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ANALOG TO DIGITAL CONVERSION Lecture 3,4 Syed M. Zafi S. Shah احسان احمد عرساڻي

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Presentation on theme: "ANALOG TO DIGITAL CONVERSION Lecture 3,4 Syed M. Zafi S. Shah احسان احمد عرساڻي"— Presentation transcript:

1 ANALOG TO DIGITAL CONVERSION Lecture 3,4 Syed M. Zafi S. Shah احسان احمد عرساڻي

2 ANALOG TO DIGITAL CONVERSION SAMPLING SAMPLING THEOREM, ALIASING QUANTIZATION QUANTIZATION NOISE In todays class

3 Need for A/D conversion We know by now the benefits of digital signals and systems But most signals of practical interest are still analog Voice, Video RADAR signals Biological signals etc So in order to utilize those benefits, we need to convert our analog signals into digital This process is called A/D conversion

4 Three step process Analog to Digital conversion is really a three step process involving Sampling Conversion from continuous-time, continuous valued signal to discrete-time, continuous-valued signal Quantization Conversion from discrete-time, continuous valued signal to discrete-time, discrete-valued signal Coding Conversion from a discrete-time, discrete-valued signal to an efficient digital data format Represent as bit?

5 SAMPLINGQUANTIZATIO N CODING CT-CV DT-CVDT-DV Analog signalBinary bits Arbitrarily, Ive chosen Differential PCM…. Can you re-create these graphs?

6 Sampling A continuous-time signal has some value defined at every time instant So it has infinite number of sample points sample every 1 sec sample every 0.1 sec sample every 1 μsec

7 It is impossible to digitize an infinite number of points because infinite points would require infinite amount of memory and infinite amount of processing power So we have to take some finite number of points Sampling can solve such a problem by taking samples at the fixed time interval If an analog signal is not appropriately sampled, aliasing will occur, where a discrete-time signal may be a representation (alias) of multiple continuous-time signals Aliasing:

8 Shannons sampling theorem The sampling theorem guarantees that an analogue signal can be in theory perfectly recovered as long as the sampling rate is at least twice as large as the highest-frequency component of the analogue signal to be sampled A signal with no frequency component above a certain maximum frequency is known as a band-limited signal (in our case we want to have a signal band-limited to ½ Fs) Some times higher frequency components are added to the analog signal (practical signals are not band-limited) In order to keep analog signal band-limited, we need a filter, usually a low pass that stops all frequencies above ½ Fs. This is called an Anti-Aliasing filter

9 In order to sample a voice signal containing frequencies up to 4 KHz, we need a sampling rate of 2*4000 = 8000 samples/second Similarly for sampling of sound with frequencies up to 20 KHz, we need a sampling frequency of 2*20000 = samples/second What is the sampling rate for CDs? Isnt it more than the one we just calculated?

10 Example 1: For the following analog signal, find the Nyquist sampling rate, also determine the digital signal frequency and the digital signal The maximum frequency component is x(t) is Therefore according to Nyquist, we need a sampling rate of The digital signal would have a frequency The digital signal can be represented as

11 Anti-aliasing filters Anti-aliasing filters are analog filters as they process the signal before it is sampled. In most cases, they are also low-pass filters unless band-pass sampling techniques are used The ideal filter has a flat pass-band and the cut-off is very sharp, since the cut-off frequency of this filter is half of that of the sampling frequency, the resulting replicated spectrum of the sampled signal do not overlap each other. Thus no aliasing occurs

12 Practical low-pass filters cannot achieve the ideal characteristics. What can be the implications? Firstly, this would mean that we have to sample the filtered signals at a rate that is higher than the Nyquist rate to compensate for the transition band of the filter Thats why the sampling rate of a CD is 44.1 KHz, much higher than the 40 KHz we calculated Go through the assignment… it has some reading task along with some problems

13 Example 2: Find the Nyquists rate for the following signal This composite signal comprises three frequencies f 1 = 25 Hz, f 2 = 150 Hz, f 3 = 50 Hz So, according to Nyquist we need to sample at 300 Hz However, for the sine term, the sampled signal has values sin( π n), meaning the samples are taken at the zero crossings, so the sine term is not counted in the process Therefore, a solution is to sample at higher than twice the max. freq component

14 Quantization Now that we have converted the continuous-time, continuous-valued signal into a discrete-time, continuous-valued signal, we STILL need to make it discrete valued This is where Quantization comes into picture The process of converting analog voltage with infinite precision to finite precision For e.g. if a digital processor has a 3-bit word, the amplitudes of the signal can be segmented into 8 levels

15 Quanitization

16 General rules for Quantization Important properties of the quantizer include Number of quantization levels Quantization resolution Note the minimum & maximum amplitude of the input signal Ymin & Ymax

17 Note the word-length of DSP m-bits Number of levels of quantizer is equal to L = 2 m The resolution of the quantizer is given as Resolution of a quantizer is the distance between two successive quantization levels More quantization levels, better resolution! Whats the downside of more quantization levels?

18 m = 4, L = 16 Ymin = 0 Ymax = 1 = 1/15 =

19 Quantization error The error caused by representing a continuous-valued signal(infinite set) by a finite set of discrete-valued levels Suppose a quantizer operation given by Q(.) is performed on continuous-valued samples x[n] is given by Q(x[n]), then the quantization error is given by

20 Lets consider the signal, which is to be quantized. In the figure (previous slide), we saw that there was a difference between the original signal and the quantized signal. This is the error produced while quantization It can be reduced, however, if the number of quantization levels is increased as illustrated on next slide

21 3-bit ADC 8-bit ADC Quant. error

22 Signal-to-Quantization-noise ratio Provides the ratio of the signal power to the quantization noise (or quanitization error) Mathematically, where P x = ¨Power of the signal x (before quantization) P q = ¨Power of the error signal x q

23 The sampled signal in the FD جُزايل سگنل ڪثرتي ميدان ۾ If Then

24 The Ideal Sampling Theorem خيالي جُزڪاري نظريئڙو Also called Uniform Sampling Theorem Nyquist Theorem If x(t) is band limited with no components at frequencies greater than f h Hz then it is completely specified by samples taken at the uniform rate fs>2f h 2f h is called the Nyquist rate

25 The Sampling Rate جُزڪاريءَ جي شرح The no. of samples per second هڪ سيڪنڊ ۾ جُزن جو تعداد If the sampling rate: f s >2f h, it is called Over-sampling f s =2f h, it is called Critical-sampling f s <2f h, it is called Under-sampling

26 لڳ ڀڳ ڪُل اوقاتي Near Continuous

27 T=0.1

28 T=?

29

30 Anti-aliasing filters Anti-aliasing filters are analog filters as they process the signal before it is sampled. In most cases, they are also low-pass filters unless band-pass sampling techniques are used The ideal filter has a flat pass-band and the cut-off is very sharp, since the cut-off frequency of this filter is half of that of the sampling frequency, the resulting replicated spectrum of the sampled signal do not overlap each other. Thus no aliasing occurs

31 Practical low-pass filters cannot achieve the ideal characteristics. What can be the implications? Firstly, this would mean that we have to sample the filtered signals at a rate that is higher than the Nyquist rate to compensate for the transition band of the filter Thats why the sampling rate of a CD is 44.1 KHz, much higher than the 40 KHz we calculated Go through the assignment… it has some reading task along with some problems

32 Example 2: Find the Nyquists rate for the following signal This composite signal comprises three frequencies f 1 = 25 Hz, f 2 = 150 Hz, f 3 = 50 Hz So, according to Nyquist we need to sample at 300 Hz However, for the sine term, the sampled signal has values sin( π n), meaning the samples are taken at the zero crossings, so the sine term is not counted in the process Therefore, a solution is to sample at higher than twice the max. freq component


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