It doesn't give resonance to a filter, but it allows you to translate an analog filter (given by a transfer function in Laplace domain) to a digital IIR filter (given in Z domain).
First: the analog world ...
For a low pass filter with resonance peak, you need a second order filter. Let's do a low pass RLC filter
Vin [1/sC] Vout
Not the best ascii art, and not even genuine electrical symbols, but you get the idea. The filter consists of a resistor R, an inductor L and a capacitor C.
Pretend all the components are just resistors, and solve for Vout:
Vout = Vin * (1/sC) / (R + sL + 1/sC)
After a bit of rearrangement, we get our transfer function in laplace domain:
H(s) = Vout/Vin = 1 / (1 + sRC + s\\^2 LC)
The cutoff frequency is w_c = 1 / sqrt(LC). Well, actually, w_c is not really a frequency measured in Hertz but in radians per second. If you have your cutoff frequency f_c measured in Hertz, you need to multiply it by 2pi: w_c = 2pi * f_c
The Q factor is given by sqrt(L / C) / R.
Given this, we can rewrite the transfer function to:
H(s) = 1 / (1 + s / (Q*w_c) + s\\^2 / w_c\\^2)
So, with this you can already play: use w_c to set the cutoff frequency and use a high Q (Q > 1) to get a resonance peak. Use Q = sqrt(2) to have a 2nd order butterworth filter.
Now you have to translate this to Z domain by substituting s by (2/T_s)*(1 - z\\^-1)/(1 + z\\^-1)
After some rearrangments, you hopefully get:
H(z) = (b_0 + b_1 * z\\^-1 + b_2 * z\\^-2) / (a_0 + a_1 * z\\^-1 + a_2 * z\\^-2)
with the following taps (best to check, cause i'm not 100% sure, but it looks ok):
b_0 = 1
b_1 = 2
b_2 = 1
a_0 = 1 + 2 / (T_s * Q * w_c) + 4 / (T_s * w_c)\\^2
a_1 = 2 * (1 - 4 / (T_s * w_c)\\^2)
a_2 = 1 - 2 / (T_s * Q * w_c) + 4 / (T_s * w_c)\\^2
This, you should be able to put in your digital IIR filter (possibly your IIR filter requires a_0 to be equal to 1, but that's easily done by dividing all taps by the a_0 you find above).
Using this approach, you can translate any linear passive analog filter to taps for your IIR digital filter.