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Chap3StudentSolutions

21 sketch these functions a g t rect 4 t b g

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Unformatted text preview: nction was defined as it was, so it could simply be the convolution of a unit rectangle with itself. This convolution can also be done analytically. For t < −1 , in the range of integration, − 1 1 < τ < , the rect function is zero and the 2 2 convolution integral is zero. For t > 1, in the range of integration, − 1 1 < τ < , the rect function is zero and the 2 2 convolution integral is zero. For −1 < t < 0 . Since the rect function is even we can say that rect ( t − τ ) = rect (τ − t ) . 1 1 This is a rectangle extending in τ from t − to t + . For t’s in the range, −1 < t < 0 , 2 2 1 1 1 t − is always less than or equal to the lower limit, τ = − , so the integral is from − to 2 2 2 1 t+ . 2 t+ g (t ) = 1 2 ∫ rect (τ − t ) dτ − 1 2 This is simply the accumulation of the area under a rectangle and therefore increases linearly from a minimum of zero for t = −1 to a maximum of one for t = 0 . Solutions 3-18 M. J. Roberts - 8/16/04 1 1 to t + . For t’s in the 2 2 1 1 range, 0 < t < 1 , t + is always greater than or equal to the upper limit, τ = , so the 2 2 1 1 integral is from t − to . 2 2 For 0 < t < 1 . This is also rectangle extending in τ from t − g (t ) = 1 2 ∫ rect (τ − t ) dτ t− 1 2 This is also the accumulation of the area under a rectangle and decreases linearly from a maximum of one for t = 0 to a minimum of zero for t = 1. t (b) g( t) = rect ( t) ∗ rect 2 This convolution is easily done graphically. t (c) g(t ) = rect(t − 1) ∗ rect 2 (d) g( t) = [rect ( t − 5) + rect ( t + 5)] ∗ [rect ( t − 4 ) + rect ( t + 4 )] Break this convolution down into the sum of four simpler convolutions. 21. Sketch these functions. (a) g( t) = rect ( 4 t) (b) g( t) = rect(4 t) ∗ 4δ ( t) (c) g( t) = rect ( 4 t) ∗ 4δ ( t − 2) (d) g( t) = rect ( 4 t) ∗ 4δ (2 t) Don’t forget the scaling property of the CT impulse. (e) g( t) = rect ( 4 t) ∗ comb( t) Convolution with a comb is relatively easy because it is simply convolution with a periodic sequence of impulses. g(t) 1 ... -2 -1 -1 1 88 ... 1 t Solutions 3-19 M. J. Roberts - 8/16/04 (f) g( t) = rect ( 4 t) ∗ comb( t − 1) This result is identical to the result of part (e). (g) g( t) = rect ( 4 t) ∗ comb(2 t) Don’t forget the scaling property of the CT impulses in the comb function. The average value of g ( t ) is 1/4. (h) g( t) = rect ( t) ∗ comb(2 t) This is the sum of multiple rectangle functions periodically repeated. 22. Plot these convolutions. (a) t + 1 t + 2 t g( t) = rect ∗ [δ ( t + 2) − δ ( t + 1)] = rect − rect 2 2 2 g(t) 1 -4 1 t -1 (b) g ( t ) = rect ( t ) ∗ tri ( t ) This is a challenging convolution because it is not so simple to do graphically (although you can get a rough idea of what it looks like that way) and it is tedious analytically. g (t ) = ∞ 1 2 ∫ rect (τ ) tri (t − τ ) dτ = ∫ tri (t − τ ) dτ −∞ − 1 2 Solutions 3-20 M. J. Roberts - 8/16/04 t < -3...
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