On the top graph, we can see a discrete-time signal $x[n]$.

  1. I don't understand how for the signal $x[3-n]$, the impulses with magnitude $1$ still are at the positive indices $n = 1, 2, 3, 4$. Why are they not at negative indices side since we have no positive $n$ this time, but have negative $n$ (see Figure 1.22(b)).

  2. Why when we take $x[3n]$ most of impulses are gone? (Figure 1.22(c))

Are the operations like time-scaling, time-reversal, time-shifting, on discrete time signals different?

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1 Answer 1


1) Reversal and time advance

$x[3-n]$ combines two operations:

  • Time reversal: $x[n] \rightarrow x[-n]$:
  • Time advance: $x[-n] \rightarrow x[-n +\tt{3}]$

Look at $x[n]$ and imagine flipping it around $n=\tt{0}$. That's time reversal. The sample at $n=\tt{-4}$, with magnitude $-1$, is now at index $n = \tt{4}$. Same goes for sample at $n=\tt{-3}$, with magnitude $-1/2$, which is now at index $n = \tt{3}$. Now advance this signal by $\tt{3}$ samples: that sample at $n=\tt{4}$ with magnitude $-1$ is now at $n=\tt{4+3} = 7$. That sample at $n=\tt{3}$ with magnitude $-1/2$ is now at $n=\tt{3+3} = 6$, etc

Write it out

You can do this for every index: let's write out the values for $x[n]$ where it is defined, i.e. for $-5 \leq n \leq 5$:

\begin{align} &n &\quad &x[n]\\ \hline -&\tt{5} &\quad &0\\ -&\tt{4} &\quad -&1\\ -&\tt{3} &\quad -&1/2\\ -&\tt{2} &\quad &1/2\\ -&\tt{1} &\quad &1\\ &\tt{0} &\quad &1\\ &\tt{1} &\quad &1\\ &\tt{2} &\quad &1\\ &\tt{3} &\quad &1/2\\ &\tt{4} &\quad &0\\ &\tt{5} &\quad &0\\ \end{align}

Now let's look at $x[3-n]$. I'll start at $n = -5$, even though for that value, $x[3-n] = x[8]$ is $\text{undefined}$: \begin{align} &n &\quad &3-n &\quad &x[3-n]\\ \hline -&\tt{5} &\quad &\tt{3} - \tt{(-5) = 8}&\quad &x[8] \text{ is undefined}\\ -&\tt{4} &\quad &\tt{3} - \tt{(-4) = 7}&\quad &\text{undefined}\\ -&\tt{3} &\quad &\tt{3} - \tt{(-3) = 6}&\quad &\text{undefined}\\ -&\tt{2} &\quad &\tt{3} - \tt{(-2) = 5}&\quad &0\\ -&\tt{1} &\quad &\tt{4}&\quad &0\\ &\tt{0} &\quad &\tt{3}&\quad &1/2\\ &\tt{1} &\quad &\tt{2}&\quad &1\\ &\tt{2} &\quad &\tt{1} &\quad &1\\ &\tt{3} &\quad &\tt{0}&\quad &1\\ &\tt{4} &\quad &\tt{-1}&\quad &1\\ &\tt{5} &\quad &\tt{-2}&\quad &1/2\\ &\tt{6} &\quad &\tt{-3}&\quad -&1/2\\ &\tt{7} &\quad &\tt{-4}&\quad -&1\\ &\tt{8} &\quad &\tt{-5}&\quad &0\\ \end{align}

2) Time Scaling

$x[Kn]$ is called decimation by the scaling factor $K>1$ (it's called expansion if $0<K<1$).

In your case, you therefore have time decimation by scaling factor $3$: $x[3n]$
This operation only keeps every 3 sample.

Let's write it out. Again, I'll start at $n = -5$, even though for that value, $x[3n] = x[-15]$ is $\text{undefined}$:

\begin{align} &n &\quad &3n &\quad &x[3n]\\ \hline -&\tt{5} &\quad &\tt{3} \times \tt{(-5) = -15}&\quad &x[-15]\text{ is undefined}\\ -&\tt{4} &\quad &\tt{3} \times \tt{(-4) = -12}&\quad &\text{undefined}\\ -&\tt{3} &\quad &\tt{3} \times\tt{(-3) = -9}&\quad &\text{undefined}\\ -&\tt{2} &\quad &\tt{3} \times\tt{(-2) = -6}&\quad &\text{undefined}\\ -&\tt{1} &\quad -&\tt{3}&\quad -&1/2\\ &\tt{0} &\quad &\tt{0}&\quad &1\\ &\tt{1} &\quad &\tt{3}&\quad &1/2\\ &\tt{2} &\quad &\tt{6} &\quad &\text{undefined}\\ &\tt{3} &\quad &\tt{9}&\quad &\text{undefined}\\ &\tt{4} &\quad &\tt{12}&\quad &\text{undefined}\\ &\tt{5} &\quad &\tt{15}&\quad &\text{undefined}\\ \end{align}


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