Question

Let $x_1(t) = u(t+1.5) - u(t-1.5)$ and $x_2(t)$ is shown in the figure below. For $y(t) = x_1(t) x_2(t)$, the $\int_{-\infty}^{\infty} y(t)dt$ is ___________ (rounded off to the nearest integer). 

 

Hint:

The "area property" of convolution is a very powerful tool. It allows you to find the total area (or DC component in the frequency domain) of a convolution result without having to perform the convolution itself. Remember: Area of $(fg)$ = (Area of $f$) $\times$ (Area of $g$).

Answer 1

Correct Answer: 15

We need to find the value of the integral of the convolution of two signals: $I = \int_{-\infty}^{\infty} y(t)dt = \int_{-\infty}^{\infty} (x_1(t) x_2(t)) dt$.
A key property of convolution states that the area under the convoluted signal is equal to the product of the areas under the individual signals.
$\int_{-\infty}^{\infty} (x_1(t) x_2(t)) dt = \left( \int_{-\infty}^{\infty} x_1(t) dt \right) \times \left( \int_{-\infty}^{\infty} x_2(t) dt \right)$.
Step 1: Calculate the area under $x_1(t)$.
The signal $x_1(t) = u(t+1.5) - u(t-1.5)$ is a rectangular pulse. It has a height of 1 and extends from $t = -1.5$ to $t = 1.5$. The width of the pulse is $1.5 - (-1.5) = 3$. Area of $x_1(t) = \text{Height} \times \text{Width} = 1 \times 3 = 3$.
Step 2: Calculate the area under $x_2(t)$.
The signal $x_2(t)$ shown in the figure is a triangular pulse. It has a base extending from $t = -1$ to $t = 1$, so the base width is $1 - (-1) = 2$. The height of the triangle is 2. Area of $x_2(t) = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2 \times 2 = 2$.
Step 3: Calculate the integral of the convolution.
$I = (\text{Area of } x_1(t)) \times (\text{Area of } x_2(t)) = 3 \times 2 = 6$.
The value of the integral is 6. This is an integer, so no rounding is required.