Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have a discrete time signal x[n]. It is said that x[n-k], where K>0, is a delayed version of x[n]. I am trying to understand this intuitively. My observation is in the signal I am subtracting time in x[n-k], by k units. Means I am doing some thing 'quickly' as compared to x[n]. So why do we call it delay instead advance?

share|cite|improve this question
If you let $y_n = x_{n-k}$, then the value of $y_n$ corresponds to the $x$ value at time $n-k$, which is before $n$ (assuming that $k>0$). – copper.hat Feb 12 '13 at 5:51

It is not an advance. You are subtracting k units from n. So if you where just calculating $x[n]$, you will be $k$ units ahead of the calculation of $x[n-k]$.

If $k = 5, n = 10$: $$ \begin{align*} &x[n] = x[10] \\ &x[n-k] = x[10-5] = x[5] \end{align*} $$

So the calculation for $x[n-k]$ lags $x[n]$ by $k = 5$ time units

share|cite|improve this answer

I hope the given answer is not satisfactory since delay means shift rightwards on time scale. But according to the explanation it's appearing at left of the x[n] sample. You just see like: n-k=0. i.e. n=k means the initiation time for signal. Practically you think the time when your class (for a student) starts as t=0 time and just assume you reached 10 minutes late. So it's 10 minutes delay on time scale. So for you the class initiation time (t=0) is t=10 or t-10=0. I hope so you understood the concept.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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