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Sampling Theorems

Sampling Theorems. Periodic Sampling. Most signals are continuous in time. Example: voice, music, images ADC and DAC is needed to convert from continuous-time signals to discrete-time signals form and vice-versa. Periodic Sampling of an analog signal is shown below:. Periodic Sampling.

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Sampling Theorems

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  1. Sampling Theorems

  2. Periodic Sampling Most signals are continuous in time. Example: voice, music, images ADC and DAC is needed to convert from continuous-time signals to discrete-time signals form and vice-versa.

  3. Periodic Sampling of an analog signal is shown below: Periodic Sampling

  4. Periodic Sampling Anti-aliasing filter S/H ADC Digital Processor Reconstruction Filter DAC The sampling process

  5. Periodic Sampling • 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

  6. Periodic Sampling • Sample-and-hold circuit (S/H) • Samples the input continuous –time signal at periodic intervals • Holds the analog sampled value constant at its output for sufficient time to allow accurate conversion by ADC

  7. Periodic Sampling • Reconstruction Filter • Smooth the staircase-like waveform of DAC output • An analog low-pass filter with cut-off frequency equal half of sampling frequency • Convert x[n] into sequence of impulses and then interpolates to form a continuous-time signal

  8. Periodic Sampling Ideal Sampler Digital processors Ideal Interpolator xa(t) x[n] A simplified representation of sampling process

  9. Periodic Sampling xa(t) x[n] • x[n] is generated by periodically sampling xa(t) • Where FT is the sampling frequency

  10. Periodic Sampling Ideal Sampler xa(t) = A cos (2πFt + φ) x[n]=A cos (2πfn + φ) xa(t) = A cos (Ωt + φ) x[n]=A cos (ωn + φ) • where f = F is the relative or normalized frequency of discrete-time signal FT ω = ΩT is the relative or normalized angular frequency for discrete-time system

  11. Periodic Sampling • If the continuous-time signal is xa(t)= A cos (Ωt + Φ) where Ω = 2πF(Angular Frequency) • After sampling, the analog signal will become discrete signal in the form of x[n] = xa[nT] = A cos (ΩnT + Φ) Since, t = nT = n FT Then, x[n] = xa[nT] = A cos (2πFnT + Φ) = A cos (2πnF/FT + Φ) = A cos (2πfn + Φ) = A cos (ωn + Φ) Where n is a time index.

  12. Periodic Sampling • Example 1: 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. 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.

  13. 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

  14. Nyquist Sampling & Aliasing Spectral of a bandlimited signal replicate itself at fs period of replication after sampling Aliasing of replicated signal results in loss of information of the original signal

  15. Nyquist Sampling & Aliasing • Sampling Theorem Let xa(t) be a band-limiting signal with Xa(jΩ) = 0 for | Ω| > Ωm. Then xa(t) is uniquely determined by its samples xa(nT), -∞ < n < ∞, if ΩT ≥ 2 Ωm (Nyquist Condition/criteria) where ΩT = 2π T

  16. Nyquist Sampling & Aliasing Example2: 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 ?

  17. Nyquist Sampling & Aliasing Solution : 1. There are 3 frequencies components in the signal which is Ω1 = 1000π, Ω2 = 1500π, Ω3 = 2500π 2. The Input frequenciesare : F1 = Ω1 / 2π = 500 Hz, F2= Ω2 / 2π = 750 Hz, F3= Ω3 / 2π =1250 Hz 3. Thus the Bandwidth Input signal is : Ω m= 1250 Hz or 1.25 kHz 4. Thus the signal should be sampled at frequency more than twice the Bandwidth Input Frequency, Ω T > 2 Ω m 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

  18. Nyquist Sampling & Aliasing Example3 : The analog signal that enters the DTS is in the form of : xa[t] = 3cos(50πt) + 10sin(300πt) - cos(100πt) 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.

  19. Nyquist Sampling & Aliasing Solutions : a. The frequencies existing in the signals are : F1 = Ω1 / 2π = 50π / 2π = 25 Hz. F2 = Ω2 / 2π = 300π / 2π = 150 Hz. F3 = Ω3 / 2π = 100π / 2π = 50 Hz. Ω m = Maximum input frequency = 150 Hz. b. The Nyquist rate is defined as : 2 Ω m = Ω T= 2(150 Hz) = 300 Hz. c. The minimum sampling rate required to avoid aliasing is Ω T ≥ 2 Ω m= 300 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)]

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