As^5.V(s)+Bs^4.V(s)+Cs^3.V(s)+Ds^2.V(s)+Es^1V(s)+F.V(s)=1, where V(s) is the transformed transient output voltage. Rearranging the 5th order transient polynomial equation above we have: V(s) = 1 / H(s) where H(s) is a polynomial is a polynomial in s to be determine. The values of the above constants: A=8,D=30,E=-33,F=9. Find an equation for H(s)in terms of A,B,C,D,E and F.
To find an equation for H(s) in terms of A, B, C, D, E, and F, we need to rearrange the given equation:
As^5.V(s) + Bs^4.V(s) + Cs^3.V(s) + Ds^2.V(s) + Es^1.V(s) + F.V(s) = 1
Let's factor out V(s) from the equation:
V(s) (As^5 + Bs^4 + Cs^3 + Ds^2 + Es + F) = 1
Now, divide both sides of the equation by V(s):
(As^5 + Bs^4 + Cs^3 + Ds^2 + Es + F) = 1/V(s)
We want to isolate V(s) on one side of the equation, so let's rearrange it:
1/V(s) = (As^5 + Bs^4 + Cs^3 + Ds^2 + Es + F)
Now, let's express it as V(s) = 1/(As^5 + Bs^4 + Cs^3 + Ds^2 + Es + F):
V(s) = 1 / (As^5 + Bs^4 + Cs^3 + Ds^2 + Es + F)
Therefore, the equation for H(s) in terms of A, B, C, D, E, and F is:
H(s) = As^5 + Bs^4 + Cs^3 + Ds^2 + Es + F