Electronic – Passive Pi Low Pass Filter LC

1. The circuit shown at the web page you linked to is only a 2-pole filter. If you try to make a 4-pole filter by cascading two such 2-pole filters, the performance is likely to be not as expected due to loading effects. Basically this means that the input impedance of this circuit is not constant across frequency, so when you cascade filters together you need to adjust the first filter to accomodate the changing load presented by the second filter.

2. It is correct that changing R does not change the filter characteristic frequency, it only affects the damping ratio. This is expected behavior.

If you want to design a 4-pole passive filter, I suggest using LTSpice and tuning the filter until you achieve good performance accounting for loading effects. If you want to just use simple calculations, then you could switch to active filters or provide a buffer between your 2-pole filter sections to eliminate loading (assuming your power handling requirements allow this).

If you really need to design this as a 4-pole passive filter, then you can go back to pre-1970's filter design textbooks or cookbooks to find the design techniques that were used before numerical analysis and active filter designs became ubiquitous.

The transfer function is: $$\small G(s)=\frac{1}{(RC)^2s^2+3RCs+1}$$

Hence \$\omega_n=\frac{1}{RC}=\small10^4\:rad/s\:(=1592\:Hz)\$, and \$\small\zeta=1.5\$, and it can be seen that the DC gain (\$\small s=0\$) is unity.

Converting this to the frequency domain, using \$ s\rightarrow j\omega\$:

$$\small G(j\omega)=\frac{1}{1-(\omega RC)^2+j3\omega RC}$$

At \$\small 10\:\small Hz\$, \$\small \omega RC=0.00628\$, hence the gain is almost unity and the phase angle is almost zero. At \$\small 1\:\small kHz\$, \$\small \omega RC=0.628\$, giving a gain of \$\small 0.505\$, and phase angle of \$\small \phi=-72^o\$.

So it seems that there's a problem with your experimental set-up. What's the input impedance of the instrument measuring Vout?

Let's do some detective work:

If the input impedance of the instrument were \$\small 3 \: k\Omega\$ resistive, then (i) the gain and phase at DC would be \$\small 0.6\$ and zero, respectively (i.e. same as your results); and (ii) the gain and phase at \$\small 1590 \:\small Hz\$ would be \$\small 0.29\$ and \$\small -79^o\$, which compares with your measurement of \$\small 0.31\$ and \$\small -73^o\$.