- #1
yungman
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- 290
I want to perform the inverse of
[tex]\frac s { [(s+α)^2-β^2](s^2+ω^2)} [/tex]
I know the conventional way is
[tex]\frac s { [(s+α)^2-β^2](s^2+ω^2)}= \frac{As+B}{[(s+α)^2-β^2]}+\frac{Ds+E}{(s^2+ω^2)} [/tex]
[tex] s= (As+B)(s^2+ω^2)+(Ds+E)[(s+α)^2-β^2][/tex]
[tex]\Rightarrow\; A+D=0,\; B+E+2\alpha A=0,\;\alpha^2A-\beta^2A+2\alpha B +D\omega^2=1,\; B\alpha^2-B\beta^2+E\omega^2=0[/tex]
You then expand and find A, B, D and E.
But this is very complicated and long. on top, α and β is a function of equations of constants. So the algebra really blow up. Is there any easy way to solve this?
Is there any site that you can enter the equation in s domain and it will transform back to the original domain.
[tex]\frac s { [(s+α)^2-β^2](s^2+ω^2)} [/tex]
I know the conventional way is
[tex]\frac s { [(s+α)^2-β^2](s^2+ω^2)}= \frac{As+B}{[(s+α)^2-β^2]}+\frac{Ds+E}{(s^2+ω^2)} [/tex]
[tex] s= (As+B)(s^2+ω^2)+(Ds+E)[(s+α)^2-β^2][/tex]
[tex]\Rightarrow\; A+D=0,\; B+E+2\alpha A=0,\;\alpha^2A-\beta^2A+2\alpha B +D\omega^2=1,\; B\alpha^2-B\beta^2+E\omega^2=0[/tex]
You then expand and find A, B, D and E.
But this is very complicated and long. on top, α and β is a function of equations of constants. So the algebra really blow up. Is there any easy way to solve this?
Is there any site that you can enter the equation in s domain and it will transform back to the original domain.
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