UNIT 6

Sampling

6.1 Sampling and reconstruction of Analog signals

Sampling is defined as, “The process of measuring the instantaneous values of continuous-time signal in a discrete form.”

Sample is a piece of data taken from the whole data which is continuous in the time domain.

When a source generates an analog signal and if that has to be digitized, having 1s and 0s i.e., High or Low, the signal has to be discretized in time. This discretization of analog signal is called as Sampling.

DT signals

The following figure indicates a continuous-time signal x (t) and a sampled signal xs (t). When x (t) is multiplied by a periodic impulse train, the sampled signal xs (t) is obtained.

Sampling Rate

To discretize the signals, the gap between the samples should be fixed. That gap can be termed as a sampling period Ts.

Sampling Frequency=1/Ts=fs

Where,

• Ts is the sampling time
• Fs is the sampling frequency or the sampling rate

Sampling frequency is the reciprocal of the sampling period. This sampling frequency, can be simply called as Sampling rate. The sampling rate denotes the number of samples taken per second, or for a finite set of values.

Nyquist Rate

Suppose that a signal is band-limited with no frequency components higher than W Hertz. That means, W is the highest frequency. For such a signal, for effective reproduction of the original signal, the sampling rate should be twice the highest frequency.

Which means,

fS=2W

Where,

• fS is the sampling rate
• W is the highest frequency

This rate of sampling is called as Nyquist rate.

Sampling theorem in time domain

The sampling theorem, which is also called as Nyquist theorem, delivers the theory of sufficient sample rate in terms of bandwidth for the class of functions that are bandlimited.

The sampling theorem states that, “a signal can be exactly reproduced if it is sampled at the rate fs which is greater than twice the maximum frequency W.”

Let us consider a band-limited signal, i.e., a signal whose value is non-zero between some –W and W Hertz.

Such a signal is represented as x(f)=0 for ∣f∣>W

For the continuous-time signal x (t), the band-limited signal in frequency domain, can be represented as shown in the following figure.

If the signal x(t) is sampled above the Nyquist rate, the original signal can be recovered, and if it is sampled below the Nyquist rate, the signal cannot be recovered.

The following figure explains a signal, if sampled at a higher rate than 2w in the frequency domain.

The above figure shows the Fourier transform of a signal xs (t).

If  fs<2W

The resultant pattern will look like the following figure.

Here, the over-lapping of information is done, which leads to mixing up and loss of information. This unwanted phenomenon of over-lapping is called as Aliasing.

Aliasing

Aliasing can be referred to as “the phenomenon of a high-frequency component in the spectrum of a signal, taking on the identity of a low-frequency component in the spectrum of its sampled version.”

Numerical:

1. The continuous-time signal x(t) = cos(200πt) is used as the input for a CD converter with the sampling period 1/300 sec. Determine the resultant discrete-time signal x[n].

Solution:

We know,

X[n] =x(nT)

= cos(200πnT)

= cos(2πn/3) , where n= -1,0,1,2……

The frequency in x(t) is 200π rad/s while that of x[n] is 2π/3.

2. Determine the Nyquist frequency and Nyquist rate for the continuous-time signal  x(t) which has the form of:

X(t) = 1+ sin(2000πt) + cos (4000πt)

Solution:

The frequencies are 0, 2000π and  4000π.

The Nyquist frequency is   4000π rad/s and the Nyquist rate is 8000π rad/s.

2.     Consider an analog signal.

Solution.

The frequency in the analog signal

The largest frequency is

The Nyquist rate is

3.     The analog signal

1. What is the Nyquist rate for this signal?
2. Using a sampling rate . What is discrete time signal obtained after sampling?
3. What is analog signal we can reconstruct from the samples if we use ideal interpolation?

Solution.

1. The frequency of the analog signal are

2.     For

For   ,the folding frequency is

Hence is not effected by aliasing

Is changed by the aliasing effect

Is changed by the aliasing effect

So that normalizing frequencies are

The analog signal that we can recover is

Which is different than the original signal

4.    

1. Find the minimum sampling rate required to avoid aliasing.
2. If , what is the discrete time signal after sampling?
3. If , what is the discrete time signal after sampling?
4. What is the frequency F of a sinusoidal that yields sampling identical to obtained in part c?

Solution. a.

The minimum sampling rate is

And the discrete time signal is

b. If   , the discrete time signal is

c. If Fs=75Hz , the discrete time signal is

d. For the sampling rate

in part in (c). Hence

So, the analog sinusoidal signal is

5.     Suppose a continuous-time signal  x(t) = cos (Ø0t) is sampled at a sampling frequency of 1000Hz to produce x[n]: x[n] = cos(πn/4)

Determine 2 possible positive values of Ø0, say, Ø1 and Ø2. Discuss if cos(Ø1t) or cos(Ø2t) will be obtained when passing through the DC converter.

Solution:

Taking T= 1/1000s

Cos(πn/4) =x[n] = x(nT) = cos (Ø0n/1000)

Ø1 is easily computed as           

Ø1 = 250π

Ø2 can be obtained by noting the periodicity of a sinusoid:

Ø2n/1000)

Ø2 = 2250π