You are working with a sinusoidal signal at 300 Hz that has a troubling noise component at 60 Hz. The unfiltered signal-to-noise ratio is 14dB.
You decide to build a simple differentiator circuit to boost the signal at the expense of the lower-frequency noise. On hand are capacitors of value 1.0 microFarad and 0.01 microFarad and numerous resistors in the range 1.0 k to 1.0 M. Design a differentiator with a gain of 100 at 300 Hz.
What is the signal-to-noise ratio at the output of the differentiator?
The gain is - Zf/Zi = - R/-jXC = - j(2pi RC) f
Solving for the RC that will give a gain of 100 at 300 Hz: RC = 53 ms.
Two values of capacitance to try:
C1 = 1.0 x 10-6 F: R = 53k
C2 = 0.01 x 10-6 F: R = 5.3M
The resistance needed to use C2 is out of range, so C1 and 53k are the design values selected.
Since the differentiator gain is directly proportional to the frequency, the gain at 60 Hz will be 20, given that the gain is 100 at 300 Hz.
The signal-to-noise ratio going into the differentiator is 5 (14dB). The differentiator adds gain of 100 for the signal of interest vs. a gain of 20 for the noise, thus another factor of 5 is added to the signal-to-noise ratio. At the output of the differentiator, the signal-to-noise ratio is 25 (28dB).