R–C Circuit as a Low-Pass Filter
Background information
Voltage divider: With increasing frequency, the capacitive reactance of capacitor $C_{\rm 1}$ decreases, which leads to a reduction in the amplitude of the output voltage $u_{\rm 2}$. The ratio of the output voltage to the input voltage as a function of frequency is called the transfer function:
$$F(j\omega) = \frac {U_{\rm 2}}{U_{\rm 1}}$$
Applying the voltage divider rule yields the following expression for the transfer function:
$$F(j\omega) = \frac {U_{\rm 2}}{U_{\rm 1}} = \frac{\frac{1}{j\omega C_{\rm 1}}}{R_{\rm 1}+\frac{1}{j\omega C_{\rm 1}}} = \frac{1}{1+j\omega C_{\rm 1}R_{\rm 1}}$$
The transfer function depends on the angular frequency ($\omega = 2\pi \cdot f$) and can be expressed in terms of magnitude and phase.
The magnitude $|\frac{U_{\rm 2}}{U_{\rm 1}}| = \frac{1}{\sqrt 1+(\omega R_{\rm 1}C_{\rm 1})^2}$ is referred to as the amplitude response.
The phase angle $\varphi$ between the input voltage $u_{\rm 1}$ and the output voltage $u_{\rm 2}$ is called the phase response. The following relation applies:
$$\varphi_{\rm U_{\rm 2}}=−\arctan\omega R_{\rm 1}C_{\rm 1}$$
The cutoff angular frequency $\omega_{\rm G}$ is defined as the frequency at which the following conditions apply to the amplitude ratio $\frac{U_{\rm 2}}{U_{\rm 1}}$ and the phase angle $\varphi$:
$$\frac{U_{\rm 2}}{U_{\rm 1}}=\frac{1}{\sqrt2}, ~~~ \varphi=−45°$$
Experimental Tasks
- Assemble the R–C measurement circuit shown in figure 1:
The circuit can be interpreted as a voltage divider.
- Calculate the cutoff angular frequency for the given filter.
$\omega_{\rm G}$ = …………………………….
From this, the cutoff frequency follows as
$f_{\rm G}$ = ………………………
- Measure the amplitude response of the low-pass filter.
The sinusoidal input voltage generated by the function generator shall have an amplitude of 6 V. The voltages $u_{\rm 1}$ and $u_{\rm 2}$ to be measured are connected to the channels of the oscilloscope.
- Draw the amplitude response on the provided logarithmic graph paper.
