HW #2 (cont'd) 2. Assume the impulse response of your circuit has the form: v.(t)=exp(-at? Ju(t) where t is time in seconds, a is a function parameter controlling the roll-off over time, and u(t) is the unit step function. Hint: Ift is a vector of time points in Matlab, the vector of parametric function values corresponding to points in t is written as v= exp(-alph*t.^2) For 5 trials and measurements at time points .01, .02, .03, .04, and .05 seconds, the data is given as (unit of volts) t=0.01 t=.02 t=.03 t=0.04 t=0.05 seconds Trial 1 0.9268 0.7855 0.5270 0.3029 0.1731 volts Trial 2 0.9909 0.8888 0.5671 0.3384 0.2305 volts Trial 3 0.9124 0.8633 0.6109 0.2904 0.1619 volts Trial 4 1.0038 0.7895 0.5753 0.2664 0.2068 volts Trial 5 0.9711 0.7383 0.5055 0.3425 0.1427 volts Write a Matlab script to estimate the value of a through a curve fit program for each trial (Hint the true value of sigma is somewhere between 100 and 1000). Compute the mean of the resulting a's, and find the 95% confidence limits (just as in the example). Hand in a hard copy of the commented script with the resulting mean a and its confidence limits and also provide a graph along with the code