EE 350 Problem Set 9 Cover Sheet Fall 2014Last Name (Print):First Name (Print):ID number (Last 4 digits):Sectio n:Submission deadl ines:• Turn in the written solutions by 4:00 pm on Friday November 21 in the homework slot outside 121 EE East.Problem Weight Score42 2043 2044 2045 2046 20Total 100The solution submitted for grading represents my own analysis of the problem, and not that of another student.Signature:Neatly print the name(s) of the students you colla bo rated with on this assignment.Reading assignment:• Lathi Chapter 6: Sections 6.1 through 6.5 and B.5 (pp. 24-33)• Priemer Chapter 13Problem 42: (20 points)This problem shows how to determine the Fourier transform of the uni t step functionf(t) = u(t).An attempt to determine F (ω) by direct integration leads to an indeterminate resultF (ω) =Z∞−∞f(t)e−ωtdt =Z∞0e−ωtdt = −1ωe−ωt∞0becauselimt→∞e−ωtyields an indeterminate answer. An alternate method is to start with the Fourier transform pairg(t) = e−atu(t) ↔ G(ω) =1ω + a,where a is a positive real-valued constant. Becausef(t) = lima→0g(t),it follows thatF (ω) = lima→0G(ω),where F (ω) is the Fourier transform of the unit step function.1. (4 points) Express G(ω) in the formG(ω) = R(ω) + X(ω)where R(ω) and X(ω) represent the real and imagina ry parts o f G(ω), respectively.2. (2 points) Show thatlima→0X(ω) =1ω.3. (8 points) Show that(a) ( 2 points) R(ω) = R(−ω ) is an even function of ω.(b) ( 4 points) The area under R(ω) is π, for any value of a > 0.(c) ( 2 points) In one or two sentences, use the results from parts (a) and (b) to conclude thatlima→0R(ω) = πδ(ω).4. (2 points) Use the results from parts (1) to (3) to obtain an expression for F (ω), the Fourier transform of theunit step f unction.5. (4 points) Use the result from part (4), along with an appropriate Fouri er transform property, to determinethe Fourier transform o fh(t) = cos(ωot)u(t)Problem 43: (20 points)1. (15 po ints) By direct integration, find the Laplace transform F (s) and the region of convergence of F (s) forthe following signals where a and b are posi tive real numbers:(a) (2 points) δ(t)(b) (2 points) u(t)(c) (3 points) e−atu(t)(d) (4 points) cos(bt) u(t)(e) (4 points) sin(bt) u(t)2. (5 points) Compare the Fourier and Laplace transforms of the signalsf(t) = δ(t)g(t) = u(t).ExplainF (ω) = F (s)|s=ω,whileG(ω) 6= G(s)|s=ω.Problem 44: (20 points)1. Let F (s) = L{f(t)} denote the unilateral L aplace transform of f(t). Prove the following properties of theLaplace transform, where to≥ 0 is a real constant and sois a complex constant.(a) (2 points) Right shift in time:L{f(t − to)u(t − to)} = F (s)e−sto, to> 0(b) (3 points) Multi plication by t:L{tf(t)} = −ddsF (s)(c) (3 points) Frequency shift:Lnesotf(t)o= F (s − so)2. Using the elementary transform pairs derived in Problem 49 and the properties derived in part 1, find theLaplace transform of the following signals where to, a, and b are positive real parameters.(a) (2 points) u(t − to)(b) (2 points) tu(t)(c) (2 points) te−atu(t)(d) (3 points) e−atcos(bt) u(t)(e) (3 points) e−atsin(bt) u(t)Note that this approach, particula r in the case of the signals considered in parts (c) a nd (d), is much easierthan finding the Laplace transform by direct integration.Problem 45: (20 points)The inverse Laplace transform of F (s) can be calculated by expanding F (s) into partial fractions whose inversetransform is known. Carefully read sections 6.1-3 and B.5 in the text. Using the basic transform pairs in Table 6.1on page 372 of the text and the techniques of partial fraction expansio n, find the inverse Laplace transforms of1. (5 points) F (s) =(s + 1)2s2− s − 62. (5 points) F (s) =2s3+ 2s2+ 17s + 15s(s + 3)(s + 5),3. (5 points) F (s) =2s3+ 15s2+ 40s + 24s(s + 3)(s2+ 4s + 8), a nd4. (5 points) F (s) =4s2+ 7s + 1s(s + 1)2.Problem 46: (20 points)1. (6 points) If the unilateral Laplace transform of f(t) is F (s), show thatLdfdt= sF (s) − f(0−)andLd2fdt2= s2F (s) − sf(0−) −˙f(0−)2. (6 points) Consider a LTI system with input f(t) and output y(t) that has the ODE representationd2ydt2+ 6dydt+ 9y(t) = 18f(t).Using the properties in part 1, determine Y (s) a nd express your answer in the formY (s) = Yzs(s) + Yzi(s),where the zero-state response Yzs(s) depends on f(t) and not the initial conditions y(0−) and ˙y(0−), while thezero-input response Yzi(s) depends on the initia l conditions y(0−) and ˙y(0−), but not the input f(t). A subtle,but yet i mportant advantage of the Laplace transform method is that the Laplace transform of the responsedepends on the initial conditio ns at time 0−and not 0+. Recall that it was first necessary to determine theinitial conditions at time 0+using the initial conditions at time 0−befo re solving an ODE using the classicalmethod (homogeneous plus particular solution).3. (8 points) In part 2, suppose thatf(t) =1 + 3e−2tu(t),y(0−) = 1 and ˙y(0−) = −2. Determine the zero-state response yzs(t), the zero-input response yzi(t), and thetotal response y (t) by computing the inverse Laplace transform of the result in part 2 using the m ethod ofpartial fractions. Compa re your result aga inst that obtained in Problem Set 3, Problem
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