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For the circuit above, I was wondering how they derived the formula

$$\omega_0 = \sqrt{\dfrac{1}{LC} - \Big(\dfrac{R}{L}\Big)^2}$$

If anyone is able to walk me through this it would be fantastic.

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To clarify the answer from Andy you need to find the frequency where the impedance is real (ie where the j part is zero). Andy's answer does this and If I were you I'd accept it. – Warren Hill Jun 27 '14 at 11:42
May I further clarify? That means: The point of resonance in a lossy parallel tank circuit is NOT always the frequency where the maximum occurs but where the phase is zero (imaginary part=0). This is the DEFINITION of resonance. – LvW Jun 27 '14 at 12:33

Impedance is \$\dfrac{R + j\omega L}{R + j\omega L +\frac{1}{j\omega C}}\cdot\dfrac{1}{j\omega C}\$ = \$\dfrac{R + j\omega L}{j\omega RC + j^2\omega^2 LC +1}\$ = \$\dfrac{R + j\omega L}{j\omega RC - \omega^2 LC +1}\$

If you were to multiply both the numerator and the denominator by the denominator's complex conjugate you'd get a complex equation on the top and a non-complex (real) equation on the bottom. Resonance is when the imaginary part of the new numerator equals zero so, working only on the new numerator we get: -

New Numerator = \$(R+j\omega L)\cdot (-j\omega RC - \omega^2 LC +1)\$

The imaginary part boils down to \$j(\omega L - \omega^3 L^2 C - \omega C R^2)\$

Equating this to zero we get, \$\omega L - \omega^3 L^2 C - \omega C R^2 = 0\$

Therefore, \$\omega^2 L^2 C = L - CR^2\$ and \$\omega^2 = \dfrac{1}{LC} - \dfrac{R^2}{L^2}\$

EDIT section that describes other "resonances"

It's also interesting to note that there is another resonance at play. Firstly, the "resonance" above is defined by the impedance looking into the LCR network being purely resistive - this is slightly different to the "natural" resonant frequency (also purely resistive) when R is ignored.

In this situation \$\omega = \sqrt{\dfrac{1}{LC}}\$

However, the real peak of the response (as would be seen on a spectrum analyser and not purely resistive) is found by equating the denominator to zero and solving for s where s = j\$\omega\$.

\$s^2LC +sRC +1=0\$ or

\$s^2 + s\frac{R}{L} + \frac{1}{LC} = 0\$

Using the general solution to a quadratic, we get: -

\$s= \dfrac{-\frac{R}{L} +/- \sqrt{\frac{R^2}{L^2} - \frac{4}{LC}}}{2}\$

If we use a trick of reversing the signs inside the square root part and bring the square root of -1 (j) outside we get: -

\$s = \dfrac{-\frac{R}{L} +/- j\sqrt{\frac{4}{LC} - \frac{R^2}{L^2}}}{2}\$

Now, divide thru by two and we get: -

\$s = -\frac{R}{2L} +/- j\sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}\$

As we should know the "jw" part is what we see in a bode plot so the peak magnitude of the response (not when the impedance is purely resistive) is when: -

\$\omega = \sqrt{\dfrac{1}{LC} - \dfrac{R^2}{4L^2}}\$

If you drew a pole zero diagram, the imaginary parts of the poles would be at

+/-\$\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{4L^2}}\$ i.e. slightly different to +/-\$\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{L^2}}\$

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So resonance is simply when the impedance is purely real? – sherrellbc Jun 27 '14 at 12:38
@sherrellbc no, resonance can mean several things but, in the question, I recognized the formula as the one that applies when the impedance is purely real - it's close to the peak of the response if you drew a bode plot - a fraction of a percent or so different and nobody would probably care except me LOL! – Andy aka Jun 27 '14 at 12:41

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