Electronic – How to implement a lowpass filter for this one zero, one pole transfer function

The problem has come from the substitution of s for 2*PI*f.

s is a substitution for j*w

thus the equation actually is: 1/(1+j)
If you resolve this back to magnitude: 1+j = sqrt(1^2 + 1^2) = sqrt(2)

resulting in: 1/sqrt(2) = 0.707 as you expect

So take a low-pass filter: RC

^{simulate this circuit – Schematic created using CircuitLab}

$$ Vout = Vin*\frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} $$ $$ Vout = Vin*\frac{1}{j\omega RC + 1} $$ $$ \frac{Vout}{Vin} = \frac{1}{j\omega RC + 1} $$ $$ \frac{Vout}{Vin} = \frac{1}{\sqrt{(\omega RC)^{2} + 1^{2}}} $$ From a magnitude point of view.

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:

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

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