Low pass filter: cutoff and peak output

[joel amos]

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 V_in = 3cos(2π(4000)t), what will the peak output voltage be?[/B]

Homework Equations

(a) f_c = 1/(2πRC)
(b) V_out/V_in = 1/√(1 + (2πfRC)²)

The Attempt at a Solution

(a) I rearranged the equation to C = 1/(2πf_cR). 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.

 

[milesyoung]

(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.

(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.

 

[joel amos]

(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.

 

[milesyoung]

(a) I rechecked my work and got 677.3 pF

Great.

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?

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.

 

[joel amos]

Ah right, so the frequency is simply 4 kHz, which means V_out would be approximately equal to V_in.

 

[LvW]

Yes - approximately. However, do you think this answers the queston? Why not do an exact calculation?

 

[joel amos]

Well, I don't know how to do the calculation. This graph makes is seem as though they'd be exactly equivalent:

 

[LvW]

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)