Signals Samppng Techniques

There are three types of samppng techniques:

Impulse samppng.

Natural samppng.

Flat Top samppng.

Impulse Samppng

Impulse samppng can be performed by multiplying input signal x(t) with impulse train $Sigma_{n=-infty}^{infty}delta(t-nT)$ of period T . Here, the ampptude of impulse changes with respect to ampptude of input signal x(t). The output of sampler is given by

$y(t) = x(t) ×$ impulse train

$= x(t) × Sigma_{n=-infty}^{infty} delta(t-nT)$

$ y(t) = y_{delta} (t) = Sigma_{n=-infty}^{infty}x(nt) delta(t-nT),...,... 1 $

To get the spectrum of sampled signal, consider Fourier transform of equation 1 on both sides

$Y(omega) = {1 over T} Sigma_{n=-infty}^{infty} X(omega - n omega_s ) $

This is called ideal samppng or impulse samppng. You cannot use this practically because pulse width cannot be zero and the generation of impulse train is not possible practically.

Natural Samppng

Natural samppng is similar to impulse samppng, except the impulse train is replaced by pulse train of period T. i.e. you multiply input signal x(t) to pulse train $Sigma_{n=-infty}^{infty} P(t-nT)$ as shown below

The output of sampler is

$y(t) = x(t) imes ext{pulse train}$

$= x(t) imes p(t) $

$= x(t) imes Sigma_{n=-infty}^{infty} P(t-nT),...,...(1) $

The exponential Fourier series representation of p(t) can be given as

$p(t) = Sigma_{n=-infty}^{infty} F_n e^{j nomega_s t},...,...(2) $

$= Sigma_{n=-infty}^{infty} F_n e^{j 2 pi nf_s t} $

Where $F_n= {1 over T} int_{-T over 2}^{T over 2} p(t) e^{-j n omega_s t} dt$

$= {1 over TP}(n omega_s)$

Substitute F_n value in equation 2

$ herefore p(t) = Sigma_{n=-infty}^{infty} {1 over T} P(n omega_s)e^{j n omega_s t}$

$ = {1 over T} Sigma_{n=-infty}^{infty} P(n omega_s)e^{j n omega_s t}$

Substitute p(t) in equation 1

$y(t) = x(t) imes p(t)$

$= x(t) imes {1 over T} Sigma_{n=-infty}^{infty} P(n omega_s),e^{j n omega_s t} $

$y(t) = {1 over T} Sigma_{n=-infty}^{infty} P( n omega_s), x(t), e^{j n omega_s t} $

To get the spectrum of sampled signal, consider the Fourier transform on both sides.

$F.T, [ y(t)] = F.T [{1 over T} Sigma_{n=-infty}^{infty} P( n omega_s), x(t), e^{j n omega_s t}]$

$ = {1 over T} Sigma_{n=-infty}^{infty} P( n omega_s),F.T,[ x(t), e^{j n omega_s t} ] $

According to frequency shifting property

$F.T,[ x(t), e^{j n omega_s t} ] = X[omega-nomega_s] $

$ herefore, Y[omega] = {1 over T} Sigma_{n=-infty}^{infty} P( n omega_s),X[omega-nomega_s] $

Flat Top Samppng

During transmission, noise is introduced at top of the transmission pulse which can be easily removed if the pulse is in the form of flat top. Here, the top of the samples are flat i.e. they have constant ampptude. Hence, it is called as flat top samppng or practical samppng. Flat top samppng makes use of sample and hold circuit.

Theoretically, the sampled signal can be obtained by convolution of rectangular pulse p(t) with ideally sampled signal say y_δ(t) as shown in the diagram:

i.e. $ y(t) = p(t) imes y_delta (t), ... , ...(1) $

To get the sampled spectrum, consider Fourier transform on both sides for equation 1

$Y[omega] = F.T,[P(t) imes y_delta (t)] $

By the knowledge of convolution property,

$Y[omega] = P(omega), Y_delta (omega)$

Here $P(omega) = T Sa({omega T over 2}) = 2 sin omega T/ omega$

Nyquist Rate

It is the minimum samppng rate at which signal can be converted into samples and can be recovered back without distortion.

Nyquist rate f_N = 2f_m hz

Nyquist interval = ${1 over fN}$ = $ {1 over 2fm}$ seconds.

Samppngs of Band Pass Signals

In case of band pass signals, the spectrum of band pass signal X[ω] = 0 for the frequencies outside the range f₁ ≤ f ≤ f₂. The frequency f₁ is always greater than zero. Plus, there is no apasing effect when f_s > 2f₂. But it has two disadvantages:

The samppng rate is large in proportion with f₂. This has practical pmitations.

The sampled signal spectrum has spectral gaps.

To overcome this, the band pass theorem states that the input signal x(t) can be converted into its samples and can be recovered back without distortion when samppng frequency f_s < 2f₂.

Also,

$$ f_s = {1 over T} = {2f_2 over m} $$

Where m is the largest integer < ${f_2 over B}$

and B is the bandwidth of the signal. If f₂=KB, then

$$ f_s = {1 over T} = {2KB over m} $$

For band pass signals of bandwidth 2f_m and the minimum samppng rate f_s= 2 B = 4f_m,

the spectrum of sampled signal is given by $Y[omega] = {1 over T} Sigma_{n=-infty}^{infty},X[ omega - 2nB]$