# Time-reversal and delay commutativity and resulting signals

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.

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?

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

• 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? – jodles Jul 22 '13 at 18:35
• @jodles See the addendum to my answer. – Dilip Sarwate Jul 22 '13 at 22:28