amplifierphasetransfer function
I have a following problem, I have to find amplitude and phase response of an amplifier whose function is $$ H(s)={\frac {1}{s^2 + s + 100}} $$
I have solved simillar problems such as for this function $$ H(s)={\frac {1000s}{s + 10}}$$.
Where I got $$ Ф(jw)=arctg(∞) – arctg(\frac{w}{10}) $$ and $$ 20log|H(jw)| = 20log100 + 20logw – 10log(1 + (\frac{w}{10})^2) $$
Does anyone know how to get amplitude and phase response for the first function?
For the zero state: Find
$$ F(s) =\frac{1} {(s-3)} $$
Which is computed by taking the Laplace transform of course. Now, multiply F(s) with your transfer function. You will have $$ Y(s) = \frac{s} { (s-3)(s^2+s+1)} $$
Now simply, use partial fraction and take Laplace inverse to find y(t)
Solution on wolframalpha:
As for the zero state response: You can find the differential equation by doing the cross multiplication As for s^2 is second derivative and "s" is first derivative: $$ y' ' + y' + y = f ' $$ Take Laplace transform again considering initial conditions: $$ s^2Y(s) - sy(0) - y'(0) + sY(s) - y(o) + Y(s) = sF(s) - f(0) $$ No as for the input F(s) = 0 (for zero input)
So, $$ Y(s) = \frac{sy(0) + y'(0) +y(0)}{(s^2 +s +1)} $$ Now sub for the initial conditions and take Laplace inverse to find y(t).
There seems to be an error in the approach of the transfer function:
$$\frac{V(s)}{V_s(s)}=\frac{\frac{1}{sC}}{R_1+\frac{1}{sC}||(R_2+sL)}$$
when the correct is:
$$ \frac{V(s)}{V_s(s)}=\frac{\frac{1}{sC}||(R_2+sL)}{R_1+\frac{1}{sC}||(R_2+sL)} $$
according to the voltage divider.
Best Answer
First equate:$$ s = j\omega$$ then: $$s^{2} = -\omega^{2}$$ That gets rid of the squared term.
Then multiply top and bottom by the complex conjugate of the divisor and proceed as normal.