Find the bandwidth and center frequency of the bandstop...

Key Concepts

Filter Design Analysis

Filter design analysis involves evaluating the frequency response of the circuit, understanding its transfer function, and identifying important parameters such as the center frequency and bandwidth. This process is crucial for ensuring that the filter meets the required performance specifications and effectively targets the undesired frequency band.

Center Frequency

The center frequency is the midpoint of the frequency range that the filter affects. In the context of a bandstop filter, it is the frequency at which the attenuation is typically strongest, and it represents the point around which the unwanted frequency band is centered.

Bandstop Filter

A bandstop filter is an electrical circuit that attenuates a specific range of frequencies while allowing frequencies outside that range to pass through with minimal attenuation. It is commonly used to eliminate unwanted noise or interference from signals that occur within a particular frequency band.

Bandwidth

Bandwidth is the measure of the range of frequencies over which the filter effectively attenuates the signal. For a bandstop filter, it is the difference between the upper and lower cutoff frequencies where the attenuation begins to fall off. This determines the width of the frequency band that is attenuated.

Transcript

00:01 Find the bandwidth and center frequency of the band stop filter and the figure is referring to the figure the capacitance is written as 1 by sc and for the inductance written by sl the combination at the output is the inductor and the capacitor which is in series this can be written as r into sl plus 1 by sc divided by r plus sl plus 1 by sc.

00:33 Here s is equals j omega.

00:36 We can write this impedance as r into 1 plus s square lc by 1 plus src plus s square lc.

00:47 Determining the transfer function h s equals v0s by vis substituting here the output voltage is the load impedance at the output by load impedance at output plus r0.

01:09 We can substitute the value here r into 1 plus s square lc by r0 plus s r r not plus s square lc plus s square lc r plus r plus s square lc r.

01:29 Substituting s equals to j omega here, then the transfer function in terms of j omega will be r into 1 plus j omega whole square lc by r not plus j omega into r r not c plus j omega whole square lc r c r r not plus r plus j omega whole square l c r plus the equivalent impedance of the circuit r not plus z s this is equals to r not plus z s this is equals to r not plus z s value is into 1 plus s square lc by 1 plus src plus s square lc the equivalent impedance will be r not plus s r not r c plus s square r not lc plus r plus r s square lc plus r s square lc whole divided by 1 plus src plus s square lc.

02:49 Substituting as equals to j omega here into this equation, z -in will be r0 plus r -c minus omega square l -c r -cr -r -c -r -plus j -o -mega -r -n -c, multiply 1 minus omega -square l -c minus j -o -mega -r -c.

03:17 This is divided by 1 -1 -1 -1 -omega -square -l -c, whole square plus omega r c whole square.

03:27 To determine the center frequency, the imaginary part of the impedance must be zero, therefore substituting r0 plus r not minus omega square lc r0 minus omega square lc r not minus r0 plus omega square r0lc equals 0.

03:55 From here the value 4, omega comes out to be 1 by root lc...

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