transfer function question

Thread Starter

Ramiel

Joined Feb 19, 2018
65
May someone explain the reason of the substitution of W0 that is circled in each picture. I understand how to find the transfer function but not the W0 substitution. Why do we substitute different variables for W0 in each of the question?
 

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Zeeus

Joined Apr 17, 2019
468
Hi..Looks like the "fundamental' or "resonant" frequency where reactance of components are equal...Dunno

Nice book ? Have that book, did not read more than chapter 5 :(
 

Thread Starter

Ramiel

Joined Feb 19, 2018
65
Hi..Looks like the "fundamental' or "resonant" frequency where reactance of components are equal...Dunno

Nice book ? Have that book, did not read more than chapter 5 :(
It is good so far. I just use it cause it is what my Uni uses as a reference.
 

danadak

Joined Mar 10, 2018
3,617
Fig 1. is where the W0 is the frequency where the response is 3 db down from
maximum, also called the half power point. Same for figure 2.

https://www.electronics-tutorials.ws/accircuits/rms-voltage.html

Figure 3 a little more comlicated. w0 is the resonant frequency in that
circuit when R = 0. But in that circuit the effective C seen by L is reduced
due to R, so resonant freq of overall circuit is raised. There is a transform
one can use to compute the effective C -

https://analog.intgckts.com/impedance-matching/series-to-parallel-conversion/

http://www.phys.ufl.edu/~majewski/nqr/reference2015/nqr_detection_educational/The Series to Parallel Impedance Transformation.pdf


Regards, Dana.
 

MrAl

Joined Jun 17, 2014
6,641
May someone explain the reason of the substitution of W0 that is circled in each picture. I understand how to find the transfer function but not the W0 substitution. Why do we substitute different variables for W0 in each of the question?
Hello,

For each drawing there would be a different w0 found and that is found by equating the response to 1/sqrt(2) and solving for w0.
The value of w0 is the value of w that forces a response that is 1/sqrt(2) times the response at some other place in the response. This other place is where the frequency causes some particular response such as maximum.

For example, for a low pass filter with response Hs where the response is 1 for w=0 we would equate:
|Hs|=1/sqrt(2)

and then solve for w0. That gives us the value of w where the response is 3db down from when w=0.
Once you place this value w0 into the equation in place of w, the response becomes 1/sqrt(2) or approximately 0.7071 .

One slight irregularity here. In the third drawing, it appears they may have not calculated the right value for w0. We can take a closer look at this later if you like.
 

Thread Starter

Ramiel

Joined Feb 19, 2018
65
Hello,

For each drawing there would be a different w0 found and that is found by equating the response to 1/sqrt(2) and solving for w0.
The value of w0 is the value of w that forces a response that is 1/sqrt(2) times the response at some other place in the response. This other place is where the frequency causes some particular response such as maximum.

For example, for a low pass filter with response Hs where the response is 1 for w=0 we would equate:
|Hs|=1/sqrt(2)

and then solve for w0. That gives us the value of w where the response is 3db down from when w=0.
Once you place this value w0 into the equation in place of w, the response becomes 1/sqrt(2) or approximately 0.7071 .

One slight irregularity here. In the third drawing, it appears they may have not calculated the right value for w0. We can take a closer look at this later if you like.
The thing is I am taking circuit theory 2 now, and this is the last subject included. We were just given this part of this chapter(transfer function) without bode plot or anything else. But from what I understood is that I equate the magnitude of the transfer function to 1/sqrt(2) and find the w value? If it is possible to tell me What is wrong with last picture that would be great because this is an answer key from a previous exam, so is my teacher wrong?
 

Zeeus

Joined Apr 17, 2019
468
But from what I understood is that I equate the magnitude of the transfer function to 1/sqrt(2) and find the w value? If it is possible to tell me What is wrong with last picture that would be great because this is an answer key from a previous exam, so is my teacher wrong?
Something is wrong with it? dont think so


Resonance when R = 0 is the w....from 2nd-to-last line to last line, he just grouped the real part in denominator and had form : (1 - x^2) ... I think you have the w value already..


But from what I understood is that I equate the magnitude of the transfer function to 1/sqrt(2) and find the w value?
Can you attempt the math?
 
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Zeeus

Joined Apr 17, 2019
468
Is this correct? read somewhere maybe

The resistor of 125 series with capacitor and we can transform the series resistor(Xs) into parallel resistor(Xp)....

Xs * Xp = (Zc)^2 , Zc reactance of that capacitor..

So little Xs then the resistance does not matter much due to high Xp... This is true or fake?
 
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MrAl

Joined Jun 17, 2014
6,641
The thing is I am taking circuit theory 2 now, and this is the last subject included. We were just given this part of this chapter(transfer function) without bode plot or anything else. But from what I understood is that I equate the magnitude of the transfer function to 1/sqrt(2) and find the w value? If it is possible to tell me What is wrong with last picture that would be great because this is an answer key from a previous exam, so is my teacher wrong?
Hi,

Thanks for the question. The answer could be different depending on what the teacher wants. This happens sometimes because the teacher sometimes takes a short cut just to make it easier on the student for the time being. In later course, this would not be acceptable at all, but in courses that lead up to that it is entirely possible that you will be given a short cut answer just to keep moving at a steady pace.
Also, we might note that it is possible he did not want to worry about the -3db point but rather just wanted to use the point where capacitive reactance equals inductive reactance. That's probably a good guess actually although it would have to be proved that can actually be done with this circuit, but i show the other view anyway.

That being said, it is possible that the teacher gave the short cut result for w0 as 1/sqrt(LC) just to avoid the complexities of the actual solution. The actual solution depends not only on the circuit but also on the application. Please try to follow along as i try to explain and how this will show possibly why the teacher chose to use a short cut result for w0 rather than the actual result.

First let's find the transfer function.
Allowing C=1 and L=1 we get:
Io=(j*w*R+1)/(j*w*R-w^2+1)

and so the amplitude is:
Ampl=sqrt(w^2*R^2+1)/sqrt(w^2*R^2+(1-w^2)^2)

and with R=3 we get:
Ampl=sqrt(9*w^2+1)/sqrt((1-w^2)^2+9*w^2)

Now when we plot this, we see we have a low pass filter with a bump on the top that could be used as a pseudo band pass filter.
[note: we would actually use three different values for R to show that the response always has this general shape but let's keep it simpler for now]
This means we have to choose what kind of filter we are really working with in the end application as well as possibly how we interpret the response curve. We will choose a low pass for this example where the bump is considered narrow relative to the application.
If we chose a band pass, we would base the -3db point on the peak of the response, but because we chose low pass and the bump is not very wide, we use the w=0 amplitude peak as reference point.
Since the above amplitude comes out to 1 for w=0, we can equate to 1/sqrt(2). If we chose bandpass we would have to first solve for the peak and then use that which would be peak/sqrt(2).
Ok so we are using 1/sqrt(2) for the low pass with narrow bump on top of the frequency response curve.
Equating, we have:
sqrt(9*w^2+1)/sqrt((1-w^2)^2+9*w^2)=1/sqrt(2)

Solving for w, we get one viable solution:
w=sqrt(5^(3/2)+11)/sqrt(2)

and thus w0 is this value of w:
w0=sqrt(5^(3/2)+11)/sqrt(2)

and numerically this is:
w0=3.330190676785561

To verify, we place that into the amplitude equation:
Ampl=sqrt(9*w^2+1)/sqrt((1-w^2)^2+9*w^2)

and doing that we get:
Ampl=1/sqrt(2)

This verifies that we did it right.


Second, let's prove that w0=1/sqrt(LC) is not the exact correct result now that we have one actual solution.
Since C=1 and L=1 that would make the simpler w0 equal to:
w=1/sqrt(1*1)=1/1=1

This is clearly not the same as the above value of 3.33019 so the supposed solution of 1/sqrt(LC) is not right.
Placing this supposed solution into the amplitude equation we get:
Ampl=sqrt(10)/3=1.05409255338946

The question then is are they using the peak as reference?
Using a little calculus and finding the peak, the value of w at the peak is:
wpeak=sqrt(sqrt(19)-1)/3

Now all we have to do is divide the amplitude with w=1 by the amplitude with w=wpeak, and we get numerically:
0.97793

and this is clearly not 1/sqrt(2) or numerically 0.7071 so this proves that w=1 is not correct for the -3db point.


Now again i want to stress that your instructor may have wanted to skip all this for now as you can see it gets a little more involved.
The more general solution is as follows...

Find the transfer function Hjw.
Calculate the amplitude Aw.
Determine the reference point and associated value of w, call it wpeak, the amplitude we'll call Apeak.
Equate:
Aw=Apeak/sqrt(2)
and solve for w. The only real positive solution of w will be w0 the -3db point frequency.


Notes:
To get a better feel for how this works, plot the response for the amplitude shown above.
See if you can visualize where the -3db point is by looking for an amplitude around 70 percent of where the amplitude is with w=0.
 
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RBR1317

Joined Nov 13, 2010
502
Did a Bode Plot for the third circuit, but could not find any particular correspondence between the defined ωo and the plot. However, reducing R by a factor of 100 does lead to a noticeable equality.
Current-Division-LRC.png BodePlot_sysLRC.png BodePlot_R_one-percent.png
 

MrAl

Joined Jun 17, 2014
6,641
Did a Bode Plot for the third circuit, but could not find any particular correspondence between the defined ωo and the plot. However, reducing R by a factor of 100 does lead to a noticeable equality.
View attachment 179133 View attachment 179134 View attachment 179135
Hi,

The closer R gets to zero (0.000) the closer w comes to 1/sqrt(LC) for the xC=xL calculation.
That is because R plays a part in the actual accurate calculation of w unless it is zero and then plays no part.

I think i got this result when R is not equal to zero:
w0=sqrt(sqrt(C^2*R^4+4*C*L*R^2+8*L^2)+C*R^2+2*L)/(sqrt(2)*sqrt(C)*L)

When R=0 we still dont get 1/sqrt(LC) unless we do the xC=xL calculation.
 
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