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: Find where $x(t)=2$.
By definition $x(t)=2$ only for $0\le t<1$, and $x(t)=0$ everywhere else.

Step 2: Find the interval where $x(-3t+5)$ is non-zero.
We need $0\le -3t+5<1$. Subtracting $5$ throughout gives $-5\le-3t<-4$. Dividing by $-3$ flips both inequalities: \[ \frac{5}{3}\ge t>\frac{4}{3} \] So $x(-3t+5)=2$ for $t$ in the interval $\left(\frac{4}{3},\frac{5}{3}\right]$, whose length is $\frac{1}{3}$.

Step 3: Find $w(t)=3x(-3t+5)$ on this interval.
On $\left(\frac{4}{3},\frac{5}{3}\right]$, $w(t)=3(2)=6$, so $[w(t)]^2=36$. Outside this interval $w(t)=0$.

Step 4: Integrate to get $E[w]$.
\[ E[w]=\int_{4/3}^{5/3}36\,dt=36\times\frac{1}{3}=12 \]

Step 5: Recompute $E[x]$ directly.
\[ E[x]=\int_0^1 (2)^2\,dt=4 \]

Step 6: Form the ratio.
\[ \frac{E[x]}{E[w]}=\frac{4}{12}=\frac{1}{3} \] \[ \boxed{\frac{1}{3}} \] This matches the result from the general scaling rule.