Lecture5_Sampling_of_Continuous-Time_Signals

Digital Signal Processing

Lecture5

Sampling of Continuous-Time Signals

Tesheng Hsiao,Associate Professor

1Periodic Sampling

Most discrete-time signals arise from periodic sampling of continuous-time signals.A discrete-time signal x[n]sampled from a continuous-time signal x c(t)can be expressed as

x[n]=x c(nT),n∈Z

where T is the sampling period,and its reciprocal,f s=1/T,is called the sampling frequency (in Hz).?s=2π/T=2πf s is the sampling frequency in radians per second.

The device that converts a continuous-time signal to a discrete-time signal is called the analog-to-digital(A/D)converter.Practical considerations in the implementation of an A/D converter include quantization error,resolution and bandwidth of the signal as well as the complexity of the electronic circuit.We temporarily leave alone the hardware implementation issues and focus ourselves on the sampling process in this lecture.It will be more convenient to express the conversion process as a”conceptual”continuous-to-discrete-time(C/D)converter,which consists of two parts:(1)impulse train modulation and(2) conversion from an impulse train to a discrete-time signal.The C/D converter is shown in Fig.(1),where s(t)is a periodic impulse train s(t)= ∞n=?∞δ(t?nT).Signals sampled by di?erent sampling rates are shown in Fig(2).

Figure1:C/D Converter

Figure 2:Signals sampled by di ?erent sampling rates.(a)x s (t )(b)x [n ]

2Frequency-Domain Representation of Sampling Preliminary Mathematics

Theorem 1(Poisson Formula)For any T >0,the following equation holds in the sense of distribution equality.

∞ n =?∞e ?jnT ?

=2πT ∞ k =?∞δ ??2πk T

where δis the Dirac delta function.

Hint of the proof:Let ?s (?)denote the series of functions on the left hand side.Appar-ently,?s is periodic with period 2π/T .It therefore su ?ces to prove that the restriction ?s to [?π/T,π/T ]is equal to 2πT δ.To do this,we have to show that for any test function ?φwhich is in?nitely di ?erentiable on a compact support included in [?π/T,π/T ],the following integration equality holds: ∞?∞?s (?)?φ(?)d ?=2πT ?φ(0)c.f.Stephane Mallat,A Wavelet Tour of Signal Processing ,2nd Ed.,Academic Press,1999Corollary 1Let s (t )=

∞n =?∞δ(t ?nT ).Then the (continuous-time)Fourier transform

of s is S (j ?)=2πT ∞ k =?∞δ ??2πk T

Proof:By the de?nition of the Fourier transform and the Poisson formula.

Sampling Theorem

From Fig.(1),we have

x s(t)=x c(t)s(t)

=x c(t)

∞ n=?∞δ(t?nT)

=

∞ n=?∞x c(nT)δ(t?nT)(1)

Since x s is the product of x c and s,the Fourier transform of x s is the convolution of X c and S.In other words,

X s(j?)=

1

2π

X c(j?)?S(j?)

=

1

2π

X c(j?)? 2πT∞ k=∞δ(??k?s)

=1

T

∞ k=∞X c(j?)?δ(??k?s)

=1

T

∞ k=?∞X c j(??k?s) (2)

where?s=2π/T is the sampling frequency in radians/sec.

Note that Eq.(2)describes the frequency-domain relationship between a continuous-time signal and the corresponding sampled signal.X s(j?)consists of periodically repeated copies of X c(j?).Each copy is shifted by an integer multiple of the sampling frequency.See Fig.(3).

Now consider the e?ect of the other block within the C/D converter which converts the impulse train to the desired discrete-time sequence.Take continuous-time Fourier transform on both sides of Eq.(1)and we get

X s(j?)=

∞ n=?∞x c(nT)e?j?T n

Since x[n]=x c(nT),the discrete-time Fourier transform of x[n]is

X(e jω)=

∞ n=∞x[n]e?jωn

It follows that

X s(j?)=X(e jω)|w=?T=X(e j?T)

Consequently,from Eq.(2),we have

Figure3:Spectrum of the sampled signals(a)the original continuous-time signal(b)the impulse train signal(c)the sampled signal(no aliasing)(d)the sampled signal(with aliasing) Remark1Converting the impulse train signal x s(t)to the discrete-time sequence x[n]has

the e?ect of frequency scaling,i.e.ω=?T.Since?s=2π

T ,the”continuous-time frequency”

(frequency of the continuous-time signal)?=?s corresponds to the”discrete-time frequency”(frequency of the discrete-time signal)ω=2π.Recall that X(e jω)is a2πperiodic function

and the”highest”frequency of X(e jω)isπ,which corresponds to?s

2in X c(j?).Frequency

components of X c(j?)outside the band[??s

2,?s

2

]will be folded into the low frequency band,

which is referred to as aliasing.

We now deal with the conditions of sampling that allow us to recover the original signal from its samples.If x c(t)is band-limited by?N(see Fig.(3a)).Suppose?s??N>?N or?s>2?N,then all shifted copies of X c(j?)do not overlap.Therefore the low-frequency band(|?|≤?N)of X s(j?)is the same as the original continuous-time spectrum(up to a scalar factor T).Consequently X c(j?)can be reconstructed from X s(j?)by?ltering out the high frequency components.In other words,let H r(j?)be an ideal lowpass?lter with cuto?frequency at?c,?N

X c(j?)=H r(j?)X s(j?)

The frequency spectrum is shown in Fig(4).

On the other hand if?s≤2?N,the copies of X c(j?)overlap and X c(j?)is no longer recoverable from X s(j?)as shown in Fig(3d).This phenomenon is called aliasing distortion, or simply,aliasing.

Figure4:Perfect reconstruction of a continuous-time signal from its samples

For example,let x c(t)=cos?0t.the Fourier transforms X c(j?),X s(j?)for?0?s/2are shown in Fig.(5(a)-(c)),respectively.When we apply the lowpass?lter

H r(j?)to reconstruct the signal from its samples,we get x r(t)=cos?0t if?0?s/2.In other words,the high frequency component of X c(j?)is”re?ected”into the low frequency part and contaminate the original signal due to the aliasing distortion.

Theorem2(Nyquist Sampling Theorem)Let x c(t)be a bandlimited signal with X c(j?)= 0for|?|≥?N.Then x c(t)is uniquely determined by its samples x[n]=x c(nT),n∈Z if

?s=2π

T≥2?N

The frequency?N is commonly referred to as the Nyquist frequency,and the frequency 2?N that must be exceeded by the sampling frequency is called the Nyquist rate.

3Reconstruction of a Bandlimited Signal from Its Sam-ples

In this section,we consider the process that converts a discrete-time sequence to a continuous-time signal.The”conceptual”device,discrete-to-continuous-time(D/C)converter is shown

Figure5:The e?ect of aliasing sampling of a cosine signal

in Fig.(6).The?rst step of the D/C converter is to convert the discrete-time sequence to an impulse train,

x s(t)=

∞ n=?∞x[n]δ(t?nT)

where T is the sampling period

The reconstruction?lter H r(j?)is an ideal lowpass?lter with cuto?frequency?c=

?s 2=π

T

and the passband gain is T.Let h r(t)be the impulse response of H r(j?).It is easy

to show that

h r(t)=sin(πt/T)πt/T

Figure 6:D/C Converter

The output of the D/C converter is

x r (t )=x s (t )?h r (t )

= ∞ n =?∞x [n ]δ(t ?nT ) ?h r (t )

=

∞ n =?∞x [n ]h r (t ?nT )(3)=∞ n =?∞x [n ]sin[π(t ?nT )/T ]π(t ?nT )/T (4)

Note that Eq.(4)is a kind of interpolation which inserts values of x r (t )between two adjacent samples.

Now consider the frequency domain relationship.Take continuous-time Fourier transform on both sides of Eq.(3),X r (j ?)=∞ n =?∞

x [n ]H r (j ?)e ?j ?nT

Factor out H r (j ?)and we get

X r (j ?)=H r (j ?)X (e j ?T )

(5)

Eq.(5)is the frequency-domain description of the D/C converter.