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### exam1_sample_solutions-s05

Course: ECE 350, Spring 2012
School: Boise State
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Word Count: 1032

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350 EE Signals and Systems Spring 2005 Sample Exam #1 - Solutions 1) a. x(t) = u(t)+u(t-1)-2u(t-2) b. x[n] = 5 k [n k ] c. k = 5 5 x[n] = 5 x(t) = u(t)+u(t-1)-2u(t-2) 2 1 2 1/2 k = 5 0 -5 1 k [n k ] 1 n 5 t 3 -4 1 t -2x(2-4t) -5 2) a - Linearity Property Additivity property: add the inputs: x3(t) = x1(t)+x2(t), so also x3(t+2) = x1(t+2)+x2(t+2) y3(t) = (t-2)x3(t+2) &lt;- from...

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350 EE Signals and Systems Spring 2005 Sample Exam #1 - Solutions 1) a. x(t) = u(t)+u(t-1)-2u(t-2) b. x[n] = 5 k [n k ] c. k = 5 5 x[n] = 5 x(t) = u(t)+u(t-1)-2u(t-2) 2 1 2 1/2 k = 5 0 -5 1 k [n k ] 1 n 5 t 3 -4 1 t -2x(2-4t) -5 2) a - Linearity Property Additivity property: add the inputs: x3(t) = x1(t)+x2(t), so also x3(t+2) = x1(t+2)+x2(t+2) y3(t) = (t-2)x3(t+2) <- from system definition y3(t) = (t-2)[x1(t+2)+x2(t+2)] add the outputs: y1(t)+y2(t)=(t-2)x1(t+2) + (t-2)x2(t+2) Scalability property: scale the input: x4(t) = ax1(t), so also x4(t+2) = ax1(t+2) y4(t) = (t-2)x4(t+2) <- from system definition y4(t) = (t-2)[ax1(t+2)] scale the output: ay1(t) = a[(t-2)x1(t+2)] because y3(t)=y1(t)+y2(t), (response to sum of inputs is sum of outputs) additivity is satisifed. because y4(t) = ay1(t), (response to scaled input is scaled output) scalability is satisfied because additivity and scalability are both satisfied, then the system is linear. <- make sure to state this final answer 2) b - Time Invariance Property y1(t) = (t-2)x1(t+2) y2(t) = y1(t-t0) y2(t) = ((t-t0)-2)x1(t-t0) (1) define system example 1 (2) shift output (3) apply time shift from (2) to system in (1) (replace ALL ts by t-t0) x2(t) = x1(t-t0) (4) shift input (5) define system example 2 y2(t) = (t-2)x2(t) y2(t) = (t-2)x1(t-t0) (6) apply input shift from (4) (inside parentheses of all x(*), insert a -t0) because y2(t) y1(t-t0) (steps 3 vs 6), (response to shifted input is NOT shifted output) time invariance is not satisifed. Therefore the system is not time invariant, or in other words it is time variant. This can be seen by inspection by observing the (t-1) outside the input signal. 2) c - Not Causal: y(0)=(-2).x(2) , y(0) depends on the input x(t) at time 2 which hasn't occurred yet. 2) d - Not memoryless: y(t) depends only on x(t+2) which isn't at the same time also - memoryless causal so not causal not memoryless. 2) e - Not stable: as t infinity, y(t) infinity for most bounded inputs , from (t-2) multiplicative factor. 3) a) y(t) = h(t) + h(t-6) y(t) 2 2 -1 t 6 b) y(t) = h(t) + h(t-2) y(t) y(t) 2 -1 2 4 t 2 -1 2 4 t c) y(t) = h(t) + 2h(t+1) 5 4 y(t) y(t) 4 2 2 2 -1 t 4 2 -1 t 4 4) x2(t) = x1(t-1) + 2x1(t-5) so y2(t) = y1(t-1) + 2y1(t-5) y1(t) 2 t 01 3 5 7 5) h(t) = etu(-1-t) a. Causality not causal. The function has non zero values for t<-1. b. Stability stable 1 e t u (1 t )dt = c. e t dt = e t 1 = e 1 e = e 1 < Memory has memory because function is not a scaled impulse at t=0: h(t ) k (t ) 6) Determine y(t) if y(t) = x(t) * h(t) . h(t) x(t) 2 +2 2 x() +2 1 -2 2 t t -2 , -2 2 h(t-) t 1 t+2 Case I +2 t t+2 2 x() +2 2 -2 t+2 < -2 t<4 y(t) = 0 Case II +2 t 2 x() t < -2, t+2 < 0, t+2 > -2 +2 2 -2 t+2 t +2 y (t ) = so -4 < t < -2 (1)( + 2)d 2 y(t) = t2/2 + 4t + 8 Case III +2 2 -2 t Case IV x() +2 t+2 2 t > -2, t < 0, and t+2 > t+2 0, < 2 so -2 <t <0 0 y (t ) = (1)( + 2)d + t 2 y(t) = -t 2t + 2 t +2 (1)( + 2)d 0 x() 2 +2 -2 t > 0, t < 2, t+2 > 2 +2 y (t ) = (1)( + 2)d so 0<t<2 2 t+2 t 2 t 2 y(t) = t /2 2t + 2 Case V x() 2 +2 2<t 2t -2 y(t) = 0 +2 t+2 t <= 4 0 t 2 + 4t + 8 y (t ) = t 2 2t + 2 t 2 / 2 2t + 2 0 4 < t <= 2 2 < t <= 0 0 < t <= 2 2<t 7) Determine y(t) if y(t) = x(t) * h(t) . -2 h(t) 2 x(t) 2 +2 t 1 2 t , -2 h(t-) x() 2 1 t-2 t-1 2 t Case I 2 t-2 t +2 t < -2 2 1 -2 y(t) = 0 Case II t > -2, t-1 < -2 2 t-2 1 -2 t t y (t ) = so -2 < t < -1 (2)( + 2)d 2 y(t) = t2 + 4t + 4 Case III t-1 > -2, t-2 < -2 2 -2 t-2 Case IV t 1 so -1 <t <0 t 1 y (t ) = t (1)( + 2)d + (2)( + 2)d 2 t 1 2 y(t) = t /2 + 3t + 7/2 t > 0, t -1< 0, t<1 2 -2 y (t ) = 0 t t 1 0 (1)( + 2)d + (2)( + 2)d + (2)(2)d t 2 so 0<t<1 1 t t-2 t 1 2 y(t) = -t + 3t + 7/2 Case V t > 1, t-1 > 0, t-2 < 0 2 -2 1 t t-2 y (t ) = t 1 1 t 2 so 1<t<2 0 0 t 1 (1)( + 2)d + (1)(2)d + (2)(2)d 2 y(t) = -t /2 2t + 8 Case VI t-2 > 0, t-1 > 1, t-2 < 1 2 -2 1 t-2 t so 2<t<3 1 y (t ) = (1)(2)d t 2 y(t) = 6 2t Case VII t-2 > 1 2 1 -2 t-2 y(t) = 0 so t>3 t 0 2 t + 4t + 4 t 2 / 2 + 3t + 7 / 2 y (t ) = t 2 + 3t + 7 / 2 t 2 / 2 2t + 8 6 2t 0 7) Determine the impulse response for the following system y[n] - 1/2y[n-1] = x[n-1] y[n] y[n-1] = x[n-1] y[n] = y[n-1] + x[n-1] y[-inf] = 0 y[0] = 0+0 = 0 y[1] = 0+1 = 1 y[2] = (1) + 0 = y[3] = () + 0 = y[n] = ()n-1u[n-1] t < 2 2 <= t < 1 1 <= t < 0 0 <= t < 0 1 <= t < 2 2 <= t < 3 t >= 3 2 x[n] 2 h [n] 1 -2 -1 8) 1 1 -1 2 -2 n 2 x[k] -1 y[n] = 0 -1 2 n h [n-k] 1 -2 1 -1 1 1 -1 n+1 < -2 2 n-2 n-1 k n n+1 n+2 k n < -3 y[-3] = -1 y[1] = 1+1+2+0 = 4 y[-2] = -1+1 = 0 y[2] = 1+2 =3 y[-1] = -1+1+1 = 1 y[3] = 2 y[0] = -1+1+1+2 = 3 y[n] = 0 y[n] n-2 > 1 n>3 4 3 2 1 -4 -3 -2 9) -1 0 1 2 3 4 n y[n] y[n-1] = x[n-1] y[n] = y[n-1] + x[n-1] n=0 n=1 n=2 n=3 n=4 y[0] = y[0-1] + x[0-1] y[1] = y[1-1] + x[1-1] y[2] = y[2-1] + x[2-1] y[3] = y[3-1] + x[3-1] y[4] = y[4-1] + x[4-1] y[0] = y[-1] + x[-1] = 0 + 0 = 0 y[1] = y[0] + x[0] = 0 + 1 = 1 y[2] = y[1] + x[1] = + 0 = y[3] = y[2] + x[2] = + 0 = y[4] = y[3] + x[3] = 1/8 + 0 = 1/8 Therefore y[n] = h[n] = (1/2)n-1u[n-1] 10) y" + 3y' + 2y = x(t) x(t) = u(t) = 1, t>0, by causality, yu(0) = 0, yu(0) = 0 yp = K, yp=0, yp =0 0+0+2k=1 k=1/2 x=-1, -2 yh: s2 + 3s + 2=0 yh = Ae-t + Be-2t yu(t) = yp + yh = Ae-t + Be-2t + yu(0) =A + B + = 0 A = -1, B = yu(t) = (-e-t + e-2t + )u(t) yu(t) = -Ae-t + -2Be-2t yu(0) = -A - 2B = 0 (chain rule) h(t) = dyu(t)/dt = (e-t - e-2t)u(t) + (-e-t + e-2t + )(t) = (e-t - e-2t)u(t) + (-e0 + e0 + )(t) = (e-t - e-2t)u(t)
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ECE 350 Signals and SystemsSpring 2011Sample Exam #2 - Solutions1. For the following signalx(t) = cos(2t) + 2sin(3t) - cos(5t) - 1T0=2, 0=2/T0=1e 2 jt + e 2 jt 2e 3 jt 2e 3 jt e 5 jt e 5 jt++-1x(t) =22j2a0=-1, a1=a-1=0, a2=a-2= , a3=1/j=-j=
New Haven - FIN - 602
ProfitPlanningChapter8PowerPoint Authors:Susan Coomer Galbreath, Ph.D., CPACharles W. Caldwell, D.B.A., CMAJon A. Booker, Ph.D., CPA, CIACynthia J. Rooney, Ph.D., CPACopyright 2012 by The McGraw-Hill Companies, Inc. All rights reserved.8- 2Learni