Exercise:
A simple band-pass filter consists of a resistor an inductor and a capacitor connected in series to the input signal. abcliste abc Show that the partial voltage amplitude across the resistor is given by V_R V_fracRsqrtR^+leftomega L - fracomega Cright where V_ is the amplitude of the input signal. vspacemm Sketch the value of V_C/V_ as a function of the angular frequency. Why is this circuit called band-pass filter? abc Explain why the voltage at the resonance frequency omega_ fracLC corresponds to the input voltage. abcliste

Solution:
abcliste abc The impedance for an RCL series circuit is Z sqrtR^+leftomega L+fracleftomega Crightright^ It follows for the partial voltage across the resistor V_R R I_ R fracV_Z V_ fracRsqrtR^+leftomega L+fracleftomega Crightright^ quad square The diagram below shows how the partial voltage deps on the frequency. The angular frequencies omega and omega_L are defined as omega_ fracLC omega_L fracRL center includegraphicswidthtextwidth#image_path:band-rejection-# center For given values of L and C i.e. a fixed resonance frequency the resistance value determines the width of the peak. It can be shown that the width at half the maximum voltage is given by the expression verify this as an additional exercise Deltaomega sqrtomega_L The output signal across the resistor only has a significant amplitude if the frequency is close to the resonance frequency omega_. Therefore we can say that this filter lets through a frequency band which is defined by the resonance frequency and the width. abc At the resonance frequency the inductive and capacitive reactance are equal: X_Lomega_ omega_ L fracLsqrtLC fracsqrtLCC fracomega_ C X_Comega_ Hence the impedance is just Zomega_ sqrtR^+leftX_L-X_Cright^ sqrtR^+ R and therefore the voltage across the resistor V_Romega_ V_ fracRZomega_ V_ fracRR V_ quad square abcliste