Let us assume x (t) = A cos(ωt + φ), on taking the partial fractions for the response we get?

Let us assume x (t) = A cos(ωt + φ), on taking the partial fractions for the response we get? Correct Answer Y(s)=k1/(s-jω)+(k1‘)/(s+jω)+Σterms generated by the poles of H(s)

By taking partial fractions, Y(s)=k1/(s-jω)+(k1‘)/(s+jω)+Σterms generated by the poles of H(s). The first two terms result from the complex conjugate poles of the deriving source.

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On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at a point A, the slope of the reversible adiabatic line $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{S}}}$$  and the slope of the reversible isothermal line $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{T}}}$$  are related as (where, $${\text{y}} = \frac{{{{\text{C}}_{\text{p}}}}}{{{{\text{C}}_{\text{v}}}}}$$  )