모두CBT
그림의 블록선도에서 C(s)R(s)\dfrac{C(s)}{R(s)}R(s)C(s)를 구하면?R(s)+−G₁(s)G₂(s)C(s)G₄(s)G₃(s)
G3(s)G4(s)1+G1(s)G2(s)G3(s)G4(s)\dfrac{G_3(s)G_4(s)}{1+G_1(s)G_2(s)G_3(s)G_4(s)}1+G1(s)G2(s)G3(s)G4(s)G3(s)G4(s)
G1(s)G2(s)1+G1(s)G2(s)+G3(s)G4(s)\dfrac{G_1(s)G_2(s)}{1+G_1(s)G_2(s)+G_3(s)G_4(s)}1+G1(s)G2(s)+G3(s)G4(s)G1(s)G2(s)
G1(s)G2(s)1+G1(s)G2(s)G3(s)G4(s)\dfrac{G_1(s)G_2(s)}{1+G_1(s)G_2(s)G_3(s)G_4(s)}1+G1(s)G2(s)G3(s)G4(s)G1(s)G2(s)
G1(s)+G2(s)1+G1(s)G2(s)+G3(s)G4(s)\dfrac{G_1(s)+G_2(s)}{1+G_1(s)G_2(s)+G_3(s)G_4(s)}1+G1(s)G2(s)+G3(s)G4(s)G1(s)+G2(s)
순방향 경로는 G1(s)G2(s)G_1(s)G_2(s)G1(s)G2(s)이고, 출력 C(s)C(s)C(s)가 G4(s)G_4(s)G4(s)와 G3(s)G_3(s)G3(s)를 거쳐 입력 측 가산점으로 되돌아오는 부(−)궤환 경로를 이루므로 궤환 전달함수는 G3(s)G4(s)G_3(s)G_4(s)G3(s)G4(s)이다.부궤환 폐루프의 전달함수 공식 CR=G1+GH\dfrac{C}{R}=\dfrac{G}{1+GH}RC=1+GHG에 대입하면 C(s)R(s)=G1(s)G2(s)1+G1(s)G2(s)G3(s)G4(s)\dfrac{C(s)}{R(s)}=\dfrac{G_1(s)G_2(s)}{1+G_1(s)G_2(s)G_3(s)G_4(s)}R(s)C(s)=1+G1(s)G2(s)G3(s)G4(s)G1(s)G2(s)가 되어 ③이 정답이다.