Question

Consider a real signal \(x(t)\), \(-\infty < t < \infty\), such that \(x(t)=0\) for \(t<0\), \(x(t)=2\) for \(0\le t<1\) and \(x(t)=0\) for \(t\ge1\).
Let \(E[x(t)]=\displaystyle\int_{-\infty}^{\infty}[x(t)]^2\,dt\).
Which of the following options correctly represents the ratio, \(E[x(t)]/E[3\,x(-3t+5)]\)?

• A. 3
• B. 1
• C. \(1/3\)
• D. \(1/9\)

Hint:

For \(w(t)=A\,x(Bt+C)\), the energy scales as \(E[w]=\dfrac{A^2}{|B|}E[x]\).

Answer 1

The Correct Option is C

Step 1: Compute E[x(t)].
The signal $x(t)$ is a rectangular pulse of height $2$ over $0\le t<1$ and zero elsewhere, so \[ E[x(t)]=\int_0^1 (2)^2\,dt=\int_0^1 4\,dt=4 \]

Step 2: Set up the scaled and shifted signal.
Let $w(t)=3x(-3t+5)$, so we need $E[w(t)]$.

Step 3: Use the general scaling rule for the energy integral.
For a signal of the form $w(t)=A\,x(Bt+C)$, substitute $u=Bt+C$, so $du=B\,dt$ and $dt=du/|B|$ (the modulus takes care of the limits flipping when $B$ is negative). Then \[ E[w]=\int [A\,x(Bt+C)]^2\,dt=A^2\int [x(u)]^2\,\frac{du}{|B|}=\frac{A^2}{|B|}E[x] \]

Step 4: Plug in the given values.
Here $A=3$ and $B=-3$, so \[ E[w]=\frac{3^2}{|-3|}E[x]=\frac{9}{3}E[x]=3E[x]=3(4)=12 \]

Step 5: Form the required ratio.
\[ \frac{E[x(t)]}{E[3x(-3t+5)]}=\frac{4}{12}=\frac{1}{3} \]

Step 6: Check the other options.
Option (A), $3$, is the reciprocal of the correct ratio and would come from dividing the wrong way round. Option (B), $1$, ignores the scaling factor $A^2/|B|$ altogether. Option (D), $1/9$, would follow only if the factor were mistakenly taken as $1/A^2$ instead of $A^2/|B|$.

Step 7: Final conclusion.
\[ \boxed{\frac{1}{3}} \]