# Time shifting and time reversal

For the problem given in the fig. how to solve it?

My approach:

Let $y(n)$ be the first output (e.g for 1st case output of system $A$ is $y_1(n)$; for 2nd case output of system $B$ is $y_2(n)$, then $$y_1(n)=x_1(-n)$$ $$V_1(n)=y_1(n+1)=x_1(-n+1)$$

And for the 2nd case:

$$y_2(n)=x_2(n+1)$$ $$V_2(n)=y_2(-n)=x_2(-n-1)$$

So if $x_1(n)=x_2(n)=x(n)$, then $$V_1(n)=x(-n+1)$$ $$V_2(n)= x(-n-1)$$

So the answer should be 'C', but the given answer is 'B'.

What is the mistake and how should I proceed?

• What if for all values of n, x1 = x2 = 0? Jan 24 '18 at 16:35

First of all, since the system A is not LTI, then the ordering of systems is critical in determining the output, and you can weakly conclude that $V_1[n] \neq V_2[n]$ in general, but is it so for all possible inputs? The following is true: if the orders of $N$ LTI systems are interchanged, the resulting outputs would be the same for all possible inputs. But the following is not (necessarily) true: if at least one system were not LTI, then outputs corresponding to different orderings must be different for all possible inputs... It's not true because there can exist some inputs for which the coprresponding outputs could be same in different orderings...
You have solved the two outputs correctly and reached the condition that $V_1[n]=x[-n+1]$ and $V_2[n]=x[-n-1]$. Can you find some inputs $x[n]$ whose transform under these two systems might yield the same output? i.,e., are there any $x[n]$ for which $x[-n-1] = x[-n+1]$ for all $n$ ?
As @loudnoises has exemplified a trivial input of the form $x[n] = 0$ for all $n$ provides a single example to the option b. Indeed all inputs of the form $x[n] = K$ (the constant signal) satisfy the relation that $x[-n+1] = x[-n-1]$ hence there exists infinitely many (for each different K) inputs of the form $x[n]=K$ for which $V_1[n] = V_2[n]$ .