Question

Problem 1: The impulse response of a system driven by a 18 order differential equation + ay = Ku(t) is given by y(t) = Ke-at where K is usually referred to as the “DC gain of the system” and 1 = = is referred to as the “time constant. Impulse responses are very important for understanding system behavior. t[sec] 0 30 se 5.887 - 3.319 +2.50 +.302 +0.3075 0.0929 40 0.6729 10 LO7 ENDA 5.487 3.319 2.307 0.8075 Use the data in the table above to identify system parameters K and a if we assume the impulse response to be an exponential fit. Please report your results using 4 significant digits.

Answer 1

- . Given, A y = ke at applying natural logarithmic on both sides 3. excy) = lnilke at) a = lnck) - at lnce) ency) = ln(k) - at It is in form Y = A + B6 Y = lncy) i A= ls(k) X-t B=-a Yes I y = lnly) 1 x 1 xy 5.487 1.7024 3.319 1.1997 5.9983 2.307 0.8359 100 8.3595 1.302 0.2639 400 5.2780 0.8075 -0.1387 900 1-4.1606 0.6729 -0.3462 -15.8433 3025 0.37111 Suma l 105 3.467 Ex = los E Y = 3.467. no. of date Linear regression EXY = -0.3711 exe - 3025 points n-6

16 los7(A) (3.467) 105 3025B] * |-0-37"} By solving the above equations A = 1.47744459 B = -0.051405887 A = ln (K) B = -a k = exp(A) a -B - exp(1.4774) Q = 0.0S1405857 (K = 4.3817 Ta-0.0514) y = ke at 4C) = 4,3817 me -0.05140

I hope this will help you.