Q2 Laplace Transforms For Circuit Analysis A First Order Butterworth Filter Is Shown In Transfer Function Form In Figure 1 (96.54 KiB) Viewed 43 times
Q2 Laplace Transforms for Circuit Analysis A first-order Butterworth filter is shown in transfer function form in Figure Q2(a). x(t) Wc y(t) s+wc Figure Q2(a) A first-order Butterworth filter with cut-off frequency rads! Answer all parts of this question. Use of mathematical software, such as MATLAB, is permitted. Enter all numbers rounded to 3 significant figures unless otherwise advised Q2(i) Impulse response and step response ) For the system shown in Figure Q2(a) a) of the cut-off frequency of the filter is to be f. = 2 kHz, what is the corresponding value of a rads? Round your answer to significant figures. Submit part 1 mark Unanswered Impulse Response The Impulse response of the filter shown in Figure 02(a) will be h(t) = w, expl-0,1). (t. (-1 This is graphed for me = 1 rads-U. = 0.159 Hz) in Figure Q2(b) = puterepare 09 D - DA 02 01- 0 1 0.5 2 25 a 25 d + Figure Q2(b) Impulse response (t) for a Butterworth filter with w. = 1 rads The time constant of a first order system can be computed from the time-response for h(r) which will be like the graph illustrated in Figure Q2(b). The time constant is defined as the value of 1 = r for which exp(-1) = exp(-1). = f () (). For the case Illustrated, = 1 rad/s and the time constant s = 1 S. This point is shown on Figure 02(b). b) For the system of Figure Q2(a) with = 2 kHz, what is the value of in milliseconds? fe 5 Round your answer to significant figures Submit part 1 mark Unanswered c) What is the value of the Impulse response h(t) when i = 0.2r? h? 11 Round your answer to significant figures. Submit part 1 mark Unanswered Q2 (ii) System Response by Convolution The response of a system is given by the time convolution of the system impulse response htt) with the Input x(t) such that hx y(t) = h(t)* x(1) y(t) ) * 1 Convolution in the time domain is multiplication in the Laplace transform (complex frequency domain): Y(s) = H (3)X(3)
Convolution in the time domain is multiplication in the Laplace transform (complex frequency domain): Y(s) = H(3)X(3) The next few questions will take you through the determination of the step response of the system of Figure Q2(a). d) The system input is x(r). = 3uo(t). Give the Laplace transform of the Input x(s). xt/). Submit part 1 mark Unanswered e) For the system of Figure Q1(a) with f. = 2 kHz, use convolution to write down an expression for y(s). Submit part 1 mark Unanswered f) Use the Inverse Laplace transform to write down the system response y(t). y(). Submit part 2 marks Unanswered Q2(iii) Step Response The unit step response of the Butterworth filter (x(t) = u(t)) is illustrated in Figure Q2(a): (1) * u(t) = (1 - exp( -1) ult). This gives the well-known first-order response Illustrated for , = 1 rad/s in Figure 02(c). Stap response 09 OR 07 00 0.4 wa 02 PW LE 16 20 Figure Q2(c)-Step response of a first-order Butterworth filter with, = 1 rad/s. As for the previous question, the time constant in the figure is r = 1 s, and the corresponding value of the step response is marked on the graph by the solid red line. At t = 1 ms, what is the value of the step response as a percentage of the final value? (Give your answer as an integer percentage. E.g. If 0.5 of the final value would be 50%.) Round your answer to the nearest integer % Submit part 1 mark Unanswered h) In control theory, rise time T, is defined as the time taken for the step response of a system to move from 10% to 90% of the final value (Indicated in Figure 22(b) by the green dashed lines). Compute the rise time in milliseconds for the system with cut-off frequency f. = 2 kHz and input x(t) = 3uo(t). Round your answer to 3 sig-figs. Round your answer to significant figures. Submit part 2 marks Unanswered Submit all parts 10 marks
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