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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?

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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

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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.

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