A simple RL circuit, getting transfer function, matching with results
First of all, the circuit drawn is a high-pass filter.

At very high frequencies, the inductor has very high impedance and thus, the output \$V_{ab}\$ should essentially equal the input.

At very low frequencies, the inductor has very low impedance and thus, the output should be essentially zero.

The frequency where the output voltage is \$\frac{1}{\sqrt{2}}\$ times the input voltage is given by

$$f_0 = \frac{R}{2\pi L} $$

So, either the circuit drawn does not reflect the circuit measured or something went terribly wrong with your measurement apparatus (or operator of said apparatus).



For completeness, the transfer function is, by inspection:

$$H(j\omega) = \frac{j\omega L/R}{1 + j\omega L/R}\quad,\quad\omega = 2\pi f$$

$$|H(f)| = \frac{2\pi fL/R}{\sqrt{1 + (2\pi fL/R)^2}} $$