EECE 290 Homework, Spring 2008

é 13.59 Find f(t) * x(t) when h(t) and x(t) are the rectangular pulses shown in Fig. P13.59(a). b) Repeat (a) when x(r) changes to the rect- angular pulse shown in Fig. P13.59(b). c) Repeat (a) when A(t) changes to the rect- angular pulse shown in Fig. P13.59(c). Figure P13.59 , A(t) x(t) 25 25 t t 0 10 0 10 (a) x(t) A(t) 25 12.5 : | 0 2° Olt ‘ (b) (c) P 13.59 [a] x(A) 25 h(t-A) 25 A (t-10) 0 t t<0: y(t) =0 [b] (A) 12.5 A 0 20 Ih (t-A) 25 A (t-10) i) t t<0 y(t) =0 [c] ot y(t) =| 312.5d\ = 3125 t —10 y(t) = / 312.5 d\ = 312.5(30 — t) t 2125 t d h(t-)) n (t-1)} Ot t<0 y(t) =0 t 0<t<1 y(t) = [625 aa = 628 ie} t 1<t<10: y(t)=/ 625a\ = 625 t-1 10 l<t<u y(t) = [| 625d = 625(11 — 8) t-1 ll<t<o: y(t) =0 ct) 625 T T t AY 13.61 The voltage impulse response of a circuit is shown in Fig. P13.61(a). The input signal to the circuit is the rectangular voltage pulse shown in Fig. P13.61(b). a) Derive the equations for the output volt- age. Note the range of time for which each equation is applicable. b) Sketch v, for -1 <1 <345, Figure P13.61 A(t) (V) 10 T T T 10 20 30 9) (a) vi(A) (V) 10 ali234 ro) (b) P13.61 [a] -l1<t<4; t+1 > |} Up = LOAd\ = 5A?) = 5t? + 10t+5V 0 0 A<t<9: t+1 5 tl Up = 1W\dA\=5\?} =50t—75V t-—4 t—4 9<t<l4: 10 t+] vo=10/ Ad\+10 10d) t-4 10 10 t+1 =5)\? +100A | = —5t? + 140¢ — 480 V t—4 10 1d<t<19: rt] vy = 100 d\ = 500V t-4 19<t< 24: +20 t+1 v= | 0d + | 10(30 — A) da J/t—4 J20 20 t+1 t+1 = 100\ + 300A —5\? t-4 20 20 = —5t? + 190t — 1305 V 24<t< 29: t+1 t+1 t+1 U=10/ (30—A)dA=300A} — 5? t—4 t-4 t-—4 = 1575 — 50tV 29<t< 34: 30 30 30 Ve=10/ (30—A)d\=300A| —5)? t—4 t—4 t—4 = 5¢? — 340¢ + 5780 V Summary: Ug = 0 -o<t<-l Uo = St? +10 +5V -1<t<4 Up = 50t — 75 V 4<t<9 Up = —5t? + 140t — 480 V 9<t<14 [b] Ug = 500 V l4<t<19 Up = —5t? + 190¢ — 1305 V 19<t<24 ty = 1575 — 50tV 24 <t < 29 Up = 5t? — 340¢ + 5780 V 29<t< 34 lo = OV 384<t< oo 600 5 v(t) (Vv) 500 + 400 4 200 4 200 3 400 no tu a Ss = a nq 26 Assume the voltage impulse response of a circuit can be modeled by the triangular waveform shown in Fig. P13.62. The voltage input signal to this circuit is the step function 10u(t) V. a) Use the convolution integral to derive the expressions for the output voltage. b) Sketch the output voltage over the interval Otol5s. c) Repeat parts (a) and (b) if the area under the voltage impulse response stays the same but the width of the impulse response nar- rows to 4s. d) Which output waveform is closer to repli- cating the input waveform: (b) or (c)? Ex- plain. : Figure P13.62 h(t) (V) 2 on | | | | | ol 5 10 t(s) 2 P 13.62 fa] h(\)= =A O<AK<5 5 2 h(A) = (4-5) 5<A< 10 r8) h(A) 2-4 ACs) oO 5 10 uy (t-A) id As) oO t O<t<5: t2 Up =10 f =A\d\ = 20? 095 5 = —100 + 40¢ — 2%? 10 <t< ow: 5D 10 2 vo =10 | =\d\ +10 (4 - =A) dd 0 5 5 5 5 + 40A 0 = 50 + 200 — 150 = 100 10 4)? 5 _ 4\? 2 4 bo Summary: Up = 2t?V 0<t<5 Up = 40¢ — 100 — 2%? V 5<t< Up = 100 V 10<t<co [b] 120 Juecv) 100 - part (c) ao 4 oa 60 5 Mpart {(b) 40 4 vw “ rae 20 4 a4 : t(s) 0 2 4 6 10 12 1 1 {c] Area = 5 (10) (2) = 10 g (4h =10 h=5 0 2 4 y(t-A)} 10 ACS) 0 6 0<t<2 5 7 5 vo =10 [ S\d\+ 10 [ (10 - 5) dd 25d? |! wo = —100 + 100t — 12.5¢? 4<t<o: 25 4 5 vo =10 | sda +10 [ (10- =a) dD 0 2 2 2 2 12 4 R\2 14 _ 254 +1002 _ 25A 2 0 2 2 2 = 50 + 200 — 150 = 100 Up = 12.5t?V 0<t<2 Up = 100¢ — 100 — 12.5%? V 2<t<4 Up = 100 V 4<t<oo [d] The waveform in part (c) is closer to replicating the input waveform because i part (c) h(A) is closer to being an ideal impulse response. That is, the area w preserved as the base was shortened.