1
$\begingroup$

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?

enter image description here

$\endgroup$

1 Answer 1

1
$\begingroup$

1) Reversal and time advance

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

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

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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