모두CBT
입력 r(t)r(t)r(t), 출력 c(t)c(t)c(t)인 제어시스템에서 전달함수 G(s)G(s)G(s)는? (단, 초기값은 000이다.)d2c(t)dt2+3dc(t)dt+2c(t)=dr(t)dt+3r(t)\dfrac{d^2c(t)}{dt^2} + 3\dfrac{dc(t)}{dt} + 2c(t) = \dfrac{dr(t)}{dt} + 3r(t)dt2d2c(t)+3dtdc(t)+2c(t)=dtdr(t)+3r(t)
s2+3s+2s+3\dfrac{s^2+3s+2}{s+3}s+3s2+3s+2
s+3s2+3s+2\dfrac{s+3}{s^2+3s+2}s2+3s+2s+3
3s+12s2+3s+1\dfrac{3s+1}{2s^2+3s+1}2s2+3s+13s+1
s+1s2+3s+2\dfrac{s+1}{s^2+3s+2}s2+3s+2s+1
초기값이 000이므로 미분방정식의 양변을 라플라스 변환하면 dndtn→sn\dfrac{d^n}{dt^n}\to s^ndtndn→sn으로 치환된다.좌변은 (s2+3s+2)C(s)(s^2+3s+2)C(s)(s2+3s+2)C(s), 우변은 (s+3)R(s)(s+3)R(s)(s+3)R(s)가 되므로 전달함수는 G(s)=C(s)R(s)=s+3s2+3s+2G(s)=\dfrac{C(s)}{R(s)}=\dfrac{s+3}{s^2+3s+2}G(s)=R(s)C(s)=s2+3s+2s+3이다.따라서 ② s+3s2+3s+2\dfrac{s+3}{s^2+3s+2}s2+3s+2s+3가 정답이다. 전달함수는 출력/입력이므로 분자·분모를 뒤집은 ① s2+3s+2s+3\dfrac{s^2+3s+2}{s+3}s+3s2+3s+2과 혼동하지 않아야 한다.