A complex exponential signal, whether in continuous time or in discrete time, is completely determined by its amplitude A, phase α, and frequency which can be F0 Hz (or Ω0 rad/sec) in continuous time or w0 radians in discrete time, as in the following expressions:

X(t) =A ejα  ejΩ0t = A ejα  ej2πF0t

X[n] = A ejα  ejw0n

When the signal is the superposition (ie sum) of two or more complex exponentials of different amplitudes and phases, we just add them in the plot and place them appropriately in frequency. So for example a sinusoidal signal, as

X(t) = A+ cos(2π F0t + α)

      = (A/2) ejα  ej2πF0t + (A/2) e-jα  e-j2πF0t

is represented by two complex exponentials with frequencies F0 and -F0 respectively, as shown in figure.  Notice the appearance of the negative frequency -F0 which corresponds to the complex exponential with the same amplitude and opposite phase.

Similarly for the discrete time case.

A discrete time sinusoid

X[n] = A cos(w0n +α) = (A/2) ejα  ejw0n + (A/2) e-jα  e-jw0n

Fig 1 Mapping of Analog Frequency and Digital Frequency

Orthogonality

Let us consider a set of n mutually orthogonal functions x1(t), x2(t)... xn(t) over the interval t1 to t2. As these functions are orthogonal to each other, any two signals xj(t), xk(t) have to satisfy the orthogonality condition. i.e.

Let a function f(t), it can be approximated with this orthogonal signal space by adding the components along mutually orthogonal signals i.e.

If are two complex functions, then can be expressed in terms of   as 

, with negligible error

Where,

Where, c0mplex conjugate of

If are orthogonal then

Basis Function

• Basic idea: If a signal can be represented by n-tuple then it can be treated in much the same way as a n-dim vector.
• Let be n signal.
• Consider a signal x(t) and suppose that

• If every signal can be written as above

basic function

•  Signal set is an orthogonal set if

• is an orthonormal set
• 

Example

Consider the following signal set:

Solution:

By inspection, the signals can be expressed in terms of the following two basis functions:

Note that the basis is orthogonal

=0

Also note that each these functions have unit energy

We say that they form an orthonormal basis.

Key takeaway

Signal set is an orthogonal set if

The general nth order difference equation is given as

a0y[n]+ay[n-1]+….+aN-1y[n-N+1]+any[n-N]=b0x[n]+b1x[n-1]+….+bMx[n-M]

The difference equation can be written as

The output equation can be given as

y[n] = { - }

The above difference equation can be solved by finding homogeneous and particular solution or by z transform method. Z transform is discussed in above section.

Homogeneous Solution:

Let x[n]=0

=0

The solution to above equation will be of the form y[n] = rn

Then solution will be of the form

y[n] = p1 + p2+ p3+……+ pn                            (For distinct roots)

y[n]= p1n + p2+ p3+……+pm+1+…….+ pnrN               (For multiple roots)

Key takeaway

1. Then solution will be of the form
2. y[n] = p1 + p2+ p3+……+ pn                            (For distinct roots)
3. y[n]= p1n + p2+ p3+……+pm+1+…….+ pnrN               (For multiple roots)

Examples

Que) Find the homogenous solution for the difference equation y[n]-y[n-1]+y[n-2]=1+3-n. Given y[-2] = 0 and y[-1]=2

Sol: For homogeneous solution

Y[n] = rn

rn-rn-1+rn-2 = 0

rn-2(r2-r+) = 0

Solving above equation we get

r= and r=1

Roots are distinct so the solution will be of the form

y[n] = p1 + p2

y[n] = p12-n +

Particular Solution:

The complete solution of any difference equation is the sum of homogeneous solution and the particular solution.

Que) For the difference equation find the particular solution x[n] = 3n

y[n] = x[n]+y[n-1]- y[n-2]?

Sol: As x[n] = 3n

From table the particular solution will be y[n] = k3n

y[n] = x[n]+y[n-1]- y[n-2]

k3nu(n) = 3nu(n)+ k 3n-1u(n-1)-k3n-2u(n-2)

Finding k for n≥2

K 32 = 32+ k 32-1- k 32-2

9k-k+k = 9

k= 27/20

y[n]= 3n

Sampling usually refers to conversion of continuous signal into short duration of pulses each pulse followed by a skip period when no signal is available. Below shown is a uniformly sampled signal.

Following are two popular sampling operations:

 1. Single rate or periodic sampling

 2. Multi-rate sampling

Figure below shows the structure and operation of a finite pulse width sampler, where 2(a) represents the basic block diagram and 2(b) illustrates the function of the same. T is the sampling period and p is the sample duration.

Figure 2 (a): Basic block diagram

Figure 2 (b): Sampler output

Finite pulse width sampler converts a continuous time signal into a pulse modulated or discrete signal. The most common type of modulation in the sampling and hold operation is the pulse amplitude modulation.

The block diagram of sampler is shown above, having a pulse train of p seconds and sampling period of T seconds.

p(t)= unit pulse train with period T

p(t)=

Us(t)=unit step function

In frequency domain p(t) can be represented as

p(t)=

= 2π/T

Cn=

p(t)=1 for 0≤ t ≤p

The output of the ideal sampler can be expressed as

f*(t)=

F*(s)=

The output of the sampler can be approximated as

Cn=

    =

Reconstruction process

Fig 3 Sampled Data

The sampled data signal is modified by the controller. The hold circuit than converts the signal to analog form. The simplest hold circuit is ZOH (zero order hold) in which the reconstructed signal acquires the same value as the last received sample for the entire sampling period.

The basic sampler is shown in above figure (a) and output in figure (b). The high frequency signal present in the reconstructed signal is filtered by the controller elements which are the low pass in frequency behaviour.

The first or higher order holds have no advantage over the ZOH. In the first order hold the last two signal samples are used to reconstruct the signal for the current sampling period.

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.”

Key takeaway

1. We can simply avoid aliasing by sampling the signal at a higher rate than the Nyquist rate (Fs>Fm). Or, we can use anti-aliasing filters. These are special low-pass filters that are usually found in the initial stages of any digital signal processing operation. The anti-aliasing filters attenuate the unnecessary high-frequency components of a signal. They band-limit the input signal by removing all frequencies higher than the signal frequencies.

Examples

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.

3) Consider an analog signal

Solution.

The frequency in the analog signal

The largest frequency is

The Nyquist rate is

4) The analog signal

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

Solution.

• The frequency of the analog signal are

• 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

5) For given signal

• Find the minimum sampling rate required to avoid aliasing.
• If , what is the discrete time signal after sampling?
• If , what is the discrete time signal after sampling?
• 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

6) 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

In sampling the signal m(t) is multiplied with periodic pulse train. Let M(ω) the spectrum of the input signal be band limited with the maximum frequency of fm as shown in figure 4.

                                               Figure: 4 Spectrum of input signal

                                                 Figure: 5 (fs>2 fm)

                                                 Figure: 6 (fs<2 fm)

The frequency spectrum of this signal when impulse sampled is plotted in figure 5 (for fs>2 fm). In figure 6 for (fs<2 fm). From figure 5 and figure 6 we can conclude that as long as fs≥2fm the original signal is preserved in the sampled signal and can be extracted from it by the low pass filter. This is known as Shannon’s Sampling theorem. This theorem states that the information contained in a signal is fully preserved in the sampled form as long as the sampling frequency is at least twice the maximum frequency contained in the signal.

1.4.2 Nyquist rate.

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.

Fig 7 Sampled Signal

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.

Key takeaway

1. Nyquist rate is given by fS=2W
2. Sampling Frequency=1/Ts=fs