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) π‘ ∞ π¦ π‘ = π₯(π‘ − π) β β π β ππ 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 ππ 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 ππ π=−∞ π‘ = ππ’(π) β (π‘ − π) β expβ‘(−2 β (π‘ − π))π’(π‘ − π) β ππ π=−∞ π’ π = 0 when π < 0. Likewise π’ π‘ − π = 0 when π‘ < π. In this case π’ π will set the limits of integration. π‘ = π β (π‘ − π) β expβ‘(−2π‘ + 2π) β‘ β ππ 0 π‘ = π β expβ‘(−2π‘) (π‘ − π) β expβ‘(2π) β ππ 0 π¦(π‘) = π β π‘ π −2π‘ β‘(π‘π 2π − π β π 2π ) β ππ 0 =πβπ −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‘ − + 2β0β − 2 4 4 4 4 =πβ 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‘ = ππ‘ − π β π‘π − ππ‘ + 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 ππππ‘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 πΆ 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 πΆ=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 + 3πΆ = − 4 3π 1 −1 −1 π΅ β = β −3 3 πΆ 4 −1 π΄ π΅ π(π ) = + π π +2 π B=− 4 π C=− 2 πΆ 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 πΆ + 1 π +2 1 π΅π −2π‘ β π’(π‘) 2 β −1 πΆ π +2 2 πΆπ‘π −2π‘ β π’(π‘) 1 1 1 −2π‘ −2π‘ π¦(π‘) = π + ππ + π‘π β π’(π‘) 4 2 4