May23,2018 [CommunicationsSystemI]Prof.Young-ChaiKo Week11(Chapter7.)SamplingTheoremandReconstructionformSamples

Week 11 (Chapter 7.)

Sampling Theorem and Reconstruction form Samples

[Communications System I]

Prof. Young-Chai Ko

EE at KU

May 23, 2018

Outline

1 Introduction to ADC

2 Sampling and interpolation Sampled signal

Interpolation Sampling rate

Signal bandwidth and sampling rate

3 Sampling theorem Preliminary Proof

Frequency-domain representation of the sampled signal

Introduction to ADC

Analog-to-Digital Conversion (ADC)

We need ”sampling theorem” and ”quantization” for analog-to-digital conversion (ADC).

We will study the sampling theorem and the quantization.

Sampling and interpolation

Sampling

Sampling of the analog signal x(t)

”Sampling” can be explained by the following expression:

x_δ(t) =

∞

X

n=−∞

x(nT_s)δ(t − nT_s)

where T_s is sampling interval.

Sampling and interpolation Sampled signal

Sampled signal: Discrete time signal

Sampling circuit

Sampled signal which is discrete-time signal

Sampling and interpolation Sampled signal

Two questions in sampling

1 Can we reconstruct x(t) from x(nT_s)?

The sampling interval may matter.

2 If so, how?

Interpolation may matter.

Sampling and interpolation Interpolation

Reconstructing x(t) from x(nT

) by interpolator

We can reconstruct x(t) (perfectly or maybe approximately) from x(nT_s) by interpolation.

Example of linear interpolation.

- Is it perfectly reconstructed?

No.

- Can we do better if we have a certain good interpolator?

Yes. Use sinc function as an interpolator which will be explained later.

Sampling and interpolation Sampling rate

Does sampling interval T

matter in reconstruction?

Consider T_s1 > T_s2, say T_s1 = T_s and T_s2 = 0.5T_s. Then the sampled (discrete-time) signal x(nT_s1) and x(nT_s2) may look like the following:

Sampling and interpolation Sampling rate

Remark

Sampling interval T_s is smaller which means the faster sampling rate, we will have better reconstruction.

However, note that the faster sampling rate means increasing the power consumption and difficulty in circuit design which increases the cost.

The question is what is the minimum sampling rate to reconstruct x(t) from x(nTs).

Sampling and interpolation Signal bandwidth and sampling rate

Relation between Signal bandwidth and sampling rate

Sampling and interpolation Signal bandwidth and sampling rate

Remark 1

x₂(t) is changing faster than x₁(t).

- It means the bandwidth of x2(t) is larger than the bandwidth of x1(t).

Remark 2

Which do you think the sampling rate should be faster between x₁(t) and x₂(t)?

- Answer is x2(t).

Remark 3

We can conjecture that the sampling rate must be related to the bandwidth of the signal.

Sampling theorem

Concept of Sampling theorem

Goal of sampling

We want to convert the analog signal to discrete time signal with a certain sampling rate.

Then we want to reconstruct perfectly x(t) from x(nT_s).

Two things to consider

1 Sampling rate

How fast do we need to sample the analog signal?

Is there any minimum sampling rate without having any problem in reconstructing the signal perfectly?

2 Interpolation

If the sampling rate is fast enough by a certain rule, can we have the interpolator for prefect reconstruction.

Sampling theorem

Sampling theorem

Rate and interpolator in sampling theorem

Consider the analog signal x(t) with the signal bandwidth W . That means, |X(f )| = 0 for

|f | > W .

Sampling interval T_s must be faster than _2W¹ for perfect reconstruction:

Ts ≤ 1 2W

The sinc function should be used for interpolator for perfect reconstruction.

b sinc(at)

where a and b are certain constants which will be explained in detail later.

Sampling theorem

Interpolating the discrete-time signal using sinc function

Sampling theorem Preliminary

Preliminary

1 Convolution property in Fourier transform

F [x₁(t) × x2(t)] = X1(f ) ∗ X2(f ) and vice versa.

2 Fourier transform of periodic signal x(t) with period T₀.

- Let {x_n} denote the Fourier series coefficients corresponding to this signal. Then x(t) =

∞

X

n=−∞

xne^{j2π nT0t}

By taking the Fourier transform of both sides and using the fact that F



e^{j2π nT0t}



= δ



f − n T0

 , we obtain

X(f ) =

∞

X

n=−∞

x_nδ



f − n T0

 .

Sampling theorem Preliminary

1 Convolution property of FT

2 FT of periodic signal x(t)

X(f ) =

∞

X

n=−∞

xnδ



f − n T₀

 .

3 FT of impulse train x(t) = P∞

n=−∞δ(t − nT₀)

X(f ) = F





∞

X

n=−∞

δ(t − nT₀)



 =

∞

X

n=−∞

x_nδ



f − n T₀



where x_n is the Fourier coefficients when x(t) is expressed in Fourier series form.

Sampling theorem Preliminary

Now let us find x_n:

xn = 1

T₀ Z

T₀

δ(t − nT₀)e^−j2π

n T0t

dt

= 1

T0

Z

T₀

δ(t)e^−j2π

n

T0(t+nT₀)

dt

= 1

T₀ Z

T₀

δ(t)e^−j2π

n T0t

e^j2πn2

| {z }

=1

dt

= 1

T₀ Z

T₀

δ(t)e^−j2π

n T0t

dt

= 1

T₀

Sampling theorem Preliminary

Sampling theorem Preliminary

1 Convolution property of FT

F [x₁(t) × x₂(t)] = X₁(f ) ∗ X₂(f )

2 FT of periodic signal x(t) with period T₀

X(f ) =

∞

X

n=−∞

x_nδ



f − n T₀

 .

where xn is the Fourier coefficients in Fourier series form of x(t).

3 FT of impulse train with period T₀

X(f ) = F





∞

X

n=−∞

δ(t − nT₀)





= 1

T₀

∞

X

n=−∞

δ



f − n T₀



Sampling theorem Proof

Proof of sampling theorem

Let us denote the sampled signal as

x_δ(t) =

∞

X

n=−∞

x(nT_s)δ(t − nT_s)

which can be rewritten as

x_δ(t) = x(t) ×

∞

X

n=−∞

δ(t − nTs)

where we make use of the property:

x(t)δ(t − t₀) = x(t₀)δ(t − t₀)

Sampling theorem Proof

Fourier transform of x_δ(t):

X_δ(f ) = F [x_δ(t)]

= F



x(t) ×

∞

X

n=−∞

δ(t − nT_s)





By convolution property of FT, we have

X_δ(f ) = F [x(t)] ∗ F





∞

X

n=−∞

δ(t − nT_s)





= X(f ) ∗ F





∞

X

n=−∞

δ(t − nT_s)





= X(f ) ∗ 1 Ts

∞

X

n=−∞

δ



f − n Ts



= 1

T_s

∞

X

n=−∞

X



f − n T_s



Sampling theorem Proof

The sampled signal x_δ(t) is given as

x_δ(t) =

∞

X

n=−∞

x(nTs)δ(t − nTs)

Then, its Fourier transform is given as

X_δ(f ) = X(f ) ∗ 1 Ts

∞

X

n=−∞

δ



f − n Ts



= 1

T_s

∞

X

n=−∞

X



f − n T_s



Sampling theorem Frequency-domain representation of the sampled signal

Frequency-domain representation of the sampled signal

Sampling theorem Frequency-domain representation of the sampled signal

What happened if

s

≥ 2W ?

Sampling theorem Frequency-domain representation of the sampled signal

We can reconstruct X(f ) from X

(f ) if

s

≥ 2W ?

We can reconstruct X(f ) from X_δ(f ) by passing through ideal lowpass filter.

Sampling theorem Frequency-domain representation of the sampled signal

Reconstruction of X(f ) from X

(f )

We can reconstruct X(f ) from X_δ(f ) such as:

X(f ) = X_δ(f )H(f ) where H(f ) is given as

H(f ) = Ts

Y

 f 2W



Then

X(f ) = X_δ(f )T_sY f 2W



Sampling theorem Frequency-domain representation of the sampled signal

Taking the inverse Fourier transform of both sides, we obtain x(t) = x_δ(t) ∗ 2W T_ssinc(2W t)

=





∞

X

n=−∞

x(nT_s)δ(t − nT_s)



 ∗ 2W T_ssinc(2W t)

=

∞

X

n=−∞

2W Tsx(nTs)sinc(2W (t − nTs))

−

∞

X

n=−∞

x(nT_s)sinc

 t Ts

− n



Sampling theorem Frequency-domain representation of the sampled signal

What happened if

s

< 2W ?

There is no way to reconstruct X(f ) from X_δ(f ).