I am reading book signal processing first Chapter 5

I am reading article of time invariance

Where author mentions example of a flip system as shown in attached photo ,i have drawn a red curve above confusion equation

I am confused, how he is doing this when he say"if we delay input and then reverse the order"

I think he is not properly reversing the order,as he reversed the sign of only "n" but he didn't reversed the sign of "n0"

Why this discrepancy in reversal/flipping?

enter image description here


The book is correct, there is no discrepancy. When we reverse a system in time, only the time-variable will get negated and not the shift. Time-reversal does not mean that the whole argument of $x[n]$ gets negated.

Take example of a sequence : $x[n] = {\hat{0},1,2,3,4,5}$. Shift this by 2 samples, so $x_{shift}[n] = {\hat{0},0,0,1,2,3,4,5}$. Now, reverse the time, what do you think should be expected sequence : $x_{final}[n] = 5,4,3,2,1,0,0, \hat{0}$.

You will get it only when $x_{final}[n] = x[-n-2]$ and not $x[-n+2]$. Check it out. Whatever operation happens, time scaling, shifting and reversal, it happens on n and not on the complete argument.

That is why when the author, reverses the time first and then delays, the shift gets negated too.

Time-reversal : $x_{reverted}[n] = x[-n]$ and then shift of $n_o$, so again shift will happen to $n$ and not to $-n$, and hence, $x_{final}[n] = x[-(n-n_o)]$.

Lets take the same sequence as an example : $x[n] = \hat{0},1,2,3,4,5$.

Revert time first : $x_{reverted}[n] = 5,4,3,2,1,\hat{0}$

Now, when delay the sequence by 2 samples, the expected sequence should look like : $x_{final}[n] = 5,4,3,\hat{2},1,0$. Check that this would happen iff $x_{final}[n] = x[-n+2]$ and not if $x_{final}[n] = x[-n-2]$. Just the put the values of n and it will be clear.

It always helps to take an example sequence and do the shifting, delaying and reversal on that sequence and then see what should you do to the argument on $x$ to get he final expected sequence.


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