## A question on passive low-pass filter

1. The problem statement, all variables and given/known data

This is a simple passive low-pass filter stated in the Wikipedia.
I am wondering why we need the resistor to allow it operates as a low pass filter.
Can we just ignore the resistor(by shorting it)? What will the consequence be?

2. Relevant equations

3. The attempt at a solution
 Recognitions: Gold Member The cut-off frequency (aka 3dB point, corner frequency or half power point) is the point in the frequency based attenuation curve where a signal strength has been halved by a filter, and this is one of the characteristics of filters that allows us to compare different designs. This cut-off frequency is determined by the time constant of the circuit, which is the multiplication of the Resistance (R) and the Capacitance (C). You may have seen any of the following formulae: RC = 1/wc OR RC = 1/(2*pi*fc) OR tau = 1/wc where: tau = RC and is the commonly used symbol wc = cutoff frequency in radians fc = cutoff frequency in hertz If we did not put a resister in the circuit, the only resistance would be the static resistance of a wire (very low) and the capacitance would be the only variable we could use to determine the cutoff frequency. e.g. assume the wire resistance is 1uR and you want a filter with a cut off frequency of 1kHz. 1uR * C = 1 / (2*pi*1000) --> C = 160 F ... yikes! practical circuit values of capacitance exist in the high pico + nano + micro + low milli range, and an average 1 F capacitor is about as large as 2 standard soft drink cans on top of each other (and costs ~$100 and has its own voltmeter). If we add a 1kR resister into this circuit the capacitance required is now 159 nF, which is much more reasonable. Using both these variables you can mix and match to get the required value of cutoff frequency. Additionally, in real circuits, if there was only the resistance of the wire, by Ohms law when you have a very low resistance you have a very high current, and this would possibly melt or burn your wire. And lastly, when looking at the resistance and capacitance, remember to look at the *total* resistance and capacitance, so combine everything in the circuit together to get the equivalent values (including the load resistance!).  Quote by Zryn The cut-off frequency (aka 3dB point, corner frequency or half power point) is the point in the frequency based attenuation curve where a signal strength has been halved by a filter, and this is one of the characteristics of filters that allows us to compare different designs. This cut-off frequency is determined by the time constant of the circuit, which is the multiplication of the Resistance (R) and the Capacitance (C). You may have seen any of the following formulae: RC = 1/wc OR RC = 1/(2*pi*fc) OR tau = 1/wc where: tau = RC and is the commonly used symbol wc = cutoff frequency in radians fc = cutoff frequency in hertz If we did not put a resister in the circuit, the only resistance would be the static resistance of a wire (very low) and the capacitance would be the only variable we could use to determine the cutoff frequency. e.g. assume the wire resistance is 1uR and you want a filter with a cut off frequency of 1kHz. 1uR * C = 1 / (2*pi*1000) --> C = 160 F ... yikes! practical circuit values of capacitance exist in the high pico + nano + micro + low milli range, and an average 1 F capacitor is about as large as 2 standard soft drink cans on top of each other (and costs ~$100 and has its own voltmeter). If we add a 1kR resister into this circuit the capacitance required is now 159 nF, which is much more reasonable. Using both these variables you can mix and match to get the required value of cutoff frequency. Additionally, in real circuits, if there was only the resistance of the wire, by Ohms law when you have a very low resistance you have a very high current, and this would possibly melt or burn your wire. And lastly, when looking at the resistance and capacitance, remember to look at the *total* resistance and capacitance, so combine everything in the circuit together to get the equivalent values (including the load resistance!).
Thanks a lot!
You cleared my question!
I am studying Analog Integrated Circuit, have been learning some amplifier designs like current-mirror and compensated amplifiers, and now comes to the topic of filters.
It looks there are much for me to study.

## A question on passive low-pass filter

If you want to you can take a look at these rf filter specs to see if they help http://www.oscilent.com/catalog/Cate...saw_filter.htm