Amplitude of Analog Filter Question

  • #1
nao113
68
13
Homework Statement
Calculate V/E of the circuit below. I tried to calculate this one, is it correct? Thank you
Relevant Equations
Vc = 1/jwC
Question
Screen Shot 2022-06-01 at 17.49.43.png

Answer;
Screen Shot 2022-06-01 at 17.49.25.png
 
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  • #2
That does not look right to me.
You have a potential divider.
Reactance; Xv = XL + XC;
But what is the sign of XC ?
Impedance; Ztotal = R + jXv
The output V = E * jXv / Ztotal
 
  • #3
Baluncore said:
That does not look right to me.
You have a potential divider.
Reactance; Xv = XL + XC;
But what is the sign of XC ?
Impedance; Ztotal = R + jXv
The output V = E * jXv / Ztotal
How about this one?
 

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  • #4
2nd guessing will not educate you, nor solve the problem.
At the frequency where L and C are resonant, there will be a deep notch with V = 0.
How can XL + XC = 0 ?
 
  • #5
Baluncore said:
2nd guessing will not educate you, nor solve the problem.
At the frequency where L and C are resonant, there will be a deep notch with V = 0.
How can XL + XC = 0 ?
thank you for the feedback, what do you mean by 2nd guessing? for `How can XL + XC = 0 ?` what should I do for that question? Did I got it wrong for my calculation?
 
  • #6
What is the reactance XL of an inductor?
What is the reactance XC of a capacitor?
 
  • #7
Baluncore said:
What is the reactance XL of an inductor?
What is the reactance XC of a capacitor?
based on my class, here it is
 

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  • #11
Both XL and XC are imaginary. R is real.
So XL + jXC does not equal Xv. Both XL and XC lie on the same axis.
Xv = ωL - 1/ωC
Then you fail to divide.
 
  • #12
Then, I remove the imaginary, how about it?
 

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  • #13
You must move more methodically and accurately.
You removed the j, but immediately ignored the negative sign, then forgot to divide.
 
  • #14
Baluncore said:
You must move more methodically and accurately.
You removed the j, but immediately ignored the negative sign, then forgot to divide.
how about this? I am sorry, I am new in this area so kinda confused
 

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  • #15
I wouldn't multiply XL by j, and then divide XC by j . That does not seem fair.
Z = R + j X;
If you work on the reactance, X only, you can ignore j for the moment.
You are having problems with fractions.
X = ωL - 1 / ωC = ( ωC·ωL - 1 ) / ωC .
 
  • #16
I am so sorry, I didn't realize it
I revised again
 

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  • #17
Resistance, real, is plotted along the x-axis, reactance, imaginary, is on the y-axis. Impedance is a point on the map, represented by a complex number. Below the x-axis is negative, so it is a capacitive reactance, above the x-axis is inductive reactance. Resonance lies on the x-axis, where reactance is zero.

You must introduce j when you write a complex impedance, because it keeps the orthogonal real and imaginary axes apart. Z = R + jX is prevented from becoming Z = R + X by the presence of j.

So in your second line you must carefully identify the reactance components with j .
 
  • #18
Screen Shot 2022-06-01 at 22.16.54.png

is it? do I need to change plus to minus after R in Z total? Then for |V/E|, did I put quadrat and roots correctly? should. I also put root on the top?
 
  • #19
Now you are going to need to propagate that R + j ( ) along the second line;
That is needed to keep the impedance orthogonal.
 
  • #20
Baluncore said:
Now you are going to need to propagate that R + j ( ) along the second line;
That is needed to keep the impedance orthogonal.i
how about this? is this what you mean as propagate?
 

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  • #22
Baluncore said:
I think you need to brush up on your complex arithmetic.
https://en.wikipedia.org/wiki/Complex_number

V = E * ( 0 + j Xv ) / ( R + j Xv )
So you need to divide an imaginary by a complex.
Thank you very much, I see, I think my answer is still not correct. I ll try to learn your suggestion.
 

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