h(0) Inverse Transform Relation

The value of a signal at time $t=0$, denoted as $h(0)$, is related to its Fourier Transform $H(j\omega)$ by the inverse Fourier Transform evaluated at $t=0$:

$h(0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} H(j\omega) d\omega$

H(jω) Trigonometric Simplification

The given Fourier Transform is $H(j\omega) = \frac{(2\cos\omega)(\sin 2\omega)}{\omega}$. Using the trigonometric identity $2\cos A \sin B = \sin(A+B) - \sin(A-B)$, with $A=\omega$ and $B=2\omega$, the numerator becomes:

$2\cos\omega \sin 2\omega = \sin(\omega+2\omega) - \sin(\omega-2\omega) = \sin(3\omega) - \sin(-\omega) = \sin(3\omega) + \sin(\omega)$

Therefore, $H(j\omega)$ simplifies to:

$H(j\omega) = \frac{\sin(3\omega) + \sin(\omega)}{\omega}$

Integral of H(jω) Calculation

We calculate the integral $\int_{-\infty}^{\infty} H(j\omega) d\omega$:

$\int_{-\infty}^{\infty} H(j\omega) d\omega = \int_{-\infty}^{\infty} \frac{\sin(3\omega) + \sin(\omega)}{\omega} d\omega$

This integral can be split: $\int_{-\infty}^{\infty} \frac{\sin(3\omega)}{\omega} d\omega + \int_{-\infty}^{\infty} \frac{\sin(\omega)}{\omega} d\omega$

Using the standard integral result $\int_{-\infty}^{\infty} \frac{\sin(ax)}{x} dx = \pi \cdot \text{sgn}(a)$:

• $\int_{-\infty}^{\infty} \frac{\sin(3\omega)}{\omega} d\omega = \pi$ (since $a=3$)
• $\int_{-\infty}^{\infty} \frac{\sin(\omega)}{\omega} d\omega = \pi$ (since $a=1$)

The total integral is $\pi + \pi = 2\pi$.

Final h(0) Value Computation

Substitute the integral result back into the formula for $h(0)$:

$h(0) = \frac{1}{2\pi} \times (2\pi)$

$h(0) = 1$