Properties of DTFT MCQ Quiz in தமிழ் - Objective Question with Answer for Properties of DTFT - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

The time-shifting property states that:

Application:

Now, we wish to calculate the DTFT of

This can be written as:

x(n) = (0.8)ⁿ [u(n) - u(n - 6)]

x(n) = (0.8)ⁿ u(n) – (0.8)ⁿ u(n - 6)

x(n) = (0.8)ⁿ u(n) – (0.8)⁶ (0.8)^{n – 6} u(n - 6)

Using time-shifting property, the DTFT of x(n) will be:

Alternate Method (Option checking):

1) Option (3) is ruled out because we can see that DTFT of x(n) will consist of two terms as the interval 0 ≤ n ≤ 5 and can be defined by:

2) Option (4) is ruled out since there is a positive sign in the denominator that comes only when (-a)ⁿ u(n) is given. But we have 0.8 which does not contain a negative sign in the given question.

3) Option (2) is also ruled out since DTFT of both terms u(n) and u(n - 6) components are getting added. Instead, they must have been subtracted.

So the correct answer is Option (1).

Tips: In the early stage of learning, try to eliminate the options. Eventually, with practice, one can get used to it to solve all the easy questions in minimum time.

Concept:

The DTFT and Inverse DTFT is defined as:

Application:

If 

The Fourier transform will be:

x²(n) → (a²)^n-1 u(n – 1)

∴ DTFT of x²(n) will be:

Concept:

Any discrete signal is said to be periodic if:

 is a rational number 

Calculation:

Given signal is , ω₀ = 1 rad / sec

Since this is not a rational number, so it is not periodic at all.

Note:

Continuous sinusoidal and complex sinusoids are periodic for every value of ‘ω₀’, but discrete-time signals are periodic only if  is a rational number .

Let the signal in frequency be represented by:

To occupy the entire region from -π to π, x[n] must be down sample by factor 14/3

Since it is not possible to down sample by non-integer factor, up sample single first by 3 (= L).

Then down sample signal by 14 (= N).

Thus, L = 3

N = 14

= g[n].q[n]

 g[n] is of form αⁿ u[n]

⇒ α = ¼

Period of q[n] is 4

Hence the product αN = ¼ × 4 = 1

We have

From properties of Fourier transform we have,

Rewriting it we have

Putting ω = 0, we have

Likewise,

Again at ω = 0

Since, for calculating α, we need value only at . So, simply, we express  mathematically for interval

So,

Thus,  

and

Now,

Substituting the Fourier equivalents we have,

Using the property that ,

 can be rewritten, as

Let, and .

Now,  can be obtained by first passing  through the filter with response  and the convolving the result with .

Now,  is periodic with period , the Fourier series coefficient  of  are

 for all

Now, output

As, the signal is periodic, taking the convolution over one period we have,

Now since all  and  are 0

 for all

Now,  for all

Thus,

DTFT (x[n]) = X(e^jω)

In first part, we take m = -n then we get

x[n] = u [n – 2] – u [n – 5]

= δ[n – 2] + δ[n – 3] + δ[n - 4] 

= e^-j2Ω + e^-j3Ω + e^-j4Ω

So that

Now,