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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 −...
View Full Document

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.

Page1 / 8

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

This preview shows document pages 1 - 4. Sign up to view the full document.

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