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

Periodic Sampling

• The sampling process

Anti-aliasing

filter

DAC

Digital Processor

ADCS/H

Reconstruction Filter

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

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

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

)()()(

)(][)(

Tjrr

nrr

eXjHjX

nTthnxtx

Periodic Sampling

• A simplified representation of sampling process

Ideal Sampler

Digital processors

Ideal Interpolato

r

xa(t) x[n]

Periodic Sampling

)()(][ nTxtxnx anTta

• x[n] is generated by periodically sampling xa(t)

• Where FT is the sampling frequencyTT

1F

xa(t) x[n]

Periodic Sampling

• wheref = 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

xa(t) = A cos (2πFt + φ)

x[n]=A cos (2πfn + φ)

xa(t) = A cos (Ωt + φ) x[n]=A cos (ωn + φ)Ideal Sampler

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.

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 :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.

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

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

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

Nyquist Sampling & Aliasing

ExampleExample 22:

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 ?

Nyquist Sampling & AliasingSolution :Solution :

1. There are 3 frequencies components in the signal which is

Ω1 = 1000π, Ω2 = 1500π, Ω3 = 2500π

2. The Input frequencies are :

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 T > 2> 2 Ω mm

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

Nyquist Sampling & Aliasing

ExampleExample 3 3 : 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.

Nyquist Sampling & AliasingSolutionsSolutions :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)]