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RC LPF Filter

timjohnson

New Member
Hello, I'm a beginner in electronics, sorry if what I'm asking is basic. I have a task for one of my assignments which I'm unsure about. It is about RC LPF filters. I am asked to modify a circuit so the cut-off frequency is doubled. How would I go about doing so? Would I simply reduce the resistor or capacitor value for the formula: F(cutoff)= 1/2piRC. Also how would I show this practically? My lecturer says to keep the amplitude of the input constant but I'm not sure what he means by that. The circuit consists of just a signal generator, supplying a resistor of 1k and capacitor of 100n and a Vout measurement between the resistor and capacitor. Any help would be great. I have also attached an image.
 

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ronsimpson

Well-Known Member
Most Helpful Member
so the cut-off frequency is doubled. …... Would I simply reduce the resistor or capacitor value
Yes you understand.
keep the amplitude of the input constant but I'm not sure what he means by that.
V_in should keep the same value. If you signal generator is set to 1 volt do not change it.

resistor or capacitor
Looking for other ways. You have stated that you can change the resistor or you can change the capacitor.
RC_old/2= RC_new (500 ohms and 100nF) or (1k ohms and 50nF)
Please tell me about (707 ohms and 70.7nF). What is the frequency?
------------
What about 2k & 25nF?
 
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JimB

Super Moderator
Most Helpful Member
Looks like this belongs in the "Homework Help" section.

Moving...

JimB
 

timjohnson

New Member
RC_old/2= RC_new (500 ohms and 100nF) or (1k ohms and 50nF)
Please tell me about (707 ohms and 70.7nF). What is the frequency?
------------
What about 2k & 25nF?
For 707 ohms and 70.7nF, the frequency would be 3184Hz
and for 2k and 25nF it would be 3183Hz. I see how you could use a combination of both or either one to get double the cut off.
I forgot to mention that the signal generator has an output impedance of 600 ohms, how would this effect the circuit?

Also I kept V_in the same amplitude (1v), but when I change frequencies from 100ohm to 1kohm the amplitude drops and is not constant. Shouldn't it be the same, as those frequencies are before the roll off?
 

crutschow

Well-Known Member
Most Helpful Member
the signal generator has an output impedance of 600 ohms, how would this effect the circuit?
Any generator impedance is in series with R1 and will thus affect both the low frequency amplitude and the cut-off frequency.
when I change frequencies from 100ohm to 1kohm
Are you talking about frequency or resistance since Ohm is a unit of resistance?
 

ronsimpson

Well-Known Member
Most Helpful Member
I forgot to mention that the signal generator has an output impedance of 600 ohms, how would this effect the circuit?
So there is a 600 ohm resistor inside the signal generator that we did not know about.
1k + 600 = 1600 ohms. (you should add the 600 ohm in) R_total=R1+600
You could take R1 down to 0 and you still have the 600 ohms.

Probably an audio signal generator. Most signal generators are more like 50 ohm.
but when I change frequencies from 100ohm to 1kohm the amplitude drops and is not constant.
" frequencies from 100ohm to 1kohm" resistance from 100 to 1kohm
From here I can not see what you are doing. But at low frequencies like 100hz the (600 ohms in the generator OR the 1k R1) should not effect you but at high frequencies the signal level should be lower with higher resistance.
 

timjohnson

New Member
Any generator impedance is in series with R1 and will thus affect both the low frequency amplitude and the cut-off frequency.
Are you talking about frequency or resistance since Ohm is a unit of resistance?
Apologies, I meant frequency; 100Hz to 1kHz
 

timjohnson

New Member
Probably an audio signal generator. Most signal generators are more like 50 ohm.
I think it is an audio generator from memory.

" frequencies from 100ohm to 1kohm" resistance from 100 to 1kohm
From here I can not see what you are doing. But at low frequencies like 100hz the (600 ohms in the generator OR the 1k R1) should not effect you but at high frequencies the signal level should be lower with higher resistance.
Apologies, I meant 100Hz to 1kHz
 

Ratchit

Well-Known Member
Hello, I'm a beginner in electronics, sorry if what I'm asking is basic. I have a task for one of my assignments which I'm unsure about. It is about RC LPF filters. I am asked to modify a circuit so the cut-off frequency is doubled. How would I go about doing so? Would I simply reduce the resistor or capacitor value for the formula: F(cutoff)= 1/2piRC. Also how would I show this practically? My lecturer says to keep the amplitude of the input constant but I'm not sure what he means by that. The circuit consists of just a signal generator, supplying a resistor of 1k and capacitor of 100n and a Vout measurement between the resistor and capacitor. Any help would be great. I have also attached an image.
The schemat you submitted does not give any voltage source resistance, so assume it is zero. This assumption also coincides with the Bode plot you submitted. If you want to increase the cut-off frequency of your low pass filter by an octave, just reduce the RC time constant by one-half. It is now 0.0001 sec. Make it 0.00005 sec.

Ratch
 
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