Question

David and Austin are trying to digitize analog pulses coming from a detector. The pulses are on average 35 mV in amplitude, with a peak width (duration) of 5 ms.

If the maximum signal they expect is 50 mV, what is the least significant bit (LSB, or q) if they use a 12-bit digitizer?

If the digitizer samples at a rate of 100 Hz (100 samples per second), can they resolve the signals as a function of time? Explain your answer in a single sentence.

Answer 1

The maximum signal that is expected = 50mV

Let the minimum signal be 0V

Thus Q = Vhi - Vlo / 2¹² since it is a 12 bit digitiser

Q = 50-0 / 4096 = 1.22*10^-2 mV

By Nquist theorem, to reproduce a signal adequately, the sampling rate should be >= 2 X highest frequency of the analog singal.

THus sampling rate for this signal should be>= 2* 1/5mS = 0.4KHz = 400 Hz

Therefore a digitiser sampling at 100 Hz cannot reproduce the signal as a function of time.