File:R-c curve3.svg
Summary
Transfer function of a simple resistor-capacitor filter.
Graph shows the input-output characteristic, | Vout / Vin |, as a function of frequency, W. Also known as the frequency response, or "what you get out of the filter at various frequencies (W)".
If W ("omega") is in radians per second, the function is valid for a filter with R = 1 Ohm and C = 1 Farad.
Other R and C values scale the frequency linearly, in inverse proportion; that is, doubling R or C will halve W.
The graph shows the amplitude response only. There is also phase shift, which is not shown in the graph. The phase shift can be briefly described as follows: At very low frequencies, the output voltage leads the input voltage by 90 degrees; at W=1, it leads by 45 degrees; at very high frequencies there is no phase shift, i.e. the output is in phase with the input.
The mathematical formula for the transfer function is:
Vout 1
------ = ------------------- where i is the square root of -1
Vin (( i * W ) + 1 )
The amplitude response (ignoring phase) is the absolute value of the above, or:
| Vout | 1 | ------ | = ------------ | Vin | (W2 + 1)^1/2
The red line in the graph shows the above response. The green line shows the response for two such filters cascaded one after the other; i.e., the output of the first filter is the input of the second. It is assumed that the second filter does not load the first filter; this might be achieved by placing a unity-gain buffer between the two, or by making the second filter with a much higher impedance level, e.g., by making the second filter's resistor 1000 times the first filter's resistor, and the second filter's capacitor 1/1000 the size of the first filter's capacitor, thus keeping the frequency the same but raising the impedance 1000 times.
Schematic diagram of the filter
C
IN __________| |__________________________ OUT
| | |
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---
| |
| |
R | |
| |
| |
---
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|
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|
___|___
GND. ___
_
Shematic of two cascaded filters
__
C1 | \ C2
| \
IN _________| |____________ ________________| \ _______________| |____________ _____________ OUT
| | | | / | | |
| | / |
| |__/ |
| |
| Buffer, |
--- gain = 1 ---
| | | |
| | | |
R1 | | R2 | |
| | | |
| | | |
--- ---
| |
| |
| |
| |
___|___ ___|___
GND. ___ ___
_ _
If using a buffer, as shown, then R2 can be the same as R1 and C2 the same as C1. Alternatively, omit the buffer and make R2 = R1 * Y and C2 = C1 / Y , where Y is a large number.
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