Frequency Response RLC circuit - Current against Frequency graph - Symmetry?

I understand that in a Frequency Response experiment dealing with an RLC circuit, the graph of Current against Frequency is supposed to be symmetrical about the resonant frequency theoretically.

However, experimentally it is not the case.

Could anyone explain why this happens?

-
Since frequency $f\geq 0$ is a non-negative quantity, and under the plausable assumption that the theoretical current $I(f)> 0$ is non-vanishing for all $f>0$, why would you expect the theoretical graph $f\mapsto I(f)$ to be a symmetric function $I(f)\stackrel{?}{=}I(2f_0-f)$ around the resonance frequency $f_0$ in the first place? – Qmechanic May 11 '13 at 12:21
@Qmechanic: Does $I(f/f_0) = I(f_0/f)$ seem more plausible? – Art Brown Aug 9 '13 at 16:02
@ArtBrown: Right. Technically, that's logarithmically symmetric $I(\ln(f))=I(2\ln(f_0)-\ln(f))$. – Qmechanic Aug 9 '13 at 16:20
Thanks, @Qmechanic, this terminology is v useful! With your permission, I would like to add it to my answer, with attribution of course. – Art Brown Aug 9 '13 at 17:41
Study the filter characteristic by plotting bode asymptotic (Frequency response) – user38344 Jan 27 '14 at 15:33

For a series LRC circuit, the magnitude of the current is indeed symmetrical about the resonant frequency, if you plot the frequency on a logarithmic scale. $$\left| Z \right| = \sqrt{R^2 +\frac{L}{C} \left[ \frac{\omega}{\omega_0}-\frac{\omega_0}{\omega}\right]^2}$$ where the resonant angular frequency $\omega_0$ is just $$\omega_0 = \frac{1}{\sqrt{L C}}$$ so $$\left|Z \left(\frac{\omega}{\omega_0} \right) \right|=\left|Z \left(\frac{\omega_0}{\omega} \right) \right|$$ Since the magnitude of the current is the applied voltage divided by $|Z|$, for a constant applied voltage (e.g. magnitude of voltage frequency-independent), the result follows.