Question

The value of the convolution of $f(x) = 3 \cos 2x$ and $g(x) = \frac{1}{3} \sin 2x$ where $x \in [0, 2\pi]$, at $x = \frac{\pi}{3}$ is ............................. (Rounded off to 2 decimal places).

Hint:

To calculate the convolution of two functions, multiply the functions for different values of $x$ and integrate. In discrete cases, this can be done via summation.

Answer 1

The convolution of two functions is defined as \[ (f g)(x) = \int_{-\infty}^{\infty} f(\tau) g(x - \tau) \, d\tau. \] In this case, we are interested in finding $(f g)(x)$ at $x = \frac{\pi}{3}$. Given: \[ f(x) = 3 \cos 2x, g(x) = \frac{1}{3} \sin 2x. \] At $x = \frac{\pi}{3}$, the convolution integral becomes: \[ (f g)\left(\frac{\pi}{3}\right) = \int_{-\infty}^{\infty} 3 \cos 2\tau . \frac{1}{3} \sin \left(2 \left(\frac{\pi}{3} - \tau\right)\right) \, d\tau. \] After simplifying the expression: \[ (f g)\left(\frac{\pi}{3}\right) = \int_{-\infty}^{\infty} \cos 2\tau . \sin \left(2 \left(\frac{\pi}{3} - \tau\right)\right) \, d\tau. \] Substituting the value, you compute the result numerically as: \[ (f g)\left(\frac{\pi}{3}\right) = 2.72. \]