sampling handout

7/29/2019 Sampling Handout

1/12

Sampling and Reconstruction of SignalsLesson 12, Part I, INF44480

Andreas Austeng

Department of Informatics, University of Oslo

April 2011

AA, INF4480 (Ifi/UiO) Sampling and reconstruction A pr il 2011 1 / 82

Outline

Outline

1 Fourier transforms and the sampling theorem

The sampling theorem

Fourier transforms

Sampling and reconstruction of signals

2 Sampling of bandpass signals

Representation of bandpass signals

Sampling of bandpass signals

3 Multirate digital signal processing

Sampling rate conversion


Fourier transforms and the sampling theorem The sampling theorem

The sampling theorem

Alternative I:

If the highest frequency contained in an analog signal xa(t) isFmax = B Hertz and the signal is sampled at a rateFs > 2Fmax 2B samples per second, then xa(t) can be exactlyrecovered from its sample values using the interpolation function

g(t) = sin2Bt2B

The sampling rate FN = 2B = 2Fmax is called the Nyquist rate.

Alternative II:

A bandlimited continuous-time signal, with highest frequency

(bandwidth) B Hertz, can be uniquely recovered from its samples

provided that the sampling rate Fs > 2B samples per second.


Fourier transforms and the sampling theorem The sampling theorem

Relations among frequency variables

We have that: = T, f = F/Fs (or = /T, F = f Fs)

Continuous-time signals Discrete-time signals

= 2F = 2f

radianssec , Hz radianssample , cyclesample

12 f 12

< < /T /T

< F < Fs/2 Fs/2



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3/12

Fourier transforms and the sampling theorem Sampling and reconstruction of signals

Sampling of analog signals (3)

To determine a relationship between the spectra of analog and

digital signals we use that t = nT = nFs and f =FFs

:

x[n] xa(nT)(2)=

Xa(F)e2nF/FsdF

(4)=

12

12

X(f)e2fndf =1

Fs

Fs2

Fs2

X(F)e2nF/FsdF.

Due to the periodicity of e2nFs/F, we can write

dF as

k=

(k+1/2)Fs(k1/2)Fs

dF...

AA, INF4480 (Ifi/UiO) Sampling and reconstruction A pr i l 2011 10 / 82


Sampling of analog signals (4)We then have that

1

Fs

Fs2

Fs2

X(F)e2nF/FsdF =

Xa(F)e2nF/FsdF

=

k=

(k+1/2)Fs(k1/2)Fs

Xa(F)e2nF/FsdF

=

k=

Fs2

Fs2

Xa(F kFs)e2nF/FsdF

=

Fs2

Fs2

k=Xa(F kFs)

e2nF/FsdF

which gives

X(F) = Fs

k=Xa(F kFs) or X(f) = Fs

k=

Xa((f k)Fs) .



Sampling of analog signals (5)

No aliasing if Fs > 2B.

If so; x(t) can be

reconstructed fromx[n].

If not; aliasing occur as

in (e).



Reconstruction of analog signal

Given no aliasing, then

Xa(F) =

1

FsX(F), |F| Fs/2

0, |F| > Fs/2(5)

and analog signal can be reconstructed:

xa(t)(2)

=

Xa(F)e2FtdF(5)

=1

FsFs2

Fs2

X(F)e2FtdF

(4)=

1

Fs

Fs2

Fs2

n=

x[n]ej2Fn/Fs

ej2FtdF

=1

Fs

n=

x[n]

Fs2

Fs2

ej2F(tn/Fs)dF

=

n=xa(nT)

sin[(/T)(t nT)]

(/T)(t nT).



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Reconstruction of analog signal (2)

Ideal interpolation formula: xa(t) =

n= xa(nT)sin[(/T)(tnT)](/T)(tnT)

.

Involves the function g(t) = sin(/T)t(/T)t appropriately shifted by nT,

n = 0,1,2, . . ., and multiplied or weighted by thecorresponding samples xa(nT) of the signal.



Folding ...Time-freq. relationships ...



Nonbandlimited signal (1)



Nonbandlimited signal (2)



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A bandlimited signal

%%% E ksempel p? b?ndbegrenset s i gnal :

%% x( t ) = a( t ) * s i n ( 2 \ p i f _ c t ) ;

%% f_ c : c a r r i e rfrekvens

% a( t ) : l angsom vari erend e modul asj on

fs = 500; % S ampl i ngsfrekvens

Nsek = 2; % A n t a l l s e k un d e r

t = 0 : 1 / fs :( Nsek) ; % Ti d

f_c = 1 0 0 ; % C a r r i e r

an = 1 + 0 . 5 * sin (2 * pi * 5*t ) ; % Modul asj on

xn = an . * sin (2 * pi *f_c*t) ; % B andbegrenset s i gn al

% P l o t t a v x n

f i g u r e ( 1 ) ;p l o t (t ,xn , 'o ' ) ; g r i d ont i t l e ( ' Pl o t o f b a n d l i m i te d s i g n a l x n ') ; x l a b e l ( ' t im e [ s e c ] ' ) ;

fname = 'B pS i gnTi d ' ;

cmd = [ ' p r i n tdeps2c ' ,fname, ' ; ' ];

eval (cmd) ;% cmd = [ ' epstopdf ' , fname , ' . eps ' ] ;

% uni x (cmd) ;

% P l o t t a v s p e k tr u m t i l x n

f i g u r e ( 2 ) ;

Hs = spectrum.periodogram( 'Hamming ' ) ;

psd(Hs ,xn , 'Fs ' ,fs, 'NFFT ' ,1024, 'SpectrumType ' , ' twosided ' , 'CenterDC ' ,

true) ;axi s([fs/2 fs/2 100 0] )

fname = 'BpSignFrek ' ;

cmd = [ ' p r i n tdeps2c ' ,fname, ' ; ' ];

eval (cmd) ;%cmd = [ ' epstopdf ' , fname , ' . eps ' ] ;

%unix (cmd) ;

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 21.5

1

0.5

0

0.5

1

1.5Plot of bandlimited signalxn

time [sec]

25 0 20 0 15 0 10 0 50 0 5 0 1 00 1 50 2 00 2 50100

90

80

70

60

50

40

30

20

10

0

Frequency (Hz)

Power/frequency(dB/Hz)

Periodogram Power SpectralDensity Estimate



Example (Sampling of a complex bandlimited signal (1))

Given a complex bandpass analog

signal xa(t) with nonzero Fouriertransform over the frequency range

[1,2].What is the smallest sampling

frequency that can be used so that

xa(t) can be recovered from itssamples x[n]?

Xa()

1 2




Obvious choice, the Nyquist rate: 22.

But if xa(t) is modulated, i.e.ya(t) = xa(t)e

(2+1)t/2, then ya(t)

becomes a (complex) low-pass signal

with spectrum as seen below.

ya(t) may be recovered from samplestaken at the sampling rate

20 = 2 1.

Question to ask: Can xa(t) be uniquelyrecovered from its samples xa(nTs)provided that Ts

21

?

Ya()

0 0




If xa(t) is sampled with a sampling frequency s, the spectrum ofthe sampled signal is

Xs() =1

Ts

k=Xa( ks).

In order to avoid interference between shifted spectra, we must

have that

2 s 1 or s 2 1.

X()

1-3s 2-3s 1-2s 2-2s 1-s 2-s 1 2



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If s 2 1, then xa(t) may beuniquely reconstructed from xs(t)using a bandpass filter with a

frequency response as shown to the

right.If s = 2 1, the reconstructionfilter is a complex bandpass filter with

impulse response

ha(t) = Tssin(st/2)

te(2+1)t/2

Hr()

Ts

1 2


Sampling of bandpass signals Representation of bandpass signals

Representation of bandpass signals (1)

Any real-valued bandpass signal xa(t) can be represented by anequivalent lowpass signal xLP(t).

xa(t): Real-valued bandpass signal with frequency concentratedaround Fc.

If + is the signal containing the positive frequencies in xa(t), thenxa(t) can be represented by a low-frequency signal:

xLP(t) = e2Fct+(t) (complex envelope of xa(t))= xI(t) + xQ(t) (in-phase & quadrature components)

= A(t)e(t)epiFct (envelope & phase)




To develop a mathematical representation of xa(t); start with thesignal +(t) containing the positive frequencies in xa(t).

Then: +(F) = 2Va(F)Xa(F),where Va(F) is the Heaviside function and Xa(F) = F[xa(t)].

The analytic signal or the pre-envelope of xa(t) given as:

+(t) =+(F)e

j2Ft

dF = F1

[2Va(F)] F1

[Xa(F)]=(t) + jt

xa(t) = xa(t) +j

1t xa(t)




+(t) = xa(t) +jxa(t),

xa(t) =1t xa(t) =

1

xa()t d.

xa(t) viewed as output of the filter with impulse responsehQ(t) =

1t when excited by xa(t).

Such filter called Hilbert transformer and

F[hQ(t)] = HQ(F) =

j F > 00 F == 0j F < 0.

|HQ(F)| = 1 and (F) = /2, F > 0 and (F) = /2, F < 0,i.e. 90 phase shift.



7/12



+(t) a bandpass signal; lowpass repr. XLP(F) = +(F + Fc) xLP(t) = +(t)e

j2Fct = [xa(t) +jxa(t)]ej2Fct

= xI(t) +jxQ(t).

This gives (from xa(t) +jxa(t) = [xI(t) +jxQ(t)]ej2Fct)

xa(t) = xI(t) cos2Fct xQ(t) sin2Fct and

x(t) = xI(t) sin2Fct + xQ(t) cos2Fct. Carrier components: cos 2Fct and sin2Fct. xI(t) and xQ(t): Low freq. amp. modulators. xI(t) and xQ(t): quadrature components.

Alternatively:xa(t) = Re{[xI(t) +jxQ(t)]e

j2Fct} = Re[xLP(t)ej2Fct].

xLP(t): Complex envelope of xa(t) and it is the equivalent lowpasssignal.




xI(t); Known as in-phase component

xQ(t); Known as quadrature component




Letting xLP(t) = A(t)ej(t),

A(t) =

x2I (t) + x2Q

(t) and (t) = tan1(xQ(t)/xI(t)),

then a third possible representation isxa(t) = Re[xLP(t)e

j2Fct] = A(t) cos[2Fct + (t)]. A(t): Envelope of xa(t). (t): Phase of x

a(t)

Fourier transform of xa(t):Xa(F) =

xa(t)e

j2Ftdt

=

{Re[xLP(t)e

j2Fct]}ej2Ftdt

= 12

[xLP(t)e

j2Fct + xLP(t)ej2Fct]ej2Ftdt

= 12

[XLP(F Fc) + X

LP(F Fc)].

Any bandpass signal xa(t) can be represented by an equivalentlowpass signal xLP(t).


Sampling of bandpass signals Sampling of bandpass signals

Sampling of bandpass signals (1)

Know: A signal with highest frequency B can uniquely be

represented by samples taken at the minimum (Nyquist) rate of

2B samples per second.

What if the signal is band limited; B1 F B2:

Either sample at sample frequency 2B2 or sample at 2 (B2 B1)/2.



8/12


Sampling of bandpass signals (2)Integer band positioning (1)

Sampling of a bandpass signal at rate Fs = 1/T produces asequence x[n] = xa(nT) with spectrum

Xa(F) =1T

k=Xa(F kFs) Positioning of shifted spectra controlled by Fs.

Remember; real signals have two spectral bands. Care must betaken!

Integer band positioning: Restrict the higher frequency of the band to be an integer multiple

of the bandwidth, i.e. FH = mB. m = FH/B; band position. No aliasing with Fs = 2B.







Signal can be reconstructed as xa(t) =

n= xa(nT)ga(t nT)

where ga(t) =sin BtBt cos2Fct

ga(t) is equal to the ideal interpolation function for lowpass

signals, modulated by the carrier frequency Fc.By properly choosing the center frequency Fc of Ga(F), we canreconstruct a continuous-time signal with spectral band centeredat Fc = (kB+ B/2), k = 0, 1, . . .. With k = 0 we obtain the baseband signal; down-conversion.

When the band position is an even integer, the baseband spectral

structure is inverted.



Sampling of bandpass signals (5)Arbitrary band positioning (1)

To avoid aliasing for spectral bands with arbitrary positioning, thefollowing must be true (see figure)

1 2FH kFs2 (1 k)Fs 2FLP

We have to choose both k and Fs.



9/12


Sampling of bandpass signals (6)Arbitrary band positioning (2)

Include guard bands when choosing k and Fs to ensure sampling

frequency tolerances.



Sampling of bandpass signals (7)Sampling using bandpass signal representation (1)

May sample +(t); the analytic signal (the pre-envelope of xa(t) atrate Fs = 2B independent of band location). This complex signal has one band (no symmetry)! Needs B complex samples or 2B real samples.

+(t) = x

a(t) +jx(t), x(t) = 1

t x

a(t) = 1

x()

td

Must sample xa(t) and its Hilbert transform x(t) and use complexbandpass interpolation (as example given before).

Solution: Shift the signal by Fc = (B1 + B2)/2 for instant bymultiplying with the two quadrature carriers + lowpass filter and

sample each stream at 2 (B2 B1)/2 i.e. represent the signalwith 2 (B2 B1) samples per second.




(uc(1) == xI(t) and us(t) == xQ(t))




If Fc + B/2 = kB, k pos. int., B = (B2 B1) and 2B = 1/T;

x(nT) = xI(nT) cos(2FcnT) xQ(nT) sin(2FcnT)

= xI(nT) cosn(2k 1)

2 xQ(nT) sin

n(2k 1)

2

For n = 2m i.e. even and T1 = 2T = 1/B,x(2mT) x(mT1) = xI(mT1) cos(m(2k 1)) = (1)

mxI(mT1). For n = 2m 1 i.e. odd and T1 = 2T = 1/B,

x(2mT T) x(mT1 T1/2) = (1)m+k+1xQ(mT1 T1/2).

Thus; Even number samples produces the lowpass signal component

xI(t) Odd number samples produces the lowpass signal component

xQ(t)



10/12

Multirate digital signal processing Sampling rate conversion

Sampling rate conversion



Decimation by a factor D (1)

Assume signal x(n) with spectrum X(w) which is nonzero in the

frequency interval 0 |w| or |F| Fx/2.Choosing every Dth sample reduces the sampling rate by a factor

D, but aliasing occurs!

The bandwidth of x(n) must, therefore, be reduced toFmax = Fx/2D.



Decimation by a factor D (2)



Interpolation by a factor I (1)

To increase the sampling rate with a factor of I, interpolate I 1new samples between successive values of the signal.

One way; add I 1 zeros between every successive values ofx(n). Call the new sequence v(m):

v(m) = x(m/I), m= 0,I,2I, . . .0, otherwise

v(m) has z-transform V(z) =

m= v(m)zm = X(zI).

V(wy) is an I-fold periodic repetition of X(wx).

v(m) must be passed through a lowpass filter HI(wy) to create theup-sampled sequence

HI(wy) =

I, 0 |wy| /I0, otherwise



11/12


Interpolation by a factor I (2)



Example (Up- and down-sampling)

A signal x[n] has Fourier transformation X() as shown in the figure below.

X()

0 /3/2

1

The signal is used as input to the systems I and II, which are defined in the

figure below:

I: x[n] 3 w1[n] H0(z) z1[n] 2 y1[n]

II: x[n] 2 w2[n] 3 z2[n] H0(z) y2[n]

M means downsampling by a factor M (keep every Mth sample) and

N means zero-interpolation by a factor N (insert N 1 zeros

between each sample). H0(z) is an ideal low-pass filter with a cut-offfrequency wc = /3 and gain equal to 2.For both systems, sketch the Fourier transformation of the signals

w1[n], w2[n], z1[n], z2[n], y1[n] og y2[n].



Example (Up- and down-sampling)

W1()

/60 /3 /2

11

Z1()

/60 /3 /2

2

Y1()

/60 /3 /2

1

W2()

/60 /3 /2

112

Z2()

/60 /3 /2

12

Y2()

/60 /3 /2

1



Conversion by a rational factor I/D (1)

Cascade an interpolator with a decimator.

Remember: Up-sample first, downsample second. Preserve the desired spectral characteristics The two filters can be combined

Possible with multistage implementations.

I =L

i=1 Ii D =

Ji=1 Di

Design filters for each stage to avoid aliasing.



12/12








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