Devre Teorisi-II Ödev-1 Cevap Anahtarı
Cevap-1
[a] 𝐿{20e−500(t−10)𝑢(t−10)}=20e−10s
(𝑠+500)
[b] 𝑓(𝑡)=(5𝑡+20)𝑢(𝑡+4)−(10𝑡+20)𝑢(𝑡+2)+(10𝑡−20)𝑢(𝑡−2)−
(5𝑡−20)𝑢(𝑡−4)
=5(𝑡+4)𝑢(𝑡+4)−10(𝑡+2)𝑢(𝑡+2)+10(𝑡−2)𝑢(𝑡−2)−5(𝑡−4)𝑢(𝑡 −4)
𝐹(𝑠)=5[𝑒4𝑠 −2𝑒2𝑠 +2𝑒−2𝑠 −𝑒−4𝑠]
𝑠2
[c] 𝐿{𝑡}=1
𝑆2 bu nedenle 𝐿{𝑡𝑒−𝑎𝑡}=1
(𝑠+𝑎)2
[d] 𝑓(𝑡)= 𝑐𝑜𝑠ℎ𝑡.𝑐𝑜𝑠ℎƟ+𝑠𝑖𝑛ℎ𝑡.𝑠𝑖𝑛ℎƟ
𝐿{𝑐𝑜𝑠ℎ𝑡}=𝑠
𝑠2−1 𝐿{𝑐𝑜𝑠ℎ𝑡}=1
𝑠2−1
𝐿{cosh(𝑡+Ɵ)} =𝑐𝑜𝑠ℎƟ[𝑠
(𝑠2−1)]+ 𝑠𝑖𝑛ℎƟ[1
𝑠2−1]=𝑠𝑖𝑛ℎƟ+𝑠[𝑐𝑜𝑠ℎƟ]
(𝑠2−1)
Cevap-2
[a] 𝑓(𝑡)=(−8𝑡−80)[𝑢(𝑡+10)−𝑢(𝑡+5)]+8𝑡[𝑢(𝑡+5)−𝑢(𝑡−5)]+
(−8𝑡+80)[𝑢(𝑡−5)−𝑢(𝑡−10)]
=−8(𝑡+10)𝑢(𝑡+10)+16(𝑡+5)𝑢(𝑡+5)−16(𝑡−5)𝑢(𝑡−5)
+8(𝑡−10)𝑢(𝑡−10)
𝐹(𝑠)=8[−𝑒10𝑠+2𝑒5𝑠 −2𝑒−5𝑠 +𝑒−10𝑠]
𝑠2
[b]
