Hi I built a simple LC low pass filter and I supply it with a 1 volt peak to peak ac voltage at 200 Hz from a function generator. My cutoff is set at 700 Hz and I have selected my circuit components based on this cutoff. However, when I hook up my circuit to the Oscilloscope both the input and output seem to be attenuated when measured on the Oscilloscope. The input and output shows up to be around 150 mv Peak to Peak instead of 1Volt Peak to Peak. I believe the capacitors have something to do with this attenuation. What are your suggestions to solve this problem ? Thanks 
LC Filter Attenuation
low passpassivepassive-networks
Related Solutions
Here is a circuit which only lets through pulses with a repetition frequency of less than 1KHz.
When triggered by the leading edge of each input pulse, Monostable IC1 generates a 1ms pulse which clocks D F/F IC3's output high. IC3 is reset when the input pulse goes low again, so the output follows the input. If the frequency is higher than 1KHz then IC1 is continuously triggered and doesn't produce any clock pulses, so IC3 stays in reset.
R2, C2 and IC2B provide a short delay to ensure that IC3 is out of reset when it receives the clock pulse from IC1.
BTW this circuit can be simplified down to just one IC by configuring the other half of the CD4528 as a basic flip-flop. Connect pins 15 and 14 to GND, pin 11 to Vdd, apply clock input (from pin 6) to pin 12, and connect the input frequency direct to pin 4 (+TR) and pin 13 (/RESET). Output is on pin 10.
The values in the circuit produce a natural resonance at about 89Hz based on this formula: -
Natural resonance = \$\dfrac{1}{2\pi \sqrt{R_1R_2C_1C_2}}\$
This circuit can also produce what to the uninitiated might think of as strange behaviour - it can actually resonate i.e. produce a significnatly higher amplitude at a range of frequencies just when you might have expected it to be attenuating those frequencies. This is component value dependent.
I believe this is what you are seeing. I think you are seeing this because you have one of the capacitors at the wrong value. For instance, if the 47n were in fact a 1nF capacitor the circuit would resonate at about 600 Hz with a significant peak: -
I used an online sallen-key calculator and plugged in the R and C numbers to get this. Here is the link.
I would encourage you to check the capacitor and resistor values you used.
Best Answer
If your function generator has a 50-ohm output impedance, I think your results make sense.
$$Z_{C1} = Z_{C2} = \frac 1 {j 2 \pi \cdot 200\ \mathrm{Hz} \cdot 28\ \mathrm{\mu F}} = -j15.9\ \mathrm \Omega$$
$$Z_{L1} = j 2 \pi \cdot 200\ \mathrm{Hz} \cdot 3.63\ \mathrm{mH} = j4.6\ \mathrm{\Omega}$$
Using series and parallel combinations, I get a total impedance of \$6.4 -j1.3\ \Omega\$, or \$~6.6 \angle -11 ^\circ\ \Omega\$ in polar notation. With a 50-ohm output impedance, your expected scope voltage on the input would be:
$$1.414\ \mathrm V \frac {6.6\ \Omega} {50\ \Omega + 6.6 \Omega} \approx 165\ \mathrm{mV}$$
That's amplitude, not peak to peak, so maybe I'm missing a factor of two somewhere in my calculations, or maybe there's some other limit on the generator output. Regardless, that is the right order of magnitude to expect. You can't hook an ~8-ohm load up to a 50-ohm generator and expect to see the full generator voltage.