Question

Problem 2. Determine whether the following signals are power or energy signals, or neither. Justify your answers. a) x(t)-Asint -00<t<oo b) x(t) = r(1)-r(1-1) c) x(t)-tu(t) d) x(t)- Aexp(bt) , b>0

Answer 1

a)
  
"a) x(t) = A sin t So, E lim_T rightarrow infinity integral^T_-T

So,
   $E=\lim_{T\rightarrow \infty}\int_{-T}^{T}|x(t)|^2~dt=A^2\lim_{T\rightarrow \infty}\int_{-T}^{T}\sin^2 t~dt$
    $\Rightarrow E=\frac{A^2}{2}\lim_{T\rightarrow \infty}\int_{-T}^{T}(1-\cos 2t)~dt$
    $\Rightarrow E={A^2}\lim_{T\rightarrow \infty}\int_{0}^{T}(1-\cos 2t)~dt$
   $\Rightarrow E={A^2}\lim_{T\rightarrow \infty}\left [ t-\frac{\sin 2t}{2} \right ]_{0}^{T}$
   $\Rightarrow E={A^2}\lim_{T\rightarrow \infty}\left [ T-\frac{\sin 2T}{2} \right ]$
   $\Rightarrow E\rightarrow \infty$
And
   $P=\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2~dt=A^2\frac{1}{2T}\lim_{T\rightarrow \infty}\int_{-T}^{T}\sin^2 t~dt$
   $\Rightarrow P=\frac{A^2}{2}\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}(1-\cos 2t)~dt$
  
   $\Rightarrow P={A^2}\lim_{T\rightarrow \infty}\frac{1}{2T}\left [ t-\frac{\sin 2t}{2} \right ]_{0}^{T}$
   $\Rightarrow P={A^2}\lim_{T\rightarrow \infty}\frac{1}{2T}\left [ T-\frac{\sin 2T}{2} \right ]$
   $\Rightarrow P=\frac{A^2}{2}\lim_{T\rightarrow \infty}\left [ 1-\frac{\sin 2T}{2T} \right ]$
   $\Rightarrow P=\frac{A^2}{2}<\infty$
So, it is a power signal.
b)
  
    $x(t)=r(t)-r(t-1)$
where,
   $r(t)=\left\{\begin{matrix} t, &t\geq 0 \\ 0, & t<0 \end{matrix}\right.$
So,
   $x(t)=\left\{\begin{matrix} 1, & t\geq 1\\ t, & 0\leq t < 1 \\ 0, & t<0 \end{matrix}\right.$
So,
   $E=\lim_{T\rightarrow \infty}\int_{-T}^{T}|x(t)|^2~dt=\lim_{T\rightarrow \infty}\left [\int_{0}^{1}t^2~dt +\int_{1}^{T}dt\right ]$
   $\Rightarrow E=\lim_{T\rightarrow \infty}\left [\frac{1}{3} +T-1\right ]$
   $\Rightarrow E=\lim_{T\rightarrow \infty}\left [T-\frac{2}{3} \right ]\rightarrow \infty$
And

$\Rightarrow P=\lim_{T\rightarrow \infty}\frac{1}{2T}\left [\frac{1}{3} +T-1\right ]$
   $\Rightarrow P=\lim_{T\rightarrow \infty}\frac{1}{2T}\left [T-\frac{2}{3} \right ]=\lim_{T\rightarrow \infty}\left [\frac{1}{2}-\frac{1}{3T} \right ]=\frac{1}{2}<\infty$
So, this is again a power signal.
  c)
   
where,
  
So,
$E=\lim_{T\rightarrow \infty}\int_{-T}^{T}|x(t)|^2~dt=\lim_{T\rightarrow \infty}\int_{0}^{T}t^2~dt$
   
   $\Rightarrow E=\lim_{T\rightarrow \infty}\left [ \frac{T^3}{3} \right ]\rightarrow \infty$
And

$\Rightarrow P=\lim_{T\rightarrow \infty}\frac{1}{2T}\left [ \frac{t^3}{3} \right ]_{0}^{T}$
   $\Rightarrow P=\lim_{T\rightarrow \infty}\frac{T^3}{6T}$
   $\Rightarrow P=\lim_{T\rightarrow \infty}\frac{T^2}{6}\rightarrow \infty$
This signal is neither power signal nor energy signal.
d)
   $x(t)=Ae^{bt}$
so,
   $E=\lim_{T\rightarrow \infty}\int_{-T}^{T}|x(t)|^2~dt=A^2\lim_{T\rightarrow \infty}\int_{-T}^{T}e^{2bt}~dt$
   $\Rightarrow E=A^2\lim_{T\rightarrow \infty}\left [ \frac{e^{2bt}}{2b} \right ]_{-T}^{T}$
   $\Rightarrow E=A^2\lim_{T\rightarrow \infty}\left [ \frac{e^{2bT}-e^{-2bT}}{2b} \right ]\rightarrow \infty$
And
   $P=\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2~dt=A^2\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}e^{2bt}~dt$
$\Rightarrow P=A^2\lim_{T\rightarrow \infty}\frac{1}{2T}\left [ \frac{e^{2bt}}{2b} \right ]_{-T}^{T}$
$\Rightarrow P=A^2\lim_{T\rightarrow \infty}\frac{1}{2T}\left [ \frac{e^{2bT}-e^{-2bT}}{2b} \right ]$
   $\Rightarrow P=\frac{A^2}{2}\lim_{T\rightarrow \infty}\frac{e^{2bT}}{2bT}\rightarrow \infty$
This is also neither power signal nor energy signal.