say the signal $x(n)$ has the z transform $X(z)$ and there is signal $x_1(n)$ that


I tried 2 different approach to get the relationship between $x(n)$ and $x_1(n)$ and the result is different without any error being noticed

Approach 1:

$x_i(n)=x(-n) \rightarrow X_i(z) = X(z^{-1})$

$x_1(n)=x_i(n+1) = x(-n-1) \rightarrow X_1(z) = zX_i(z)=zX(z^{-1})$

Approach 2:

$x_i(n)=x(n-1) \rightarrow X_i(z) = z^{-1}X(z)$

$x_1(n)=x_i(-n)=x(1-n) \rightarrow X_1(z) = X_i(z^{-1}) = zX(z^{-1})$

Hence in the end, I don't know $x_1(n) = x(-n-1)$ or $x_1(n) =x(1-n)$ and I can't find any mistake. I know that different function can have same z-transform but different ROC but in this case, no any info of ROC is given


1 Answer 1


The first result is correct, the second is wrong. You're right that $x_i[n]=x[n-1]$ corresponds to $z^{-1}X(z)$. And if you replace $z$ by $1/z$ you need to replace $n$ by $-n$. But $x_i[-n]$ is simply $x[-n-1]$, and not $x[-n+1]$. Note that you don't replace the whole argument of $x[n-1]$ by its negative value, but you just replace $n$ by $-n$.

  • $\begingroup$ can you explain further, for $x_i[n] = x[n-1]$ then isn't $x_i[-n]=x[-(n-1)]=x[-n+1]$. I don't get why it's wrong because I apply the same rule for approach 1, that is $x_i[n]=x[-n]$ so $x_i[n+1]=x[-(n+1)].=x[-n-1]$? $\endgroup$
    – aukxn
    Nov 18, 2015 at 14:50
  • $\begingroup$ @aukxn: In the first case, you replace $n$ by $n+1$, so $-n$ becomes $-(n+1)=-n-1$. In the second case you replace $n$ by $-n$, so $n-1$ becomes $-n-1$. It's just straightforward substitution. $\endgroup$
    – Matt L.
    Nov 18, 2015 at 15:05
  • $\begingroup$ So in some case I can't replace entirely the time index like what I did in 1st case. That's strange, is there any rule for what case the replace must take entirely and what is not? I was taught that for time reversal, any shifting in input in one direction result in shifting in opposing direction of output but what you just said is new to me. $\endgroup$
    – aukxn
    Nov 18, 2015 at 15:15
  • $\begingroup$ @aukxn: There is no strange rule, it's just plain substitution. Please re-read my previous comment, it's really straightforward. In the first case you replace $n$ by $n+1$, in the second $n$ by $-n$. If you do that you get the correct result. $\endgroup$
    – Matt L.
    Nov 18, 2015 at 16:46

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