www.Hobby-Electronics.info - Electronics Course

Frequency filters

Take a look on the diagram above. Assume that the voltage source supplies a 1V/10kHz signal (this means: the amplitude is 1V and the frequency is 10kHz = 10000Hz).

The impedance of capacitor C will be X_C = 1/(2∙π∙10⁴∙10^-6) = 15,9Ω. The output voltage (voltage across capacitor C) will be 1V∙(X_C/Z_R+C), where Z_R+C is the total impedance of R and C. Because a capacitor causes a phase shift in the current flow, we cannot just state that Z_R+C = R + X_C. Using some complex math^[2] we can prove that:

Z_R+C = √(R²+X_C²).

In our case Z_R+C = √(1k²+15.9²) = 1000.13Ω. So the output voltage becomes 1V∙(15.9/1000.13) = 0.0159V.

Now assume that the voltage source supplies a 1V/10Hz signal. The impedance of capacitor C will then be X_C = 1/(2∙π∙10∙10^-6) = 15,9kΩ. The output voltage will be 1V∙(X_C/(Z_R+C)) = 1V∙(15.9k/√(1k²+15.9k²)) = 0.998V. So we created a very simple frequency filter with just a resistor and a capacitor.

In this case we created a so called low pass filter (LPF) since it passes low frequency signals and suppresses high frequency signals. If you swap R and C, you create a high pass filter (HPF).

Let's calculate the cut-off frequency of our filter. The cut-off frequency is the frequency at which R=X_C => R = 1/(2∙π∙f∙C) =>

In our case f = 1/(2∙π∙10³∙10^-6) = 159Hz.^[2]

^[2]We already know how to calculate the impedance of a capacitor:

At 10kHz a 1nF capacitor has an impedance of 15.9k.

When we connect a 10k resistor in series with this capacitor, we may expect that the total impedance will be 25.9k is. But that's not the case, because a capacitor causes a -90 degrees phase shift in the current flow. The equation

doesn't show this. But what if we plot the impedance as a vector? The length will be the (absolute) impedance and the angle the phase shift:

This component has an absolute impedance of 1 ohm and causes a 45 degrees phase shift between voltage and current.

A resistor doesn't cause a phase shift; this vector will be on the x-axis. a capacitor causes a -90 degrees phase shift and will be on the (negative) y-axis. The total impedance of R and C is R + X_C. However, we have to add vectors instead of plain numbers. Since we have 90 degrees angles, it's easy to calculate the (absolute) impedance: we can just use Pythagoras' theorem. Z_t = √(R²+X_C²). The phase shift is equal to arctan(X_C/R)

Wouldn't it be nice if we had a more simple way for saying: the impedance is x ohms and causes a y degrees fase shift? A 180 degrees phase shift is easy; in that case we could say: the impedance equals -x ohms. A 180 degrees phase shift equal to multiplying by -1. Now suppose that a phase shift of 90 degrees is equal to multiplying by j. A 180 degrees phase shift will be the same as multiplying by j². This means that j² = -1. A negative square is only possible in so called 'complex math'. (Mathematicians among us may be accustomed to use i instead of j. But we already use i as a symbol for current, so that's confusing.)

Every impedance can be written as: a + bj. Number a is called the real part and is plotted on the x-axis. Number b is called the imaginary part and is plotted on the y-axis.

A resistor doesn't cause a phase shift and is therefore purely real.

We know that a capacitor causes a -90 degrees phase shift; its impedance is therefore purely imaginary, and can be written as:

As an example, we'll look again at our 10k resistor in series with a 15.9k capacitor. The total (complex) impedance is 10k - 15.9kj. The absolute value (|Z|) equals √(10k² + 15.9k²) = 18.78k. The phase shift it causes in the current (arg(Z)) is arctan(-15.9k/10k) = -57.8 degrees.

So, when we connect a resistor and capacitor in series: Z_R+C = R - X_Cj.

The absolutie value is: |Z_R+C| = √(R²+X_C²).

The phase shift is: arg(Z_R+C) = arctan(-X_C/R).