if you have a transfer function like,

        b2·s² + b1·s + b0
H(s) = ———————————————————
        a2·s² + a1·s + a0

whereas s would be (1 - z⁻¹)∕(1 + z⁻¹)∕e^(j·ω) in the bilinear transform,
you can find its magnitude response with this equation:

             (b2·x)² - (2·b2·b0 - b1²)·W·x·y + (b0·W·y)²
|H(j·ω)|² = —————————————————————————————————————————————
             (a2·x)² - (2·a2·a0 - a1²)·W·x·y + (a0·W·y)²

(analog)        x = ω²
                y = 1
                W = ω0²

(digital)       x = sin(ω∕2)²
                y = cos(ω∕2)²
                W = tan(ω0∕2)²

whereas         ω is the physical frequency in rads/sec
                ω0 is the center frequency in rads/sec

and the phase? maybe some other time
note: I'm no math genius and there's probably an error in here