The other way to make an infinite-length signal from a finite-length one is to
periodize it, which means replicating a finite-length signal over and over to create an infinite-length periodic version. Mathematically, that means defining the new infinite-length periodic signal like this:
\begin{eqnarray*}
y[n]&=&\sum_{m=-\infty}^{\infty} x[n-mN], \quad n\in\Integers \\&=&\cdots + x[n+2N] + x[n+N]+ x[n] + x[n-N]+ x[n-2N] + \cdots\end{eqnarray*}
Graphically, we can see that this amounts to repeating the signal over and over, before and after the original portion:
Finite-length signalOriginal signal periodized to create an infinite-length signal.(a) finite-length signal
is periodized to create (b) an infinite-length signal
(only a portion of it is shown).
Periodization and modular arithmetic
It turns out that, as we consider periodization and periodic signals, the notion of modular arithmetic will be helpful. In modular arithetic, integers do not lie on a line stretching from negative infinity to infinity, but rather on a circle of a defined size $N$. In modulo-8, for example, the numbers are 8 "hours" on a "clock." Our convention will be for the numbers to traverse from 0 to $N-1$, counterclockwise:
CAPTION.
Consider a finite-length signal of size $N$. We can align the time-dependent values of the signal on the modulo circle:
CAPTION. When we travel around the clock once, from time index 0 to 7, we express the finite-length signal. But if we keep traveling, in one direction or the other, then that amounts to periodizing the signal. Using our modulo notation, we can periodize a finite-length signal
to be an infinite-length periodic signal
like this: $y[n]=x[(n)_N]$.
Finite/periodic signals relationship
We have seen that we can take an $N$-length finite-length signal and periodize it to make an infinite-length periodic signal with a period of $N$. By the same token, we can also work in reverse and extract one period worth of signal from any periodic signal to create a finite-length signal.
What this means is we can consider periodic signals and finite-length signals to be essentially equivalent : we can consider just one period of a periodic signal (the rest of the signal is redundant, by definition), or periodize a finite-length signal. They are two ways of looking at the same thing, a phenomenon we will often see in our study of signals and systems, and we will choose the perspective that best suits our needs for particular applications.
Shifting infinite-length signals
Given some signal $x[n]$, it will often be necessary for us to consider that signal shifted in time. We denote such a shift mathematically with an expression like $x[n-m]$, where $m$ is some integer. If $m$ is greater than zero, then $x[n-m]$ will be just like $x[n]$, except it will be shifted to the right by $m$ time units. If $m$ is less than zero, it will be shifted to the left. Here is who that might look for a couple values of $m$:
Original signal $x[n]$.$x[n]$ shifted to the right.$x[n]$ shifted to the left.A (a) signal $x[n]$ shifted according to the expression $x[n-m]$ where $m$ is (b) positive, and (c) negative. This type of shifting works the same with periodic signals:
Original periodic signal $y[n]$.$y[n]$ shifted to the right.A periodic signal shifted one value in time. Of course, for periodic signals, certain shifts actually do not have any effect on them. If a signal repeats with a period of $N$, then shifting that signal by any integer multiple of $N$ will yield the original signal. Take a look at the signal above, which was shifted to the right by a time unit of 1. If we keep on shifting it until it is shifted 8 time units the result will be identical to the original signal.
Shifting finite-length signals
Can finite-length signals be shifted, as well? There does not seem to be any reason why not. Suppose we have some finite-length signal $x[n]$ (of length $N$) and we define another finite-length signal $v[n]$ to be $v[n]=x[n-1]$. So we have $v[1]=x[0]$ and $v[2]=x[1]$ and so on until $v[N-1]=x[N-2]$. But what about $v[0]$, what do we put there? And how about $x[N-1]$, where is that supposed to go? We do not want to invent information to put in $v[0]$, nor lose the information of $x[N-1]$. An elegant solution is to periodize $x[n]$, and then consider the relation $v[n]=x[n-1]$. In this case, we now have a value for $v[0]$: $v[0]=x[-1]$. Since $x[n]$ is periodic with period $N$, it also happens that $x[-1]=x[N-1]$, so we do not lose that information.
This kind of operation, for finite-length signals, is called a
circular shift , and we can express it mathematically with the help of our modular arithmetic operator. Circularly shifting a finite-length signal $x[n]$ by $m$ time units is expressed as $x[(n-m)_N]$. It can also be visualized by turning $x[n]$ about the circle on which it resides:
Original finite-length signal $x[n]$.$x[(n-3)_8]$Circularly shifting a signal by $m$ amounts to turning it counter-clockwise $m$ steps.
Time reversing finite-length signals
For infinite length signals, the transformation of reversing the time axis $x[−n]$ is obvious: just flip the signal about $n=0$. But things are not quite so obvious for finite-length signals; if a signal is defined for, say, $n$ between 0 and N, then what gets flipped across the $n=0$ from the negative side? Once again, it turns out the modular arithmetic opererator can be called in for help. We reverse the time axis, modulo N: $x[(-n)_N]$. Below is an image of a finite-length ($N=8$) signal, time-reversed:
We can time reverse a finite-length signal $x[n]$ with the mathematical expression $x[(-n)]_8$.
Signal causality
A signal $x[n]$ is
causal if $x[n]=0$ for all $n \lt 0$, it is
anti-causal if $x[n]=0$ for all $n \geq 0$, and it is
acausal if is neither causal nor anti-causal.
A causal signal.An anti-causal signal.
Even and odd signals
A signal $x[n]$is defined as
even if $x[-n]=x[n]$, and
odd if $x[-n]=-x[n]$.
An even signal.An odd signal.
Even/odd signal decomposition
MM: Is it really necessary to include this? Usage of the concept (wavelets?) is outside the scope of the class.
Digital signals
Digital signals are a special sub-class of discrete-time signals. While the independent time variable for discrete-time signals is integer-valued, the dependent variable (i.e., the value the signal takes at any given time) can take on any value. However, for digital signals, both of these variables are discrete-valued. Rather than take any value on a continuum, discrete signals take only a limited number of values, or
levels . Typically, the number of levels is expressed as $D = 2^q$, and each possible value of $x[n]$ is represented as a digital code with $q$ bits.
A digital signal with $q = 2$ bits, so $D = 2^2= 4$ levels.
