The full question is asking: Derive the operation transfer function & the sinusoidal transfer function for an RC high-pass filter. Plot the step-response and the complete frequency response (magnitude and phase).
I have a conceptual question first about circuits in series: Because I am looking at an RC circuit that is in series would the placement of my resistor or capacitor matter? My understanding is that it would not matter because they are in series.
I used x(t)=i(t)R + y(t) I used the relationship between current and capacitance and plugged it into i(t) and took the took the laplace transform and ended up getting my transfer function $H(S)$ to be $H(S) = \frac{\frac{1}{RC}}{S+\frac{1}{RC}}$ which simplifies to $H(S)= \frac{1}{RC}* \frac{1}{S+\frac{1}{RC}}$
However the solution manual has the answer being $H(S)=\frac{D}{D+\frac{1}{RC}}$
I am confused on why my transfer function is not the same.
Update: I realize I am supposed to do voltage division here but I don't quite understand how to accomplish that. I have looked at other sources for voltage division but many are either too complex or simply have resistors and no capacitor in the system.
Well, in general, we can solve this using the following circuit:
The input voltage can be written as:
$$\text{V}_\text{in}\left(t\right)=\hat{\text{v}}\cos\left(\omega t+\varphi\right)\tag1$$
The complex form of the input voltage is given by:
$$\underline{\text{V}}_{\space\text{in}}=\hat{\text{v}}\exp\left(\varphi\text{j}\right)\tag2$$
Using the voltage divider formula, we get:
$$\underline{\text{V}}_{\space\text{Z}_2}=\frac{\underline{\text{Z}}_2}{\underline{\text{Z}}_1+\underline{\text{Z}}_2}\cdot\underline{\text{V}}_{\space\text{in}}\tag3$$
So, the transfer function is given by:
$$\underline{\mathcal{H}}\left(\omega\text{j}\right):=\frac{\underline{\text{V}}_{\space\text{Z}_2}}{\underline{\text{V}}_{\space\text{in}}}=\frac{\underline{\text{Z}}_2}{\underline{\text{Z}}_1+\underline{\text{Z}}_2}\tag4$$
Now, you can use: