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I've created a first-order RC filter with R = 68 Ohm, C = 5.0 uF, and I am studying the transfer function of this filter. I have calculated the transfer function to be 1/(1+jwRC), based on the relationship between the output and input voltage (see image): Circuit Diagram.

I measured the magnitude and phase of the output waveform using an oscilloscope, and plotted it in a Bode diagram against the magnitude and phase of the transfer function above, and this is what I get:

enter image description here

It appears from doing some research that the experimental data is following the curve that is the sum of the Bode pole and Bode zero functions. How do I calculate what the zero and pole equations are from my transfer function, or how do I model the actual response I measured?

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  • \$\begingroup\$ What is "ca" in your question title? \$\endgroup\$ – Transistor May 22 '18 at 19:33
  • \$\begingroup\$ the title got cut off - should be fixed now. \$\endgroup\$ – sidney13 May 22 '18 at 19:46
  • \$\begingroup\$ What kind of capacitor did you use? \$\endgroup\$ – mkeith May 22 '18 at 20:21
  • \$\begingroup\$ a 4.7 uF nominal (5.0 uF actual, measured with LCR meter) non-polar ceramic capacitor \$\endgroup\$ – sidney13 May 22 '18 at 20:56
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    \$\begingroup\$ maybe related to the esl of the capacitor. You can try higher frequency, like 25MHz, i once also do this kind of experiment with my function generator. You may have positive gain with higher frequency, if i remember right. \$\endgroup\$ – iouzzr May 22 '18 at 23:15
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What you observe is normal and you could even have a more twisted response if the sweep frequency were increased. This is because the capacitor (and the resistor also but we can neglect them so far) host parasitic terms. The ac model of the capacitor includes an equivalent series resistance (ESR) noted \$r_C\$ and an equivalent series inductance if you are interested in the upper frequency response. For a low-frequency model, the below circuit is a good start:

enter image description here

Using the FACTs, you can determine the transfer function without writing a line of algebra. For \$s = 0\$ the gain is 1, then turning the excitation off (replace the source by a short circuit), the time constant is \$\tau_2=(r_C+R)C\$ in which \$R\$ is your 68-ohm element. This induces a pole located at \$\omega_p = \frac{1}{(r_C+R)C}\$. Now, what you see is the zero contributed by the series connection of \$r_C\$ and the capacitor. If you null the output - \$V_{out} = 0\$ - the time constant becomes \$\tau_1=r_CC\$ with a zero located at \$\omega_z = \frac{1}{r_CC}\$. The transfer function is thus \$H(s) = \frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$

At low frequency, the magnitude is 1 and the phase 0°. As you increase the frequency, the transfer function magnitude drops with a -1-slope (20 dB/decade) and the phase gently goes towards -90°. When the zero contributed by the ESR kicks in, it breaks the slope to 0 and the phase rises up again towards 0°. A pole lags the phase (-1-slope, 90° lag) while a zero leads the phase (+1-slope, 90° lead). Combining the two implies what you've observed. As \$s\$ approaches infinity, you end up with a gain equal to \$\frac{r_C}{R+r_C}\$.

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For an RC, the pole is at 1/RC rad/sec, or 1/2piRC Hz. You'll be doing this calculation your whole career, so the thing to remember is:

1K 1uF 160Hz

Since they're linear, you can adjust any two...as above, 1/15K and 5uF makes for 15/5*160Hz, or 480Hz. Multiply by 2pi to get rad/s.

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  • \$\begingroup\$ Is that the same as the corner frequency in this case? I have the corner frequency displayed as the vertical red line at w = 1/RC. After the corner, the theory and experiment diverge... \$\endgroup\$ – sidney13 May 22 '18 at 19:50
  • \$\begingroup\$ Yes, it's the same as the corner frequency. I think the difference between theory and experiment in this case is largely noise, leakage, tolerance, and/or measurement error. When the attenuation is 20dB, your output is 1/10 your input, and an imperfection which would have made a 0.2dB difference on the full scale input will change the output by about 1dB. It's typical for experimental filter results to appear noisy or inaccurate in the stopband, because all noise and error is proportionately higher as your signal diminishes. \$\endgroup\$ – Cristobol Polychronopolis May 23 '18 at 12:38

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