Quiz5Soln

Quiz5Soln - “copies” of X ω-kω s(signifying aliasing...

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Name: SOLUTION ECE 301: Quiz 5 Purdue University, Summer 2009 Consider a signal x ( t ) with a Fourier transform X ( ω ) = u ( ω + 2) - u ( ω - 2). Let x ( t ) be sampled by the system pictured below, with H ( ω ) = u ( ω + π T s ) - u ( ω - π T s ) 1. What is the largest value of T s which guarantees that x R ( t ) = x ( t )? Solution: We ±nd that X ( ω ), as plotted below, is band limited to ω m = 2 (i.e. X ( ω ) = 0 for | ω | > ω m ). The Nyquist Sampling Theorem states that we must have the sampling frequency ω s > 2 ω m in order to perfectly reconstruct x ( t ) from x p ( t ). Therefore the sampling period T s = 2 π ω m < 2 π 2 ω m = π 2 , so that π 2 is the upper bound on T s . -3 -2 -1 0 1 2 3 -0.5 0 0.5 1 1.5 ω X( ω ) 1
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2. Consider that T s = 2 π 3 . (a) Plot X p ( ω ) = F{ x p ( t ) } . Solution: Since T s is larger than the upper bound found in 1, we know that there will be aliasing . We have that X p ( ω ) = 1 T s s k = -∞ X ( ω - s ) , where ω s = 2 π T s = 3. We see that there is overlap between adjacent
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Unformatted text preview: “copies” of X ( ω-kω s ) (signifying aliasing), as shown in the left Fgure below. X p ( ω ) is shown in the right Fgure. X p ( ω ) is plotted below:-3-2-1 1 2 3 1/Ts-3-2-1 1 2 3 1/Ts 2/Ts ω X p ( ω ) (b) Plot X R ( ω ) = F{ x R ( t ) } . Solution: We have that X R ( ω ) = H ( ω ) X p ( ω ), where the Flter, H ( ω ) ampliFes everything in the range ω ∈ [-3 2 , 3 2 ), and atten-uates everything else to zero. The left Fgure below shows H ( ω ) and X p ( ω ) on the same axis; the Fgure to the right plots X R ( ω ).-3-2-1 1 2 3 1/Ts 2/Ts Ts ω-3-2-1 1 2 3 0.5 1 1.5 2 ω R 2...
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Quiz5Soln - “copies” of X ω-kω s(signifying aliasing...

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