ECE3123_Chapter-01-02__Sampling - ECE3221 Digital Signal Processing Sampling Prof Dr Othman O Khalifa Electrical and Computer Engineering Kulliyyah of

ECE3123_Chapter-01-02__Sampling - ECE3221 Digital Signal...

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Unformatted text preview: ECE3221 Digital Signal Processing Sampling Prof. Dr. Othman O. Khalifa Electrical and Computer Engineering Kulliyyah of Engineering International Islamic University Malaysia Sampling Def. A digital signal can be obtained from an analogue signal by the following steps: 1. Sampling 2. Quantization 3. Coding The process of converting a continuous-time signal to a sequence of numbers is called Sampling. Event Name Basic ADC Analog to Digital Conversion ADC is an acronym for Analog to Digital Converter which converts the analog signal x(t) into the digital signal sequence x(n). Analog-to-digital conversion or Analog- todigitization consists of the sampling and quantization processes. The sampling process depicts a continuously varying analog signal as a sequence of values. The quantization process approximates a waveform by assigning an actual number for each sample. An ADC consists of two fundamental blocks; an ideal sampler and a quantizer. A/D converter Quantizer Ideal Sampler x(t) Event Name x(nT) x(n) Analog to Digital Conversion Analog-to-digital conversion carries out the following steps: Analog- to1. The bandlimited signal x(t) is sampled at uniformly spaced instants of time , nT, where n is a positive integer, and T is the nT, sampling period in seconds. This sampling process converts an analog signal into a discrete-time signal, x(nT), with discretex(nT), continuous amplitude value. 2. The amplitude of each discrete-time sample is quantized discreteinto one of the 2B levels, where B is the number of bits the 2B ADC has to represent for each sample. The discrete amplitude levels are represented (or encoded) into distinct binary words x(n) with a fixed wordlength B. This binary sequence, x(n), is x(n), the digital signal for DSP hardware. Sampling Process Use A-to-D converters to turn x(t) into numbers x[n] Take a sample every sampling period Ts – uniform sampling Continuous- x(t) time to Discrete-time x[n] x[n]=x(nTs) fs=2 kHz f=100 Hz fs=500 Hz Event Name Sampling Theorem Bridge between continuous-time and discrete-time continuousdiscreteTell us HOW OFTEN WE MUST SAMPLE in order not to loose any information Sampling Theorem A continuous-time signal x(t) with frequencies no higher than fmax (Hz) can be reconstructed EXACTLY from its samples x[n] = x(nTs), if the samples are taken at a rate fs = 1/Ts that is greater than 2fmax. For example, the sinewave on previous slide is 100 Hz. We need to sample this at higher than 200 Hz (i.e. 200 samples per second) in order NOT to loose any data, i.e. to be able to reconstruct the 100 Hz sinewave exactly. fmax refers to the maximum frequency component in the signal that has has significant energy. Consequence of violating sampling theorem is corruption of the signal signal in digital form. Whose theorem is this ? The sampling theorem is usually known as the Shannon Sampling Theorem due to Claude E. Shannon’s paper “A mathematical theory of communciation” in 1948. The minimum required sampling rate fs (i.e. 2xB) is known as the Nyquist sampling rate or Nyquist frequency because of H. Nyquist’s work on telegraph transmission in 1924 with K. Küpfmüller. The first formulation of the sampling theorem precisely and applied it to communication is probably a Russian scientist by the name of V. A. Kotelnikov in 1933. However, mathematician already knew about this in a different form and called this the interpolation formula. E. T. Whittaker published the paper “On the functions which are represented by the expansions of the interpolation theory” back in 1915! Event Name Whose theorem is this ?... The discovery of the sampling theorem is attributed to Harry Nyquist and Claude Shannon. In 1928, Nyquist referenced the existence of the theorem in his paper, "Certain Topics in Telegraph Transmission Theory," but he did not explicitly explore it. It remains a mystery why Nyquist is considered one of the founders of the principal, except for the fact that the company he worked for--Bell Labs--referred to the concept as the Nyquist Sampling Theorem in their texts. In 1949, credit for the theorem was also given to Shannon, a mathematical engineer, based on the results outlined in his published work, "Communication in the Presence of Noise." This altered the official name to the Nyquist-Shannon Sampling Theorem. This stick even though other scientists--including E. T. Whittaker and V. A. Kotelnikov--had published similar findings in 1915 and 1933 respectively. Sampling An ideal sampler can be considered as a switch that is, it is periodically open and closed every T seconds which is expressed as: T = f1s x[n] = xa (t ) t = nT = xa (nT ) xa(t) x[n] where fs is the sampling frequency (or sampling rate) in hertz (Hz, or cycles per second). Event Name Sampling … In order to represent an analog signal x(t) by a x(t) discrete-time signal x(nT) accurately, the following discretex(nT) conditions must be met: fs ≥ 2 fM − Shannon’s sampling theorem 1. The analog signal, x(t), must be bandlimited by x(t) the bandwidth of the signal fM 2. The sampling frequency, fs, must be at least twice the maximum frequency component fM in the analog signal x(t). That is, x(t). Sampling … Shannon’s Sampling Theorem: This states that when Shannon’ the sampling frequency is greater than twice the highest frequency component contained in the analog signal, the original signal x(t) can be perfectly reconstructed from the discrete signal x(nT). nT). The minimum sampling frequency fs=2fM is the fs= Nyquist rate while fN= fs / 2 is the Nyquist frequency fN= (or folding frequency). The frequency interval [−fs / 2, [− fs / 2] is called the Nyquist interval. When an analog signal is sampled at sampling frequency, fs, frequency components higher than fs / 2 fs, fold back into the sampling range [0, fs / 2]. This [0 undesired effect is known as aliasing. An anti-aliasing antifilter is an analog lowpass filter with the cut-off cutfrequency of f fc ≤ s 2 Event Name Sampling The intermediate signal, x(nT), is a discrete-time nT), discretesignal with a continuous value (a number has infinite precision) at discrete time nT, n = 0, 1, …, ∞ as nT, illustrated in the figure below. The signal x(nT) is an nT) impulse train with values equal to the amplitude of x(t) at time nT. The analog input signal x(t) is continuous in nT. both time and amplitude while the sampled signal x(nT) nT) is continuous in amplitude, but defined only at discrete points in time. Thus the signal is zero except at sampling instants t = nT. nT. x(nT) 0 T 2T 3T Time, t 4T Impulse Sampling Sampled waveform Signal waveform 0 0 1 201 1 201 Impulse sampler 0 1 Event Name 201 Sampling proof Consider a bandlimited signal x(t) and is x(t) spectrum X(ω): X(ω Ideal sampling = multiply x(t) with x(t) impulse train: train: * Therefore the sampled signal has a spectrum: Sampling proof … Therefore, to reconstruct the original signal x(t), we can use an ideal lowpass filter on the sampled spectrum: This is only possible if the shaded parts do not overlap. This means that fs must be more than TWICE that of B. Event Name Sampling Theorem: Mathematical proof The sampled version can be expressed as: x(t ) = x(t )δTs (t ) = ∑ x(nT )δ (t − nTs ) n We can express the impulse train as a Fourier series: δ T (t ) = T1 [1 + 2 cos ωs t + 2 cos 2ω s t + ....] where ω s = Therefore: s s 2π Ts x(t )δ Ts (t ) = T1s [ x(t ) + 2 x(t ) cos ωs t + 2 x(t ) cos 2ω s t + ....] Since 2 x(t ) cos ω s t X (ω ) = ⇔ 1 Ts X (ω − ω s ) + X (ω + ω s ) ∞ ∑ X (ω − nω ) n = −∞ s Which is essentially the spectrum shown in the previous slide. What happens if we sample too slowly? What are the effects of sampling a signal at, above, and below the Nyquist rate? Consider a signal bandlimited to 5Hz: Sampling at Nyquist rate of 10Hz give: Event Name What happens if we sample too slowly? Sampling at higher than Nyquist rate at 20Hz makes reconstruction much easier. Sampling below Nyquist rate at 5Hz corrupts the signal. ALIASING Anti-aliasing filter Anti-aliasing filter To prevent aliasing effect A low-pass analog filter with cut-off frequency less than half of sampling frequency Pre-filtering to ensure all frequency components outside band-limited signal sufficiently attenuated Event Name Anti-aliasing filter ,,,, To avoid corruption of signal after sampling, one must ensure that that the signal being sampled at fs is bandlimited to a frequency B, where B < fs/2. Consider this signal spectrum: After sampling: After reconstruction: Nyquist Sampling & Aliasing Given a sequence of number representing a sinusoidal signal, the original waveform of the signal (continuous-time signal) cannot be determined Ambiguity caused by spectral replicating effect of sampling Event Name Practical Sampling Impulse train is not a very practical sampling signal. Let us consider a train of pulses pT(t) of pulse width t=0.025 sec. * Ideal Signal Reconstruction Use ideal lowpass filter: That’s why the sinc function is also known as the interpolation That’ function: Event Name Practical Signal Reconstruction Ideal reconstruction system is therefore: In practice, we normally sample at higher frequency than Nyquist rate: Example of sample and hold Event Name Example 1 : If the analog signal is in the form of : xa[t] = 3cos(1000πt-0.1π)- 2cos(1500πt+0.6π) + 5cos(2500πt+0.2π) Determine the signal bandwidth and how fast to sample the signal without losing data ? Solution : 1. There are 3 frequencies components in the signal which is w1 = 1000π, w2 = 1500π, w3 = 2500π 1000π 1500π 2500π 2. The Input frequencies are : F1 = w1 / 2π = 500 Hz, F2 = w2 / 2π = 750 Hz, F3 = w3 / 2π =1250 Hz 2π 2π 2π 3. Thus the Bandwidth Input signal is : fmax = 1250 Hz or 1.25 kHz 4. Thus the signal should be sampled at frequency more than twice the Bandwidth Input Frequency, F T > 2 fm Thus the signal should be sampled at 2.5 kHz in order to not lose the data. In other words, we need more than 2500 samples per seconds in order to not lose the data Event Name Example 2 : The input continuous signal which have frequency of 2kHz enter the DTS system and being sampled at every 0.1ms. Calculate the digital and normalized frequency of the signal in Hz and rad. rad. Solution : 1. Calculate the Sampling Rate : FT = 1 / T = 1 / (0.1ms) = 10 kHz. 2. Now, calculate the digital frequency. f = F / FT = 2 kHz / 10 kHz = 0.2 3. The digital frequency in radian, ω = 2πf = 2π (0.2) = 0.4π rad. 4. The normalized digital frequency in radian, ω = ΩT = 2πFT = 2π(2kHz)(0.1ms) = 0.4. Event Name Example : The analog signal that enters the DTS is in the form of : xa[t] = 3cos(50πt) + 10sin(300πt) - cos(100πt) [t] 3cos(50π 10sin(300π cos(100π a. Determine the input signal bandwidth. b. Determine the Nyquist rate for the signal. c. Determine the minimum sampling rate required to avoid aliasing. d. Determine the digital (discrete) frequency after being sampled at sampling rate determined from c. e. Determine the discrete signal obtained after DTS. Solutions : a. The frequencies existing in the signals are : F1 = w1 / 2π = 50π / 2π = 25 Hz. 2π 50π 2π π F2 = w2 / 2π = 300π / 2π = 150 Hz. 2 300π 2π F3 = w3 / 2π = 100π / 2π = 50 Hz. 2π 100π 2π f m = Maximum input frequency = 150 Hz. Hz. b. The Nyquist rate is defined as : 2 f m = f T = 2(150 Hz) = 300 Hz. Hz. c. The minimum sampling rate required to avoid aliasing is f T ≥ 2 f m = 300 Hz. Hz. d. f1 = F1 / FT = 25 Hz / 300 Hz = 1/12 f2 = F2 / FT = 150 Hz / 300 Hz = 1/2 f3 = F3 / FT = 50 Hz / 300 Hz = 1/6 e. The discrete signal after DTS is : x[n] = xa[nTs] = 3cos[2πn(1/12)] + 10sin[2πn(1/2)]- cos[2πn(1/6)] 3cos[2π 10sin[2π (1/2)]- cos[2π Event Name Quantizing and Encoding The fundamental distinction between discrete-time discretesignal processing and DSP is the wordlength. The former assumes that discrete-time signals values x(nT) discretenT) have infinite wordlength while the latter assumes that digital signal values x(n) only have a limited B-bit. The quantizing and encoding process is a method of representing the sampled discrete-time signal x(nT) as discretenT) a binary number that can be processed with DSP hardware. To process or store the discrete-time signal discretewith DSP hardware, the signal must be quantized to a digital signal x(n) with a finite number of bits. If the wordlength of an ADC is B bits, there are 2B different 2B values (levels) that can be used to represent a sample. Quantization is therefore a process that represents an analog-valued sample x(nT) with its nearest level analognT) that correspnds to the digital signal x(n). Quantizer (Quantization) The real-valued signal has to be stored as a code for realdigital processing. This step is called quantization. quantization. ˆ x[n] = Q( x[n]) Event Name Quantizer (Quantization) Quantizer (Quantization) In general, if we have a (B+1)-bit binary two’s (B+1)two’ complement fraction of the form: a0◊ a1a2 ...aB then its value is − a0 20 + a1 2 −1 + a2 2 −2 + ... + aB 2 − B The step size of the quantizer is Δ = 2 X m / 2 B +1 = X m / 2 B where Xm is the full scale level of the A/D converter. The numerical relationship beween the code words and the quantizer samples is ˆ ˆ x[n] = X m xB [n] Event Name Example of quantization Analysis of quantization errors ˆ e[n] = x[n] − x[n] Quantization error In general, for a quantizer with step size Δ, the quantization error satisfies that when − Δ / 2 < e[n] ≤ Δ / 2 (− X m − Δ / 2) < x[n] ≤ ( X m − Δ / 2) If x[n] is outside this range, then the quantization x[n] error is larger in magnitude than Δ/2, and such samples are saided to be clipped. clipped. Event Name Analysis of quantization errors Analyzing the quantization by introducing an error source and linearizing the system: The model is equivalent to quantizer if we know e[n]. e[n]. Example of quantization error original signal 3-bit quantization result 3-bit quantization error Event Name Example of quantization error 8-bit quantization error In a heuristic sense, the assumptions of the statistical model appear to be valid if the signal is sufficiently complex and the quantization steps are sufficiently small, so that the amplitude of the signal is likely to traverse many quantization steps from sample to sample. Quantization Noise Event Name Quantization error analysis The mean value of e[n] is zero, and its variance is e[n] σ = 2 e Δ/2 ∫ e2 −Δ / 2 Since Δ= Δ2 1 de = Δ 12 Xm 2B For a (B+1)-bit quantizer with full-scale value Xm, the (B+1)fullnoise variance, or power, is σ e2 = 2 2 −2 B X m 12 Quantization error analysis A common measure of the amount of degradation of a signal by additive noise is the signal-to-noise ratio (SNR), signal- to(SNR), defined as the ratio of signal variance (power) to noise variance. Expressed in decibels (dB), the SNR of a (B+1)(B+1)bit quantizer is 2 ⎛σ 2 ⎞ ⎛ 12 ⋅ 2 2 B σ x ⎞ ⎟ SNR = 10 log10 ⎜ x ⎟ = 10 log10 ⎜ 2 ⎜σ 2 ⎟ ⎜ ⎟ Xm ⎝ e⎠ ⎝ ⎠ ⎛X ⎞ = 6.02 B + 10.8 − 20 log10 ⎜ m ⎟ ⎜σ ⎟ ⎝ x ⎠ Hence, the SNR increases approximately 6dB for each bit added to the world length of the quantized samples. Event Name ...
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