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Unformatted text preview: 1 EE102 Spring 200910 Lee Systems and Signals Homework #3 Solutions Due: Tuesday, April 27, 2010 at 5 PM. 1. It is often useful to represent operations on signals as convolutions. For each of the follow ing, find a function h ( t ) such that y ( t ) = ( x * h )( t ) . (a) y ( t ) = Z t∞ x ( τ ) dτ Solution: We want to find an h ( t ) so that y ( t ) = Z t∞ x ( τ ) dτ = Z ∞∞ x ( τ ) h ( t τ ) dτ From the lecture notes, if h ( t ) was causal, the upper limit of the convolution would be t . In addition, we want h ( t ) to be of unit amplitude, so that we get x ( τ ) when we multiply by h ( t τ ) . This means that h ( t ) = u ( t ) , since h ( t ) is one for positive time, and zero for negative time. Then y ( t ) = Z ∞∞ u ( τ ) u ( t τ ) dτ = ( x * u )( t ) . where u ( t ) is the unit step. We conclude that y ( t ) = ( x * h )( t ) with h ( t ) = u ( t ) . 1 t h ( t ) = u ( t ) (b) y ( t ) = Z t t T x ( τ ) dτ Solution: y ( t ) = Z t t 1 x ( τ ) dτ = Z t∞ x ( τ ) dτ Z t T∞ x ( τ ) dτ = ( x * h )( t ) ( x * h T )( t ) 2 where h ( t ) = u ( t ) is the unit step, as in (a), and h T ( t ) = u ( t T ) is the unit step delayed by T . Then y ( t ) = ( x * ( h h T ))( t ) = ( x * h )( t ) where h ( t ) = h ( t ) h T ( t ) = u ( t ) u ( t T ) . This is a square pulse of of amplitude 1, tha goes from time zero to T . t h ( t ) = u ( t ) u ( t T ) T T 2 T 1 (c) y ( t ) = x ( t ) Solution: Here we want a convolution system that simply returns the input y ( t ) = Z ∞∞ x ( τ ) h ( t τ ) dτ If h ( t ) is an an impulse function, y ( t ) = Z ∞∞ x ( τ ) δ ( t τ ) dτ = x ( t ) by the sifting property. So h ( t ) = δ ( t ) . (d) y ( t ) = x ( t 1) Solution: This will be similar to (c), with an addition that the output should be delayed, y ( t ) = Z ∞∞ x ( τ ) h ( t τ ) dτ If we let h ( t ) be a delayed impulse δ ( t 1) , then y ( t ) = Z ∞∞ x ( τ ) δ (( t τ ) 1) dτ = Z ∞∞ x ( τ ) δ (( t 1) τ ) 1) dτ = x ( t 1) again by the sifting property. So h ( t ) = δ ( t 1) . 3 2. Graphically compute the convolution of these two functions: 31 1 2 1 2 f ( t ) t 31 1 2 1 2 t g ( t ) Solution: The convoltion can be graphically performed by flipping f about the origin, shifting it right by t , multiplying point by point with g ( t ) , and then evaluating the area. At time t = 0 there is no overlap, and the output is zero. The first overlap begins at t = 2 , so this is the first output point. The area of the overlap is Z t 2 4( τ 2) dτ = (1 / 2)( t 2)(4( t 2)) = 2( t 2) 2 for 2 < t < 3 . This is a smoothly increasing quadratic. over this interval, which reaches a peak amplitude of 2 at t = 3 . From t = 3 to t = 4 , the area of the overlap starts to decrease slowly, as the initial part of the triangle no longer overlaps the rectangle. The rate at which we loose area is exactly the same as the rate we gained area from t = 2 to t = 3 , so the result is the same shape, but inverted. We get a smooth, quadratic signal, concave down,result is the same shape, but inverted....
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This note was uploaded on 10/21/2010 for the course EE ee102 taught by Professor Levan during the Spring '09 term at UCLA.
 Spring '09
 Levan

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