# High pass filter with given attenuation

Choose a frequency $f_0$.We have to project a circuit such as:

1. all the $f \ge 10f_0$ can pass without attenuation
2. all the $f \le \frac{1}{10} f_0$ pass with an attenuation of a factor of 10.

The solution is giving by

simulate this circuit – Schematic created using CircuitLab

but I can't understand why. Could you help me?

In this instance, think of the capacitor as a frequency-dependent resistor.

At high frequencies, you can think of it as a dead short relative to the fixed resistance. In that case, the input is simply connected to the output, so the gain is 1.

At low frequencies, the capacitor is a open circuit. Now you only have the two resistors forming a voltage divider with a gain of 1/11. Whoever said this had a low frequency gain of 1/10 was wrong, unless they consider 1/11 within their error margin. Note that the gain of the voltage divider is:

gain = R / (10R + R) = R / 11R = 1/11

To find the crossover frequency of a R-C filter, you set the impedance magnitude of the capacitor equal to the resistance:

R = 1 / (2πfC)

When f is in units of Hz, C in Farads, then R is in Ohms.

• Could you explain me why, at low frequencies, the capacitor is an open circuit? – sunrise Jul 3 '13 at 12:57
• If I consider a frequency like 10 Hz and a C=47nF, according to $Z=\frac{1}{\omega * C}$, I obtain a very heigh impedence.. where am I wronging? Many thanks for your help! – sunrise Jul 3 '13 at 13:09
• @sunri: Look at the equation I showed for the impedance magnitude of a capacitor. R becomes infinite as C goes to 0. – Olin Lathrop Jul 3 '13 at 13:10
• "Open circuit" means that the resistence is infinite.. isn't it? – sunrise Jul 3 '13 at 13:20
• If, at low frequencies, the capacitor is like a very high resistor, the current doesn't pass through it and passes through the two resistors. Is this step correct? Now, I know that if I consider a "normal" high pass filter, all low frequencies are attenuated, sometimes they are attenuated a lot. Is this strong attenuation due to the fact that a very little current passes through the capacitor? And, in our circuit, is the attenuation of a factor of 10 due to the fact that much more current passes through the resistor R1=10R than the current that passes through the capacitor? – sunrise Jul 3 '13 at 13:37