Amplitude of Analog Filter Question

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Discussion Overview

The discussion revolves around the amplitude of an analog filter, focusing on the calculations involving reactance, impedance, and the behavior of circuits containing inductors and capacitors. Participants explore various aspects of circuit analysis, including potential dividers and resonance effects.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants question the calculation of reactance, particularly the sign of capacitive reactance (XC) and its role in the total impedance.
  • There are discussions about the implications of resonance in circuits, with mentions of a deep notch in voltage at resonant frequencies.
  • Participants express confusion regarding the treatment of imaginary and real components in impedance calculations, emphasizing the need for careful handling of complex numbers.
  • Some participants suggest that the removal of imaginary components and the handling of signs in equations need to be approached methodically.
  • There are repeated references to the need for understanding the orthogonal nature of real and imaginary axes in complex impedance.
  • Several participants provide links to external resources for further clarification on reactance and complex numbers.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are multiple competing views on the correct approach to calculating reactance and impedance. Confusion and uncertainty persist regarding the treatment of complex numbers and the implications of resonance.

Contextual Notes

Limitations include potential misunderstandings of complex arithmetic and the need for clarity on the definitions of reactance and impedance. Some calculations remain unresolved, and assumptions about the treatment of signs and components are not fully agreed upon.

nao113
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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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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
 
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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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 ?
 
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?
 
What is the reactance XL of an inductor?
What is the reactance XC of a capacitor?
 
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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  • #21
  • #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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