Analog Filter Circuits NCEES Ref Handbook p206 - Understanding R1 & R2 in Fig.2

[xconwing]

NCEES Reference Handbook p.206Hi,

My question is: In Fig.2, how can R1 and R2 be parallel? If anything, it should be in series because capacitor in the long run is an open.

Thanks ahead.

 

[sophiecentaur]

Hi and welcome.
"Series" and "parallel" are just names. They are not relevant to the analysis of circuits - except for the most elementary cases. But, if you bear in mind that a voltage source (v1) has zero impedance and if you replace it with a piece of wire, then, from the point of view of the C, they are in parallel and the equivalent resistance can be calculated with that formula.

 

[zoki85]

Consider form of the transfer function in the s-domain for that circuit. Introducing Rp gives a shorter way to write it down and calculate.

 

[xconwing]

Hi and welcome.
"Series" and "parallel" are just names. They are not relevant to the analysis of circuits - except for the most elementary cases. But, if you bear in mind that a voltage source (v1) has zero impedance and if you replace it with a piece of wire, then, from the point of view of the C, they are in parallel and the equivalent resistance can be calculated with that formula.

thanks for that answers (and others too )

how did they come to the equation H(s)?
I tried to do the circuit analysis but it didn't come out the same, will put it up here when i find a decent equation editor

 

[zoki85]

how did they come to the equation H(s)?
I tried to do the circuit analysis but it didn't come out the same, will put it up here when i find a decent equation editor

Let i₁ be current through R₁, i_c current through C, and i₂ current through R₂.
For sine wave voltage input it holds:

i₁ = i_c+ i₂
v₁ = i₁·R₁ + i_c/(jωC)
0 = i₂·R₂ - i_c/(jωC)

From this you find (express) i₂.
Then, H(s)=H(jω)=v₂/v₁= (i₂·R₂)/v₁.
Finally, when you introduce R_p=R₁·R₂/(R₁+R₂), you'll get the result.

 

[xconwing]

Let i₁ be current through R₁, i_c current through C, and i₂ current through R₂.
For sine wave voltage input it holds:

eq.1 i₁ = i_c+ i₂
eq.2 v₁ = i₁·R₁ + i_c/(jωC)
eq.3 0 = i₂·R₂ - i_c/(jωC)

From this you find (express) i₂.
eq.4 Then, H(s)=H(jω)=v₂/v₁= (i₂·R₂)/v₁.
eq.5 Finally, when you introduce R_p=R₁·R₂/(R₁+R₂), you'll get the result.

Is eq.2 missing: v₁ = i₁·R₁ + i_c/(jωC) + i₂·R₂?
Is eq.3 suppose to be: 0 = i₁·R₁ - i₂·R₂ - i_c/(jωC)?
Also, how does eq.5 implement onto eq.4?

 

[zoki85]

Nothing is missing. 2nd eq. is for the first closed loop: v₁ - R₁ - C and 3rd eq. is for second closed loop: C-R₂.
Do you understand this?

 

[xconwing]

Nothing is missing. 2nd eq. is for the first closed loop: v₁ - R₁ - C and 3rd eq. is for second closed loop: C-R₂.
Do you understand this?

Okay, I understand eq.1-3, 2&3 are from KVL. i₂ should express from which eq.? Why v₁ become a short?

 

[zoki85]

i₂ to be expressed as one of the solutions to the given system of 3 linear equations (other two solutions are i₁ and i_c but you don't need them). What "v₁ become a short"? I don't understand you

 

[NascentOxygen]

Okay, I understand eq.1-3, 2&3 are from KVL. i₂ should express from which eq.? Why v₁ become a short?

Eqn 3 comes from saying "because C and R₂ are connected in parallel, then the voltage across C equals the voltage across R₂."

Eqn 1 comes from saying "the voltage across C equals V₁ minus the voltage lost across R₁".

 

[xconwing]

i₂ to be expressed as one of the solutions to the given system of 3 linear equations (other two solutions are i₁ and i_c but you don't need them). What "v₁ become a short"? I don't understand you

That is how they get the R_P to be R₁ and R₂ parallel right?

As for finding H(s), the path using system of equation is kind of tedious. My i₂ came out to be pretty ugly, maybe the algebra is wrong somewhere or it just simply ugly. I try to use the voltage divider approach, where R_x is C and R₂ parrallel.

 

[NascentOxygen]

H(s) tells you Vout/Vin. I'm wondering why you are needing to find I2 ?

For a complex impedance, you should be using Zx not Rx. Rx should be used only where the element is known to be a pure resistance.

 

[xconwing]

H(s) tells you Vout/Vin. I'm wondering why you are needing to find I2 ?

For a complex impedance, you should be using Zx not Rx. Rx should be used only where the element is known to be a pure resistance.

I was trying out the method suggested by zokia85 in post #5

 

[zoki85]

That is how they get the R_P to be R₁ and R₂ parallel right?

As for finding H(s), the path using system of equation is kind of tedious. My i₂ came out to be pretty ugly, maybe the algebra is wrong somewhere or it just simply ugly. I try to use the voltage divider approach, where R_x is C and R₂ parrallel.

Oho, looks like you know how to use a voltage divider principle in complex impedance domain:)
Then, it is obvious you obtained what you wanted:

I don't know what the heck you were doing with the system of equations, but i₂ turns up quickly and quite nicely in a form:

H(s) tells you Vout/Vin. I'm wondering why you are needing to find I2 ?

He doesn't need to, but when someone asks questions wether R₁ and R₂ are in series or parallel it is more pedagogical to show approach with fundamentals of circuit analysis.

 

[xconwing]

Okay, so far the algebra checked off their final equation against my. Though I still don't understand how they see R₁ and R₂ as being parallel, the v₁ is in the way. If you say, turn v₁ into a short, why?

 

[zoki85]

Okay, so far the algebra checked off their final equation against my. Though I still don't understand how they see R₁ and R₂ as being parallel, the v₁ is in the way. If you say, turn v₁ into a short, why?

Not 100% sure, but my guess is they prefer approach to the problem by the method of Thevenin's theorem
In that case the filter circuit can be replaced by equivalent Thevenin's circuit with:

in order to find current and voltage of capacitor's impedance X_c :

 

[xconwing]

Ugh, I see. They want to be fancy schmancy.

Okay, one last thing. That omega, the way they set it up seem like R_P will ALWAYs be equal to capacitor impedance. Is that the case?

 

[zoki85]

That omega, the way they set it up seem like R_P will ALWAYs be equal to capacitor impedance. Is that the case?

Are you drunk?

 

[xconwing]

No, I'm sober, are you?

 

[zoki85]

ω_c is the critical angular frequency. ω can vary between 0 and ∞ rad/s . Just have a look at Fig.1 . And stop trolling.

 

[xconwing]

Nah brah, I'm extremely lazy so "trolling" is not something I do. However, I am curious about every detail of any particular concept. With your word and mine, shall we call this whole thing "curious" "trolling"

Thanks for the inquired infos
-X