Let's solve **Q1(a)** step by step: --- ## Q1(a) \( x(t) = e^{-8t} u(t) \) Where: - \( u(t) \) is the unit step function (\( u(t) = 1 \) for \( t \geq \), otherwise). ### **Step 1: Energy Calculation** Energy of a signal \( x(t) \) is, \[ E = \int_{-\infty}^{\infty} |x(t)|^2 dt \] For \( x(t) = e^{-8t} u(t) \): \[ |x(t)|^2 = (e^{-8t} u(t))^2 = e^{-16t} u(t) \] So, \[ E = \int_{-\infty}^{\infty} e^{-16t} u(t) dt \] Since \( u(t) = \) for \( t < \), \[ E = \int_{}^{\infty} e^{-16t} dt \] Integrate: \[ E = \left[ -\frac{1}{16} e^{-16t} \right]_{}^{\infty} \] \[ E = - \left( -\frac{1}{16} \right) \] \[ E = \frac{1}{16} \] --- ### **Step 2: Power Calculation** Power of a signal \( x(t) \) is, \[ P = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T} |x(t)|^2 dt \] But for energy signals (finite energy), power is zero. Let's check: \[ P = \lim_{T \to \infty} \frac{1}{2T} \int_{}^{T} e^{-16t} dt \] \[ = \lim_{T \to \infty} \frac{1}{2T} \left[ -\frac{1}{16} e^{-16t} \right]_{}^{T} \] \[ = \lim_{T \to \infty} \frac{1}{2T} \left( -\frac{1}{16}e^{-16T} + \frac{1}{16} \right) \] As \( T \to \infty \), \( e^{-16T} \to \): \[ P = \lim_{T \to \infty} \frac{1}{2T} \cdot \frac{1}{16} = \] --- ## **Final Answer (a):** - **Energy:** \( \boxed{\frac{1}{16}} \) - **Power:** \( \boxed{} \) --- Would you like the step-by-step solution for part **(b)** next?