Currently I am learning control systems.
I specifically find the mathematics very hard to understand.
For my exam I am practicing on this question:
"Determine the transfer function of this circuit in the S-domain."
I came to here:
I don't know if it's correct and I don't know how to simplify this.
Could you please explain me in steps how to solve this mathematically?
Thank you in advance for your help!
Best Answer
Edit: In order to address your question more completely, the step you have missed in completing the transfer function is to substitute in the circuit element impedances for the components in the circuit. Also, you you did calculate the parallel impedance of \$R_2\$ and \$C\$ incorrectly.
In order to find the transfer function \$H(s)\$ of this circuit we use the voltage divider rule, that is: \begin{equation} H(s) = \frac{U_{out}}{U_{in}}(s) = \frac{Z_2}{Z_1 + Z_2} \end{equation} where, \begin{equation} Z_2 = R_2||\frac{1}{sC} \qquad Z_1 = R_1 \end{equation} so we now have, \begin{equation} Z_2 = \left(\frac{1}{R_2} + sC\right)^{-1} = \left(\frac{sCR_2 + 1}{R_2}\right)^{-1} = \frac{R_2}{sCR_2 + 1}. \end{equation}
From here we can derive the transfer function using the previously stated voltage divider rule, substituting in the impedances for \$Z_1\$ and \$Z_2\$, \begin{equation} \begin{split} H(s) &= \frac{\frac{R_2}{sCR_2 + 1}}{\frac{R_2}{sCR_2 + 1} + R_1} = \frac{R_2}{R_2 + R_1 + sCR_1R_2} \\ &=\frac{\frac{1}{R_1C}}{s + \frac{R_1 + R_2}{R_1R_2C}}= \frac{\frac{1}{R_1C}}{s + \frac{1}{R_1||R_2C}}. \end{split} \end{equation} Note how the denominator of the transfer function is normalized to the coefficient of the largest power of \$s\$. From this transfer function if we look at the limiting behavior as \$s \rightarrow 0\$ or \$s \rightarrow \infty\$ we see that the circuit exhibits a "low-pass" behavior.
Since often in controls the impulse response \$H(s)\$ is not what you will be interested in, to find the time domain response to say a step (Heaviside) function you would multiply the impulse response function by the Laplace transform of the time domain voltage input, \begin{equation} U_{out}(s) = H(s)U_{in}(s), \end{equation} and taking the inverse Laplace transform of the result you arrive at the time domain response of your circuit to the given input. \begin{equation} U_{out}(t) = \scr{L}^{-1}\lbrace H(s) U_{in}(s) \rbrace \end{equation}