The waveform of a periodic signal $\mathrm{x}\left(\mathrm{t}\right)$ is shown in the figure.



A signal g(t) is defined by $\mathrm{g}\left(\mathrm{t}\right)=\mathrm{x}\left(\frac{\mathrm{t}-1}{2}\right)$. The average power of $\mathrm{g}\left(\mathrm{t}\right)$ is __________.

In the range: 2 - 2

Concept:

The instantaneous power of a signal is calculated as:

$p\left(t\right)=\int s^2\left(t\right)dt$

The average power will be:

$P_{avg}=\frac{1}{T}\underset{0}{\overset{T}{\int }}|s^2\left(t\right)|dt$

Calculation:

$x\left(\frac{t-1}{2}\right)=x\left(\frac{1}{2}\left(t-1\right)\right)$

So, we first scale x(t) by 1/2.and then shift the result by 1 unit right. The resulting waveform is obtained as:







Now, the average power is given by:

$P_{av}=\frac{1}{T}\underset{T}{\int }{\left|g\left(t\right)\right|}^{2}dt$

We have the magnitude of waveform g(t) as shown below.



Taking the sample period -3 ≤ t ≤ 3, we obtain:

$P_{av.}=\frac{1}{6}\underset{-3}{\overset{3}{\int }}{\left|g\left(t\right)\right|}^{2}dt$

$P_{av.}=\frac{1}{6}\underset{-3}{\overset{3}{\int }}{\left|g\left(t\right)\right|}^{2}dt$

$=\frac{1}{6}\left[\underset{-1}{\overset{1}{\int }}{\left(\frac{-3}{2}t+\frac{3}{2}\right)}^{2}dt+\underset{1}{\overset{3}{\int }}{\left(\frac{3}{2}t-\frac{3}{2}\right)}^{2}dt\right]$

$=\frac{3}{8}\left[\underset{-1}{\overset{1}{\int }}{\left(1-t\right)}^{2}dt+\underset{1}{\overset{3}{\int }}{\left(t-1\right)}^{2}dt\right]$

$=\frac{3}{8}\left\{{\left[-\frac{{\left(1-t\right)}^3}{3}\right]}_{-1}^1+{\left[\frac{{\left(t-1\right)}^3}{3}\right]}_1^3\right\}$

$=\frac{3}{8}\left\{\frac{8}{3}+\frac{8}{3}\right\}$

= 2