Question

Let continuous-time signals \( x_1(t) \) and \( x_2(t) \) be defined as:

\[ x_1(t) = \begin{cases} 1, & t \in [0, 1] \\ 2 - t, & t \in [1, 2] \\ 0, & \text{otherwise} \end{cases} \quad \text{and} \quad x_2(t) = \begin{cases} t, & t \in [0, 1] \\ 2 - t, & t \in [1, 2] \\ 0, & \text{otherwise} \end{cases} \]

Consider the convolution \( y(t) = x_1(t) * x_2(t) \). Then

\[ \int_{-\infty}^{\infty} y(t)\,dt =\ ? \]

• A. 1.5
• B. 2.5
• C. 3.5
• D. 4

Hint:

The integral of the convolution of two signals equals the product of their individual integrals: \[ \int_{-\infty}^{\infty} (x_1 * x_2)(t)\,dt = \left( \int_{-\infty}^{\infty} x_1(t)\,dt \right) \left( \int_{-\infty}^{\infty} x_2(t)\,dt \right) \]

Answer 1

The Correct Option is A

The integral of a convolution of two signals is equal to the product of their individual integrals: \[ \int_{-\infty}^{\infty} y(t)\,dt = \left( \int_{-\infty}^{\infty} x_1(t)\,dt \right) \cdot \left( \int_{-\infty}^{\infty} x_2(t)\,dt \right) \] Calculate: \[ \int_{0}^{1} 1\,dt + \int_{1}^{2} (2 - t)\,dt = 1 + \left[ 2t - \frac{t^2}{2} \right]_1^2 = 1 + \left[(4 - 2) - (2 - 0.5)\right] = 1 + (2 - 1.5) = 1 + 0.5 = 1.5 \] So: \[ \int x_1(t)\,dt = 1.5, \quad \int x_2(t)\,dt = 1 \Rightarrow \int y(t)\,dt = 1.5 \times 1 = 1.5 \]