The document discusses periodic sampling and reconstruction of continuous-time (analog) signals, detailing the relationship between discrete-time signals and the original signals through mathematical transformations. It explains the Nyquist sampling theorem, emphasizing the importance of sampling frequency in avoiding aliasing when reconstructing signals. Various examples illustrate how appropriate sampling rates preserve signal integrity, while insufficient rates lead to loss of information.

1. Chapter Two
Sampling and Reconstruction
Lecture #5
Rediet Million
AAiT, School Of Electrical and Computer Engineering
rediet.million@aait.edu.et
March,2018
(Rediet Million) DSP-Lecture #5 March,2018 1 / 22

2. Introduction
Under reasonable constraints,a continuous -time(analog) signals can
be quite accurately represented by samples taken at discrete point in
time.
Basic digital processing of continuous-time signals perform in three
stages:
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3. 2.1 Periodic sampling
Typically,discrete-time signals,x(n), are formed by periodically
sampling of a continuous-time signals, xc(t), according to the relation
x(n) = xc(nTs) ,−∞ < n < ∞
where Ts is the sampling period(interval) and fs =
1
Ts
or Ωs =
2π
Ts
is the sampling frequency in sample/sec (Hz) or in rad/sec.
It is convenient to represent an ideal sampling process mathematically
in two steps:
First, the CT signal, xc(t) , is multiplied by a periodic sequence
of infinite impulse train, s(t), to form the sampled signal xs(t).
xs(t) = xc(t)s(t)
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4. Periodic sampling
Mathematically,
s(t) =
∞
n=−∞
δ(t − nTs)
xs(t) = xc(t)s(t) = xc(t)
∞
n=−∞
δ(t − nTs)
xs(t) =
∞
n=−∞
xc(nTs)δ(t − nTs)
Then the sampled signal, xs(t), is converted into a discrete-time
signal by mapping the impulse that are spaced in time by Ts into a
sequence x(n).
x(n) = xs(nTs) = xc(nTs)
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5. Periodic sampling
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6. Periodic sampling
Example
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7. Periodic sampling
Frequency domain representation of sampling
The effect of continuous to discrete time conversion maybe analyzed
by considering the Fourier transform of xs(t).
Since xs(t) is the product of xc(t) & s(t) ,the Fourier transform of
xs(t) is the convolution of the Fourier transforms of Xc(jΩ) & S(jΩ).
xs(t) = xc(t)s(t) ⇒ Xs(jΩ) =
1
2π
Xc(jΩ) ∗ S(jΩ)
s(t) is a periodic function of time and it is possible to compute its
Fourier series representation
s(t) =
∞
n=−∞
δ(t − nTs)
S(jΩ) = 2π
∞
k=−∞
Ckδ(Ω − kΩs)
Where Ckis Fourier series coefficient and Ωs =
2π
Ts
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8. Periodic sampling
Frequency domain representation of sampling
Ck is determined from exponential Fourier series
Ck =
1
Ts
Ts /2
−Ts /2
δ(t)e−jkΩs tdt =
1
Ts
Thus,
S(jΩ) =
2π
Ts
∞
k=−∞
δ(Ω − kΩs)
⇒ The Fourier transform of a periodic infinite impulse train,s(t), is a
periodic infinite impulse train ,S(jΩ) ,in frequency domain.
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9. Periodic sampling
Frequency domain representation of sampling
Since Xs(jΩ) =
1
2π
Xc(jΩ) ∗ S(jΩ) , it follows that
Xs(jΩ) =
1
2π
Xc(jΩ) ∗
2π
Ts
∞
k=−∞
δ(Ω − kΩs)
Xs(jΩ) =
1
Ts
∞
k=−∞
Xc(jΩ − jkΩs)
Finally the discrete Fourier transform of x(n) can be obtained as follows:
- We know that x(ejw ) =
∞
n=−∞
x(n)e−jnw =
∞
n=−∞
Xc(nTs)e−jnw and
Xs(jΩ) =
1
Ts
∞
k=−∞
Xc(jΩ − jkΩs) =
∞
n=−∞
Xc(nTs)e−jnΩTs
-Combining the two equations:
x(ejw ) = Xs(jΩ)|Ω=w/Ts
=
1
Ts
∞
k=−∞
Xc(
jw
Ts
−
2πk
Ts
)
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10. Periodic sampling
Frequency domain representation of sampling
Example: Suppose that Xc(jΩ) is strictly band-limited so that Xc(jΩ) = 0
for |(Ω)| > ΩN as shown in figure bellow.
If xc(t) is sampled with a sampling frequency Ωs > 2ΩN , the Fourier
transform of xs(t) is formed by periodically replicating Xc(jΩ) as
illustrated bellow.
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11. Periodic sampling
Frequency domain representation of sampling
However ,if Ωs < 2ΩN ,the shifted spectra Xc(jΩ − jkΩs) overlap, and
when these spectra are summed to form Xs(jΩ) ,the result is as shown
bellow.
This overlapping of spectral components is called aliasing.When aliasing
occurs,the frequency of xc(t) is corrupted, and Xc(jΩ) cannot recovered
from Xs(jΩ).
Aliasing can be avoided by using Nyquist sampling theorem.
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12. Periodic sampling
Nyquist sampling theorem
If xc(t) is strictly band-limited,
Xc(jΩ) = 0 for |Ω| > ΩN
then xc(t) may be uniquely recovered from its sample xc(nTs) if
Ωs ≥ 2ΩN =
2π
Ts
The frequency ΩN = Ωs/2 is called the Nyquist frequency, and the
minimum sampling frequency, Ωs = 2ΩN, is called the Nyquist rate.
Thus ,in order to avoid aliasing, the operating frequency of the signal
should be in Nyquist interval [−
Ωs
2
,
Ωs
2
].
This is done by analog low-pass pre-filter known as an anti-aliasing
pre-filter. The cutoff frequency of this pre-filter,Ωc, is taken to be the
Nyquist frequency,
Ωs
2
.
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13. Periodic sampling
Aliased frequency components
The unique frequency components that appear within the Nyquist
interval maybe obtained by reducing the original frequency f modulo fs, i.e
fa = f mod (fs)
-This modulo-fs operation is obtained by adding or subtracting from f
enough multiples of fs until it lies within the Nyquist interval [−fs/2, fs/2].
Thus, the sinusoid that is extracted by the analog reconstructor will be
xa(t) = ej2πfat fa = f if |f | ≤
fs
2
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14. Periodic sampling
Aliased frequency components
Example:
Let x(t) = 4 + 3cos(πt) + 2cos(2πt) + cos(3πt).
Determine the minimum sampling rate that will not cause any aliasing
effects, If x(t) is sampled at half the Nyquist rate determine xa(t) that
would be aliased with x(t).
-The frequencies in x(t) are
f1 = 0Hz f2 =
1
2
Hz f3 = 1Hz f4 = 1.5Hz
Thus,fmax = f4 = 1.5Hz and the minimum sampling rate is
fs = 2fmax = 3Hz
- If x(t) is sampled at half of this rate i.e fs =
fs
2
= 1.5Hz ,then aliasing
will occur.
The Nyquist interval of this case is [−
fs
2
,
fs
2
] i.e [−0.75, 0.75].
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15. Aliased frequency components
-The aliased frequencies will be only due to f3 & f4 since they are outside
the Nyquist interval.
- The aliased components will have frequencies
f3a = f3mod(fs ) = 1mod(1.5) = 1 − 1.5 = −0.5Hz
f4a = f4mod(fs ) = 1.5mod(1.5) = 1.5 − 1.5 = 0Hz
- Thus,x(t) will be aliased with
xa(t) = 4cos(2πf1t) + 3cos(2πf2t) + 2cos(2πf3at) + cos(2πf4at)
= 4 + 3cos(πt) + 2cos(−πt) + cos(0)
= 5 + 5cos(πt)
-The signal x(t)& xa(t) are shown below.They agree only at their sampled
values i.e x(nT) = xa(nT)
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16. Sampling and reconstruction
The above example shows that the sample values x(nT) don’t uniquely
determine the continuous-time signal they can form.
Hence, an ideal reconstructor is supposed to extract from a sampled
signal,all the frequency components that lie within the Nyquist interval,
[−
Ωs
2
,
Ωs
2
], & removes all frequencies outside the interval.
This shows that an ideal reconstructor,Hr (jΩ), acts as an ideal low-pass
filter with cutoff frequency equal to the Nyquist frequency
Ωs
2
.
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17. Sampling and reconstruction
Example: Sampling and reconstruction of sinusoidal signal
Suppose we sample the CT signal xc(t) = cos(4000πt) with sampling
period Ts =
1
6000
sec.
-We can obtain x(n) = xc(nTs) = cos(4000πTsn) = cos(ω0n)
,where ω0 = 4000πTs =
2π
3
- The highest frequency of xc(t) is Ω0 = 4000π and the sampling
frequency is Ωs =
2π
Ts
= 12000π
Ωs = 12000π ≥ 2Ω0 = 8000π ⇒ there is no aliasing.
- The Fourier transform of xc(t) is
Xc(jΩ) = πδ(Ω − 4000π) + πδ(Ω + 4000π)
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18. Sampling and reconstruction
and the sampled signal for sampling frequency of Ωs = 12000π is
Xs(jΩ) =
1
Ts
∞
k=−∞
Xc(jΩ − jkΩs)
=
π
Ts
∞
k=−∞
δ(Ω − 4000π − kΩs) + δ(Ω + 4000π + kΩs)
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19. Sampling and reconstruction
the above figure(a) shows Xc(jΩ) is a pair of impulse at Ω = ±4000π
and the shifted copies of is FT centered on ±Ωs, ±2Ωs , etc
Plotting x(ejω) = xs(
jω
Ts
) as a function of the normalized frequency
ω = ΩTs gives in figure(b).
The above figure also shows the frequency response of the ideal
reconstructor filter Hr (jΩ) for Ωs = 12000π
⇒ the signal that would be reconstructed would have frequency
Ω0 = 4000π which is the frequency of the original signal.
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20. Sampling and reconstruction
Example: Aliasing in the reconstruction of under-sampled sinusoidal
signal
Suppose the signal is again xc(t) = cos(4000πt).However now the
sampling period is increased to Ts =
1
1500
sec
- This sampling period fails to satisfy the Nyquist criterion ,since
Ωs = 3000π 2Ω0 = 8000π ⇒ there is will be aliasing.
- the sampled signal for sampling frequency of Ωs = 3000π is
Xs(jΩ) =
π
Ts
∞
k=−∞
δ(Ω − 4000π − kΩs) + δ(Ω + 4000π + kΩs)
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21. Sampling and reconstruction
This time the impulse is located at Ω = −1000π is from
δ(Ω + 4000π − kΩs) and the impulse at Ω = 1000π is from
δ(Ω + 4000π + kΩs).
It is clear that the signal would be reconstructed using sampling rate
Ts =
1
1500
sec would have frequency Ω0 = 1000π and not 4000π.
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22. Periodic sampling
(#1 ) Class exercises & Assignment
1) The sequence x(n) = cos(
πn
4
) was obtained by sampling a CT signal
xc(t) = cos(Ω0t) at a sampling rate of 1000 sample/sec. Determine the
two possible values of Ω0 that could have resulted in sequence x(n).
2 Consider the discrete-time sequence x(n) = sin(
πn
8
) .Find the two
different continuous-time signals that would produce this sequence when
sampled at a frequency of fs = 20Hz.
3) For the following continuous signals xc(t) and the corresponding
sampling frequency fs
a.xc(t) = 10sin(2πt) + 10sin(8πt) + 5sin(12πt) ,fs = 5Hz
b.xc(t) = sin(6πt)[1 + 2cos(4πt)] ,fs = 4Hz
i.Determine xa(t) aliased with xc(t) and show that the two signals have
the same sample values i.e xc(nTs) = xa(nTs).
ii. Find the condition for Ts to correctly sample xc(t) and avoid aliasing.
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