Question

The continuous time signal \(x(t)\) is described by \[ x(t) = \begin{cases} 1, & 0 \leq t \leq 1 \\ 0, & \text{elsewhere} \end{cases} \] If \(y(t)\) represents \(x(t)\) convolved with itself, which of the following statements is/are TRUE?

• A. \(y(t) = 0 \; \text{for all } t<0\)
• B. \(y(t) = 0 \; \text{for all } t>1\)
• C. \(y(t) = 0 \; \text{for all } t>3\)
• D. \(\int_{0.1}^{0.75} \frac{dy(t)}{dt}\, dt \neq 0\)

Hint:

- Convolution of two rectangular pulses yields a triangular waveform. - The length of the output is the sum of the lengths of the two input signals. - Derivatives and integrals can be simplified using the Fundamental Theorem of Calculus: \(\int f'(t) dt = f(b) - f(a)\).

Answer 1

The Correct Option is A, D

Step 1: Convolution setup. We have \(y(t) = x(t) * x(t)\). Since \(x(t)\) is a rectangular pulse of duration 1, convolving it with itself gives a triangular signal of duration 2: \[ y(t) = \begin{cases} 0, & t<0 \\ t, & 0 \leq t \leq 1 \\ 2-t, & 1<t \leq 2 \\ 0, & t>2 \end{cases} \] Step 2: Check each option. - (A) For \(t<0\), clearly \(y(t) = 0\). True. - (B) For \(t>1\), \(y(t)\) is not zero immediately; in fact, for \(1<t \leq 2\), \(y(t) = 2 - t \neq 0\). Hence this is False. - (C) For \(t>3\), of course \(y(t) = 0\). But the maximum support of \(y(t)\) is only up to \(t=2\). So the statement "for all \(t>3\)" is trivially true. - (D) Evaluate: \[ \int_{0.1}^{0.75} \frac{dy(t)}{dt} dt = y(0.75) - y(0.1) \] Since \(0 \leq t \leq 1\), we have \(y(t) = t\). Thus \(y(0.75) = 0.75, \; y(0.1) = 0.1\). So the difference is \(0.75 - 0.1 = 0.65 \neq 0\). True. Final Answer: (A) and (D)