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.

A2) 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.

A3) 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

A4)

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.

A6)

The frequency in the analog signal

The largest frequency is

The Nyquist rate is

A7)

• 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

A8)

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

A9)

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

A10) 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 1.

Figure: 1 Spectrum of input Signal

Figure: 2 (fs>2 fm)

                                                 Figure: 3 (fs<2 fm)

The frequency spectrum of this signal when impulse sampled is plotted in figure 2 (for fs>2 fm). In figure 3 for (fs<2 fm). From figure 2 and figure 3 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.