High Pass Filter
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Exercise:
A simple high pass filter consists of a resistor and an inductor connected in series to the input signal. abcliste abc Show that the partial voltage amplitude across the inductor is given by V_C fracV_sqrtleftfracomega_comegaright^ + where V_ is the amplitude of the input signal and omega_c the cutoff frequency: omega_c sqrtfracRL What is the partial voltage at the cutoff frequency? Sketch the value of V_L/V_ vs frequency. abc For a typical high pass filter used in audio circuits the input signal has an amplitude of VO and the resistance and inductance are RO and LO respectively. Calculate the cutoff frequency and the frequency at which the output voltage i.e. the partial voltage across the inductor is VLO. abcliste
Solution:
abcliste abc The impedance for an RL series circuit is Z sqrtR^+leftomega Lright^ It follows for the partial voltage across the capacitor V_L X_L I_ omega L fracV_Z fracV_ omega LsqrtR^+leftomega Lright^ fracV_sqrtleftfracRomega Lright^+ With fracRL omega_c it follows V_L fracV_leftfracomega_comegaright^+ quad square For omegaomega_c we find V_L fracV_sqrtleftfracomega_comega_cright^+ fracV_sqrt i.e. the voltage across the inductor is sqrt times smaller than the input voltage. vspacemm The diagram below shows how the partial voltage deps on the frequency. center includegraphicswidthtextwidth#image_path:high-pass-filter-# center abc The cutoff angular frequency is omega_c omcF fracRL omc This corresponds to a frequency f_c fracomega_cpi fc approx resultfcP The partial voltage across the inductor is V_L fracV_sqrtleftfracomega_comegaright^+ fracV_sqrtleftfracRpi f Lright^+ Solving for the frequency yields f fF fracRpitimesLtimessqrtleftfracVVLright^- f approx resultfP abcliste
A simple high pass filter consists of a resistor and an inductor connected in series to the input signal. abcliste abc Show that the partial voltage amplitude across the inductor is given by V_C fracV_sqrtleftfracomega_comegaright^ + where V_ is the amplitude of the input signal and omega_c the cutoff frequency: omega_c sqrtfracRL What is the partial voltage at the cutoff frequency? Sketch the value of V_L/V_ vs frequency. abc For a typical high pass filter used in audio circuits the input signal has an amplitude of VO and the resistance and inductance are RO and LO respectively. Calculate the cutoff frequency and the frequency at which the output voltage i.e. the partial voltage across the inductor is VLO. abcliste
Solution:
abcliste abc The impedance for an RL series circuit is Z sqrtR^+leftomega Lright^ It follows for the partial voltage across the capacitor V_L X_L I_ omega L fracV_Z fracV_ omega LsqrtR^+leftomega Lright^ fracV_sqrtleftfracRomega Lright^+ With fracRL omega_c it follows V_L fracV_leftfracomega_comegaright^+ quad square For omegaomega_c we find V_L fracV_sqrtleftfracomega_comega_cright^+ fracV_sqrt i.e. the voltage across the inductor is sqrt times smaller than the input voltage. vspacemm The diagram below shows how the partial voltage deps on the frequency. center includegraphicswidthtextwidth#image_path:high-pass-filter-# center abc The cutoff angular frequency is omega_c omcF fracRL omc This corresponds to a frequency f_c fracomega_cpi fc approx resultfcP The partial voltage across the inductor is V_L fracV_sqrtleftfracomega_comegaright^+ fracV_sqrtleftfracRpi f Lright^+ Solving for the frequency yields f fF fracRpitimesLtimessqrtleftfracVVLright^- f approx resultfP abcliste