Combined signal folding and shifting

Let $g[n] = f[-n]$(n is the time index). Then what will be expression for $g[n+k]$.

Is $g[n+k] = f[-n + k]$ or $g[n+k] = f[-n - k]$. Which one is true and Why?

• This isn't really a signal-processing question at all; instead, this is a simple algebraic substitution problem. Sep 30 '13 at 13:09
• @JasonR Yes, it is a simple algebraic substitution problem, but far too many people get it wrong because of poor notation which leads to wrong understanding of the material. See, for example, this question Sep 30 '13 at 16:48

2 Answers

The statement $g[n]=f[-n]$ is meaningless unless you assign some numerical value to $n$, e.g. $g[3]=f[-3]$, and if you do say that $n=3$, then the equality $g[3] = f[-3]$ tells you nothing about the relationship between $g[n+k]=g[3+k]$ to the value taken on $f[\cdot]$ for any choice of argument.

What you really meant to say is that

$$\text{For all integers } n, g[n]=f[-n]$$

which asserts that the equality holds for all choices of $n$, and in this case, it follows by definition that $g[n+k] = f[-n-k]$; you just need to negate the argument $n+k$ of $g$ (to get $-n-k$) and use it (it meaning $-n-k$) as the argument of $f$ exactly as the definition says you should. The complicated arguments in another answer that claim that $g[n+k]=f[-n+k]$ are wrong, as discussed in the comments following that answer.

The answer is f[-n+k].

I'll try to explain this with example. Let the signals be discrete & k=2. The bold numbers indicate sample at time 0(I am not well versed with editing here)

Let f[n] = {1,1,1,1}
Thus f[-n] = {1,1,1,1} = g[n]
Now g[n+2] = {1,1,1,1,0,0}

Now how can you obtain the same answer from f[n]
f[n-2] = {0,0,1,1,1,1}
And if we invert this signal ie f[-(n-2)], we get {1,1,1,1,0,0} which is g[n+2].
Thus g[n+2] = f[-(n-2)] = f[-n+2]
Hence proved.
The theoretical answer are the rules followed in Shift & Scale
y[n] = x[an+b]
If you shift first & then scale, then
1) y[n] = x[n+k]
2) Then scale y[n] by a

If you scale first & then shift, then
1) y[n] = x[a(n+(b/a))] ie scale x[n] by a first
2) Shift y[n] by (b/a)

• No, this is wrong. The definition $g[n] = f[-n]$ means that to find the value of $g$ at any time index, take the negative of whatever appears in the square brackets following $g$, and use that in the square brackets following $f$. Thus, $g[n+2] = f[-n-2]$, not $g[n+2]=f[-n+2]$ as KharoBangdo claims. Note that $g[n+2]=f[-n+2]$ implies, upon setting $n=0$ that $g[2]=f[2]$ which is patently false; by definition, $g[2] = f[-2]$. Sep 30 '13 at 13:00
• @DilipSarwate So what did i do wrong in my example? Sep 30 '13 at 14:28
• Denote the signal $\{1,1,1,1,0,\mathbf 0\}$ as $h[\cdot]$, that is, $h[-2]=1, h[-1]=0, h[0] = 0$, etc. Thus, for all integers $n$, $h[n] = g[n+2]$. Denote the signal $\{\mathbf 0, 0, 1,1,1,1\}$ as $e[\cdot]$ so that $e[0] = 1$, $e[1]=0$, $e[2]=1$, etc., that is, for all integers $m$, whether positive or negative, $e[m] = f[m-2]$. In particular, we can substitute any number, even a negative number such as $(-n)$, as the argument $m$ for $e[\cdot]$ and find $e[(-n)] = f[(-n)-2] = f[-n-2]$. Then, we have that for all choices of $n$, $$g[n+2] = h[n] = e[-n] = f[-n-2].$$ Sep 30 '13 at 20:53
• That should have been $e[0] = 0$, not $e[0]=1$. Sep 30 '13 at 21:04