Low pass filter: cutoff and peak output

AI Thread Summary
To determine the cutoff frequency for a low-pass filter with R1 at 4.7kΩ, a capacitor value of 677.3 pF is calculated using the formula C = 1/(2πfcR). The input voltage is given as Vin = 3cos(2π(4000)t), indicating a frequency of 4 kHz. Since this frequency is below the cutoff of 50 kHz, the output voltage (Vout) will be approximately equal to Vin, but a small voltage division occurs due to the capacitor's impedance. A more precise calculation is needed to account for phase shift and magnitude using the complex transfer function. Understanding these concepts is crucial for accurate analysis of low-pass filter behavior.
joel amos
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Homework Statement


(a) If R1 = 4.7kΩ, what value for C1 will give a cutoff frequency of 50kHz?
(b) If the input voltage is described by the equation Vin = 3cos(2π(4000)t), what will the peak output voltage be?[/B]
669es3.png


Homework Equations


(a) fc = 1/(2πRC)
(b) Vout/Vin = 1/√(1 + (2πfRC)2)

The Attempt at a Solution


(a) I rearranged the equation to C = 1/(2πfcR). I got 71.2 pF but am uncertain.
(b) Not sure if this is the relevant equation. If it is, than I am uncertain how to get the frequency of the input voltage.
 
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joel amos said:
(a) I rearranged the equation to C = 1/(2πfcR). I got 71.2 pF but am uncertain.
The equation is fine, but you got the wrong number out of it. Can't tell you what went wrong if you don't show your calculation.

joel amos said:
(b) Not sure if this is the relevant equation. If it is, than I am uncertain how to get the frequency of the input voltage.
Recall the general expression ##A\cos(\omega t + \phi)##, where ##A##, ##\omega##, ##\phi## is the amplitude, angular frequency and phase, respectively, of the sinusoid. Match that to what you're given.

After you've done that, think about what the purpose of a low-pass filter is and how it affects signals below and above its cut-off frequency.
 
(a) I rechecked my work and got 677.3 pF
(b) So if without the low pass filter the peak voltage would be 3V. Would the frequency be 2π4000 = 25132.7? What about phase? If I'm assuming the values from part (a), then wouldn't the output voltage match the input voltage as 25.1 kHz < 50kHz and this is a low pass filter? If this is wrong and the frequency is really higher than the cutoff, I know that output lessens as frequency increases, but I'm not sure of the equation that would give me the voltage output.
 
joel amos said:
(a) I rechecked my work and got 677.3 pF
Great. :smile:

joel amos said:
Would the frequency be 2π4000 = 25132.7?
You have ω = 2π4000 rad/s, which is the angular frequency. What is the relationship between frequency and angular frequency?

joel amos said:
If I'm assuming the values from part (a), then wouldn't the output voltage match the input voltage as 25.1 kHz < 50kHz and this is a low pass filter?
Approximately, if the frequency of the input waveform was lower. There'll be a difference in phase.
 
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Ah right, so the frequency is simply 4 kHz, which means Vout would be approximately equal to Vin.
 
Yes - approximately. However, do you think this answers the queston? Why not do an exact calculation?
 
Well, I don't know how to do the calculation. This graph makes is seem as though they'd be exactly equivalent:
lowpassfreqdomain.png
 
No - it`s impossible. Because at 4 kHz the capacitor has an impedance other than infinite, you have a (small) voltage division between R and C.
In addition, there is a (small) phase shift, Thus, you have to use the complex transfer function for separately calculating magnitude and phase.
Don`t you remember - in your first post you have given already the correct magnitude expression. Insert the frequency - and you have the result.

(Remark: The shown graph is not very realistic for a first-order lowpass)
 
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