h\sum_{i=0}^m \alpha_{i} q_{n-1-iN_{s}} } $ like this. (I have not written the division by 2 here) $\endgroup$ – S. Khan Oct 31 '17 at 14:14

In continuous time, we can, though it is a mathematical curiosity of little practical use. Suppose we have a smooth continuous-time signal $x$. Using the Taylor expansion,

$$x (t-1) = x(t) - \dot x (t) + \frac 12 \ddot x (t) - \cdots = \left( 1 - \frac{\mathrm d}{\mathrm d t} + \frac 12 \left(\frac{\mathrm d}{\mathrm d t}\right)^2 - \cdots \right) x(t) = \color{blue}{\exp \left( - \frac{\mathrm d}{\mathrm d t} \right) \, x (t)}$$

| improve this answer | |
  • $\begingroup$ +1: Not only continuous. You can use the natural map between bandlimited continuous time signals and discrete time sequences to "push forward" a suitable discrete time derivative operator $D$ from the continuous time derivative operator. This turns out to be the convolution with the derivative of the sinc function. So the corresponding discrete time expression for the time shift operator is $\exp(-D)$ where $D=\mathrm{sinc}'\star$. $\endgroup$ – Jazzmaniac Nov 3 '17 at 12:52

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.