Can you explain, please, step-by-step how an electronic filter does work? For example, high pass filter. I know It's a trivial things, but I can't get it completely. Don't bring me formula and etc. Just explanation in three words.
Let's start with the idea of a voltage divider. Connect two resistors in series, and apply an input voltage across the two of them. Then take the point between the two resistors as your output voltage. That output voltage will be less than your input voltage; the two resistors have divided the input voltage into two parts, each part proportional to the resistance of the two resistors. When most people refer to a "voltage divider" they mean one built out of two basic passive resistors. The resistance of these resistors doesn't depend on the voltage input, so the voltage divider works the same for all frequencies of input voltage.
But you can make a simple high-pass or low-pass filter by building a voltage divider out of a resistor and a capacitor. The resistance of a capacitor depends on the frequency of the input voltage. For very low frequencies, the capacitor has a nearly infinite resistance. As the frequency increase, the resistance of the capacitor decreases. For very high frequencies, the capacitor has nearly zero resistance. If you connect the resistor and the capacitor in series, with the resistor connected to the high of your input voltage, and take the output voltage as the point between the resistor and the capacitor, then you get a low-pass filter. Low frequencies see the capacitor as having a much larger resistance than the resistor, and therefore output a voltage close to the input voltage. High frequencies see the capacitor as having a much lower resistance than the resistor, and therefore output a voltage close to zero. To build a high-pass resistor, swap the resistor and capacitor, so that the capacitor is connected to the high of your input voltage. The same reasoning holds, with high and low frequencies switched.
As JKL's answer notes, inductors also have a resistance that depends on the frequency of the input voltage. In this case, the resistance increases with increasing frequency. So if you replace a capacitor with an inductor, you turn a low-pass filter into a high-pass filter, and vice versa.
A SIMPLE HIGH PASS FILTER: T-CONFIGURATION - no maths
For a high pass filter you can have a resistor with resistance $R$, and two capacitors with capacitance $C$, with the capacitors connected to the resistor which is grounded at one end. When a signal is entering the filter from the left, the high harmonics in the signal will pass through the capacitor on the left and close circuit through the resistor, no problem. The high frequencies will also pass the capacitor on the right. However the low frequencies that are below the lower cut-off frequency, $1/2\pi RC$, will encounter a very large resistance from the two capacitors and the circuit will be as if it is an open circuit for these low frequencies. Similarly, one could use two resistors and one inductance instead. The principle is very similar. I hope this helps you a bit.
In very rough words: electronic filters are built of "storage tanks" of electromagnetic energy (namely inductors, which store energy in a magnetic field and capacitors, which store energy in an electric field) and they work on the principle that it takes nonzero time to shuttle energy between these stores - ultimately this time is bounded by the speed of light, but in practice the time lag is much longer than this and it is the "sizes of the tanks" (which tell you how much energy is stored given the "voltage" across or the current into the tank) which sets this shuttling time. So we've essentially built a lagging system with inertia, which can only respond to slowly changing inputs. As the frequency rises, the lag means that the response is weaker and weaker owing to the delays. I've just described a low pass filter.
For a high pass filter, we arrange for the output to be the input minus the output of a lowpass filter. At low frequencies, the lowpass filter's output is significant and cancels the component of the input fed straight to the output. At high frequencies, the lowpass filter is sluggish and the cancellation is weak, so the input gets to the output almost unchanged.
protected by Qmechanic♦ Aug 16 '13 at 6:01
Thank you for your interest in this question.
Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.
Would you like to answer one of these unanswered questions instead?