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PASSIVE AND ACTIVE FILTERS

Passing and Rejecting Signals Based on Their Frequency

by James Antonakos

Start ý A Filter by Any Other Name ý Whatýs the Frequency, Kenneth? ý A Closer Look ý Itýs Just a Phase ý Whatýs Load Got to Do with It? ý Activate That Filter! ý Filters to the Rescue ý Sources and PDF

A CLOSER LOOK

Take, for example, a low-pass filter with a corner frequency of 1591 Hz and apply a 5-V_RMS, 2-KHz input signal to it. What is the output amplitude?

In order to determine the answer, you must know the relationship between the gain and frequency in the filter. Figure 3 shows how the gain formula for the passive low-pass filter is derived.

Figure 3ýHere is the derivation of the gain formula for the passive low-pass filter. j indicates that complex numbers are involved (and that the gain has real and imaginary components).

Think of the passive low-pass filter in Figure 1 as a voltage divider. Some of the input voltage will drop across the resistor, and the rest will drop across the capacitor. Though the resistor value is fixed, the capacitive reactance will vary with frequency (getting smaller as frequency increases). So, the action of the voltage divider will vary, as will the output amplitude.

Starting with the basic voltage divider formula in Figure 3, substitute the formula for capacitive reactance and then use algebra to simplify things. Notice that the resulting formula shows the gain as a function of the input frequency (f) and the corner frequency (f_c). The ratio of these two frequencies is what drives the gain and phase shift of the filter.

To check the formula, note that as f increases, the ratio of f to f_C gets larger. This makes the denominator of the gain fraction larger, which makes the overall fraction smaller. Thus, as f increases, the gain gets smaller, as it should in a low-pass filter.

Photo 1 shows a screen shot of Excel providing a suitable range of input frequencies to the low-pass filter and the associated output values. The gain in dB is easily found by multiplying the base-10 log of the filter gain by 20. For example, 20 ý (base-10 log of 0.707) = ý3 dB.

Do you see that almost the full range of gain (from one down to zero) is contained in a frequency range that goes from a decade below the corner (159 Hz) to a decade above the corner (15.9 KHz)? This is important to remember.

Also, do you see that that gain has fallen by 20 dB in the second decade of frequencies (1591 Hz to 15.9 KHz)? This is a characteristic of a first-order filter. A second-order filter has a roll-off of ý40 dB/decade. A third-order filter would be ý60 dB/decade, and so on.

Photo 1ýHere you can see Excel performing gain and phase calculations on the low-pass filter.

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