1. (20 points) Comparison of theoretical solution vs numerical simulation The current passing through a capacitor (with capacitance C) is given by ic = CdV This can be re-written as dt ic d - dt (1) To obtain V(t), you will be performing integration on both sides from time O'to time 't. This is mathematically written as {"av(a)ds Sticker de dV = () -dz (2) where x is just a dummy variable of integration. Equation 1 can also be converted to a discrete system by doing the following: V(e) v(m) = [1029 ) 7:) ic(: (3) 89 where x follows {0,T., 275,3T5, ...}. a) If C = 10F, V(0) = 0.2V and ic(x) = sin(x), compute V(2) by theoretical integration using Equation 2 (Hint: use t = 2s) b) If C = 10F, V(0) = 0.2V, ic(2) sin(x) and Tg = 0.2, compute V(2) using MATLAB and Equation 3 c) If C = 10F, V(0) = 0.2V, ic(1) = sin(.1) and T, = 0.1, compute V(2) using MATLAB and Equation 3 d) If C = 10F, V(0) = 0.2V, ic(2) = sin(x) and T, = 0.01, compute V(2) using MATLAB and Equation 3 e) Compare results from parts b), c) and d) to part a). Note: please include your MATLAB code for parts b), c) and d) in your solution