Exercise SS-C1: Understanding Convolution, part I [20pts| Consider the RC circuit shown to the right, with R-10 and C-2 mF. The "input" voltage is x(t), and the "output" voltage is y(t) For now, the input voltage is given by: x(t) C y(t) X(t)- rect- where Vo-12V and the time constant -Rc. we are going to solve this circuit in two ways. First, without using convolution, we can show that the corresponding output voltage is: Vo (1-e-t/r) %"(1-e-2)'e-(t-2r)/t O2 (discharging; pulse is "off") We will use this analytic solution to check to make sure our other (numerical) solution is correct. Second, using convolution, we start with the impulse response: h(t)- e-/r and write the output voltage as the convolution of h and x: y(t) =k(t)" x(t)-J hir)' x(t-r)' dr We will solve for this output voltage numerically, by performing the convolution integral using MATLAB, at different values of time t. Because x(t) is a pulse, its value is either 0 or Vo, and we only need to integrate over a relatively short period of time. However, because the argument is "t-r", and the integral begins at r = 0, you will need to think about what the limits of r should be as a function of t. Exercise SS-C1: Understanding Convolution, part I [20pts| Consider the RC circuit shown to the right, with R-10 and C-2 mF. The "input" voltage is x(t), and the "output" voltage is y(t) For now, the input voltage is given by: x(t) C y(t) X(t)- rect- where Vo-12V and the time constant -Rc. we are going to solve this circuit in two ways. First, without using convolution, we can show that the corresponding output voltage is: Vo (1-e-t/r) %"(1-e-2)'e-(t-2r)/t O2 (discharging; pulse is "off") We will use this analytic solution to check to make sure our other (numerical) solution is correct. Second, using convolution, we start with the impulse response: h(t)- e-/r and write the output voltage as the convolution of h and x: y(t) =k(t)" x(t)-J hir)' x(t-r)' dr We will solve for this output voltage numerically, by performing the convolution integral using MATLAB, at different values of time t. Because x(t) is a pulse, its value is either 0 or Vo, and we only need to integrate over a relatively short period of time. However, because the argument is "t-r", and the integral begins at r = 0, you will need to think about what the limits of r should be as a function of t