# Help with understanding resonance...

Discussion in 'Homework Help' started by blazedaces, Aug 13, 2008.

1. ### blazedaces Thread Starter Active Member

Jul 24, 2008
130
0
I want to be able to find the resonant frequency and bandwidth of any circuit if I know the values of its components.

For example, let's say I have a standard parallel RLC circuit (a resistor in parallel with a capacitor in parallel with an inductor) with resistance R, capacitance C, and inductance L.

The admittance of the circuit can be calculated easily by adding the admittance of each individual parallel component:

Ytotal(s) = YR(s) + YC(s) + YL(s) = 1/R + C*s + 1/(L*s)

For some reason that I can not understand since the teacher skipped this in class and just assumed we knew it I'm now supposed to combine the terms and the equation eventually becomes

YTotal = C/s * (s^2 + s*1/(C*R) + 1/(L*C))

I know that the resonant frequency of a parallel circuit is Sqrt(1/LC) and the bandwidth is 1/(R*C) for this circuit.

On the other hand, if I look at a standard series RLC circuit the total impedance boils down to

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

I know that the resonant frequency of a series circuit is Sqrt(1/LC) and the bandwidth is R/L for this circuit.

My question is the following:

Is it a rule of thumb that the equation of total admittance or total impedance is of some form K/(s+c)*(s^2+s*BW+w0^2) where c and K are constants, BW is the bandwidth, and w0 is the resonant frequency in radians / second?

Why is it admittance for a parallel circuit and impedance for a series circuit? I would rather understand why these things are true than just memorize the equations...

If this is not the case, how does one calculate BW and w0 from a given circuit. I read the contents of this site: http://www.allaboutcircuits.com/vol_2/chpt_6/6.html

I know w0 is given when XL = XC, but I'm still confused on the topic and I don't understand how BW is calculated so much.

-blazed

Last edited: Aug 13, 2008
2. ### yubyub Member

Aug 13, 2008
19
6
from the two equations

1/R + C*s + 1/(L*s) = C/s * (s^2 + s*1/(C*R) + 1/(L*C))

this can be done in a few steps.
Take the LHS and put it in brackets, (1/R + C*s + 1/(L*s))
then multiply outside the brackets by C, and divide inside the brackets by C (essentially the equation is still the same but looks different) C * (1/RC + s + 1/(LC*s)
then do the same with the s and you get the result C/s * (s^2 + s*1/(C*R) + 1/(L*C))

As for the rule of thumb with admittance stuff... im having a read through my textbook.. so i might be wrong, but that rule of thumb works for second order systems. If you really need to know this stuff I reccomend getting some kind of textbook on it.

3. ### blazedaces Thread Starter Active Member

Jul 24, 2008
130
0
Firstly thank you for your help. Now, perhaps I was misleading and I'm sorry, but my issue is not with the algebra of rearranging the equation but rather with why the equation is being put in that format I guess.

If that rule of thumb works for second order systems I will keep that in mind, thank you.

The textbook covering this subject is from a previous class. Unfortunately that is tucked away somewhere in my house (back at home while I'm living here in philly) so I don't have access to it or I might not be asking this question on the forums.

Thanks again,

-blazed

4. ### Ratch New Member

Mar 20, 2007
1,068
3
blazedaces,

Resonance occurs when the the phases of the currents or voltages reinforce or hinder each other on a periodic basis. In a series electrical circuit, this occurs at one frequency only. In a parallel circuit, this can occur at three different frequencies, which are almost equal to each other if the Q of the circuit is greater than 10. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/parres.html You can also design a parallel circuit that it is resonant at all frequencies.

There is plenty of material about this on the web, and that includes the complex and involved equations. A good search engine is your friend.

Ratch

5. ### blazedaces Thread Starter Active Member

Jul 24, 2008
130
0
Just wanted to say thanks for the help.

That site helped explain how to calculate w0. But I still don't understand how to calculate BW well.

Most sites refer to BW as = w0/Q. Which implies that you have to find Q first (from X/R) and then get BW. Is that really the case?

-blazed

6. ### Ratch New Member

Mar 20, 2007
1,068
3
blazedaces,

Yer welcome.

Certainly. A high Q will allow a narrower bandwidth and vice versa. They are intimately related. Make sure you know the definition of BW. Then you can relate to how it is calculated. Here is another site that can help. It shows both series and parallel resonant circuits. http://en.wikipedia.org/wiki/RLC_circuit

Ratch