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 318 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 319 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 320 M. J. Roberts  8/16/04 t < 3...
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 Spring '09
 ATOUSA
 LTI system theory, Impulse response, M. J. Roberts

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