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. The sampling process Anti- aliasing filter DAC Digital Processor ADCS/H Reconstruction Filter

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

9. Periodic Sampling x[n] is generated by periodically sampling x a (t) Where F T is the sampling frequency x a (t)x[n]

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

11. Periodic Sampling If the continuous-time signal is ) x a (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] = x a [nT] = A cos (ΩnT + Φ) Since, t = nT = n F T Then, x[n] = x a [nT] = A cos (2πFnT + Φ) = A cos (2πnF/F T + Φ) = A cos (2πfn + Φ) = A cos (ωn + Φ) Where n is a time index.

12. Periodic Sampling Example 1Example 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 : F T = 1 / T = 1 / (0.1ms) = 10 kHz. 2. Now, calculate the digital frequency. f = F / F T = 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 x a (t) be a band-limiting signal with X a (jΩ) = 0 for | Ω| > Ω m. Then x a (t) is uniquely determined by its samples x a (nT), -∞ < n < ∞, if Ω T ≥ 2 Ω m (Nyquist Condition/criteria) where Ω T = 2π T

16. Nyquist Sampling & Aliasing Example2 Example 2: If the analog signal is in the form of : x a [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 Ω 1 = 1000π, Ω 2 = 1500π, Ω 3 = 2500π 2. The Input frequencies are : F 1 = Ω 1 / 2π = 500 Hz, F 2 = Ω 2 / 2π = 750 Hz, F 3 = Ω 3 / 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, T > 2 m 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 Example 3 : The analog signal that enters the DTS is in the form of : x a [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 Solutions : a. The frequencies existing in the signals are : F 1 = Ω 1 / 2π = 50π / 2π = 25 Hz. F 2 = Ω 2 / 2π = 300π / 2π = 150 Hz. F 3 = Ω 3 / 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. f 1 = F 1 / F T = 25 Hz / 300 Hz = 1/12 f 2 = F 2 / F T = 150 Hz / 300 Hz = 1/2 f 3 = F 3 / F T = 50 Hz / 300 Hz = 1/6 e. The discrete signal after DTS is : x[n] = x a [nT s ] = 3cos[2πn(1/12)] + 10sin[2πn(1/2)]- cos[2πn(1/6)]