102_1_hw03_sol

102_1_hw03_sol - 1 EE102 Spring 2009-10 Systems and Signals...

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1 EE102 Spring 2009-10 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, ±nd a function h ( t ) such that y ( t )=( x * h )( t ) . (a) y ( t )= ± t -∞ x ( τ ) Solution: We want to ±nd an h ( t ) so that y ( t ± t -∞ x ( τ ) = ± -∞ x ( τ ) h ( t - τ ) 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 ± -∞ u ( τ ) u ( t - τ ) =( 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 ± t t - T x ( τ ) Solution: y ( t ± t t - 1 x ( τ ) = ± t -∞ x ( τ ) - ± t - T -∞ x ( τ ) x * h 0 )( t ) - ( x * h T )( t )
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2 where h 0 ( 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 0 - h T ))( t ) =( x * h )( t ) where h ( t h 0 ( 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 0 1 (c) y ( t x ( t ) Solution: Here we want a convolution system that simply returns the input y ( t ± -∞ x ( τ ) h ( t - τ ) If h ( t ) is an an impulse function, y ( t ± -∞ x ( τ ) δ ( t - τ ) = 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 ± -∞ x ( τ ) h ( t - τ ) If we let h ( t ) be a delayed impulse δ ( t - 1) , then y ( t ± -∞ x ( τ ) δ (( t - τ ) - 1) = ± -∞ x ( τ ) δ (( t - 1) - τ ) - 1) = x ( t - 1) again by the sifting property. So h ( t δ ( t - 1) .
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3 2. Graphically compute the convolution of these two functions: 3 -1 0 1 2 1 2 f ( t ) t 3 -1 0 1 2 1 2 t g ( t ) Solution: The convoltion can be graphically performed by Fipping 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 ±rst overlap begins at t =2 , so this is the ±rst output point. The area of the overlap is ± t 2 4( τ - 2) = (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 . ²rom t 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 to t , so the result is the same shape, but inverted. We get a smooth, quadratic signal, concave down, that starts at an amplitude of 2 at t , and goes to 0 at t , when the trangle and the rectangle no longer overlap. These breakpoints are illustrated below: 3 -1 0 1 2 1 2 t f ( 0 - ! ) g ( ! ) 3 -1 0 1 2 1 2 t g ( ! ) f ( 2 - ! ) 3 -1 0 1 2 1 2 t g ( ! ) f ( 2 . 5 - ! ) t = 0 t = 2 t = 2 . 5 3 -1 0 1 2 1 2 t g ( ! ) 3 -1 0 1 2 1 2 t g ( ! ) f ( 3 - ! ) f ( 3 . 5 - ! ) 3 -1 0 1 2 1 2 t g ( ! ) f ( 4 - ! ) t = 3 t = 3 . 5 t = 4 The result is 3 -1 0 1 2 1 2 t 4 5 ( f * g )( t )
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4 3. Analytically compute the convolution ( f * g )( t ) , where f ( t ) and g ( t ) are f ( t )= u ( t ) g ( t ) = rect( t - 1) and plot the result.
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102_1_hw03_sol - 1 EE102 Spring 2009-10 Systems and Signals...

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