Creating a Bode Plot for a Series RC High Pass Filter

[AdkinsJr]

Homework Statement

Make a blode plot of a series RC high pass filter (log(gain) vs. f)...

Homework Equations

log(Gain)=log(Vout/Vin)
I'm pretty sure the "Gain" is just the transfer function, for a highpass it should be the V_in = Vsupply, and V_out = V_capacitor.

The Attempt at a Solution

I have an RC circuit hooked up and I'm going to make the bode plot taking data from the oscilloscope, I'm just going to vary the frequency and collect voltages, but I'm not sure how to go about doing it because of the phase difference between the input voltages (Ch. 2 blue line) and the output voltage (over the capacitor, Ch.1 yellow line).

Can I just measure peak-to-peak? I'm not 100% sure if that's correct because of the phase difference between peaks. Do I need to take voltages over the same point on the "time" axis or is peak-to-peak ok even if at different points with respect to time axis?

 

[rude man]

There are two kinds of Bode plots - |gain| and phase. If all you need is the gain, pick the amplitudes of input & output voltages and disregard the phase shift.

 

[gneill]

Peak-to-peak voltages will work fine. You can always convert them to peak values (divide by 2).

You might want to think about where you take the Vout measurement. That is, what's the placement of the resistor and capacitor for a high-pass filter? It's essentially a voltage divider. How does the impedance of a capacitor vary with frequency?

A Bode plot generally displays the power gain in dB. When you work with voltage ratios (Vout/Vin) take 20 Log(Vout/Vin) to get the equivalent power gain in dB.

 

[AdkinsJr]

Actually what I have hooked up is a low pass filter, I made a mistake there... I'm asked to make a high pass and a low pass and put them in series to see what happens. But the circuit I have now is just the low pass, ...a resistor and capacitor in series, it's the capacitor connected to the low end (ground) so this should be low pass.

So now I need to measure the voltages over the resistor and the supply (entire circuit). So in my excel spreadsheet I should just make a column of "V_supply - V_capacitor" to have my " V_resistor" and use that as my V_out when I make the Bode plot for the low pass.

For a high pass I'll swap the capacitor and resistor and measure the voltage over the capacitor.

 

[AdkinsJr]

Also I think I see why it doesn't matter if the peak values are at different "times" because the transfer function is a function of "s" which is a complex variable dependent on frequency, so there is no time dependency.

 

[rude man]

A Bode plot generally displays the power gain in dB. When you work with voltage ratios (Vout/Vin) take 20 Log(Vout/Vin) to get the equivalent power gain in dB.

No, Bode gain plots are typically voltage gain plots with dB = 20log₁₀(V_out/V_in).

 

[rude man]

Also I think I see why it doesn't matter if the peak values are at different "times" because the transfer function is a function of "s" which is a complex variable dependent on frequency, so there is no time dependency.

You could have used a voltmeter in lieu of your oscilloscope in which case you would not have been aware of time at all. Time has nothing to do with the gain measurement.

Just to be sure: your Vin is hooked up to the generator high and low sides and your 'scope (Vout) to the capacitor high and low (low being ground) for the lowpass. Yes, for the high-pass you swap the R and C.

 

[gneill]

No, Bode gain plots are typically voltage gain plots with dB = 20log₁₀(V_out/V_in).

There are 10 decibels in a bel.

$$dB = 10 log_{10}\left( \left(\frac{V_{out}}{V_{in}}\right)^2\right) = 20 log_{10}\left( \frac{V_{out}}{V_{in}}\right)$$

 

[rude man]

There are 10 decibels in a bel.

$$dB = 10 log_{10}\left( \left(\frac{V_{out}}{V_{in}}\right)^2\right) = 20 log_{10}\left( \frac{V_{out}}{V_{in}}\right)$$

Thanks g I had that part figured out already :)
A Bode plot graphs 20log(Vout/Vin) = 20log (voltage gain).