Question

3. A continuous-time system with impulse response h'n - 5 rect' is excited by x(1) = 4 rect(21). (a) Find the response y(t) at time 1 = 1/2. (b) Change the excitation from part (a) to x. //\= xlt-1and keep the same impulse response. What is the new responsey.(t) at time 1 =1/2?

Answer 1

This question belongs to Signals and systems.The output is found by the convolution of the two signals.here both signals are square wave so the output convolution will lead to
SOLUTION Notes - Axect (u) = st 1. Giveni, nitasrect't alt=4rect (27) = 4 rect (4/12) hittar 1 41 Ylt=h(H& ILHA convolution. > t a t u How when two square pulse are convoluted the result is trapezoidal wave with following property: ' c) starting value; sum of starting value of a signals. - i) Ending value: sum of ending value of a signals in Intermediate value - a) min of wave 1 + max of wave 2 b) mas of wave I t min of wave 2 iwla slope ope amplitude 1 x Amplitude 2 1) Starting value=(-2)+(-44) = -3/4 ii) Ending value = (1/2) + (114) = 314 iii) Intermediate value = a) Ź+(tal=-7 py Lt b) Ł +(-+) = ť (- Lain) 110 value @tot . write equation of t line: T t k 4 -> 4-41= m(0-21) → 4-10.5-2016-5) a 9=5 . Amp 1x Amp 2 20 = 40 @ t-1/2 CS) Scannéd wită -(-3) - Camsca M210)

b) h(t=srectlt' dlalte alt-1) phlt) 뇌 Am tt 1) starting value =) + 3 = ţ ii) Ending value = (%) + (Sal = 7 (ii) Intermediate value a a) Ź + s = 3 b) £ + 24 =1514) Yalth value @ tak > write equation of m=2-41 I t Yaz 20 (at-1) 44 4 के - live not az-21 Tas - (209 1910 Cams
trapezoidal wave.