Exercise:
A simple low pass filter consists of a resistor and a capacitor connected in series to the input signal. abcliste abc Show that the partial voltage amplitude across the capacitor is given by V_C fracV_sqrtleftfracomegaomega_cright^ + where V_ is the amplitude of the input signal and omega_c the cutoff frequency: omega_c fracsqrtRC Sketch the value of V_C/V_ vs frequency and show that the steepest decrease is at omegaomega_c/sqrt. abc For a typical low pass filter used in audio circuits the input signal has an amplitude of VO and the resistance and capacitance are RO and CO respectively. Calculate the cutoff frequency and the output voltage i.e. the partial voltage across the capacitor at fO. abcliste

Solution:
abcliste abc The impedance for a RC series circuit is Z sqrtR^+fracleftomega Cright^ It follows for the partial voltage across the capacitor V_C X_C I_ fracomega C fracV_Z fracV_omega C sqrtR^+fracleftomega Cright^ fracV_sqrtleftRomega Cright^+ With RC fracomega_c it follows V_C fracV_leftfracomegaomega_cright^+ quad square labeleq:partial The diagram below shows how the partial voltage deps on the frequency. center includegraphicswidthtextwidth#image_path:low-pass-filter-# center The slope is given by the derivative of refeq:partial with respect to omega. For the relevant part of this expression we find using the chain rule fractextdtextdomegaleftleftfracomegaomega_cright^ + right^-/ -fracleftleftfracomegaomega_cright^ + right^-/ fracomegaomega_c^ -leftleftfracomegaomega_cright^ + right^-/ fracomegaomega_c^ labeleq:slope At an extremal po of the slope the derivative of the slope with respect to omega is zero. Deriving refeq:slope again we find using the product rule fractextdtextdomega &left-leftleftfracomegaomega_cright^ + right^-/ fracomegaomega_c^ right & -fracomega_c^ leftleftfracomegaomega_cright^ + right^-/ &quad + frac leftleftfracomegaomega_cright^ + right^-/ fracomegaomega_c^ fracomegaomega_c^ & leftleftfracomegaomega_cright^ + right^-/ leftfracomega^omega_c^-fracleftfracomegaomega_cright^+omega_c^right & leftleftfracomegaomega_cright^ + right^-/ leftfracomega^omega_c^-fracomega^+omega_c^omega_c^right & leftleftfracomegaomega_cright^ + right^-/ fracomega^-omega_c^omega_c^ This expression is zero for omega^ omega_c^ i.e. the slope has its greatest magnitude for omega fracomega_csqrt quad square abc The cutoff angular frequency is omega_c omcF fracRtimes C omc This corresponds to a frequency f_c fracomega_cpi fc approx resultfcP The partial voltage across the capacitor is V_C fracV_sqrtleftfracomegaomega_cright^+ VcF fracVsqrtleftpitimesftimesRtimesCright^+ Vc approx resultVcP- abcliste