power and energy

[logistic_guy]

Determine the values of $\displaystyle P_x$ and $\displaystyle E_x$ for each of the following signals:

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

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

$\displaystyle \bold{(c)} \ x_3(t) = \cos t$

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

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

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

 

[logistic_guy]

Let us first show the formulas for the power $\displaystyle P_x$ and energy $\displaystyle E_x$.

For a continuous-time signal $\displaystyle \textcolor{red}{\text{x(t)}}$, we have:

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

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


For a discrete-time signal $\displaystyle \textcolor{red}{\text{x[n]}}$, we have:

$\displaystyle P_x = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\sum_{n=-N}^{N} |x[n]|^2$

$\displaystyle E_x = \sum_{n=-\infty}^{\infty} |x[n]|^2$

 

[logistic_guy]

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

$\displaystyle P_x = \lim_{T \rightarrow \infty}\frac{1}{2T}\int_{-T}^{T} |x_1(t)|^2 \ dt = \lim_{T \rightarrow \infty}\frac{1}{2T}\int_{-T}^{T} |e^{-2t}u(t)|^2 \ dt$


$\displaystyle = \lim_{T \rightarrow \infty}\frac{1}{2T}\int_{0}^{T} e^{-4t} \ dt = \lim_{T \rightarrow \infty}\frac{1}{2T} \frac{e^{-4t}}{-4}\bigg|_{0}^{T}$


$\displaystyle = \lim_{T \rightarrow \infty} \left(-\frac{e^{-4T}}{8T} + \frac{1}{8T}\right) = -0 + 0 = \textcolor{blue}{0}$

 

[logistic_guy]

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

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


$\displaystyle = \int_{0}^{\infty} e^{-4t} \ dt = \frac{e^{-4t}}{-4}\bigg|_{0}^{\infty} = 0 + \frac{1}{4} = \textcolor{blue}{\frac{1}{4}}$

 

[logistic_guy]

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

$\displaystyle P_x = \lim_{T \rightarrow \infty}\frac{1}{2T}\int_{-T}^{T} |x_2(t)|^2 \ dt = \lim_{T \rightarrow \infty}\frac{1}{2T}\int_{-T}^{T} |e^{j(2t + \pi/4)}|^2 \ dt$


$\displaystyle = \lim_{T \rightarrow \infty}\frac{1}{2T}\int_{-T}^{T} \ dt = \lim_{T \rightarrow \infty}\frac{1}{2T} t\bigg|_{-T}^{T} = \lim_{T \rightarrow \infty}\frac{2T}{2T} = \textcolor{blue}{1}$

 

[logistic_guy]

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

$\displaystyle E_x = \int_{-\infty}^{\infty} |x_2(t)|^2 \ dt = \int_{-\infty}^{\infty} \left|e^{j(2t + \pi/4)}\right|^2 \ dt$


$\displaystyle = \int_{-\infty}^{\infty} \ dt = t\bigg|_{-\infty}^{\infty} = \infty + \infty =\textcolor{blue}{\infty}$

 

[logistic_guy]

$\displaystyle \bold{(c)} \ x_3(t) = \cos t$

$\displaystyle P_x = \lim_{T \rightarrow \infty}\frac{1}{2T}\int_{-T}^{T} |x_3(t)|^2 \ dt = \lim_{T \rightarrow \infty}\frac{1}{2T}\int_{-T}^{T} |\cos t|^2 \ dt$


$\displaystyle = \lim_{T \rightarrow \infty}\frac{1}{2T}\int_{-T}^{T} \cos^2 t \ dt = \lim_{T \rightarrow \infty}\frac{1}{2T}\int_{-T}^{T} \left[\frac{\cos 2t + 1}{2}\right] \ dt$


$\displaystyle = \lim_{T \rightarrow \infty}\frac{1}{2T}\left[\frac{\sin 2t}{4} + \frac{t}{2}\right]_{-T}^{T} = \lim_{T \rightarrow \infty}\frac{1}{2T}\left[\frac{\sin 2T}{4} + \frac{T}{2} - \frac{\sin -2T}{4} - \frac{-T}{2}\right]$


$\displaystyle = \lim_{T \rightarrow \infty}\frac{1}{2T}\left[\frac{\sin 2T}{4} + \frac{T}{2} + \frac{\sin 2T}{4} + \frac{T}{2}\right] = \lim_{T \rightarrow \infty}\frac{1}{2T}\left[\frac{\sin 2T}{2} + T\right]$


$\displaystyle = \lim_{T \rightarrow \infty}\left[\frac{\sin 2T}{4T} + \frac{1}{2}\right] = 0 + \frac{1}{2} = \color{blue} \frac{1}{2}$

 

[logistic_guy]

$\displaystyle \bold{(c)} \ x_3(t) = \cos t$

$\displaystyle E_x = \int_{-\infty}^{\infty} |x_3(t)|^2 \ dt = \int_{-\infty}^{\infty} \left|\cos t\right|^2 \ dt$


$\displaystyle = \int_{-\infty}^{\infty} \cos^2 t \ dt = \int_{-\infty}^{\infty} \bigg[\frac{\cos 2t + 1}{2}\bigg] \ dt = \bigg[\frac{\sin 2t}{4} + \frac{t}{2}\bigg]_{-\infty}^{\infty} = \frac{R}{4} + \infty = \color{blue} \infty$

where $\displaystyle -1 \leq R \leq 1$

 

[logistic_guy]

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

$\displaystyle P_x = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\sum_{n=-N}^{N} |x_1[n]|^2 = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\sum_{n=-N}^{N} \bigg|\left(\frac{1}{2}\right)^n u[n]\bigg|^2$


$\displaystyle = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\sum_{n=0}^{N} \left(\frac{1}{2}\right)^{2n} = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\sum_{n=0}^{N} \left(\frac{1}{4}\right)^{n}$


$\displaystyle = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\bigg[1 + \sum_{n=1}^{N} \left(\frac{1}{4}\right)^{n}\bigg] = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\bigg[1 + \frac{\frac{1}{4}\left(1 - \left(\frac{1}{4}\right)^N\right)}{1 - \frac{1}{4}}\bigg]$


$\displaystyle = \frac{1}{3}\lim_{N \rightarrow \infty}\frac{1}{2N + 1}\left[4 - \left(\frac{1}{4}\right)^N\right] = \frac{1}{3}\left(\frac{4}{\infty} - \frac{0}{\infty}\right) = 0$

 

[logistic_guy]

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

$\displaystyle E_x = \sum_{n=-\infty}^{\infty} |x_1[n]|^2 = \sum_{n=-\infty}^{\infty} \bigg|\left(\frac{1}{2}\right)^n u[n]\bigg|^2$


$\displaystyle = \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^{2n} = \sum_{n=0}^{\infty} \left(\frac{1}{4}\right)^{n} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{4}{4} - \frac{1}{4}}$


$\displaystyle = \frac{4}{4 - 1} = \textcolor{blue}{\frac{4}{3}}$

 

[logistic_guy]

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

$\displaystyle P_x = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\sum_{n=-N}^{N} |x_2[n]|^2 = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\sum_{n=-N}^{N} \bigg|e^{j(\pi/2n + \pi/8)}\bigg|^2$


$\displaystyle = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\sum_{n=-N}^{N} 1 = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\bigg[2N + 1\bigg] = \textcolor{blue}{1}$

 

[logistic_guy]

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

$\displaystyle E_x = \sum_{n=-\infty}^{\infty} |x_2[n]|^2 = \sum_{n=-\infty}^{\infty} \bigg|e^{j(\pi/2n + \pi/8)}\bigg|^2$


$\displaystyle = \sum_{n=-\infty}^{\infty} 1 = \textcolor{blue}{\infty}$

 

[logistic_guy]

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

$\displaystyle P_x = \lim_{N \rightarrow \infty}\frac{1}{2N + 1} \sum_{n = -N}^{N} |x_3[n]|^2 = \lim_{N \rightarrow \infty}\frac{1}{2N + 1} \sum_{n = -N}^{N} \left|\cos \frac{\pi}{4}n\right|^2$


$\displaystyle = \lim_{N \rightarrow \infty}\frac{1}{2N + 1} \sum_{n = -N}^{N} \cos^2 \frac{\pi}{4}n = \lim_{N \rightarrow \infty}\frac{1}{2N + 1} \sum_{n = -N}^{N} \frac{1}{2}\bigg[1 + \cos \frac{\pi n}{2}\bigg]$


$\displaystyle = \lim_{N \rightarrow \infty}\frac{1}{2N + 1} \bigg[\frac{2N + 1}{2} + \frac{1}{2}\sum_{n = -N}^{N} \text{value between} \ \{-1,1\}\bigg]$


$\displaystyle = \lim_{N \rightarrow \infty}\bigg[\frac{1}{2} + \frac{\text{finite number}}{4N + 2}\bigg] = \frac{1}{2} + 0 = \textcolor{blue}{\frac{1}{2}}$

 

[logistic_guy]

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

$\displaystyle E_x = \sum_{n = -\infty}^{\infty} |x_3[n]|^2 = \sum_{n = -\infty}^{\infty} \left|\cos \frac{\pi}{4}n\right|^2$


$\displaystyle = \sum_{n = -\infty}^{\infty} \cos^2 \frac{\pi}{4}n = \sum_{n = -\infty}^{\infty} \frac{1}{2}\bigg[1 + \cos\frac{\pi n}{2}\bigg] = \textcolor{blue}{\infty}$