MEEN 364 ParasuramLecture 4 April 2, 2003

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EXAMPLES ON SAMPLING AND ALIASING PHENOMENA

Example 1:

Consider the following two analog signals

,)50(2cos)(

,)10(2cos)(

2

1

ttxttx

π

π

=

=

which are sampled at a rate of Fs = 40 Hz. The corresponding discrete time signals are

.2

5cos)4050(2cos)(

,2

cos)4010(2cos)(

2

1

nnnx

nnnx

ππ

ππ

==

==

However,

.2

cos)2

2cos(2

5cos nn

nn ππππ

=+=

Hence )()( 21 nxnx = . Thus the signals are identical and consequently indistinguishable.

If we are given the sampled values generated by n2

cosπ , there is some ambiguity as to

whether these sampled values correspond to x1(t) or x2(t). Since x2(t) yields exactly thesame values as x1(t) when the two are sampled at Fs = 40 samples per second, we say thatthe frequency F2 = 50 Hz is an alias of the frequency F1 = 10 Hz at the sampling rate of40 samples per second.

Example 2:

Consider the analog signal

.100cos3)( ttxa π=

a) Determine the minimum sampling rate required to avoid aliasing.b) Suppose that the signal is sampled at the rate of Fs = 200 Hz, what is the discrete-

time signal obtained after sampling?c) Suppose that the signal is sampled at the rate of Fs = 75 Hz, what is the discrete-

time signal obtained after sampling?d) What is the frequency of a sinusoid that yields samples identical to those obtained

in part (c)?

MEEN 364 ParasuramLecture 4 April 2, 2003

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a) The frequency of the analog signal can be calculated as

.50

,1002

1

1

=⇒

=

FF ππ

Hence the minimum sampling rate required to avoid aliasing is Fs = 100 Hz.

b) If the signal is sampled at Fs = 200 Hz, then the discrete-time signal is

.)2

cos(3)200

100cos(3)( nnnx ππ==

c) If the signal is sampled at Fs = 75 Hz, then the discrete-time signal is

.)3

2cos(3

)3

22cos(3

)3

4cos(3)75

100cos(3)(

n

n

nnnx

π

ππ

ππ

=

−=

==

d) For the sampling rate of Fs = 75 Hz, we have

ffFF s 752 ==

The frequency of the sinusoid in part (c) is .31

=f Hence

.25375752 === fF

Clearly, the sinusoidal signal

ttFty

π

π

50cos3

2cos3)( 2

=

=

sampled at Fs = 75 samples per second yields identical samples. Hence F1 = 50 Hz is analias of F2 = 25 Hz for the sampling rate Fs = 75 Hz.

MEEN 364 ParasuramLecture 4 April 2, 2003

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Example 3:

Consider the analog signal

.100cos300sin1050cos3)( ttttxa πππ −+=

What is the Nyquist rate for this signal?

The frequencies present in the signal above are

.50,150

,25

3

2

1

===

FFF

Thus Fmax = 150 Hz and according to the sampling theorem

HzFFs 3002 max => .

The Nyquist rate is FN = 2Fmax. Hence

FN = 300 Hz.

Discussion

It should be observed that the signal component tπ300sin10 , sampled at the Nyquistrate FN = 300 Hz, results in samples nπsin10 , which are identically zero. In otherwords, we are sampling the analog signal at its zero-crossing points, and hence we missthe signal component completely. This situation would not occur if the sinusoid is offsetin phase by some amount θ. In such a case we have )300sin(10 θπ +t sampled at theNyquist rate FN = 300 samples per second, which yields the samples

.sin10)1(

cossin10

)sincoscos(sin10)sin(10

θ

πθ

θπθπθπ

n

nnnn

−=

=

+=+

Thus if 0≠θ or π, the samples of the sinusoid taken at Nyquist rate are not all zero.However, we still cannot obtain the correct amplitude from the samples when the phase θis unknown. A simple remedy that avoids this potentially troublesome situation is tosample the analog signal at a rate higher than the Nyquist rate.

MEEN 364 ParasuramLecture 4 April 2, 2003

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Example 4:

Consider the analog signal

.12000cos106000sin52000cos3)( ttttxa πππ ++=

a) What is the Nyquist rate for this signal?b) Suppose that the signal is sampled at the rate of Fs = 5000 samples per second,

what is the discrete-time signal obtained after sampling?c) What is the analog signal we can reconstruct from the samples?

a) The frequencies existing in the above signal are

.6000

,3000

,1000

3

2

1

HzFHzFHzF

=

=

=

Thus Fmax = 6000 Hz and according to the sampling theorem,

HzFFs 120002 max => .

The Nyquist rate is

FN = 2Fmax = 12000 Hz.

b) The discrete-time signal of the signal sampled at 5000 samples per second is

nnn

nnn

nnn

nnnnx

)51(2cos10)

52(2sin5)

51(2cos3

)511(2cos10)

521(2sin5)

51(2cos3

)56(2cos10)

53(2sin5)

51(2cos3

)5000

12000cos(10)5000

6000sin(5)5000

2000cos(3)(

πππ

πππ

πππ

πππ

+−+=

++−+=

++=

++=

Finally, we obtain

nnnx )52(2sin5)

51(2cos13)( ππ −= .

Since Fs = 5000 Hz, the folding frequency is Fs/2 = 2500 Hz. This is the maximumfrequency that can be represented uniquely by the sampled signal.

MEEN 364 ParasuramLecture 4 April 2, 2003

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c) Since only the frequency components at 1000 Hz and 2000 Hz are present in thesampled signal, the analog signal that can be recovered is

ttty )2000(2sin5)1000(2cos13)( ππ −=

which is obviously different from the original signal x(t). This distortion of the originalanalog signal was caused by the aliasing effect, due to the low sampling rate used.

Assignment

1) Consider the analog signal

tttx ππ 1800cos2600cos3)( += .

a) Determine the minimum sampling rate required to avoid aliasing.b) What is the Nyquist rate for the signal x(t)?c) Suppose that the signal is sampled at Fs = 2000 Hz, what is the discrete-time

signal obtained after the sampling? What are the frequencies in the resultingdiscrete-time signal?

d) If the sampling rate is Fs = 600 Hz, then what is the maximum frequency that canbe recovered from the discrete-time signal?

2) An analog signal contains frequencies up to 10 kHz.

a) What range of sampling frequencies allows exact reconstruction of the signalfrom its samples?

b) What is the Nyquist rate for the signal?c) If the sampling rate Fs is chosen as 10 kHz, then what is the maximum frequency

that can be represented uniquely by the sampled signal?