3 7 b x f fxt fsin2 500t f sinc1000t 1 f 1 rect

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Unformatted text preview: 2 part a. 3 (7) b) X (f ) = F{x(t)} = F{sin(2π (500t))} ∗ F {sinc(1000t)} 1 f 1 rect( )] = [ (δ (f − 500) − δ (f + 500))] ∗ [ 2j 1000 1000 1 f − 500 f + 500 = [rect( ) − rect( )] 2000j 1000 1000 X(f) j (8) The problem statement asks for a sketch of X(f), not |X(f)|. Since x(t) is real and odd function of t, X(f) is an imaginary and odd function of f. I have corrected the plot to show in red what X(f) looks like. Figure 5: Sketch of problem 2 part b. c) Xs (f ) = F{xs (t)} = F{combT (x(t))} 1 1 = rep T {X (f )} T f − 500 f + 500 = rep2000 [−j [rect( ) − rect( )] 1000 1000 ∞ f − 500 − 2000k f + 500 − 2000k = [−j [rect( ) − rect( )] 1000 1000 (9) k=−∞ Again, the problem statement asked for X_s(f), not |X_s(f)|. The red over-plot shows what X_s(f) looks like. X_s(f) j ... ... Figure 6: Sketch of problem 2 part c. 4 It is important to include these ellipses to show that the function repeats periodically d) 1 = 2000 Ts 2πf 2πf = ω= fs 2000 2πf X (ω ) = Xs ( ) 2000 fs = (10) (...
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