Consider a real signal $x\left(t\right)$, $-\infty <t<\infty$, such that $x\left(t\right)=0$ for $t<0$, $x\left(t\right)=2$ for $0\le t<1$ and $x\left(t\right)=0$ for $t\ge 1$. Let $E\left[x\right(t\left)\right]={\int }_{-\infty }^{\infty }{\left|x\right(t\left)\right|}^{2}dt$. Which of the following options correctly represents the ratio, $\frac{E\left[x\right(t\left)\right]}{E\left[3x\right(-3t+5\left)\right]}$ ?

The correct answer/option is 1/3.

To understand why this is the correct answer, let us analyze the energy of a signal under amplitude scaling and time transformation.
Let $x\left(t\right)$ be a real-valued signal. The energy of the signal, denoted as $E\left[x\right(t\left)\right]$, is defined as:
$E\left[x\right(t\left)\right]={\int }_{-\infty }^{\infty }{\left|x\right(t\left)\right|}^{2}dt$

Given:
$x\left(t\right)=2$ for $0\le t<1$, and $x\left(t\right)=0$ otherwise.
We can calculate the energy of $x\left(t\right)$ directly:
$E\left[x\right(t\left)\right]={\int }_0^1{2}^{2}dt={\int }_0^{1}4dt=4$

Now, let us find the energy of the transformed signal $y\left(t\right)=3x(-3t+5)$.
By definition, the energy of $y\left(t\right)$ is:
$E\left[y\right(t\left)\right]={\int }_{-\infty }^{\infty }{\left|3x\right(-3t+5\left)\right|}^{2}dt=9{\int }_{-\infty }^{\infty }{\left|x\right(-3t+5\left)\right|}^{2}dt$

To evaluate this integral, we perform a change of variable. Let:
$u=-3t+5$
Differentiating both sides gives:
$du=-3dt\Rightarrow dt=-\frac{1}{3}du$

As $t\to -\infty$, $u\to \infty$, and as $t\to \infty$, $u\to -\infty$. Substituting these into the energy equation:
$E\left[3x\right(-3t+5\left)\right]=9{\int }_{\infty }^{-\infty }{\left|x\right(u\left)\right|}^2\left(-\frac{1}{3}\right)du$

We can use the negative sign to swap the limits of integration:
$E\left[3x\right(-3t+5\left)\right]=\frac{9}{3}{\int }_{-\infty }^{\infty }{\left|x\right(u\left)\right|}^{2}du=3E\left[x\right(\left)\right]$

In general, for any signal transformed as $Ax(at+b)$, the energy scales according to:
$E\left[Ax\right(at+b\left)\right]=\frac{A^2}{\left|a\right|}E\left[x\right(t\left)\right]$
Here, $A=3$ and $a=-3$, which gives:
$E\left[3x\right(-3t+5\left)\right]=\frac{3^2}{|-3|}E\left[x\right(\left)\right]=3E\left[x\right(\left)\right]$

Finally, we compute the required ratio:
$\frac{E\left[x\right(t\left)\right]}{E\left[3x\right(-3t+5\left)\right]}=\frac{E\left[x\right(t\left)\right]}{3E\left[x\right(t\left)\right]}=\frac{1}{3}$