-1
\$\begingroup\$

I have this anti-aliasing low pass filter where R = 100Ω and C = 3.62μF.

enter image description here

When the input has a frequency of 50Hz, Vin(t) = sin(2π*50t), I have worked out the dynamic error and found that the maximum error is 0.1123.

When I add a 100Ω resistor load at the output of the filter and re-work out the dynamic error, I find that the maximum error is now 0.1105.

Question

What is the reason for the error in the system decreasing after applying a load to the output?

\$\endgroup\$
2
  • 2
    \$\begingroup\$ How do you define dynamic error? In both cases the capacitor dynamics have little effect. \$\endgroup\$ – copper.hat Apr 11 at 21:49
  • \$\begingroup\$ @copper.hat E(t) = Vout(t) - Vin(t) \$\endgroup\$ – Jared Apr 12 at 9:15
3
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Your circuit. (b) The equivalent circuit which makes it clear that you've created a 2:1 potential divider. (c) The Thevenin equivalent has half the signal level feeding in and 50 Ω source resistance.

By adding the load you have overloaded the filter to the point of altering its response.

\$\endgroup\$
2
\$\begingroup\$

T=RC

Case 1 RC = 100 * 3.62 uF f-3dB ~ 440 Hz A(50Hz) = -0.05 dB
Case 2 RC = 50 * 3.62 uF . f-3dB ~ 880 Hz thus less Attenuation at 50 Hz.

This is not really an anti-aliasing filter and no specs.

  • no sampling rate was given nor BW ripple, band-reject attenuation and SNR from Nyquist noise.

You should give a spec for any design:

e.g.

  • attenuation error vs frequency tolerance like 0 to 0.02 dB .
  • 3rd harmonic rejection ? > 30 dB
  • Sampling resolution, and acceptable alias noise above Nyquist rate thus attenuation value at f1, f2 f3

Here is my random 50 Hz BPF with 0.02 dB ripple over +/-10 Hz

enter image description here

\$\endgroup\$
2
\$\begingroup\$

The cutoff frequency of an RC filter is determined by the total resistance and capacitance at the output.

With R = 100 Ω and C = 3.62 μF the cutoff frequency is ~440 Hz. When you put 100 Ω across the capacitor the total (Thevenin equivalent) resistance is now 50 Ω, so the cutoff frequency rises to ~880 Hz. You see less dynamic error because the filter is attenuating the higher frequency components of your signal less (but it also attenuates higher aliasing frequencies less, so the sampled signal may be less accurate).

This loading effect is one of the down-sides of passive RC filters. To minimize it the load resistance and capacitance needs to be ~10 times higher than the filter's resistor and capacitor values. For a 2 stage filter they need be ~100 times higher and so on, which quickly becomes impracticable.

One way around this is to follow each filter stage with a 'buffer' amplifier which has high input and low output impedance. The buffer amp's output can be fed back into the filter to sharpen the response. A popular configuration is the Sallen Key 'active' filter, which provides 2 stages with a sharper cutoff transition than 2 passive filter stages.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.