# Discrete Time Signals - Time Scaling and Time Reversal

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?

## 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).

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}