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
• Periodic Sampling of an analog signal is
shown below:
Periodic Sampling
• The sampling process
Antialiasing
filter
S/H
ADC
Digital
Processor
Reconstruction
Filter
DAC
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
xr (t ) 

 x[n]h (t  nT )
n  
r
X r ( j)  H r ( j) X (e jT )
Periodic Sampling
• A simplified representation of sampling
process
xa(t)
Ideal
Sampler
x[n]
Digital
processors
Ideal
Interpolator
Periodic Sampling
• x[n] is generated by periodically sampling xa(t)
x[n]  xa (t ) t nT  xa (nT )
xa(t)
x[n]
1
FT 
T
• Where FT is the sampling frequency
Periodic Sampling
xa(t) = A cos (2πFt + φ)
xa(t) = A cos (Ωt + φ)
Ideal Sampler
x[n]=A cos (2πfn + φ)
x[n]=A cos (ωn + φ)
• where
f = F is the relative or normalized frequency of discretetime signal
FT
ω = ΩT is the relative or normalized angular frequency
for discrete-time system
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 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.
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
Example 2:
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 & Aliasing
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,
ΩT >2Ωm
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
Example 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 & 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)]