모두CBT
의 라플라스 함수를 시간 함수로 고치면 어떻게 되는가?
F(t)=e-1-2e-2t
F(t)=e-1+te-2t
F(t)=e-1+e-2t
F(t)=2t+e-t
부분분수 전개:F(s)=2s+3s2+3s+2=2s+3(s+1)(s+2)=As+1+Bs+2F(s)=\dfrac{2s+3}{s^2+3s+2}=\dfrac{2s+3}{(s+1)(s+2)}=\dfrac{A}{s+1}+\dfrac{B}{s+2}F(s)=s2+3s+22s+3=(s+1)(s+2)2s+3=s+1A+s+2B.A=2(−1)+3−1+2=1A=\dfrac{2(-1)+3}{-1+2}=1A=−1+22(−1)+3=1, B=2(−2)+3−2+1=1B=\dfrac{2(-2)+3}{-2+1}=1B=−2+12(−2)+3=1이므로F(s)=1s+1+1s+2F(s)=\dfrac{1}{s+1}+\dfrac{1}{s+2}F(s)=s+11+s+21.라플라스 역변환 공식 L−1[1s+a]=e−at\mathcal{L}^{-1}\left[\dfrac{1}{s+a}\right]=e^{-at}L−1[s+a1]=e−at을 적용하면 f(t)=e^{-t}+e^{-2t}가 된다.
부분분수 전개:F(s)=2s+3s2+3s+2=2s+3(s+1)(s+2)=As+1+Bs+2F(s)=\dfrac{2s+3}{s^2+3s+2}=\dfrac{2s+3}{(s+1)(s+2)}=\dfrac{A}{s+1}+\dfrac{B}{s+2}F(s)=s2+3s+22s+3=(s+1)(s+2)2s+3=s+1A+s+2B.A=2(−1)+3−1+2=1A=\dfrac{2(-1)+3}{-1+2}=1A=−1+22(−1)+3=1, B=2(−2)+3−2+1=1B=\dfrac{2(-2)+3}{-2+1}=1B=−2+12(−2)+3=1이므로F(s)=1s+1+1s+2F(s)=\dfrac{1}{s+1}+\dfrac{1}{s+2}F(s)=s+11+s+21.
라플라스 역변환 공식 L−1[1s+a]=e−at\mathcal{L}^{-1}\left[\dfrac{1}{s+a}\right]=e^{-at}L−1[s+a1]=e−at을 적용하면 f(t)=e^{-t}+e^{-2t}가 된다.
내용에 오류가 있거나 최신 법령·기준과 다른 부분이 보이면 알려주세요. 확인 후 반영하겠습니다.
