19.49Ω seems a disastrously low parallel resistance equivalent. I think it will wreak havoc on a filter's response. Are you sure that isn't meant to be a series resistance of 19.49Ω?

But I doubt it makes any difference. Like NascentOxygen wrote, that RLC circuit will be extremely lossy, i.e. its Q factor will be very low/bandwidth will be very high.

Yes, I think 19.49 Ω may be the series resistance of the inductor (I measured it using an RCL meter, I will double check this).

But according to my book the parallel resistance R_{p} may be calculated from the graph using the quality factor relationships:

##Q_p = \frac{R_p}{\omega_0 L}=\omega_0 R_p C##

##Q= Q_p \sqrt{1-2|T_{min}|^2}##

Q_{p} being the quality factor of the parallel inductor-capacitor combination, and Q is the actual quality factor of the filter (a measure of the sharpness of the peak).

|T_{min}| is the minimum value of the transfer function (minimum value of |T|), which occurs when |z_{p}| is a maximum (at resonant frequency ##\omega_0 =1/\sqrt{LC}##).

In my plot T_{min} is my last data point where it is equal to -0.080 dB (or 1.01 in linear magnitude), so

##z_p=\frac{R_p}{1+jR_p(\omega C - \frac{1}{\omega L})}##

But then how do I solve for ##R_p## from this impedance? How can I get rid of the j's in the denominator? (I'm looking for a way to solve for R_{p} without actually having to measure it experimentally).

Yes, when I vary the resistance to to R_{p}≈1 kΩ, the theoretical graph starts to look right.