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)
출력 C(s)는 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)이다.가산점 앞뒤 관계를 세우면C=G(s)[R−H(s)C]+DC=G(s)\left[R-H(s)C\right]+DC=G(s)[R−H(s)C]+D정리하면C[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+D이므로C(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.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 핵심이다.
출력 C(s)는 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)이다.
가산점 앞뒤 관계를 세우면C=G(s)[R−H(s)C]+DC=G(s)\left[R-H(s)C\right]+DC=G(s)[R−H(s)C]+D
정리하면C[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+D이므로C(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.
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 핵심이다.
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