On taking the partial fractions for the response in the question 4, we get?

Question

TuteeHUB · Accepted Answer

Y(s)=k₁/(s+j‚âà√¨‚àö¬¢)+(k₁^{‚Äö√Ñ√∂‚àö√ë‚àö‚â§})/(s+j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s)

TuteeHUB · Answer

Y(s)=k1/(s-j‚âà√¨‚àö¬¢)+(k1‚Äö√Ñ√∂‚àö√ë‚àö‚â§)/(s+j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s)

TuteeHUB · Answer

Y(s)=k1/(s+j‚âà√¨‚àö¬¢)+(k1‚Äö√Ñ√∂‚àö√ë‚àö‚â§)/(s+j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s)

TuteeHUB · Answer

Y(s)=k1/(s+j‚âà√¨‚àö¬¢)+(k1‚Äö√Ñ√∂‚àö√ë‚àö‚â§)/(s-j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s)

TuteeHUB · Answer

Y(s)=k1/(s-j‚âà√¨‚àö¬¢)+(k1‚Äö√Ñ√∂‚àö√ë‚àö‚â§)/(s-j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s)

1.	On taking the partial fractions for the response in the question 4, we get?
A.	Y(s)=k<sub>1</sub>/(s-j‚âà√¨‚àö¬¢)+(k<sub>1</sub><sup>‚Äö√Ñ√∂‚àö√ë‚àö‚â§</sup>)/(s+j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s)
B.	Y(s)=k<sub>1</sub>/(s+j‚âà√¨‚àö¬¢)+(k<sub>1</sub><sup>‚Äö√Ñ√∂‚àö√ë‚àö‚â§</sup>)/(s+j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s)
C.	Y(s)=k<sub>1</sub>/(s+j‚âà√¨‚àö¬¢)+(k<sub>1</sub><sup>‚Äö√Ñ√∂‚àö√ë‚àö‚â§</sup>)/(s-j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s)
D.	Y(s)=k<sub>1</sub>/(s-j‚âà√¨‚àö¬¢)+(k<sub>1</sub><sup>‚Äö√Ñ√∂‚àö√ë‚àö‚â§</sup>)/(s-j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s)
Answer» B. Y(s)=k<sub>1</sub>/(s+j‚âà√¨‚àö¬¢)+(k<sub>1</sub><sup>‚Äö√Ñ√∂‚àö√ë‚àö‚â§</sup>)/(s+j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s)

Discussion