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In the picture below, what is the purpose of R1? Compared to a high pass filter with 1 resistor, how does this affect Fc?

Thank you.

enter image description here

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This is basically a voltage divider between \$R_2\$ and \$R_1, C_1\$ in series, so the transfer function is the following: $${V_{out} \over V_{in}}={R_2 \over{R_2+R_1+{1\over j\omega C_1}}}={j\omega R_2C_1 \over 1+j\omega C_1(R_1+R_2)}$$

The first resistor limits the current into the capacitor, therefore it can charge slower, with a time constant of \$\tau=(R_1+R_2)C_1\$.

The cutoff frequency is smaller than without \$R_1\$: instead of $$f_c={1\over 2\pi R_2C_1}$$ it is $$f_c={1\over 2\pi(R_1+R_2)C_1}$$

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    \$\begingroup\$ I gotta learn how to write equations for Stackexchange. That's a nice answer. \$\endgroup\$ – Alan Campbell Nov 16 '14 at 1:32
  • \$\begingroup\$ @AlanCampbell It's LaTeX. You can learn it from online tutorials and references or you can click the "edit" button on a question or answer with equations to see how the equations were typed in. \$\endgroup\$ – Null Nov 16 '14 at 1:35
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Your voltage source is an "ideal" source. In practice, voltage sources vary, depending on how much current they need to source. This is modelled with an external series resistor - \$R1\$ in your diagram.
When you draw a lot of current, the the output voltage will drop by \$V = IR\$. If \$R_2\$ is more than \$10 \times R_1\$, you can generally ignore \$R_1\$ for ballpark measurements. However, since \$R_2\$ and \$C\$ form a high pass filter, changing your selection for \$R_2\$ affects your value for \$C\$.

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    \$\begingroup\$ This answers a slightly different (equally relevant) question than hryghr's answer, but it's also a very good point. \$\endgroup\$ – krs013 Nov 16 '14 at 3:22
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Purpose of 2 resistors in high pass filter?

This is a (1st order) high-pass attenuator. That is to say, the high-frequency asymptotic gain is less than one (unity).

An ordinary RC high-pass filter has a high-frequency gain that approaches unity. What if, instead, one desires that the high-frequency voltage gain be less than unity?

One approach would be to use the circuit in your question. At high enough frequencies, where the impedance of the capacitor is insignificant, the voltage gain is approximately

$$\frac{V_{out}}{V_{in}} \approx \frac{R_2}{R_1 + R_2} < 1$$

Of course, one must take into account the resistance of \$R_1\$ in the calculation of the corner frequency:

$$\omega_c = \frac{1}{(R_1 + R_2)C}$$

Note that, for a consistency check, as \$R_1 \rightarrow 0\$, we recover the equations for the ordinary RC high-pass filter.

In summary, this high-pass filter circuit gives an additional degree of freedom: the high-frequency attenuation can be now be specified as a constraint.

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I want to show another method without math:

In high frequency, the C is shorted, so the transfer function should have "high frequency" gain

$$ \frac{R_{2}}{R_{1}+R_{2}} $$

So, the high frequency gain now won't be 1, and will be smaller, dependents on the ratio of \$\frac{R_{1}}{R_{2}}\$.

The circuit's time constant basically is $$ \tau = R_{total} \times C_{1}=(R_{1}+R{2}) \times C_{1} $$

And the well known first-order high-pass filter's transfer function is

$$ H(s)=\frac{1}{1+1/(s\tau)} $$

Combine the high frequency gain and substitute the \$\tau\$, we get the whole

$$ H(s)=\frac{R_{2}}{R_{1}+R_{2}} \frac{1}{1+\frac{1}{s(R_{1}+R{2})C_{1}}} $$

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