Finding frequency and phase?

Question

a low pass first order instrument has a time constant of 20 ms. Find the frequency, in hertz, of the input at which the output will be 93% of the dc output. Find the phase angle at this frequency.

I'm stumped here. I know how to get the frequency but I cant do the rest. All I need is the formulas.

Answer 1

The key here is to note that you've been provided with the general form of the frequency response (1st order low pass), and the corner frequency (w = 1/.02).

An equivalent circuit is a voltage divider consisting of a resistor R and capacitor C in series.  The input voltage is applied across the whole RC, while the output voltage is measured across just the C.

So, we start writing the frequency-dependent formulas:

1.  For the resistor, its impedance is just R.

2.  For the capacitor, its impedance is Zc = -j/(w*C), where w = 2*pi*f.

Frequency can be expressed as f in Hertz or w (omega) in radians/second.  Both are correct, but be careful to apply the 2*pi appropriately.

So, the voltage divider becomes:

Vout = Vin * (Zc/(R+Zc)

But, we don't really care about the exact values of Vin and Vout, only their relative values.  This is the 93% point asked for.

Vout / Vin = 0.93 = Zc/(R+Zc)

Let's expand this to show it in terms of R, C, and f:

0.93 = (-j/(2*pi*f*C)) / (R + -j/(2*pi*f*C))

Hmm...it is still complicated...and still complex numbers.

With a bit of algebra, this will reduce to:

0.93 = | 1/(1+j*2*pi*f*R*C) |, and eventually become

f = 1/(2*pi*R*C) * sqrt(1/.93^2 -1)

From the textbook (and your differential equations class), you know that the time constant tau = R*C = 0.02 seconds.

Now it reduces to:

0.93 = | 1/(1+j*2*pi*f * 0.02) |

This is to the point that you can calculate f.

To find phase, go back to the original voltage divider circuit:

Vout/Vin = (1/(1+j*2*pi*f*R*C))

Given the frequency of interest, just solve this equation in complex form and put it in polar form using your choice of sine, cosine, or tangent functions.