NRI INSTITUTE OF TECHNOLOGY
UNIT-6
SIGNALS & SYSTEMS
UNIT – 6
SAMPLING
Sampling :
Sampling theorem is significant in communication systems because it provides the basis
for transmitting analog signals by use of digital techniques. The sampling theorem can be stated
in two different ways.
1. A band limited signal having no frequency components higher than fm Hz. is completely
1
described by its sample values at uniform intervals less than or equal to
sec apart.
2 fm
1
This is frequency domain statement. Time interval between samples is
sec 
2 fm
“Nyqluist period”.
2. A band limited signal having no frequency components beyond fm Hz. may be
completely recovered from its samples taken at a rate of at least 2fm samples/sec. 2fm is
“Nyqluist rate”.
This is called uniform sampling theorem as the samples are taken at uniform intervals.
Proof :
-
A band limited has zero value of F.T. beyond the frequency fm Hz.
No useful signal is band limited in the mathematical sense because its F.T. extends from
- to . But after a particular frequency the transform diminishes to such as extent that it
can be neglected and the F.T. provides a definite band width.
Such signals are considered band limited signals for all practical purposes.
The above statement can be proved by using convolution theorem.
Proof :
- Let us consider band limited signal f(t) having no frequency components beyond fm Hz.
i.e. F() = 0 for || > m where m = 2fm.
-
Where the signal is multiplied by a periodic impulse train T (t) the product yields a
sequence of impulse located at uniform intervals of T secs.
-
The strength of the resulting impulses is equal to the value of f(t) at the corresponding
instants.
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SIGNALS & SYSTEMS
Figure shows :
f(t)  F()
T (t)  o wo ()
1
[F() * o ()]
T
The F.T. of fS (t) can be obtained by using frequency convolution of
F() & o o ()
1
Thus F.T. of f(t) fS (t) =
[F() * o o ()]
2

1
Putting o 
2 T
1
We get fS (t) =
[F() * o ()]
T
FS (t) = f(t) . T (t) 
-
Thus the spectrum of fS (t) can be obtained by convolving F() & o o ().
Convolution can be obtained by folding (flipping) T (t) around the vertical axis. Since T
(t) is an even fn. of time folding yields the same T (t).
The operation of convolution yields F() repeating itself every o radians/sec.
F() repeating periodically forms the spectrum of fS (t) and is denoted by FS ().
The spectrum FS () can be achieved analytically also the periodic fn. 0 0 () can be
written as the sum of impulses located at  = 0,  0,  20,  …..  n0.

Thus 0 () =

m  
Thus
fS (t) 
1
T
 (  m0 )
where m = 1, 2, 3, …..



F
(

)
*   (  m0 ) 

m  


1 
 F ( ) *  (  m0 )
T m
By using sampling property of “” function
1 
FS () =
 F (  m0 )
T m
FS () =
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NRI INSTITUTE OF TECHNOLOGY
UNIT-6
SIGNALS & SYSTEMS
The summation represents F() repeating every 0 radians / sec. as the one obtained by
graphical convolution.
From the fig. above we can see that F() repeats periodically w/o overlapping provided
1
2
 2 fm
0  2 m
Or
Or
T
sec.
T
2 fm
Where T is the uniform sampling interval.
1
Sampling rate f0 =
; i.e. f0 =  2 fm samples / sec.
T
Thus the sampling theorem is proved.
Recovery of the original spectrum :
-
We find that as long as the signal is sampled at interval ‘T’ or at a sampling rate “f0” the
spectrum will repeat w/o overlap.
The spectrum extends upto  and the ideal B.W. of the sampled signal is infinite.
However the desired spectrum F() centered at  = 0 can be recovered by passing the
sampled signal with spectrum FS () through an LPF with cut off frequency m. The
transfer fn. of the ideal LPF is rectangular i.e. a gate fn.
The sampled signal after filtering
yields all the frequency components
present in the desired signal. f(t)
having spectrum F().
if (a) 0 - m = m they touch each other
(b) 0 - m > m they do not touch.
(c) 0 - m < m they overlap.
Nyqluist interval :
-
-
The maximum sampling interval is
1
T=
secs. for complete recovery of the sampled signal.
2 fm
This is known as the “Nyqluist sampling interval”.
Similarly minimum sampling rate f0 = 2fm is called “Nyqluist rate”.
When the band limited signal is sampled at Nyqluist sampling interval (i.e. Nyqluist
sampling rate) the spectrum FS () will contain non-overlapping F() repeating periodically
but each spectrum F() will be touching the neighbouring ones in the FS ().
Aliasing :
When a band limited signal is sampled at a rate lower than Nyqluist rate i.e. f0 < 2fm
1
(or T >
) then periodically repeating F() in the spectrum FS () overlap with the
2 fm
neighbouring over as shown in the picture.
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-
UNIT-6
SIGNALS & SYSTEMS
The signal is under sampled and is said to produce aliasing in the under sampled process.
Because of the overlapping due to aliasing phenomenon, it is no longer possible to
recover f(t) from fS (t) by LPF.
Since the spectral components in the overlap regions, add up the signal is distorted.
To combat the aliasing effect we may use prior to sampling an LPF called anti-aliasing
filter and then the filtered signal is sampled at a rate slightly higher than the Nyqluist rate.
1
fS =
number of samples / sec.
T
fS  2 fm ; fS = 2fm is the minimum rate of sampling is called “Nyqluist sampling rate”.
To explain Aliasing let us take an examples :
x(t) = cos ot
F.T. of cos ot
We want to sample this are reconstruct by passing the
sampled o/p through an LPF of B.W. “o”.
  [ ( + o) + ( - o)]
Case 1 – When s = 4o
In this case s - o = 3o
LPF characteristics allows o to pass
through. Hence we get the
reconstructed o/p as x(t) = cos ot.
Case 2 – When s = 3o
Signal at s - o = 1.5 o
by missing of the signals o & s.
In this case we still get
x(t) = cosot at the o/p.
Case 3 – When s = 1.5 o
We get in this case when s < 2o.
x(t) = cos ot + cos (s - o) t
Which is not same as the original
signal hence distorted.
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SIGNALS & SYSTEMS
Signal recovery from its sampled version :
We have seen that the original signal f(t) can be recovered in frequency domain by
passing its sampled version through an LPF. With a cut off frequency of m (fm).
Let us see now that we can reconstruct the f(t) in time domain from its sampled version.
For fn. f(t) sampled at Nyqluist rate 0 we have 0 = 2m.
1 
And FS () =
 F (  no )
T n
1 
because o = 2m
FS () =
 F (  2mm )
T n
FS ()
F()
LPF
fS (t)
f(t)
H() = T G2m ()
Recovery of a fn. from its sampled version :
The LPF has a cut off frequency of m radians.
The transfer fn. of an LPF is a Gate fn.
Hence the base band spectrum can be recovered from FS () by multiplying FS () with
G2m ().
1
F ( )
T
F() = FS () . T G2m ( )
FS () . G2m () =
-
The gate fn. G2m ( ) is representing an LPF with a cut off frequency m. In other words
the action of the LPF is equivalent to multiplying the sampled signal FS () with a gate
fn. T G2m ( ) and the action yields the base band spectrum F().
-
The gate fn. has ht. T and width 2m.
The fn. f(t) can be obtained by evaluating the time domain equation of (Inverse F.T. of)
F().
The time domain equation of RHS can be evaluated by using time convolution theorem.
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f(t)
= fS (t) * T
m
Sa (mt)

= fS (t) * Sa (m t)
SIGNALS & SYSTEMS
i.e. if the frequency domain
as T =
2
o
is a gate fn. its time domain
Equation is a Sa fn.
The sampled fn. fS (t) can be considered as a
sum of impulses located at sampling instants (nT)
m
having strength equal to fn at that instant.
T


m

2 f m
 2 fm  f0

 T f 0 1
Hence the sampled fn. fS (t) can be expressed as
fS (t) =  f n  (t  nT ) where fn is the nth sample of f(t).
n
Substituting for fS (t) the fn. f(t) is given as [i.e. fS (t) = Sa (m t)
f(t) =  f n  (t  nT ) * Sa mt
n
Using sampling property of “” fn we have.
f(t) =  f n Sa m (t  nT )
n
Putting m T =  we get ; 2fm T = mT
2  f m 2 f m
1

  because T =
2 f0
2 fm
f0
We get f(t) =

f n Sa (mt  n)
n
-
This equation represents that the fn. f(t) can be constructed by multiplying the sample fn
by a sampling fn. Sa(mt - n) and adding the multiplied values.
Here Sa(mt - n) represents the sampling fn. at the sampling instants T = nT.
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NRI INSTITUTE OF TECHNOLOGY
UNIT-6
SIGNALS & SYSTEMS
Sampling of Band pass signals :
-
We have discussed so far band limited signals centered around the origin i.e. low pass
signals.
Let us consider band pass signals which is a more general case. Let the band be
represented by H upper cut off and L lower cut off frequencies.
The case we have considered so far was C = 0. i.e. the sampling of band limited signals.
Case I : If either H or L is a harmonic of the sampling frequency S.
i.e. S = 2(H - L) = 2  2m = 4 m.
-
Let us consider L is an integral multiple of (harmonic of) S i.e. (L = n S).
Then the spectrum of the corresponding band pass signal contains the entire pass band
and the original pass band signal can be recovered from its sampled version by passing the signal
through a band pass filter with sharp cut off having pass band L to H.
Case II :
If L or H not harmonic of S.
Then a more general sampling condition is given as follows.
2(c  m )
  m
S 
; where m is the largest integer not exceeding c
m
2m
Actually case I is a special case of Case II.
Where the frequency band occupied by the signal is located between adjacent multiples
 f  fm 
 here the minimum sampling rate is twice the band width of the signal
of 2 m i.e. m =  c
2
f
m


i.e. 2  2m = 4 m. This is same as Case I.
-
In the case of BP sampling x(t) can be recovered.
With out any error from its samples x(KTS) taken at regular intervals of TS if the sampling rate fS
is such that
1 2 f2

fS =
TS
m
f
Where m is the largest integer not exceeding 2
B
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NRI INSTITUTE OF TECHNOLOGY
UNIT-6
SIGNALS & SYSTEMS
Where f2 is the highest frequency contained in the Band Pass Signal and B is (f2 – f1).
2 f2
If must be noted that sampling frequencies higher than
may not always permit recovery of
m
x(t) without distortion (i.e. we may not be able to avoid aliasing) unless fS > 2f2. If x(t) is ideally
sampled using impulses the signal x(t) can be recovered from its samples by an ideal BPF whose
transfer fn. H(f) is given by
1 for f1  | f |  f 2
H(f) = 
0 otherwise
In fact as stated above in the sampling theorem, the required sampling rate for a Band
Pass Signal depends on ‘m’ i.e. on “f2/B”.
2 f2
fS =
m
m = largest integer of (f2/B).
Flat Top Sampling :
-
-
-
The circuitry needed for natural sampling is very complicated because the pulse shape at
the top has to be maintained. These complications can be reduced by Flat Top Sampling.
Here the pulses have a constant amplitude with in the pulse interval.
The constant amplitude of the pulse can be chosen at any value of f(t) within the pulse
interval. Say the value at the beginning of the pulse is chosen in the diagram below.
The Flat Top Sampled Signal fm (t) is shown in the figure (e) and it is the convolution of
the impulse sampled signal fS (t) of (a) and the non-periodic pulse P(t) of width  and of
ht. 1 shown in figure (c).
The spectrum of fS (t) and P(t) are shown in figs. (b) & (d).
The spectrum of fm (t) is shown at (f) which is obtained by multiplying FS () and P().
As the P() value is different at different frequencies the shape of Fm () is not similar to
FS () which shows that a distortion will be introduced if the signal is recovered by an
ideal LPF of cut off frequency m.
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UNIT-6
SIGNALS & SYSTEMS
1
.
P( )
Thus when the Flat Top Sampled Signal is passed through an LPF its o/p is P(). F()
1
and when this is passed through equalizer whose transfer fn. is
we get the original
P( )
signal f(t) as F().
To ward of this difficulty the signal is passed through a filter whose transfer fn. is
-
The amplitude of the sampling pulse train is adjusted to 1.
-
The duration of the pulses is .
-
They are separated by TS.
-
The F.S. of a aperiodic train of pulses is given by
A 2 A 
2 nt
V(t) =

Cn cos

To
To n1
To
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Where
UNIT-6
SIGNALS & SYSTEMS
A = Amplitude of the pulse
 = Duration of the pulse
To = Period of the pulse train and
 2 n  

sin 
T
 o 
Cn is given by
Cn =
 2 n  


T
 o 
Here
So V(t) = fC (t) ; A = 1 & To = Ts
Hence

2 
2t
4t
 C2 Cn
 .....
C1 cos
TS TS 
TS
TS

The constant Cn is given by
sin n2  / TS
Cn =
for n = 1, 2, 3, ……
(n 2  / TS )
The o/p of the multiplier is then


2 
2t
2t
f(t) . fc(t) =
f (t ) 
 f (t ) C2 cos 2 
 ....... 
 f (t ) C1 cos
Ts
Ts 
Ts
Ts

1
Let us choose Ts =
2 fm
Where fm = maximum frequency comp. in f(t).
We then get

2
 f (t ) C1 cos 2 (2 f m )t  ....... 
f (t ) 
f(t) . fc(t) =
Ts
Ts
If we don’t consider the multiplying factor of the first term in the RHS it is the f(t) itself.
fc (t) =


If f(t) . fc(t) is passed through an LPF, f(t) is obtained at the o/p, the cut off frequency of
the LPF = fm.
Ex : The signal x(t) = 2 cos 200 t + 6 cos 180 t is ideally sampled at a frequency of 150
samples / sec. The sampled version X0(t) is passed through an ideal LPF with a cut off frequency
of 110 Hz.
What frequency components will be present in the o/p of the LPF ?
Write down the expressions for its o/p signal.
x(t) = 2 cos 200 t + 6 cos 180 t
= 2 cos 2 (100) t + 6 cos 2 (90)t
Hence taking F.T. of both side
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NRI INSTITUTE OF TECHNOLOGY
UNIT-6
SIGNALS & SYSTEMS
= [ (f + 100) +  (f – 100)] + 3 [ (f + 90) +  (f – 90)]
X(f)
Spectrum of sampled version of x(t)
fs =
1
T

  ( f
n  
f1 = 100
250 = 100 + f0
-50 = 100 – f0
1
 f 0 )   ( f1  f 0 )   ( f 2  f 0 )   ( f 2  f 0 ) 
f1 = -100
-100 + f0 = 50
-100 – f0 = -250
f2 = 90
f2 + f0 = 240
f2 – f0 = -60
f2 = -90
f2 + f0 = 60 = (-90+150)
f2 – f0 = -90 – 150
= - 240
The output of the filter will be
x(t) = 2 [cos 2 (50) t + cos 2(100) t]
+ 6 [cos 2(60) t + cos 2 (90)t]
Ex. 1 :
The signal x(t) = 10 cos 150 t is ideally sampled at a frequency.
fs = 200 samples/sec. Sketch the spectrum of x (t)
x(t) = 10 cos 2 (75)t
X(f) = 5 [(f-75) + (f+75)]
Since the spectrum X (f) of x(t) is given by
1 
X (f) = 5 .
 X ( f  nf s )
T n  
The sketch is as follows :
1
 fs
T
f = 75
f = -75
f + fs
f + fs = 75 + 20 = 125
75+200=275 f-fs = -75–200 = -275
f – fs = 75-200
= -125
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SIGNALS & SYSTEMS
Ex. 2 : For the above probability if sampling is done at 100 SPs. Find the spectrum of sampled
o/p.
fs = 100 SPs
f = 75
f – fs = 75-100 = -25
f + fs = 75+100 = 175
T (t) =


n  
f = -75
f + fs = -75+100 = 25
f–fs =-75–100 = -175
a k e Jnot
T /2
1
1
 (t ) e  Jot dt 

T T / 2
T
1
=
T

1 Jnot
e
T (t) = 
n   T
ax =

F.T. [T (t)] =
1
T

n  
=
=
1
T





e Jnot . e  Jt dt
  [1] e
 J (  no ) t
dt
n    

2
T

n  
 (  no )
Ex. 3 : How many minimum no. of samples are required to exactly describe the following
signal.
(a)
(b)
x(t)
= 10 cos (6 t) + 4 sin (8t)
= 10 cos 2 [3] t + 4 sin 2 [4] t
1
1
T1 of (a) = & T2 of (b) =
3
4
T1 1 4 4
  
T2 3 1 3
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T = 3T1 = 4T2 = 1
SIGNALS & SYSTEMS
Period = 1 of the combined signal.
The highest frequency in the combined signal is 4 Hz.
Hence the required sampling frequency = 8 CPS.
Ex. 4 : Determine the minimum sampling frequency to be used to sample the signal
x(t) =100 sin2c (100 t) if the signal x(t) is to be recovered w/o any distortion.
Solution x(t) = 100 sin2 c (100 t) = 10 sinc (+100 t). 10 sinc (100 t)
We know that 10 sin c (100 t)  0.1  [f/100]
 F.T. [100 sin c2 100 t] = 0.1  [f/100] * 0.1  [f/100]
Convolution of two identical rectangular pulses results in a triangular pulse whose base
width in twice that of each rectangular pulse.
 F [100 sinc2 100 t] = [0.1  (f/100)] * 0.1 [ (f/100)]
= 0.01  100  (f/200)
Where  [f/200] denotes a triangular pulse as shown.
Thus the signal x(t) = 100 sin c2 100 t
is a low pass signal, band limited to 100 Hz.
Hence the Nyquist rate is 200 CPS.
Triangular spectrum of
the x[t] = 100 sinc2 100t.
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