# The tutor gives background along with a basic explanation of how a high-pass filter works.

A high-pass filter will send along high frequency signals but block low frequencies. It can do so because the impedance, X_{c}, of a capacitor of capacitance C, at frequency f, is

X_{c}=1/(2πfC)

At very high frequency, therefore, X_{c}≈0

The total impedance, Z, of a resistor R and capacitor C in series is given by

Z=(R^{2} + X_{c}^{2})^{0.5}

Therefore, with input voltage V, and instantaneous current i, Ohm’s Law gives

V=i(R^{2} + X_{c}^{2})^{0.5}

The voltage across the resitor is

V_{R}=iR

Therefore, the ratio of the voltage across the resistor to the source is

V_{R}/V = iR/i(R^{2} + X_{c}^{2})^{0.5} = R/(R^{2} + X_{c}^{2})^{0.5}

As mentioned earlier, for high frequency, X_{c}≈0, giving

V_{R}/V ≈ R/(R^{2})^{0.5} = 1

With a high-pass filter, the maximum output voltage for a high frequency signal equals the input voltage.

Convention says that the critical frequency, f_{c}, is that at which V_{R}/V = 1/(2)^{0.5} = 0.707, which occurs when X_{c} = R:

V_{R}/V = R/(R^{2} + X_{c}^{2})^{0.5} = R/(R^{2} + R^{2})^{0.5} = R/(2R^{2})^{0.5}

which leads to

V_{R}/V = 1/2^{0.5} = 0.707

Therefore, a high-pass filter will pass any frequency higher than the critical frequency f_{c}, where f_{c} is calculated from

X_{c} = R

1/(2πf_{c}C) = R

1 = 2πf_{c}CR

1/(2πCR) = f_{c}

By that reasoning, a 10kΩ resistor, in series with a 12pF (picoFarad) capacitor, placed in series, should produce a high-pass filter with critical frequency f_{c} = 1.3MHz. The output would be read across the resistor R.

Source:

www.electronics-tutorials.ws

ecee.colorado.edu

Serway, Raymond A. __Physics for Scientists and Engineers__ with modern physics. Toronto: Saunders College Publishing, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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