List of top Signals and Systems Questions

A continuous time transfer function \( H(s) = \dfrac{1 + s/10^6}{s} \) is converted to a discrete time transfer function \( H(z) \) using a bilinear transform at a sampling rate of 100 MHz. The pole of \( H(z) \) is located at \( z = \) \(\underline{\hspace{2cm}}\).

Let $X(j\omega)$ denote the Fourier transform of $x(t)$. If $X(j\omega) = 10 e^{-j\pi f} \left( \frac{\sin(\pi f)}{\pi f} \right)$, then $ \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) d\omega = \underline{\hspace{1cm}}. $ (where $\omega = 2\pi f$)