Question

2) If I need to sample an analog signal into a digital representation of it, what resolution and sampling rate would I need given the following: a) The maximum frequency I'm interested in is 38kHz. b) The analog signal is bounded between -5V and 5V. c) I need no more than 500uV of digitization error. 3) Write a ladder logic program for our PLC that will sense when one button (button A) is pressed and will start a fan. The fan should continue to run after button A is released, but stop after 5 seconds when another button (button B) is pressed. 4) Write a ladder logic program for out PLC that will use a drum sequencer to light up an array of 10 LED's like the hood display on the Pontiac Firebird in the 1980s TV show Knight Rider (https://www.youtube.com/watch?vEONyXYPhnUls) 5) Describe the steps needed to scale the conversion of an analog input such that the minimum input of OV would map to a value of 50.0 and the maximum input of 5V would map to a value of 500.0 and would be stored in the internal data register "DF10" for use in a ladder logic program for the CLICK PLC.

Answer 1

2) According to the Nyquist Sampling Theorem, A bandlimited continuous-time (analog) signal can be sampled and perfectly reconstructed from its samples if the waveform is sampled over twice as fast as it's highest frequency component.

In this case the maximum frequency of interest is 37 kHz. [Ans]

The minimum sampling rate for perfect reconstruction = 2x38 kHz = 76 kHz [Ans]

The resolution of digital representation of an analog signal is the smallest amount of change between two consecutive quantization levels.

A change of analog input equal to resolution causes a change is the least significant bit of the digital output; A change in analog input with value less than resolution does not change the digital output.

Resolution = (Full scale range of analog value)/(no. of possible binary outputs)

                   = (Maximum possible analog value - minimum possible analog value)/2^M

Where M is the number of bits in the digital output.

In terms od digital representation, resolution is the number of bits in the digital output, i.e. M.

Digitization error is the difference between analog input and it's corresponding quantization level. This error ranges between -1/2*resolution and +1/2*resolution, since the difference of < -1/2*resolution or +1/2*resolution would cause the analog input to be represented by a different quantization level.

Qn Digitization error Resolution n-1

In figure The analog signal (shown in blue) at the point marked with red has digitized value Q_n . The digitization error is Analog value - Q_n. If this error was < -1/2*resolution the analog value would be digitized as the previous quantization level Q_n-1 and if this error was > +1/2*resolution the analog value would be digitized as the next quantization level Q_n+1.

In given question, analog value range = 5V - (-5 V) = 10 V

maximum allowed digitization error = 500 $\small \mu$ V

Analog Resolution for 500 $\small \mu$ V digitization error = 2*500 $\small \mu$ V = 1mV

=> 10 V/ 2^M$\small \leq$ 1 mV [ M = resolution in terms of digital representaion]

=>2^M$\small \geq$ 10V/1mV = 10000

=> M $\small \geq$ log₂(10000) = 13.2877

=> M = 14

=> Analog resolution = 10 V / 2¹⁴=0.61 mV

=> To have digitization error $\small \leq$ 500 $\small \mu$ V, the analog resolution is 0.61 mV and digital resolution is 14 bits. [Ans]