signal processing fundamentals – part i spectrum …enpklun/eie327/sampling2.pdf · 2 signal...

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Signal Processing Fundamentals – Part ISpectrum Analysis and Filtering

4. Sampling and Aliasing

4. 4. Sampling and AliasingSampling and Aliasing

Sampler Ideal Low Pass Filter

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Sampling• Most real signals are continuous-time (analogue)

signals• E.g. speech, audio, etc.

• Computers have much difficulty in handling continuous-time signals

• Need sampling⇒ Extract samples of the signal at some

particular time instants

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Continuous-to-Discreteor

Analogue-to-Digital

x(t) x[n] = x(nTs)

Ts = 1/fs

TsWhat is the value of Ts?

Normal CD musicfs = 44.1kHz

Sampled atfs = 16kHz

Sampled atfs = 8kHz

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f = 100Hz

f = 100Hzfs = 2000Hz

f = 100Hzfs = 500Hz

Sampling Sinusoids

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( )φπ += ftAtx 2cos)(( )

( )φω

φπ

φπ

+=

+=

+==

nAffnA

fnTAnTxnx

s

ss

ˆcos

2cos

2cos)(][



sffwhere πω 2ˆ = is the so-called discrete-

time radian frequency

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Spectrum of Sampled Sinusoids• Assume f = 100Hz, fs = 300Hz, A = 1 and φ= 0

( )( )300/1002cos][

1002cos)(nnxttx

ππ

==

• From Fourier series, it is known that the spectrum of x(t), i.e. Xk is as follows:

f(Hz)0 100-100

1/21/2

Original sinusoid

Sampled sinusoid

Spectrum of original sinusoid

X-1 = X1 =

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• If x(t) is sampled to x(nTs)

∑

∑−

=

−

−

=

−

=

=

1

0

/2

1

0

/2

0

)(2

)(2 0

N

n

Nknjs

N

n

TknTjs

sPk

enTxN

enTxTTX s

π

π

• From Fourier series, we know that

dtetxT

X TktjTk

00 /200

)(2 π−∫=

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T0

Ts

N is the number of samples in one periodN is the number of samples in one period

Ts N = T0 / Ts= 20

N = T0 / Ts= 5

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( )

( )∑

∑

∑

=

+−−

=

−−

=

−

+=

+=

=

2

0

3/)1(23/)1(2

2

0

3/23/23/2

2

0

3/2

31

31

)3/2cos(32

n

nkjnkj

n

knjnjnj

n

knjPk

ee

eee

enX

ππ

πππ

ππ

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0866.05.0866.05.01

)3/22sin()3/22cos()3/2sin()3/2cos(1

1 3/223/22

0

3/2

=−−+−=

++++=

++=∑=

jjjj

eee jj

n

nj

ππππ

πππ

• Let’s consider a particular k, e.g. k = 0

In fact0

1

0

/2 =∑−

=

N

n

Nmnje π if m is not a multiple of N

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• In general

==∑

−

= NofmultiplemforNotherwise

eN

n

Nmnj 01

0

/2π

since

1

/)3(2

/)2(2

/)(2/)0(2

==

=

=

− NnNj

NnNj

NnNjNnj

e

e

ee

π

π

ππ

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( )

=+=−

=

+= ∑=

+−−

otherwiseofmultiplekif

orofmultiplekif

eeXn

nkjnkjPk

03)1(3)1(1

31 2

0

3/)1(23/)1(2 ππ

Magnitude Spectrum for sampled sinusoid

0

0.2

0.4

0.6

0.8

1

1.2

-8 -6 -4 -2 0 2 4 6 8

k

Ampli

tude

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4. Sampling and AliasingMagnitude Spectrum for sampled sinusoid

0

0.2

0.4

0.6

0.8

1

1.2

-8 -6 -4 -2 0 2 4 6 8

k

Ampli

tude

Magnitude Spectrum for original sinusoid

0

0.2

0.4

0.6

0.8

1

1.2

-800 -600 -400 -200 0 200 400 600 800

Frequency (Hz)

Ampli

tude

Spectrum of sampled sinusoid

Spectrum of original sinusoid

Ideal low pass filter

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Sampler Ideal Low Pass Filter

fs = 1/Ts

A/D and D/A conversions

A/D converter

D/A converter

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0

0.2

0.4

0.6

0.8

1

1.2

-8 -6 -4 -2 0 2 4 6 8

2 pi k

Ampl

itude

sfkf02ˆ πω =

• Very often the discrete-time signal spectrum is expressed using discrete-time radian frequency

0 2π 4π-2π-4π

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Aliasing• Assume f = 100Hz, fs = 200Hz, A = 1 and φ= 0


0

0.5

1

1.5

2

2.5

-6 -4 -2 0 2 4 6

k

Ampli

tude

Ideal low pass filter

sfkf02ˆ πω =

0 2π 4π-2π-4π

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• Assume f = 100Hz, fs = 100Hz, A = 1 and φ= 0Magnitude Spectrum for sampled sinusoid

0

0.5

1

1.5

2

2.5

-5 -4 -3 -2 -1 0 1 2 3 4 5

k

Ampli

tude

sfkf02ˆ πω =

0 2π 4π-2π-4π 6π-6π 8π-8π

Only get a DC when an ideal low-pass filter is used

Only get a DC when an ideal low-pass filter is used

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Ts

Just a DC

100 Hz cosine wavef0 = 100Hz

Samples with fs = 100Hz

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Shannon Sampling Theorem

A continuous-time signal x(t) with frequencies no higher than fmax can be reconstructed exactly from its samples x[n] = x(nTs) if the samples are taken at a rate fs = 1/Ts that is greater than 2fmax

Nyquist FrequencyNyquist Frequency

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Discrete-to-Continuous Conversion

• Achieve by low pass filtering⇒ Smooth out the sharp changes in the signal as

much as possible

The simplest low pass filter is a capacitor, which works like a reservoir to store the voltage of the samples

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x[n] x(t)

)3(]3[)2(]2[)(]1[)(]0[)(

s

ss

TtpxTtpxTtpxtpxtx

−+−+−+=

• Let p(t) =

• Such low pass filter operation can be mathematically expressed as

0 tTs

1

0 1 2 3 n t0 Ts 2Ts 3Ts

rectangular pulse

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• Let p(t) =

0 tTs

1 x[n]

0 1 2 3 nx[0].p(t)

0 tx[1].p(t-Ts)

0 tTs

t

x[2].p(t-2Ts)

0 Ts 2Tsx[3].p(t-3Ts)

0 tTs 2Ts 3Ts

x(t)t0 Ts 2Ts 3Ts

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• In general, if we have N samples,

)(][)(1

0s

N

nnTtpnxtx −= ∑

−

=

))1((]1[))2((]2[

)2(]2[)(]1[)(]0[)(

s

s

ss

TNtpNxTNtpNx

TtpxTtpxtpxtx

−−−+−−−++

−+−+=K

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• Rectangular pulse in general cannot give smooth output

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• Let p(t) =

0 t

1

• Next try triangular pulse

)(][)(1

0s

N

nnTtpnxtx −= ∑

−

=

x[n]0 1 2 3 n

x(t)t0 Ts 2Ts 3Ts

Low Pass Filter

sT− sT

x(t)t0 Ts 2Ts 3Ts

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• Triangular pulse in general gives better but not the best output

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• The best pulse is sinc pulse

( )s

sTtTttp

//sin)(

ππ= )4(]4[)(]0[)( sTtpxtpxtx −+=

Ts 5Ts

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• Sinc pulse gives the best result• However, the length of a sinc pulse is infinitely long• Cannot be implemented exactly• Low pass filter using sinc pulse is the so-called ideal

low pass filter, it has rectangular bandwidth



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ExerciseA signal can be represented by the following formulation

( )[ ] ( )tttx )10000(2cos)2000(2cos410)( ππ+=

• Sketch the two-sided spectrum of this signal• Is that signal periodic? If so, what is the period?• What is the Nyquist sampling frequency of this

signal?

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Solution

0

55

11

10000 120008000-10000-12000 -8000

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The signal is periodic since x(t) can be expressed as a sum of sinusoids with the same fundamental frequency. By Fourier series analysis, we know that the resulted signal is periodic.

The period is in fact the inverse of the fundamental frequency, in this case, it is equal to, i.e. 1/2000 sec

The Nyquist frequency of this signal is 2*fmax = 24,000Hz

signal processing fundamentals – part i spectrum …enpklun/eie327/sampling2.pdf · 2 signal...

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