Ps2soln - MIT OpenCourseWare http/ocw.mit.edu 2.161 Signal Processing Continuous and Discrete Fall 2008 For information about citing these

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Solution of Problem Set 2 Assigned: Sept. 18, 2008 Due: Sept. 25, 2008 Problem 1: Y ( j Ω) = H ( j Ω) F ( j Ω) and it is shown in the below figure Y ( jw ) 10 1 ∞ y ( t ) = Y ( j Ω) e j Ω t dt 2 π −∞ 10 400 − 200 = ( e j Ω t dt + e j Ω t dt ) 2 π 200 − 400 10 = ( e 400 jt − e 200 jt − e − 400 jt + e − 200 jt ) 2 πjt 10 = (sin(400 t ) − sin(200 t )) πt It is not a casual filter (since y ( t ) = negationslash 0 for some t < 0). h ( t ) / 5 and h ( t ) = Problem 2: We define H ∗ ( j Ω) = 1 − H ( j Ω). Then H ∗ ( j Ω) is a low pas filter, matching Prob. 5 in PS 1, which we have already found it’s impulse response: H ( j Ω) = 1 − H ∗ ( j Ω) F − 1 ( H ( j Ω)) = F − 1 (1) −F − 1 ( H ∗ ( j Ω)) h ( t ) = δ ( t ) − h ∗ ( t ) h ( t ) = δ ( t ) − sin(Ω c t ) πt 1 Problem 3: ∞ ∞ x ( t ) = A n sin( n Ω t + φ n ) = A n (sin( n Ω t ) cos φ n + cos( n Ω t ) sin φ n ) n =0 n =0 ∞ X ( j Ω) = A n ( F (sin( n Ω t )) cos φ n + F (cos( n Ω t )) sin φ n ) n =0 ∞ X ( j Ω) = A n (cos φ n ( − jπ ( δ (Ω − n Ω ) − δ (Ω + n Ω ))) + sin φ n ( π ( δ (Ω − n Ω ) + δ (Ω + n Ω )))) n =0 ∞ X ( j Ω) = − jπ A n ((cos φ n + j sin φ n ) ( δ (Ω − n Ω ) + ( − cos φ n + j sin φ n ) ( δ (Ω + n Ω )) n =0 ∞ X ( j Ω) = − jπ A n e jφ n δ (Ω − n Ω ) − e − jφ n δ (Ω + n Ω ) n =0 Problem 4: (a) The ideal multiplicative filtering operation is a low pass filtering with the pass-band Ω c = N Ω : 1 | n | ≤ N, | Ω | ≤ Ω c H n = 0 | n | > N, | Ω | > Ω c X n = X n H n (b) It’s a convolution in this specific form: 1 T 2 x ( t ) = x ( t ) ⊗ h ( t ) = x ( τ ) h ( t − τ ) dτ T − T 2 We can prove that why convolution is in this specific integral form for our Periodic Exponential Fourier Transform: 1 T ∞ ∞ X n H n e jn Ω t = 2 − jn Ω τ dτ e jn Ω t x ( t ) = X n h ( τ ) e T − T n = −∞ n = −∞ 2 ∞ jn Ω t ∞ jn Ω ( t − τ ) 1 1 T T X n e X n e 2 2 − jn Ω τ x ( t ) = h ( τ ) dτ = h ( τ ) dτ e T T T T −...
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This note was uploaded on 02/27/2012 for the course MECHANICAL 2.161 taught by Professor Derekrowell during the Fall '08 term at MIT.

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Ps2soln - MIT OpenCourseWare http/ocw.mit.edu 2.161 Signal Processing Continuous and Discrete Fall 2008 For information about citing these

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