모두CBT
그림과 같은 블록선도에서 출력 C(s)는?R(s)+-G(s)++C(s)D(s)H(s)
G(s)1+G(s)H(s)R(s)+11+G(s)H(s)D(s)\dfrac{G(s)}{1+G(s)H(s)}R(s)+\dfrac{1}{1+G(s)H(s)}D(s)1+G(s)H(s)G(s)R(s)+1+G(s)H(s)1D(s)
11+G(s)H(s)R(s)+11+G(s)H(s)D(s)\dfrac{1}{1+G(s)H(s)}R(s)+\dfrac{1}{1+G(s)H(s)}D(s)1+G(s)H(s)1R(s)+1+G(s)H(s)1D(s)
11+G(s)H(s)R(s)+G(s)1+G(s)H(s)D(s)\dfrac{1}{1+G(s)H(s)}R(s)+\dfrac{G(s)}{1+G(s)H(s)}D(s)1+G(s)H(s)1R(s)+1+G(s)H(s)G(s)D(s)
G(s)1+G(s)H(s)R(s)+G(s)1+G(s)H(s)D(s)\dfrac{G(s)}{1+G(s)H(s)}R(s)+\dfrac{G(s)}{1+G(s)H(s)}D(s)1+G(s)H(s)G(s)R(s)+1+G(s)H(s)G(s)D(s)
가산점 앞뒤 관계를 세우면 C=G(s)[R−H(s)C]+DC=G(s)\left[R-H(s)C\right]+DC=G(s)[R−H(s)C]+D 이다.정리하면 C[1+G(s)H(s)]=G(s)R+DC\left[1+G(s)H(s)\right]=G(s)R+DC[1+G(s)H(s)]=G(s)R+D 이므로 C(s)=G(s)1+G(s)H(s)R(s)+11+G(s)H(s)D(s)C(s)=\dfrac{G(s)}{1+G(s)H(s)}R(s)+\dfrac{1}{1+G(s)H(s)}D(s)C(s)=1+G(s)H(s)G(s)R(s)+1+G(s)H(s)1D(s) 가 된다.즉 외란 D(s)D(s)D(s) 는 G(s)G(s)G(s) 의 뒤쪽에서 가산되므로 분자에 G(s)G(s)G(s) 가 붙지 않는 것이 핵심이며, 정답은 ①이다.