Cut-off frequencies, Quality for a given Transfer Function

In summary, the given network transfer function is for an active band pass filter, with a center frequency of 10^3 rad/s, lower and upper half power frequencies of 9.9x10^2 rad/s and 1.01x10^3 rad/s respectively, and a quality factor Q of 1.005. By multiplying the transfer function by (30/30), we can see that it is equivalent to an ideal band pass function with a quality factor of 30. This allows us to easily determine the parameters of the given transfer function.
  • #1
mrwest09
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Homework Statement



Given the network transfer function

H(s)=10s/(s^2+300s+10^6)

Find the center frequency, lower and upper half power frequencies and the quality factor Q.

Homework Equations



Beta=R/L

(omega naught)^2=1/(LC)

BW=hi-lo=R/L

H(s)=(R/L)s/(s^2+(R/L)s+1/(LC))

The Attempt at a Solution



Well I am truly stumped. Our prof has given us an equation for the transfer function of RLC band pass filter and the equation in the problem nearly fits the format (located above), however the value for (R/L) differs in the numerator and denominator suggesting it is an active band pass filter. However our prof has not mentioned anything about active band pass filters yet in class (I was reading ahead) and therefore I was thinking that couldn't be it.

At this point I don't even know where to begin. Should I attempt to make it fit into an active band pass filter? Or solve for the cut off frequencies by finding the magnitude of the transfer function and setting it equal to 1/sqrt(2).

I think I just need a nudge the right direction.

Thanks for any help!
 
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  • #2
So I finally figured it out. Turns out if we multiply the transfer function by (30/30) we get a transfer function of
H(s)=[300s/(s^2+300s+10^6)](1/30) which fits our definition for an ideal band pass function. By simply comparing it with the format for the ideal band pass function we can determine all of the above parameters.
 

1. What is a cut-off frequency?

A cut-off frequency is a specific frequency at which a filter or system begins to attenuate or block certain signals or frequencies. It is often used in the context of electronic filters to describe the point at which the filter starts to reduce the amplitude of a signal.

2. How is cut-off frequency related to the quality of a transfer function?

The cut-off frequency is directly related to the quality of a transfer function. In fact, the cut-off frequency is often used as a measure of quality for filters. A higher quality transfer function will have a sharper cut-off frequency, meaning that it will more effectively filter out unwanted signals or frequencies.

3. Can the cut-off frequency be adjusted?

Yes, the cut-off frequency can be adjusted in certain systems or filters. This can be done by changing the components or parameters of the system, such as the resistor or capacitor values in an electronic filter. By adjusting the cut-off frequency, the filter can be customized to better suit the specific needs of the system.

4. What is the relationship between cut-off frequency and bandwidth?

Cut-off frequency and bandwidth are closely related. The cut-off frequency is the point at which the filter begins to reduce the amplitude of a signal, while the bandwidth is the range of frequencies that are allowed to pass through the filter. The cut-off frequency is typically located at the -3dB point of the bandwidth.

5. How does the shape of the transfer function affect the cut-off frequency?

The shape of the transfer function can greatly affect the cut-off frequency. A steeper slope in the transfer function will result in a sharper cut-off frequency, allowing for more precise filtering. On the other hand, a more gradual slope in the transfer function will result in a wider cut-off frequency, allowing for a broader range of frequencies to pass through the filter.

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