Use this book analog and digital communication systems 5th e by Martin S. Roden

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Use this book analog and digital communication systems 5th e by Martin S. Roden chapter 1 for more information
The question again if it is not clear in the photo: 1.15 Design a counting analog-to-digital converter to operate on the signal s(t)= 2sint.The design requires four-bit analog-to-digital conversion?.
* Use example 1.6 with it solution 1.15 Design a counting analog-to-digital converter to operate on the signal s(t) 2sin1tt. The design requires four-bit analog-to-digital conversion Example 1.6 Design a counting ADC to convert s() sin2000nt into a 4-bit digital signal. Solution: The frequency ofsay is 1000 Hz, so to comply with the sampling theorem, the sampling rate must be greater than 2000 samplessec. Let us arbitrarily choose a rate 25% above this value, or 2500 samples/sec. There are many practical tradeoff considerations which go into this choice of sampling rate. If we use a rate close to the minimum, precise lowpass filtering is required at the receiver to reconstruct the original signal. On the other hand, if we use a much higher rate, the bandwidth of the transmitted waveform increases The individual sample values range between -1 V and +1 v. The counting quantizer discussed in this section operates on positive samples. We therefore shift the signal by 1 volt to assure that samples never go negative. The shifted samples range from 0 volt to 2 volt. The ramp must be capable of reaching the maximumsample value within the sampling period, 0.4 msec. The slope must therefore be at least 2/0.4x103-5000 volts/sec. In prac Section 1.6 A/D and DIA Conversion tice, we would choose a value larger than this to account for slight jitter in the timing of the system and also to give the ramp to return to zero prior to the next sampling point. We time might choose a much value if we wanted to convert the sample in a small fraction of the period. larger slope a number of This would apply if a converter is being shared among signals. At the minimum slope, it takes the ramp function o 4 msec to reach the maximum sample value. The counter should therefore count from 0000 to 1111 in 0.4 msec. This requires a counting rate of 40,000 counts/sec

Explanation / Answer

The general formula for sine wave is s(t) = Asin(wt) and .w = 2/T = 2f

For given equation, 2sint = 2sin(2*0.5*t)

So here Amplitude, A = 2V and Frequency f = 0.5Hz

As per sampling theorem, Sampling freq, Fs= 2*f = 2 * 0.5 = 1Hz, So Sampling time Ts = 1s

Again, Total Sample value range, Rs= 4V (-2V to 2V), So Minimum Slope value = Rs/Ts= 4/1 = 4 Volts/Sec

Counting rate of 4 bit counter = 16 * 1 = 16 counts/sec

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