Basic Operations
Introduction Energy and Power Basic Operations Practice Problems Transformation of signals defined piecewise Even and Odd Signals Commonly encountered signals

A variety of operations can be carried out on signals to obtain new signals. These operations can be classified into two categories - operations that are performed on the dependent variable and operations that can be performed on the independent variable. Recall that for a signal $x(t)$, $t$ is the independent variable and $x(t)$ is the dependent variable.

Operations performed on the dependent variable

Amplitude Scaling: Signals are often amplified or attenuated, i.e., the amplitude of the signal is scaled. Mathematically, this corresponds to multiplying the signal $x(t)$ by a (possibly complex constant) $c$ to obtain a new signal, say $y(t)$ given by

\begin{align} \hbox{For CT signals}: & y(t) = c \ x(t) \nonumber \\ \hbox{For DT signals}: & y[n] = c \ x[n] \end{align}

Addition, Subtraction, Multiplication, and Division
Take two signals $x_1(t)$ and $x_2(t)$. We can add, subtract, multiply and divide these signals. These operations are performed for every value of $t$. This is similar to performing operations on functions of $t$. Thus we can obtain a new signal $y(t)$ from two signals $x_1(t)$ and $x_2(t)$ according to,

\begin{align} \hbox{Addition:} & y(t)=x_1(t)+x_2(t) \\ \hbox{Subtraction:}& y(t)=x_1(t)-x_2(t) \\ \hbox{Multiplication:}&y(t)=x_1(t).x_2(t) \\ \hbox{Division:}&y(t)=x_1(t)/x_2(t) \end{align}

Derivatives and Integrals of CT signals

For continuous-time signals (CT) signals, we define the derivative and integral of the signal as follows. It is important to note the notation used in defining the integral. Notice that the variable of integration is $\tau$ and $t$ appears in the limit of the integral making the result of the integral a function of $t$.

\begin{align} \hbox{Derivative:} & y(t)=\frac{d}{dt}x(t) \\ \hbox{Integral:} & y(t)=\int_{-\infty}^{t}x(\tau)d\tau. \end{align}

Difference and Partial sums of DT signals
For discrete-time signals, the operation of derivative and integrals are not well defined. The equivalent notion of derivative and integral for DT signals are given by backward difference and partial sum as given below

\begin{align} \hbox{Backward difference:} & y[n]=x[n]-x[n-1] \\ \hbox{Partial sum:} & y[n] = \sum_{m=-\infty}^{n} x[m]. \end{align}

Operations performed on the Independent Variable

The various operations are:
1. Time Scaling:

Given a signal $x(t)$,Time scaling is defined as-

\begin{equation} y(t)=x(at) \end{equation}
  • If $a > 1$,

Let $\hspace{0.5cm}a=2$

\begin{equation} y(t)=x(2t) \end{equation}
\begin{equation} y(0)=x(0),y(0.5)=x(2*0.5)=x(1),y(1)=x(2)... \end{equation}
  • If $a < 1$,
\begin{align} Let \hspace{0.5cm} a=0.5 \end{align}
\begin{equation} y(0)=x(0),y(1)=x(0.5),y(2)=x(1) \end{equation}

Suppose we have a DT signal $x[n]$,

\begin{align} y[n]=x\bigg[\frac{n}{2}\bigg] \nonumber \end{align}

2. Reflection:

Given a signal $x(t)$,Time scaling is defined as-

\begin{equation} y(t)=x(-t) \end{equation}

3. Time Shifting:

Given a signal $x(t)$,Time shifting is defined as-

\begin{equation} y(t)=x(t-t_0) \end{equation}
If $t_0 > 0 \Rightarrow x(t-t_0)$ is a right shift and if $t_0$ < 0 $\Rightarrow x(t-t_0)$ is a left shift.
Example: Cellular Communication System

For a DT signal $x[n]$,

\begin{align} y[n]=x[n-n_0] \nonumber \end{align}

If $n_0 > 0 \Rightarrow x[n-n_0]$ is a right shift and if $n_0 < 0 \Rightarrow x[n-n_0]$ is a left shift.

4. Multiple Operations on a Signal:

  • Given a signal $x(t)$, $y(t)=x(at-b)$
  • Given a signal $x(t)$, $y(t)=x\bigg(\frac{t-b}{a}\bigg)$
  • Given a signal $x(t)$, $y(t)=x(-at-b)$


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