Why is an RC circuit a low pass filter? Yes, this is a homework question that I'm working on, but hopefully someone can help me. I can't seem to figure it out.
Low Pass Filter in an RC Circuit
- Thread starter Zincq
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An RC circuit could be a high pass filter, too. Just swap the R and C positions.
But, I assume you have in mind a series resistor from a voltage source connected to a shunt capacitor, thence to the load.
Since a capacitor has reactance, it loads down the voltage from the resistor it is connected to. This loading increases more and more as the freq increases. So, output voltage will drop as the freq increases. Since low freqs pass with less drop than high freqs, presto, low pass filter.
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Supplemental:
This RC filter cut-off freq is defined as the freq at which C's reactance = R's resistance. At that point, the output will be 70.7% of the input. This is the half power point, -3db.
R's voltage and C's voltage are 90 degrees out of phase, so don't expect the 2 voltages to add up to the input voltage. The 2 voltages actually add the way 2 sides of a right triangle add to equal the hypotenuse. Ea= sqr(Ec^2 + Er^2)
But, I assume you have in mind a series resistor from a voltage source connected to a shunt capacitor, thence to the load.
Since a capacitor has reactance, it loads down the voltage from the resistor it is connected to. This loading increases more and more as the freq increases. So, output voltage will drop as the freq increases. Since low freqs pass with less drop than high freqs, presto, low pass filter.
--------
Supplemental:
This RC filter cut-off freq is defined as the freq at which C's reactance = R's resistance. At that point, the output will be 70.7% of the input. This is the half power point, -3db.
R's voltage and C's voltage are 90 degrees out of phase, so don't expect the 2 voltages to add up to the input voltage. The 2 voltages actually add the way 2 sides of a right triangle add to equal the hypotenuse. Ea= sqr(Ec^2 + Er^2)
highwire is correct. Here's another way to think about it:
If you have the case he talked about, with a source connected to a resistor, then capacitor in series, with the load connected across the capacitor, then the output voltage will definitely depend on frequency, since the reactance of a capacitor is 1/(omega*C). As the frequency (omega) increases, the capacitor's reactance decreases. If you let the frequency approach infinity, then the capacitor will look like 0 ohms, or a short circuit. Now, your load is connected across the capacitor. Since V=IR, there is 0 volts across the capacitor, meaning as the frequency goes to infinity, the output voltage goes to zero. At any frequency, you can find the reactance of the capacitor, then use a simple voltage divider to find the actual output voltage.
If you're into Laplace transforms and such, you can obtain a transfer function relating the output to the input for any value of s (frequency). In this case, I believe the transfer function is (1/RC)/(s+1/RC). Plug in j*omega for s, and take the magnitude of the resulting complex number, and you have your ratio of output to input.
I now realize that this is poorly written and I probably messed up somewhere, but I hope this helps somewhat.
If you have the case he talked about, with a source connected to a resistor, then capacitor in series, with the load connected across the capacitor, then the output voltage will definitely depend on frequency, since the reactance of a capacitor is 1/(omega*C). As the frequency (omega) increases, the capacitor's reactance decreases. If you let the frequency approach infinity, then the capacitor will look like 0 ohms, or a short circuit. Now, your load is connected across the capacitor. Since V=IR, there is 0 volts across the capacitor, meaning as the frequency goes to infinity, the output voltage goes to zero. At any frequency, you can find the reactance of the capacitor, then use a simple voltage divider to find the actual output voltage.
If you're into Laplace transforms and such, you can obtain a transfer function relating the output to the input for any value of s (frequency). In this case, I believe the transfer function is (1/RC)/(s+1/RC). Plug in j*omega for s, and take the magnitude of the resulting complex number, and you have your ratio of output to input.
I now realize that this is poorly written and I probably messed up somewhere, but I hope this helps somewhat.
could someone explain reactance? (I have a guess but i'm not sure)
wow... laplace. so you actually do use diffEq in reality
wow... laplace. so you actually do use diffEq in reality
Reactance is an easier way to characterize energy storage elements (inductors and capacitors) in AC circuits. In DC circuits, the relationship for an inductor is v = L*di/dt, where v is the voltage across the inductor and i is the current through the inductor, and the equation for a capacitor is i = C*dv/dt, where i is the current through the capacitor and v is the voltage across it. These equations are also valid in AC circuits, but some simplifications can be made. We can make generalizations for the sinusoidal steady state response, which is the output response to an input sinusoid (think the voltage from a wall outlet), after all transients have died out.
Using some ideas from the Laplace domain (the s-domain), it can be shown that we can represent the impedance (reactance) of a capacitor is 1/(s*C), and the impedance (reactance) of an inductor is s*L. For the AC case, we know that we're dealing with sinusoids, which allows us to substitute j*omega in for s, where j is the imaginary unit. That's right, since I is used for current, EE's decided to use j instead. This is part of a technique called phasor analysis, which is very handy when dealing with sinusoids in differential equations.
This leaves us with the impedance of an inductor as j*omega*L, and the impedance of a capacitor as 1/(j*omega*C). The advantages of these quantities is that we can then treat inductors and capacitors much like we do resistors, using the relation V=IZ, where Z now is the impedance of the circuit element. Ohm's law is better than you thought, eh?
Using these quantities, we can get the output as a function of s, which is the Laplace transform of the output. Then, we apply the inverse Laplace transform to get the solution back into the time domain, where we all live. Or, if we use phasor analysis, then the output will be some complex number. The magnitude of the complex number will be the magnitude of the resulting sinusoid, and the phase angle of the complex number's phasor will be the phase angle of the output. Either way works very well for solving such problems, and have different times where they are preferable. For instance, using the s-domain to get a transfer function between the output and input is useful in obtaining the system's frequency response, often represented by a Bode plot.
There's a brief explanation on how the impedance is calculated and used. I hope this helps. And also, what you said about differential equations actually being used in the real world, you would be amazed at how much in engineering relies on differential equations. I'm only finishing up my sophomore year, and it is very surprising how important DE's are to engineering problems. I'm not sure if you plan on entering the engineering field, but the mathematics is sometimes pretty intensive, but rather interesting.
P.S. I'm sure some of this is not quite right, so if I was wrong anywhere, someone who is more knowledgeable than I am, please feel free to correct me.
Using some ideas from the Laplace domain (the s-domain), it can be shown that we can represent the impedance (reactance) of a capacitor is 1/(s*C), and the impedance (reactance) of an inductor is s*L. For the AC case, we know that we're dealing with sinusoids, which allows us to substitute j*omega in for s, where j is the imaginary unit. That's right, since I is used for current, EE's decided to use j instead. This is part of a technique called phasor analysis, which is very handy when dealing with sinusoids in differential equations.
This leaves us with the impedance of an inductor as j*omega*L, and the impedance of a capacitor as 1/(j*omega*C). The advantages of these quantities is that we can then treat inductors and capacitors much like we do resistors, using the relation V=IZ, where Z now is the impedance of the circuit element. Ohm's law is better than you thought, eh?
Using these quantities, we can get the output as a function of s, which is the Laplace transform of the output. Then, we apply the inverse Laplace transform to get the solution back into the time domain, where we all live. Or, if we use phasor analysis, then the output will be some complex number. The magnitude of the complex number will be the magnitude of the resulting sinusoid, and the phase angle of the complex number's phasor will be the phase angle of the output. Either way works very well for solving such problems, and have different times where they are preferable. For instance, using the s-domain to get a transfer function between the output and input is useful in obtaining the system's frequency response, often represented by a Bode plot.
There's a brief explanation on how the impedance is calculated and used. I hope this helps. And also, what you said about differential equations actually being used in the real world, you would be amazed at how much in engineering relies on differential equations. I'm only finishing up my sophomore year, and it is very surprising how important DE's are to engineering problems. I'm not sure if you plan on entering the engineering field, but the mathematics is sometimes pretty intensive, but rather interesting.
P.S. I'm sure some of this is not quite right, so if I was wrong anywhere, someone who is more knowledgeable than I am, please feel free to correct me.
Just explain it mathematically... much simpler to accept and understand.
For a simple LP filter, the transfer function is 1/(1+jwRC) where w is supposed to be the frequency in rad/s. As w is varied between 0 and infinity, you can see that the transfer function starts at 1 and goes to 0. Remember that a transfer function value of 1 means it passes all of the signal input. As w gets larger, the transfer function value decreases. Plotted on a log graph, you can see the plotted curve passes most of the lower frequency range (depending on R and C values of course).
For a simple LP filter, the transfer function is 1/(1+jwRC) where w is supposed to be the frequency in rad/s. As w is varied between 0 and infinity, you can see that the transfer function starts at 1 and goes to 0. Remember that a transfer function value of 1 means it passes all of the signal input. As w gets larger, the transfer function value decreases. Plotted on a log graph, you can see the plotted curve passes most of the lower frequency range (depending on R and C values of course).
or to explain it in plain english.... a capacitor is basically an open circuit at DC, and a short circuit at high frequency.
so in a low pass RC circuit close to DC the cap is an insignificant part. at high frequencies the cap "shorts" the output to ground effectively blocking the high frequencies.
this type of generalization can be misleading at RF levels, but is a great help when just trying to understand what a circuit does.
a lot easier than punching through 5 equations to get some number that doesn't mean anything unless you're just dying to know the operating points of the circuit. so many of my EE comrades at school couldn't say what a circuit meant unless they spent 2 hours with their HP calculator and a spice program and after that they'd have a hard time coming up with a reason anybody would even use a circuit as simple as an RC filter. They sure knew how to tell you how much current was flowing to the load at 344.849038243820843902849023 Hz though.
jt
so in a low pass RC circuit close to DC the cap is an insignificant part. at high frequencies the cap "shorts" the output to ground effectively blocking the high frequencies.
this type of generalization can be misleading at RF levels, but is a great help when just trying to understand what a circuit does.
a lot easier than punching through 5 equations to get some number that doesn't mean anything unless you're just dying to know the operating points of the circuit. so many of my EE comrades at school couldn't say what a circuit meant unless they spent 2 hours with their HP calculator and a spice program and after that they'd have a hard time coming up with a reason anybody would even use a circuit as simple as an RC filter. They sure knew how to tell you how much current was flowing to the load at 344.849038243820843902849023 Hz though.
jt
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