Natural frequency, q factor and damping factor

hey guys, i got set this question and have some trouble with it:

i think i got part a:

s=wj (w is omega)

Vout/Vin = (10^9/s)/(10^9/s +80 +s810^-3)

=10^12/(s^2+80 000 s + 10^12)

However, i've learnt that the natural frequency is where the total imaginary resistance would be equal to 0, so it would be equal to 0. However the answer is 159 kHz.

I'm not certain about the quality factor either, so some help with that would be appreciated as well

(a) The frequency response function you can find by calculating the complex impedance of the circuit:

$\displaystyle V_{in} = IZ_T$, where $Z_T$ is the total impedance of the circuit.

$\displaystyle V_{out} = IZ_C$, where $Z_C$ is the impedance of the capacitor.

Subbing in for I in the 2nd equation using the 1st:
$\displaystyle V_{out} = V_{in}\frac{Z_C}{Z_T}$

$\displaystyle \Rightarrow \frac{V_{out}}{V_{in}} = \frac{Z_C}{Z_T}$

This is exactly analogous to DC circuits, where the voltage over a particular component is the corresponding ratio of resistances.

Now we know:
$\displaystyle Z_T = R + j \omega L + Z_C$
$\displaystyle Z_C = (j\omega C)^{-1}$

So substitute these in and find the magnitude of the resulting complex function - this is your frequency response curve.


The Q-factor is not a rigidly defined quantity, but can be taken as the ratio of response curve width to its peak height (give or take factors of $2\pi$).

First of all, thanks for helping me. However, i tried to do that, which gave me (10^9/s)/(10^9/s +80 +s810^-3) where s is equal to omega times j. I then simplified this to 10^12 / (s^2 + 8s*10^4 + 10^12). However the natural frequency is not 10^6. Any ideas?

Please write the bits of your working from where I finished. Leave out the values until the end, please.

ok, in that case i got to V out / V in = 1 / ( jw^2 * L * C + jwRC + 1 ), so w0 (the natural frequency) is = (L*C)^0.5 = (10^-9 * 10^-3)^-0.5 = 10^6

Well you're almost right, but it's (jw)^2 L C, don't forget.

How are you getting that formula for the natural frequency? The natural frequency is the value where the frequency response is a maximum. Find this from your expression for the frequency response (i.e. find the maximum magnitude of your V_out / V_in expression).

HINT: The undamped natural frequency is simply $\omega_0 = \sqrt{\frac{1}{LC}}$.

i'm sorry, i forgot to put the jw it into brackets and a minus in
"w0 = (L*C)^-0.5 = (10^-9 * 10^-3)^-0.5 = 10^6"

To get w0, i compared my solution with (wj)^2 / (w0)^2 + 2*(C thingy)*wj/w0 + 1 so therefore took w0 as the number squared dividing the (wj)^2.

However I seem to get the same answer as you. Could it be something due to a difference in the natural frequency and the undamped natural frequency (if there is a difference)?

Do you have an 'answer' for this that disagrees with yours, by the way? I don't see the problem, to be honest.

Try using $\displaystyle \omega = \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}$ ?

yea, it should be 159 kHz following the answer on the back of my problem sheet. It could be wrong, but I have only spotted a mistake once in about 160 questions...

See the edit to my previous post. Sorry, I should have spotted that earlier.