I'm reading Schaum's DSP book, and in Fig 1-3 they demonstrate why shifting and reversal are order-dependent, showing a couple of simple systems, 1) delay followed by reversal, and 2) reversal followed by delay.

enter image description here

The equations they present as follows:

1) (Time delay -> Reversal)

$x(n) \rightarrow x(n-n_0) \rightarrow x(-n-n_0)$

2) (Reversal -> Time delay)

$x(n) \rightarrow x(-n) \rightarrow x(-n+n_0)$

I'm fine with shifting and reversal not being commutative due to the different results above, but shouldn't the results be opposite? I.e.:

1) (Time delay -> Reversal)

$x(n) \rightarrow x(n-n_0) \rightarrow x(-(n-n_0)) = x(-n + n_0)$

2) (Reversal -> Time delay)

$x(n) \rightarrow x(-n) \rightarrow x(-n - n_0)$

I didn't find this in the errata for the book which is why I'm thoroughly confused. Which is correct?

  • $\begingroup$ Looks like you are right and the book is wrong. $\endgroup$ – Hilmar Jul 22 '13 at 12:06
  • $\begingroup$ @Hilmar Errr no, the book is correct. The OP is confusing two different meanings on $n$. $\endgroup$ – Dilip Sarwate Jul 22 '13 at 13:41

Your counterexample to the book's assertion is confusing between two different uses for $n$. There was a question earlier in which some user (endolith? datageist?) gave an answer containing a detailed description of what exactly this confusion is and how to interpret the results correctly. My cursory search has not found this great answer, and so I will just exhibit a simple example.

Suppose that $x[n] = \begin{cases}1, &n=0,\\2, & n = 1,\\0, &\text{otherwise,}\end{cases}$

  • Delaying $x[n]$ by one unit gives $x_d[n] = \begin{cases}1, &n=1,\\2, & n = 2,\\0, &\text{otherwise,}\end{cases}$

and time-reversing this gives $x_{d,r}[n] = \begin{cases}1, &n=-1,\\2, & n = -2,\\0, &\text{otherwise,}\end{cases}$ which equals $x[-n-1]$ exactly as the book says it does. Don't believe this? Here is a table of values for the two functions: $$\begin{array}{|l|c|c|c|c|c|c|c|c|c|} \hline\\n&\cdots&-3&-2&-1&0&1&2&3&\cdots\\ \hline\\ x[n]&0&0&0&0&1&2&0&0&0\\ \hline\\x_{d,r}[n]&0&0&2&1&0&0&0&0&0\\ \hline\\ -n-1&\cdots&2&1&0&-1&-2&-3&-4&\cdots\\ \hline\\ x[-n-1]&\cdots&0&2&1&0&0&0&0\\ \hline\\ \end{array}$$

OK, thus far? On the other hand,

  • Time-reversing $x[n]$ gives $x_r[n] = \begin{cases}1, &n=0,\\2, & n = -1,\\0, &\text{otherwise,}\end{cases}$
    and delaying this by one time unit gives $x_{r,d}[n] = \begin{cases}2, & n = 0,\\1, &n=1,\\0, &\text{otherwise,}\end{cases}$
    which equals $x[-n+1]$ exactly as the book says it does. Don't believe this either? Here is the table for it. $$\begin{array}{|l|c|c|c|c|c|c|c|c|c|} \hline\\n&\cdots&-3&-2&-1&0&1&2&3&\cdots\\ \hline\\ x[n]&0&0&0&0&1&2&0&0&0\\ \hline\\x_{r,d}[n]&0&0&0&0&2&1&0&0&0\\ \hline\\ -n+1&\cdots&4&3&2&1&0&1&2&\cdots\\ \hline\\ x[-n+1]&0&0&0&0&2&1&0&0&0\\ \hline\\ \end{array}$$

So how can we establish these results without setting up tables?

Well, $x_d$ is a sequence whose $n$-th term $x_d[n]$ equals $x[n-n_0]$ for all choices of $n$; in other words, take whatever is inside the square brackets after $x_d$, subtract $n_0$ from it, and stick it in as the argument/index for $x$. Now, for any sequence $y$, its time-reversal $y_r$ is a sequence for which $y_r[k] = y[-k]$ for all choices of $k$ (negate the argument of $y_r$ and stick it in as the argument for $y$. Thus, the time-reversal of $x_d$ is given by

$$x_{d,r}[k] = (x_d)_r[k] = x_d[-k] = x[(-k)-n_0] = x[-k-n_0]$$ as the book says, and not $x[-k+n_0]$ as the OP claims it should be.

Similarly, $x_r[n] = x[-n]$ and so $$x_{r,d}[k] = (x_r)_d[k] = x_r[k-n_0] = x[-k+n_0]$$ as the book says it should be, and not $x[-k-n_0]$ as the OP claims it should be.

  • 1
    $\begingroup$ Thanks Dilip for the good examples! They make perfect sense. Is there a way of reaching the same conclusion algebraically? Clearly my attempt was wrong; assuming I can not include n0 the way I did, what's the correct way of going about it? $\endgroup$ – jodles Jul 22 '13 at 18:35
  • $\begingroup$ @jodles See the addendum to my answer. $\endgroup$ – Dilip Sarwate Jul 22 '13 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.