How does a low pass harmonic filter work and why is the resistor necessary?

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A low pass harmonic filter allows low frequencies to pass while blocking high frequencies, utilizing components like inductors and capacitors to manage reactance based on frequency. Inductors increase reactance at higher frequencies, while capacitors decrease reactance, effectively shunting high frequencies to ground. The inclusion of a resistor in an RC low-pass filter alters the transfer function, changing the circuit's response and defining the break frequency. Understanding the transfer function is crucial for analyzing filter behavior, as it dictates the relationship between input and output voltages. Mastery of these concepts is essential for effective filter design and analysis.
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Can someone explain to me how a low pass harmonic filter works in laymens terms? I would really appreciate it?
 
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low pass filters pass low frequencies, which means they block high frequencies.

If you look at the reactance of circuit components which change their impedance based on frequency, it will be relatively easy to understand, assume you have a 5 ohm load across a 24 volt source. If you add an inductor in series with the load, the inductive reactance is low at low frequencies (since X = 2pifL) as frequencies increase, so too does the reactance.

Alternatively, you can add a capacitor in parallel, does the same, since the capacative reactance decreases with frequency, the capacitor in parallel looks like a shunt to high frequency (this is why capacitors (decoupling capacitors) are added in parallel with loads in the most common form of bridge rectifier circuit, they short high frequencies to ground as frequency gets high (ie, transients)

You can use math to tune the filter, and add capacitors . inductors as you need to, in order to create the circuit whereby only specific ranges of frequencies can be applied to a load..
 
Thank You that helps!
 
Why is the resistor necessary for the RC low-pass filter?
 
fonz said:
Why is the resistor necessary for the RC low-pass filter?

Once again...the key lies in the transfer function of your circuit!

If you just have the capacitor...your transfer function is 1/jωc. Looks a lot like an integrator.

If you add in the resistor and are taking the voltage across the cap...your transfer function is now 1/(jωRC+1)...big difference! (simple voltage division)

Clearly two different bode plots. When ω= 1/RC in your low pass filter...that's when your break frequency occurs. Your transfer function will be 1/(j+1) at this point. Exact mathematical equation for your bode plot.

In regards to filters, unless you are talking about transfer functions (V0/VI) you are essentially wasting time! Learn it! TRANSFER FUNCTION.

In other words...when you come across any filter...the first question in your mind should be...

WHAT'S THE TRANSFER FUNCTION!

The filter follows the mathematical equation of your transfer function. Plug in values for omega equalling zero and infinity...then plug in values in between.

When you have a computer do your bode plot for you...the program is following the exact mathematical equation found in your transfer function. It's not magic!

Because of the "j" in your transfer functions...you will also have the phase change...again it just follows the math.
 
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