ECE308_F08_Sample_Problems_2-Solution

ECE308_F08_Sample_Problems_2-Solution - Problem 2-7 (3)...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 2-7 (3) Linear. Consider the responses to two arbitrary inputs: ' m) = an?) yztr) = x202) Multiply first by a and the second by bland add to get ayiU) + [73:20) = M102) + M202) That is, for the input ax1(t)+ lax/2(1), we replace 1 by :2 to get the new output which is the right-hand side of the above equation. (b) Time varying. Consider the response to the delayed input: 3’00) = x02 - 1:) Now consider the delayed output due to the undelayed input: y(r — 1:) = x[(r - 1?] Clearly the two are not the same. ((2 ) Noncausal. Consider t = 2 which gives y(2) = x(4); i.e., the output depends on a future value of the input. (d) Not zero memory. This follows from (c J where it was found that the output does not depend only on values of the input at the present time only. Problem 2-10 Using Kirchoff’s voltage equation and Ohm’s law, the appropriate equations are fl0=Lfim+fl0 d: fln=mm £0; 2 mm d! R d! Substitute the last equation in the first and rearrange to obtain 6%) + E = .6. dr Ly“) Lxm (b) The proof is similar to those of Problems 2-8 and 2-9. (c ) Consider _ r r dyU—Tl : where [I d: d,” d: ll As I 1" Thus L“ L r) + £311- 1) = £x(t — 1:) dr L L which shows that the system is fixed. (d) Note that the solution to the homogeneous equation is yHU) = Ae 'R'”: t > 0 Assume a complete solution of this form where A is time varying. Substitute into the differential equation of part (a) to obtain I A(:) = f—E-xflkkmdl +140 0 Since the inductor current is assumed 0 at 1’ = 0, this gives A0: 0, so the solution to the differential equation is ‘ R R y(t) = -—x(}L)exp[F—(r - 1)}Ak it L Problem 2-12 (a) Linear; first order; causal; time invariant. (b) Linear; first order; causal; time varying. (c ) Nonlinear; second order; causal; time invariant. ((1) Nonlinear; zero order; causal; time invariant. W (a) By KVL and Ohm's law: 51 (tho) R dt + h(:) = 6(1) where the forcing function being a delta function means that the response is the impulse response. For t < 0. the impulse resp0nse is zero because the input is 0 and the initial conditions are assumed 0. For t > 0, we solve the homogeneous differential equation to get the solutiou h(t) : Ae'Rm‘, r > 0 To find the condition required to fixA, integrate the ditfereutial equation (with impulse forcing function) through I = 0: mg dh(t) R a: o— 0? o. d: + h(r)d: = 60M: : 1 l l Fromthe form of Mr), we see that it has a step discontinuity at r = 0 and therefore its derivative has an impulse at r = 0. Thus the second term on the left-hand side integrates to 0 (it only has a step discontinuity). The first term on the left-hand side integrates to (UR)[h(0+) - h(0-)]. Thus, the above equation becomes h(0+) = RIL = A. and the impulse response becomes h(t) = {—e “Lam (b) Let the voltage across the resistor be visa) and the voltage across the inductor be vL(t). Thus vL(r) + vR(r) = 6(t) But the voltage across the resistor cannot be proportional to an impulse because then the current around the loop would be proportional to an impulse and this means the inductor voltage (L times the derivative of the current) w0uld be proportional to the derivative of an impulse. Since there is no derivative of an impulse on the right-hand side to balance it, this cannot be the case. Therefore, it must be true at time 0 that vL(t) = 50) and the Current possesses a step at time 0. In particular, 0v . _ 1 F 1 40+) u Elana: _ L For t > 0, the current around the resistor-inductor loop must satisfy dim dt L +Rt’(r) = 0 or £0) = Ae'R‘IL, t > 0 The constant A can be fixed by setting t'(0) = i(0+) = 1/L. Thus, the same result is obtained for the impulse response as obtained in part (a). W By KVL around the loop, I x(r) = Rita) + % fi(l)d}. +R2i(r) —w But i0) = 3’0)er Substitute this into the integro-differential equation and differentiate once to get R1+R2dy(t) + 3’“) = E R2 at: ch d: One way to find the impulse response is to find the step response and differentiate it. The solution to the homogeneous equation is v R390 I a(t) = Ae'm‘ I > 0 With a step input, the right-hand side of the differential equation is an impulse. To get the required initial condition, we integrate the differential equation through I = 0: 0+ 1131+]?2 da(t)dr+ 1 R2 0_ d: RZC Dr 0e a(r)dr = 6(r)dr : 1 l i To match the right-hand side, the integrand of the first term on the left-hand side must contain a unit impulse and, therefore, the second term on the left-hand side is proportional to a unit step. Hence the integral on the secmrd term through I = 0 is 0 (a step discontinuity). The first term is a perfect differential. Thus, we obtain a(0+) = Rim?!1 + R,) as the required condition, and the step response is R _ i R - + 2 6 "(RI Wan) and h(t) : La“) = 2 an) — —1—6 “‘R' MW) R1 +R2 d‘ R1+R2 (R1+R2)C a(t) : W (a) The step response is 030‘) = I MAMA = exp(-Rt/L)u(t) (b) The ramp response is a,(:) = fa,(l)dl = %[1~ cxp(—Rr/L)]u(r) .1 WM Write the input as x0) = r0) — r(r — 1) — u(t # 2) Thus. by superposition, the output is y(t) = a,(:) - a,(z — 1) — (1,0 - 2) = in — exp(—R:/L)]u(r) - fi {1 — cxp[-R(t - 1)/L)]u(r - 1) — exp[—R(t _ 2)/L]u(t — 2) Sketches of the input and output are given below: 1.5 ‘u' a...’ K 0.5 Fromm (a) The frequency response function is given in terms of the impulse response by How) 2 f h(t)e 'h'dr From the impulse response given in Problem 2-29, this gives jw HUto) = f[6(t) ~ £exp(erlL)u(t)]exp(-jmt)dt = L (b) In terms off = mfln, the frequency l‘CSpOnSC function is Jff/f3 where f = R 3 1 + mf3 Err—L H0): Taking the magnitude, the amplitude response function is [fl/fa «1 + (flfg)2 AU): (c ) The phase response function is the argument of the frequency response function. It is given by 60‘) = g—m'v/fg) Plots of the amplitude and phase msp0nsc functions are given below: phase resp.; radians ...
View Full Document

Page1 / 11

ECE308_F08_Sample_Problems_2-Solution - Problem 2-7 (3)...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online