Signals & Systems : Signal Energy & Power

Signal Energy & Power :

The total Energy over the time interval t1 ≤ t t2 in a Continuous time Signal x(t) is defined as $\displaystyle \int_{t_1}^{t_2} |x(t)|^2 dt$ , and during − ∞ < t < ∞ , The Total Energy is given as :

$\displaystyle E_{\infty} = Lim_{T \rightarrow \infty} \int_{-T}^{T} |x(t)|^2 dt $

$\displaystyle E_{\infty} = \int_{-\infty}^{\infty} |x(t)|^2 dt $

In case of Discrete time signal x[n] , the Total Energy during -∞ < n < ∞ is expressed by ,

$\displaystyle E_{\infty} = Lim_{N \rightarrow \infty} \Sigma_{-N}^{N} |x[n]|^2 $

$\displaystyle E_{\infty} = \Sigma_{- \infty}^{\infty} |x[n]|^2 $

The Time Average Power over an infinite interval is given as :

In continuous time ,

$\displaystyle P_{\infty} = Lim_{T \rightarrow \infty} \frac{1}{2 T} \int_{-T}^{T} |x(t)|^2 dt $

In discrete time ,

$\displaystyle P_{\infty} = Lim_{N \rightarrow \infty} \frac{1}{2N + 1} \Sigma_{-N}^{N} |x[n]|^2 $

Periodic Signals :

A Periodic continuous time signals x(t) has the property that :

x(t) = x(t + T) for all values of T

The fundamental period To of x(t) is the smallest positive value of T for which above equation holds .

A discrete time signal x[n] is periodic with period N , if it is unchanged by time shifting of N i.e.

x[n] = x[n + N] for all values of n

The fundamental period No of x[n] is the smallest positive value of N for which above equation holds .

Even & Odd Signals :

In continuous time a signal is even if

x(-t) = x(t)

A discrete time signal is even if ,

x[-n] = x[n]

A Signal is referred to as Odd if ,

x(-t) = – x(t)

also , x[-n] = – x[n]

Even & Odd part of signal x(t) can be Expressed as :

$\displaystyle Even \; x(t) = \frac{1}{2}[x(t) + x(-t)] $

$\displaystyle Odd \; x(t) = \frac{1}{2}[x(t) – x(-t)] $

In case of discrete time x[n] can be Expressed as :

$\displaystyle Even \; x(n) = \frac{1}{2}[x(n) + x(-n)] $

$\displaystyle Odd \; x(n) = \frac{1}{2}[x(n) – x(-n)] $

Solved Example: Determine the values of P and E for each of the following Signals :

(a) $\displaystyle x_1(t) = e^{-2t} u(t)$

(b) $\displaystyle x_2(t) = e^{j(2t + \pi/4)} $

(c) $\displaystyle x_3(t) = cos( t )$

(d) $\displaystyle x_1[n] = (\frac{1}{2})^n u[n]$

(e) $\displaystyle x_2[n] = e^{j(\pi/2n + \pi/8)} $

(f) $\displaystyle x_3[n] = cos(\frac{\pi}{4}n) $

Solution:

(a) $x_1(t) = e^{-2t} u(t)$

The value of Total Energy in the interval from -∞ to ∞

$\displaystyle E_{\infty} = \int_{-\infty}^{\infty} |x_1(t)|^2 dt $

Since , u(t) = 0 for t < 0

$\displaystyle E_{\infty} = \int_{0}^{\infty} |e^{-2t}|^2 dt $

$\displaystyle E_{\infty} = \int_{0}^{\infty} e^{-4t} dt $

$\displaystyle E_{\infty} = [\frac{e^{-4t}}{-4}]_{0}^{\infty} = \frac{1}{4} $

Average Power , $\displaystyle P_{\infty} = lim_{T \rightarrow \infty} \frac{E_{\infty}}{2T} $

P = 0

(b) $x_2(t) = e^{j(2t + \pi/4)} $

|x2(t)| = 1

$T_0 = \frac{2 \pi}{\omega} = \frac{2 \pi}{2}= \pi $

Now , Total energy over one period ,

$\displaystyle E_{period} = \int_{0}^{T_0} |x_2(t)|^2 dt $

$\displaystyle E_{period} = \int_{0}^{T_0} 1^2 dt $

$\displaystyle E_{period} = T_0 = \pi $

E→ ∞ , there are infinite number of Periods .

Now , Power of one period ,

$\displaystyle P_{period} = \frac{E_{period}}{T_0} $

$\displaystyle P_{period} = \frac{T_0}{T_0} = 1 $

(c) $x_3(t) = cos( t )$

Total Energy over one Period ,

$\displaystyle E_{period} = \int_{0}^{T_0} |x_3(t)|^2 dt $

$\displaystyle E_{period} = \int_{0}^{T_0} |cos t|^2 dt $

$\displaystyle E_{period} = \int_{0}^{T_0} |\frac{1 + cos2t}{2}| dt $

$\displaystyle E_{period} = [\frac{t}{2} + \frac{sin 2t}{4}]_{0}^{T_0} = \frac{T_0}{2} $

Average Power over this Period ,

$\displaystyle P_{period} = \frac{E_{period}}{T_0} = \frac{T_0 /2}{T_0}$

$\displaystyle P_{period} = \frac{T_0 /2}{T_0} = \frac{1}{2}$

Since , there are infinite number of Periods , the total energy integrated over all time is infinite . However , each period of the signal looks exactly the same . Since the average power of the signal equals 1 over each period , averaging over multiple periods always yields an average power of 1 , i.e.

$\displaystyle P_{\infty} = Lim_{T_0 \rightarrow \infty} \frac{1}{2T_0} \int_{-T_0}^{T_0}|cos^2 t|dt = 1 $

(d) Given $\displaystyle x_[n] = (\frac{1}{2})^n u[n] $

= 0 , n < 0

$= (\frac{1}{2})^n , n \ge 0 $

Total Energy in the given discrete signal equals to ,

$\displaystyle E_{\infty} = Lim_{N \rightarrow \infty} \Sigma_{n=-N}^{+N} |x[n]|^2 $

$\displaystyle = \Sigma_{n = -\infty}^{+ \infty} |x[n]|^2 $

$\displaystyle = \Sigma_{n = -\infty}^{+ \infty} (\frac{1}{2})^{2n} $ (Since , u[n] = 0 for n < 0)

$\displaystyle = 1 + \frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + ….$

$\displaystyle = \frac{1}{1 – \frac{1}{4}} = \frac{4}{3}$

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