# Solving for Q factor

Joined Oct 18, 2011
19
The problem asks to calculate the Q factor of the filter. I have read up on solving for the Q factor using the inductance. With Zin = Vin/Iin = ZL = jωL.

After doing some circuit evaluation I came up with my Zin being
s^2*C2*C1*nR^2

after solving it two times I came up with the same answer twice. Then setting s = jω and then Zin = jωL I do not get all of the "s" to cancel. The problem does not offer me a frequency only the values of R,m,n,C1,C2.

I am unsure of what to do from this point, or if I have even came at the problem the right way from the beginning. Any help would be appreciated, I also can assume I probably did not calculate the right Zin.

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#### steveb

Joined Jul 3, 2008
2,436
The problem asks to calculate the Q factor of the filter. I have read up on solving for the Q factor using the inductance. With Zin = Vin/Iin = ZL = jωL.

After doing some circuit evaluation I came up with my Zin being
s^2*C2*C1*nR^2

after solving it two times I came up with the same answer twice. Then setting s = jω and then Zin = jωL I do not get all of the "s" to cancel. The problem does not offer me a frequency only the values of R,m,n,C1,C2.

I am unsure of what to do from this point, or if I have even came at the problem the right way from the beginning. Any help would be appreciated, I also can assume I probably did not calculate the right Zin.
You don't want to calculate Zin. Instead you want to calculate the transfer function of output voltage over input voltage. You will find it is a second order system. You then put the transfer function into standard form to match up the Q factor, and identify it.

EDIT: Thinking about it further, your expression for input impedance doesn't look right. I say this because I just noticed that Iin=(V2-Vo)/R, hence the expression for Vo (which relates to the transfer function T=Vo/V2) is included in the input impedance.

Hence, the input impedance can be found by Iin=V2(1-T)/R, which leads to Zin=R/(1-T). Note that if T is a second order system, then Zin will be much more complicated than what you wrote. It should look something like the following form.

$$Y_{in}={{1}\over{Z_{in}}}={{1}\over{R}}{{s^2+\omega_B s+\omega_o/(m+1)}\over{s^2+\omega_B s+\omega_o}}$$

where the omegas are parameters with units of radians/s, which can be related to the circuit component values. Note that the Q can be related to the $$\omega_B$$ value.

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Joined Oct 18, 2011
19
I think I understand what you are saying. I am still really confused where you got your Zin equation. if Zin = R/(1-T) how did you come up with

$image=http://forum.allaboutcircuits.com/mimetex.cgi?Z_%7Bin%7D=%7B%7B1%7D%5Cover%7BR%7D%7D%7B%7Bs%5E2+%5Comega_B%20s+%5Comega_o/%28m+1%29%7D%5Cover%7Bs%5E2+%5Comega_B%20s+%5Comega_o%7D%7D&hash=1f36f88fd91d7072a565719ab2375517$

I can see the general form of a second order transfer function but I am at a loss as to how R went from being in the numerator to the denominator. Another question I have is when writing the transfer function am I solving for Vo or Iin?

Joined Oct 18, 2011
19
The more I work on this problem, the most lost I become. I would really appreciate a more indepth description. Thanks!

#### steveb

Joined Jul 3, 2008
2,436
The more I work on this problem, the most lost I become. I would really appreciate a more indepth description. Thanks!
I apologize for causing more confusion. I made a mistake above, and you are correct to note that the impedance shouldn't have units of 1/Ohms. I accidentally wrote the formula for input admittance Yin=1/Zin. I have now corrected it above.

So let's go through a few steps to clarify.

First, how did I get Zin=R/(1-T)? Note that the input current will equal the input voltage minus the opamp negative input terminal voltage divided by the resistance R.

$$I_{in}={{V_2-V_m}\over{R}}$$

But, the Vm voltage is about equal to the Vlp voltage on the other input terminal of the opamp. Hence, we get ...

$$I_{in}={{V_2-V_{lp}}\over{R}}$$

Now, lets define the voltage gain transfer function to be T=Vlp/V2, and substitute this in to the above equation.

$$I_{in}={{V_2-T\;V_2}\over{R}}$$

Factor out the V2 to get ...

$$I_{in}={{V_2\;(1-T)}\over{R}}$$

Now input impedance is defined to be Zin=V2/Iin which leads to

$$Z_{in}={{R}\over{1-T}}$$

OK, Let's end this post and continue in the next post.

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#### steveb

Joined Jul 3, 2008
2,436
The more I work on this problem, the most lost I become. I would really appreciate a more indepth description. Thanks!
Next, lets consider how to determine the transfer function for voltage gain. There are a number of ways to do this, but it all amounts to writing out some linear circuit equations and then solving for Vlp/V2.

I prefer to start at the output and work backwards, so I do the following.

$$V_{lp}={{-I_2}\over{sC_2}}$$, where I2 is the current through C2.

$$I_2={{V_{bp}}\over{nR}}$$ since the negative terminal of A2 is near ground potential.

$$V_{bp}=V_{lp}-{{I_1}\over{sC_1}}$$, where I1 is the current through C1.

$$I_1={{V_2-V_{lp}}\over{R}}-{{V_o}\over{mR}}$$

Now you have the task of solving 4 linear equations with 4 unknowns (Vlp, I2, Vbp and I1).

When I solve this I get.

$$T={{1/(nC_1C_2R^2)}\over{s^2+s/(nC_2R)+(m+1)/(mnC_1C_2R^2)}}$$

Now we can use Zin=R/(1-T) to get

$$Z_{in}=R\Biggl({{s^2+s/(nC_2R)+(m+1)/(mnC_1C_2R^2)}\over{s^2+s/(nC_2R)+1/(mnC_1C_2R^2)}}\Biggr)$$

Double check my work and keep an eye out for more mistakes ... I'm old!

#### t_n_k

Joined Mar 6, 2009
5,447
I derived the same transfer function as steveb shows. I simplified the math somewhat by first reducing the source and the R, mR voltage divider to its Thevenin equivalent.

I would finally calculate the Q as being

$$Q=\sqr{\frac{n(m+1)C_2}{mC_1}}$$

That also needs checking as I'm also in the older age bracket.

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