Signal manipulations

Signal manipulations are generally decomposed of a few basic transformations. These transformations may be classified either as those that are transformations of the independent variable $n$ or those that are transformations of the amplitude of $x[n]$, i.e. the dependent variable.

The most common types of amplitude transformations are addition, multiplication and scaling: \begin{eqnarray*} \color{blue}{\textit{Addition}} &:& y[n]=x_1[n]+x_2[n] \newline \color{blue}{\textit{Multiplication}} &:& y[n]=x_1[n] \cdot x_2[n] \newline \color{blue}{\textit{Scaling}} &:& y[n] = c \cdot x[n] \end{eqnarray*} Performing these operations is straightforward and involves only point-wise operations on the signal.

Sequences are often altered and manipulated by modifying the independent function variable $n$, where $f[n]$ is some function of the index $n$. If, for some value of $n$, $f[n]$ is not an integer, $y[n] = x[f[n]]$ is undefined. The most common transformations include shifting (delaying or advancing), reversal, and time-scaling as summarized in the following equations: \begin{eqnarray*} \textit{Time shifting (delay or advance)} &:& f[n]=n-n_0 \newline \textit{Time reversal} &:& f[n]=-n \newline \textit{Time scaling = Down- or Up- sampling} &:& f[n]=M \cdot n \text{ or } f[n]=n/M \end{eqnarray*} Determining the effect of modifying the index $n$ may always be accomplished using a simple tabular approach of listing, for each value of the index $n$, the value of $f[n]$ and then setting $y[n] = x[f[n]]$. However, for many transformations this is not necessary, and the sequence may be determined or plotted directly. Examples of delaying, time-reversal and time scaling, such as down sampling and up-sampling are illustrated in Fig. 1.

Examples of transformation of the function variable.

Note that shifting, reversal, and time scaling operations are order dependent. Therefore, one needs to be careful in evaluating compositions of these operations.

Shifting and reversal are order dependent operations.

Fig. 2 shows an example in which a signal $x[n]$ is first delayed over $n_0$ samples, resulting in $x[n-n_0]$. Reversing this signal, that is replacing the running index $n$ by $-n$, results in the signal $x[-n-n_0]$. Switching the order of these two operations, that is first reverse and then delay results in another signal namely $x[-n+n_0]$.

Finally the delta pulse may be used to decompose an arbitrary signal $x[n]$ into a, possible infinite, sum of weighted and shifted delta pulses: $$ x[n] = \sum_{k=- \infty}^{\infty} x[k] \delta[n-k] $$ Each term in the sum, $x[k] \delta[n-k]$, is a signal that has amplitude $x[k]$ at index $n=k$ and a value of zero for all other values of the index $n$.

Example

Express the signal $$ x[n] = \begin{cases} 3 & n=0 \newline 2 & n=1 \newline 1 & n=2 \newline 0 & \text{elsewhere} \end{cases} $$ as a sum of scaled and shifted unit step functions.

First express $x[n]$ as a sum of weighted and shifted delta pulses: $$ x[n]=3 \delta[n] + 2 \delta[n-1] + \delta[n-2]. $$ With $\delta[n] = u[n] - u[n-1]$ this becomes: \begin{eqnarray*} x[n] &=& 3u[n] - 3u[n-1] + 2u[n-1] - 2u[n-2] + u[n-2] - u[n-3] \newline &=& 3u[n]-u[n-1] -u[n-2] - u[n-3] \end{eqnarray*}

Example

The power in a real valued signal $x[n]$ is defined as the sum of squares of the sample values: $P= \sum_{n=-\infty}^{\infty} x^2[n]$. Suppose that a signal $x[n]$ has an even part $x_e[n]= (\frac{1}{3})^{|n|}$. If the power in $x[n]$ is $P=2$, find the power in the odd part $x_o[n]$ of $x[n]$.

\begin{eqnarray*} P &=& \sum_{n=-\infty}^{\infty} x^2[n] = \sum_{n=-\infty}^{\infty} (x_e[n] + x_o[n])^2[n] \newline &=& \sum_{n=-\infty}^{\infty} x_e^2[n] + \sum_{n=-\infty}^{\infty} x_o^2[n] + 2 \sum_{n=-\infty}^{\infty} x_e[n] \cdot x_o[n]. \end{eqnarray*} The product $x_e[n] \cdot x_o[n]$ is odd and the sum of this odd signal is zero, thus we have: $$ P = \sum_{n=-\infty}^{\infty} x_e^2[n] + \sum_{n=-\infty}^{\infty} x_o^2[n] = P_e + P_o $$ with $$ P_e = \sum_{n=-\infty}^{\infty} (\frac{1}{3})^{2|n|} = -1 + 2 \sum_{n=0}^{\infty} (\frac{1}{3})^{2n} = \frac{5}{4}. $$ From this we obtain the result: $$ P_o = P - P_e = 2 - \frac{5}{4} = \frac{3}{4}. $$

Last updated on Mar 7, 2020