# Time Domain Convolution ECE 3100 Western Michigan University John Stahl

```Time Domain Convolution
ECE 3100
Western Michigan University
John Stahl
Probabilistic Methods Signal &amp; System Analysis , 3rd edition, George R. Cooper, 1999
Time Domain
x(t)
h(t)
y(t)
π‘
∞
π¦ π‘ =
π₯(π‘ − π) β β π β ππ
or
π=0
Laplace Domain
X(s)
H(s)
π π  = π»(π ) β X(s)
Y(s)
π¦ π‘ =
π₯(π) β β π‘ − π β ππ
π=−∞
Time Domain Convolution
π₯ π‘ = π β π’(π‘)
Where M is a DC value.
β π‘ = π‘ β expβ‘(−2π‘) β π’(π‘)
π‘
π¦ π‘ =
π₯(π) β β π‘ − π β ππ
π=−∞
π‘
=
ππ’(π) β (π‘ − π) β expβ‘(−2 β (π‘ − π))π’(π‘ − π) β ππ
π=−∞
π’ π = 0 when π &lt; 0. Likewise π’ π‘ − π = 0 when π‘ &lt; π. In this
case π’ π will set the limits of integration.
π‘
=
π β (π‘ − π) β expβ‘(−2π‘ + 2π) β‘ β ππ
0
π‘
= π β expβ‘(−2π‘)
(π‘ − π) β expβ‘(2π) β ππ
0
π¦(π‘) = π β
π‘
π −2π‘
β‘(π‘π 2π − π β π 2π ) β ππ
0
=πβπ
−2π‘
π‘ 2π π 2π
β π − 2 2π − 1 |π‘0
2
2
Slowly and carefully apply the
limits to each term. Be careful of
sign errors.
2π‘
2π‘
0
0
π‘
π
π
π
π
= π β π −2π‘ β (π 2π‘ −π 0 ) − 2π‘
−
+ 2β0β −
2
4
4
4
4
=πβ
π −2π‘
π‘ 2π‘
π 2π‘ π 2π‘
1
β (π −1) − 2π‘
−
+ 0−
2
4
4
4
2π‘
2π‘
π‘
π
π
1
= π β π −2π‘ β (π 2π‘ −1) − 2π‘
−
+ 0−
2
4
4
4
1
1
1
1
1
−2π‘
= ππ‘ − π β π‘π
− ππ‘ + π + 0 − π β π −2π‘
2
2
2
4
4
π¦ π‘ =
1
1
1
π − π β π −2π‘ − π β π‘π −2π‘
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 π  = π»(π ) β X(s)
π₯ π‘ = π β π’(π‘)
β π‘ = π‘ β expβ‘(−2π‘) β π’(π‘)
Use the Laplace table.
π
π
π π  =β π₯ π‘
=
π» π  =β β π‘
1
=
π +2
2
π π  = π»(π ) β X(s)
1
π π  =
π +2
π
β
2 π
ππππ‘π :
π =0
π  = −2
Partial Fraction Expansion with repeated roots
1
π +2
2β
π π΄
π΅
= +
π
π
π +2
π΄ π +2 π +2
2
πΆ
π +2
1+
+ Bs π  + 2
@π  = 0
2
2
+ Cs π  + 2 = π π  + 2
@π  = −2
8π΄ + 0 = 2π
0+0= 0
π΅=0
π
π΄=
4
Fails for repeated
roots
πΆ=0
With repeated roots:
Eq 1. @π  = −1
π΄ + B(−1) 1
Eq 2. @π  = −3
π΄ −1 −1
2
2
+ C(−1) 1 = π 1
+ B(−3) −1
2
+ C(−3) −1 = π −1
Eq 1.
π
−B−C=π
4
3π
−B − C =
4
Eq 2.
−
π
− 3B + 3C = −π
4
3π
−3B + 3πΆ = −
4
3π 1
−1 −1 π΅
β
=
β
−3 3
πΆ
4 −1
π΄
π΅
π(π ) = +
π
π +2
π
B=−
4
π
C=−
2
πΆ
1+ π +2
2
Take Y(s) back into the time-domain.
π΄
π΅
π(π ) = +
π
π +2
β −1
π΄
π
β −1
π΄ β π’(π‘)
π΅
π +2
πΆ
+
1
π +2
1
π΅π −2π‘ β π’(π‘)
2
β −1
πΆ
π +2
2
πΆπ‘π −2π‘ β π’(π‘)
1
1
1 −2π‘
−2π‘
π¦(π‘) =
π + ππ
+ π‘π
β π’(π‘)
4
2
4
```