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RC LPF Filter

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timj:
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.

bson:
It doesn't matter which you change; the cutoff frequency follows their product, R*C.  Reducing either will increase the cutoff frequency since it's reciprocal.  The choice of values is mainly practical; smaller values for R increase current consumption and power dissipation, large values for R increase noise, and larger capacitors are physically bigger and cost more.  Since the current values are presumably well-chosen, you could reduce both enough to double the cutoff frequency.  Of you could reduce the capacitor alone; usually smaller capacitors are a win, at least down to 100pF or so.

todorp:
To keep the loading of the source similar only the cap C should be reduced (without touching the R).

timj:
So if the audio generator has an output impedance of 600ohm, does that mean I'll need to include that in my calculation for R1?

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

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