Energy and Power of Signals

Introduction

Energy and Power

Basic Operations

Practice Problems

Transformation of signals defined piecewise

Even and Odd Signals

Commonly encountered signals

Motivating question

Consider a simple circuit where a DC voltage source with $v$ volts is connected to a 1 Ohm resistor as shown in the figure in the left panel. How much power is dissipated by the resistor? The power dissipated is given by $v^2/R = v^2$ watts or joules/sec. How much energy is dissipated in the resistor over a 1 minute time duration? Since energy is the integral of power, the energy dissipated is $v^2 \times 60 = 60 v^2$ Joules.

Now consider the same circuit but with a voltage source whose voltage varies with time as shown in the panel on the right, i.e., the voltage at time $t$ is $x(t)$. Let us now consider the question of how much energy is dissipated from the resistor over the entire time interval from $-\infty,\infty$. At any given time $t$, the (instantaneous) power is given by $x^2(t)$ and the overall energy is given by $\int_{-\infty}^{\infty} x^2(t) \ dt$. Notice that we said that the power dissipated at time $t$ is $x^2(t)$, but can we define one value for the power dissipated when the voltage source is $\underline{x}(t)$? This would represent the average energy dissipated per unit time when the voltage signal is $x(t)$. Indeed, we define such a quantity below. We refer to the energy and power dissipated by the resistor as the energy and power associated with the voltage or signal $x(t)$.

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Definition of Energy and Power

The energy and power of a CT signal $\underline{x}(t)$ and DT signal $\underline{x}[n]$ are defined as

(1)

\begin{align} \hbox{Energy of a CT signal} : & E_x=\int_{-\infty}^{\infty} |x(t)|^2 \ dt \\ \hbox{Energy of a DT signal} : & E_x=\sum_{n=-\infty}^{\infty} |x[n]|^2 \\ \hbox{Power of a CT signal} : & P_x= \lim_{T \rightarrow \infty} \frac{1}{2T} \int_{-T}^{T} |x(t)|^2 \ dt \\ \hbox{Power of a DT signal} : & P_x=\lim_{N \rightarrow \infty} \frac{1}{2N+1} \sum_{n=-\infty}^{\infty} |x[n]|^2 \\ \end{align}

These defintions apply to both real and complex signals $x(t)$ and $x[n]$.

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Power of periodic signals

Consider a periodic CT signal $\underline{x}(t)$ with time period $T_0$ such as the example shown in the figure below.

For such a signal, the energy of the signal given by $\int_{-\infty}^{\infty} |x(t)|^2 dt$ is infinite. Since the power is the average energy
per period, the power is given by

(2)

\begin{align} \hbox{Power}: P_{x}=\frac{1}{T_0} \int_{t_0}^{t_0+T_0} |x(t)|^2 dt. \end{align}

where $t_0$ is any arbitrary time instant starting from which we measure the time period. Since the signal is periodic, $t_0$ can be arbitrary i.e., regardless of which time interval we choose to measure the energy over, as long as we are measuring the energy over a time interval equal to one time period, the result is identical since the signal is periodic.

If $x(t)$ is a periodic DT signal with time period $N_0$, the power of a periodic signal is defined as

(3)

\begin{align} \hbox{Power}: P_x=\frac{1}{N_0} \sum_{n_0}^{n_0+N_0-1} |x[n]|^2 \end{align}

Just like in the CT case, $n_0$ can be arbitrary and the choice of $n_0$ does not affect the result.

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Energy as the strength of a signal

Even though we used the circuit example as a motivation to define the energy of a signal, the definition of energy is not confined only to signals which can be interpreted as a voltage waveform. Rather, the energy of a signal can be used as a measure of the strength of a signal. Often, we encounter situations where we would like to measure the strength of a signal or compare the strengths of two signals and the energy of the signal provides a quantitative measure of the strength of the signal. The above definition of energy to measure the strength of a signal is indeed only one of many possible choices and there are other ways to define the energy or strength of a signal. For example, one can look at the maximum value taken by the signal as one measure of strength, one can look at the sum of the absolute values of a DT signal as another choice. All these measures are meaningful and depending on the decision that we would like to make, we must choose the measure. The energy of a signal defined as in (1) is commonly used and in this course, this will be our default definition of energy. The definition of energy is closely related to what is called in mathematics as the norm of a vector

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Energy type and power type signals

1. $\underline{x}(t)$ is an energy type signal if $0<E_x<\infty$
2. $\underline{x}(t)$ is a power type signal if $0<P_x<\infty$

Clearly, for any periodic signal $E_x$ is not bounded and hence, periodic signals cannot be energy type signals. If the energy within one period of the signal is bounded, then the power will be bounded and hence, such signals will be power type signals.

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Example Problems

Example 1: Consider the signal given below. Is this power or energy type signal?

Solution

(4)

\begin{align} x(t)= \begin{cases} 3 \left(1-\frac{t}{4}\right) & \text{if} \ 0 \leq t \leq 4 \nonumber \\ 3 \left(1+\frac{t}{4}\right) & \text{if} \ -4\leq t \leq 0 \nonumber\\ 0 & \text{otherwise} \nonumber \end{cases} \end{align}

Therefore,

(5)

\begin{aligned} E_x & = \int_{-4}^{4} |x(t)|^2 dt \\ & = \int_{-4}^{4} x^2(t) dt \nonumber \\ & = 9\int_{-4}^{0} (1+\frac{t}{4}))^2 dt + 9\int_{0}^{4} (1-\frac{t}{4}))^2 dt \nonumber \\ & = 9\frac{(1+\frac{t}{4})^3}{\frac{3}{4}} \Bigg|_ {-4}^0 + 9\frac{(1-\frac{t}{4})^3}{\frac{-3}{4}} \Bigg|_ {0}^4 \nonumber \\ & = 9 \frac{4}{3} + 9 \frac{4}{3} \nonumber \\ & = 24 \\ \end{aligned}

So, $x(t)$ is an Energy Signal.

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Example 2: Let $x(t) = A \cos\left(\omega_0t+ \theta\right)$? Is this a power signal or energy signal?

Solution

$E_x$ is unbounded. So,

(6)

\begin{aligned} P_x & =\frac{1}{T_0} \int_{0}^{\frac{2\pi}{\omega_0}} A^2 cos^2(\omega_0t+ \theta)dt \nonumber \\ & = \frac{1}{T_0} \int_{\frac{-\theta}{\omega_0}}^{\frac{-\theta+2\pi}{\omega_0}} A^2 cos^2(\omega_0t+ \theta) dt \nonumber \\ & = \frac{A^2}{T_0} \int_{\frac{-\theta}{\omega_0}}^{\frac{-\theta+2\pi}{\omega_0}} \frac{1+cos(2\omega_0t+2\theta)}{2}dt \nonumber\\ & = \frac{A^2}{T_0} \bigg[ \frac{t}{2} + \frac{sin(2\omega_0t+2\theta)}{4\omega_0}\bigg]_{\frac{-\theta}{\omega_0}}^{\frac{-\theta+2\pi}{\omega_0}} \nonumber\\ & = \frac{A^2}{T_0} \bigg[ \frac{1}{2} \bigg(\frac{-\theta+2\pi}{\omega_0} - \frac{-\theta}{\omega_0} \bigg) \bigg] \nonumber \\ & = \frac{1}{2}*\frac{A^2}{T_0}*\frac{2\pi}{\omega_0}\nonumber\\ & = \frac{A^2}{2} \end{aligned}

So, $x(t)$ is a Power Signal.

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Example 3: What is the power of the signal $x(t) = e^{j\omega_0t}$? where $(T_0=\frac{2\pi}{\omega_0})$

Solution

Given that $T_0=\frac{2\pi}{\omega_0}$ $\Rightarrow$ $e^{j\omega_0t}=e^{j\omega_0(t+\frac{2\pi}{\omega_0})}=e^{j(\omega_0t+2\pi)}$.
We know that $e^{j\theta}=e^{j(\theta+2\pi)}$ $\Rightarrow$ $x(t)$ is periodic with a period of $\frac{2\pi}{\omega_0}$.Therefore,

(7)

\begin{aligned} P_x & =\frac{1}{T_0}\int_{0}^{T_0} |e^{j\omega_0t}|^2 dt \nonumber \\ & =\frac{1}{T_0}\int_{0}^{T_0} 1.dt \hspace{1cm} \big[ |e^{j\theta}|=1 \hspace{0.2cm}\forall \hspace{0.2cm}\theta \Rightarrow |e^{j\omega_0t}|=1\big]\nonumber \\ & = 1 \end{aligned}

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Example 4: Compute the energy of the signal $x[n]$ given by

(8)

\begin{align} x[n]= \begin{cases} \big( \frac{1}{3}\big)^{n} & \text{if}\ n\geq0 \\ 0 & \text{otherwise} \end{cases} \end{align}

Solution

(9)

\begin{aligned} E_x & =\sum_{-\infty}^{\infty}x^2[n] =\sum_{0}^{\infty}x^2[n] =\sum_{0}^{\infty}\bigg(\bigg(\frac{1}{3}\bigg)^{n}\bigg)^2\nonumber\\ & =\sum_{0}^{\infty}\bigg(\frac{1}{3}\bigg)^{2n}\nonumber =\sum_{0}^{\infty}\big(\frac{1}{9}\big)^{n}\nonumber =\frac{1}{1-\frac{1}{9}}\nonumber \\ & =\frac{9}{8} \end{aligned}

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