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ps9 - EE 350 PROBLEM SET 9 DUE 3 December 2007 Reading...

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EE 350 PROBLEM SET 9 DUE: 3 December 2007 Reading assignment: Lathi Sections 6.1 through 6.5 and B.5 (pp. 24-33) Recitation sections meet the week of November 26. In problem 44 you will find the Laplace transform of several elementary signals. Problem 45 shows how these basic transform pairs, along with several Laplace transform properties, can be used to quickly find the Laplace transform of more complicated functions without evaluating the Laplace transform integral. The basic transform pairs obtained in Problem 44 will also allow you to find the inverse Laplace transform using the partial fraction expansion method in Problem 46. Problem 44: (15 points) By direct integration, find the Laplace transform F ( s ) and the region of convergence of F ( s ) for the following signals where a and b are positive real numbers: 1. (2 points) δ ( t ) 2. (2 points) u ( t ) 3. (3 points) e - at u ( t ) 4. (4 points) cos( bt ) u ( t ) 5. (4 points) sin( bt ) u ( t ) Problem 45: (20 points) 1. Let F ( s ) = L{ f ( t ) } denote the unilateral Laplace transform of f ( t ). Prove the following properties of the Laplace transform, where t o 0 is a real constant and s o is a complex constant. (a) (2 points) Right shift in time: L{ f ( t - t o ) u ( t - t o ) } = F ( s ) e - st o , t o > 0 (b) (3 points) Multiplication by t: L{ tf ( t ) } = - d ds F ( s ) (c) (3 points) Frequency shift: L e s o t f ( t ) = F ( s - s o ) 2. Using the elementary transform pairs derived in Problem 51 and the properties derived in part 1, find the Laplace transform of the following signals where t o , a , and b are positive real parameters. (a) (2 points) u ( t - t o ) (b) (2 points) tu ( t ) (c) (2 points) te - at u ( t ) (d) (3 points) e - at cos( bt ) u ( t ) (e) (3 points) e - at sin( bt ) u ( t ) Note that this approach, particular in the case of the signals considered in parts (c) and (d), is much easier than finding the Laplace transform by direct integration.
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Problem 46: (15 points) The inverse Laplace transform of F ( s ) can be calculated by expanding F ( s ) into partial fractions whose inverse transform is known. Carefully read sections 6.1-3 and B.5 in the text. Using the basic transform pairs in Table 6.1 on page 372 of the text and the techniques of partial fraction expansion, find the inverse Laplace transforms of 1. (5 points) F ( s ) = 2 s 3 + 2 s 2 + 17 s + 15 s ( s + 3)( s + 5) , 2. (5 points) F ( s ) = 2 s 3 + 15 s 2 + 40 s + 24 ( s )( s + 3)( s 2 + 4 s + 8) , and 3. (5 points) F ( s ) = 4 s 2 + 7 s + 1 s ( s + 1) 2 . Problem 47: (12 points) For a causal and real-valued function f ( t ), a comparison of the unilateral Laplace transform F ( s ) = 0 - f ( t ) e - st dt to the Fourier transform F ( ω ) = -∞ f ( t ) e - ωt dt = 0 f ( t ) e - ωt dt suggests that F ( ω ) = F ( s ) | s = ω .
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