Time Domain Convolution ECE 3100 Western Michigan University John Stahl

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Time Domain Convolution
ECE 3100
Western Michigan University
John Stahl
Probabilistic Methods Signal & System Analysis , 3rd edition, George R. Cooper, 1999
Time Domain
x(t)
h(t)
y(t)
𝑑
∞
𝑦 𝑑 =
π‘₯(𝑑 − πœ†) βˆ™ β„Ž πœ† βˆ™ π‘‘πœ†
or
πœ†=0
Laplace Domain
X(s)
H(s)
π‘Œ 𝑠 = 𝐻(𝑠) βˆ™ X(s)
Y(s)
𝑦 𝑑 =
π‘₯(πœ†) βˆ™ β„Ž 𝑑 − πœ† βˆ™ π‘‘πœ†
πœ†=−∞
Time Domain Convolution
π‘₯ 𝑑 = 𝑀 βˆ™ 𝑒(𝑑)
Where M is a DC value.
β„Ž 𝑑 = 𝑑 βˆ™ exp⁑(−2𝑑) βˆ™ 𝑒(𝑑)
𝑑
𝑦 𝑑 =
π‘₯(πœ†) βˆ™ β„Ž 𝑑 − πœ† βˆ™ π‘‘πœ†
πœ†=−∞
𝑑
=
𝑀𝑒(πœ†) βˆ™ (𝑑 − πœ†) βˆ™ exp⁑(−2 βˆ™ (𝑑 − πœ†))𝑒(𝑑 − πœ†) βˆ™ π‘‘πœ†
πœ†=−∞
𝑒 πœ† = 0 when πœ† < 0. Likewise 𝑒 𝑑 − πœ† = 0 when 𝑑 < πœ†. In this
case 𝑒 πœ† will set the limits of integration.
𝑑
=
𝑀 βˆ™ (𝑑 − πœ†) βˆ™ exp⁑(−2𝑑 + 2πœ†) ⁑ βˆ™ π‘‘πœ†
0
𝑑
= 𝑀 βˆ™ exp⁑(−2𝑑)
(𝑑 − πœ†) βˆ™ exp⁑(2πœ†) βˆ™ π‘‘πœ†
0
𝑦(𝑑) = 𝑀 βˆ™
𝑑
𝑒 −2𝑑
⁑(𝑑𝑒 2πœ† − πœ† βˆ™ 𝑒 2πœ† ) βˆ™ π‘‘πœ†
0
=π‘€βˆ™π‘’
−2𝑑
𝑑 2πœ† 𝑒 2πœ†
βˆ™ 𝑒 − 2 2πœ† − 1 |𝑑0
2
2
Slowly and carefully apply the
limits to each term. Be careful of
sign errors.
2𝑑
2𝑑
0
0
𝑑
𝑒
𝑒
𝑒
𝑒
= 𝑀 βˆ™ 𝑒 −2𝑑 βˆ™ (𝑒 2𝑑 −𝑒 0 ) − 2𝑑
−
+ 2βˆ™0βˆ™ −
2
4
4
4
4
=π‘€βˆ™
𝑒 −2𝑑
𝑑 2𝑑
𝑒 2𝑑 𝑒 2𝑑
1
βˆ™ (𝑒 −1) − 2𝑑
−
+ 0−
2
4
4
4
2𝑑
2𝑑
𝑑
𝑒
𝑒
1
= 𝑀 βˆ™ 𝑒 −2𝑑 βˆ™ (𝑒 2𝑑 −1) − 2𝑑
−
+ 0−
2
4
4
4
1
1
1
1
1
−2𝑑
= 𝑀𝑑 − 𝑀 βˆ™ 𝑑𝑒
− 𝑀𝑑 + 𝑀 + 0 − 𝑀 βˆ™ 𝑒 −2𝑑
2
2
2
4
4
𝑦 𝑑 =
1
1
1
𝑀 − 𝑀 βˆ™ 𝑒 −2𝑑 − 𝑀 βˆ™ 𝑑𝑒 −2𝑑
4
4
2
Laplace Domain
X(s)
H(s)
Convert the x(t) and
h(t) into X(s) and H(s).
Y(s)
π‘Œ 𝑠 = 𝐻(𝑠) βˆ™ X(s)
π‘₯ 𝑑 = 𝑀 βˆ™ 𝑒(𝑑)
β„Ž 𝑑 = 𝑑 βˆ™ exp⁑(−2𝑑) βˆ™ 𝑒(𝑑)
Use the Laplace table.
𝑀
𝑠
𝑋 𝑠 =β„’ π‘₯ 𝑑
=
𝐻 𝑠 =β„’ β„Ž 𝑑
1
=
𝑠+2
2
π‘Œ 𝑠 = 𝐻(𝑠) βˆ™ X(s)
1
π‘Œ 𝑠 =
𝑠+2
𝑀
βˆ™
2 𝑠
π‘Ÿπ‘œπ‘œπ‘‘π‘ :
𝑠=0
𝑠 = −2
Partial Fraction Expansion with repeated roots
1
𝑠+2
2βˆ™
𝑀 𝐴
𝐡
= +
𝑠
𝑠
𝑠+2
𝐴 𝑠+2 𝑠+2
2
𝐢
𝑠+2
1+
+ Bs 𝑠 + 2
@𝑠 = 0
2
2
+ Cs 𝑠 + 2 = 𝑀 𝑠 + 2
@𝑠 = −2
8𝐴 + 0 = 2𝑀
0+0= 0
𝐡=0
𝑀
𝐴=
4
Fails for repeated
roots
𝐢=0
With repeated roots:
Eq 1. @𝑠 = −1
𝐴 + B(−1) 1
Eq 2. @𝑠 = −3
𝐴 −1 −1
2
2
+ C(−1) 1 = 𝑀 1
+ B(−3) −1
2
+ C(−3) −1 = 𝑀 −1
Eq 1.
𝑀
−B−C=𝑀
4
3𝑀
−B − C =
4
Eq 2.
−
𝑀
− 3B + 3C = −𝑀
4
3𝑀
−3B + 3𝐢 = −
4
3𝑀 1
−1 −1 𝐡
βˆ™
=
βˆ™
−3 3
𝐢
4 −1
𝐴
𝐡
π‘Œ(𝑠) = +
𝑠
𝑠+2
𝑀
B=−
4
𝑀
C=−
2
𝐢
1+ 𝑠+2
2
Take Y(s) back into the time-domain.
𝐴
𝐡
π‘Œ(𝑠) = +
𝑠
𝑠+2
β„’ −1
𝐴
𝑠
β„’ −1
𝐴 βˆ™ 𝑒(𝑑)
𝐡
𝑠+2
𝐢
+
1
𝑠+2
1
𝐡𝑒 −2𝑑 βˆ™ 𝑒(𝑑)
2
β„’ −1
𝐢
𝑠+2
2
𝐢𝑑𝑒 −2𝑑 βˆ™ 𝑒(𝑑)
1
1
1 −2𝑑
−2𝑑
𝑦(𝑑) =
𝑀 + 𝑀𝑒
+ 𝑑𝑒
βˆ™ 𝑒(𝑑)
4
2
4
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