How To Find Center Frequency Of Bandpass Filter Resonance

The following circuit is an example of a band pass filter:

Commencement we will consider a qualitative analysis of the circuit. Recall that the impedance of the inductor and capacitor are:

and .

Hence if the frequency is cypher (i.eastward. D.C.) the impedance of the inductor is nil (i.e. a short excursion) and the impedance of the capacitor is infinite (i.e. an open excursion), this is shown in the circuit below:

At present if the frequency is infinite, the impedance of the inductor is infinite (i.eastward. an open circuit) and the impedance of the capacitor is nothing (i.e. a short circuit), this is shown in the excursion beneath:

Now we will consider the quantitative analysis.

Using Kirchoffs' voltage law gives:

and ohm's law:

we can calculate the gain of the circuit by:

The post-obit graph is of the gain of the band pass filter excursion shown above:

The gain of the excursion is:

and the post-obit graph shows the phase every bit a office of frequency:

A bandpass filter has 5 characteristic parameters. These are listed in the following table:

Name of Variable	Description	Symbol
Center Frequency	This is the frequency at which the transfer role is at a maximum
Cut off frequency 1	This is the lower frequency at which the transfer part equals of the maximum value
Cut off frequency 2	This is the higher frequency at which the transfer office equals of the maximum value
Bandwidth	This variable is the width of the pass band. (meet graph below)
Quality cistron	This parameter is the ratio of the eye frequency to the bandwidth. This gives a measure of the pass band and can be used to describe the shape of the transfer role graph	Q

Calculation of the heart frequency

The middle frequency is when the impedance of the whole circuit is existent, hence:

Calculation of the cut off frequencies

By definition, the cut off frequency is when the transfer function is of the maximum value. Hence to detect the cut off frequencies we gear up the proceeds to equal to and solve for:

which and so gives:

The solution of this yields four values for the cutoff frequencies. Only two are positive and take physical significance; they place the laissez passer pand of the filter: