Test: Signals & Systems - 2 - Electronics and Communication Engineering (ECE) MCQ

The Fourier Series Coefficient of a discrete time Periodic sequence is given by:

Where N = Period of the sequence.

Calculation:

Given, x (n) = {1, 1, 0, 0} and N = 4

We are asked to calculate C₃*

Putting k = 3.

y[n] + 3y[n – 1] + 2y[n – 2] = 2x[n] – x[n – 1]
By applying Z-transform,

Y[z] + 3(z-1Y(z) + y[-1]) + 2(z-2Y(z) + y[-2] + z-1y[-1]) = 2X(z) – z-1X(z) – x[-1]

By applying partial fractions,

The steady state response of the output is

The Fourier series representation of the discrete-time periodic sequence is given by:

ak = Fourier series coefficient periodic by N.

ω0 = Fundamental frequency.

Application:

Now, we can rewrite the given sequence as

Given signal is, x[n] = r[n] (u[n] – u[n – 11])
Energy of x[n] is

Given that f(t) = v(t) = e^-2t u(t)

The total energy dissipated by the resistor is

The energy in the frequency band 0 < ω < 10 is,

The percentage of the total energy is 

y [n] is the first difference of x [n]
y(n) = x(n) – x(n - 1)

By applying z-transform.

⇒ Y(z) = X(z) – z-1 X(z)

⇒ Y(z) = 3 + 4z – 6z^-1 + 2z^-2 – z^-1 (3 + 4z – 6z^-1 + 2z^-2)

= 3 + 4z – 6z^-1 + 2z-2 – 3z^-1 – 4 + 6z^-2 – 2z^-3

= -1 + 4z – 9z^-1 + 8z^-2 – 2z^-3

1. X(s) converges to  with ROC: Re (s) > 0.
For limit  tends to zero for σ = Re(s) > 0 as the exponent term dominates and its integral from 0 to +∞ converges.

Therefore, the given statement is false.

2.  does not tend to zero as t → +∞, which means the integral from 0 to +∞ is unbounded for all s. Thus the Laplace transform cannot converge for any value of s.

Therefore, the given statement is true.

3. For  does not tend to zero for t → -∞.

For Re(s) < 0, it does not converge for t → +∞.

Hence for no value of s does the Laplace transform converge.

Therefore, the given statement is true.
 

4. |t| = t u(t) – t u(-t).

For t u(t), the ROC is Re(s) > 0

For t u(-t), the ROC is Re(s) < 0

Hence, there is no value of s common to both the ROCs.Therefore, the given statement is true.    

If the DTFT of a signal f(n) ↔ F(ω), then from the frequency shifting property

Calculation:

Given, f(n) = {a, b, c, d}

If f(n) ↔ F(ω)

Concept:
z-transform of a discrete signal can be written as;

Calculation:

We have to calculate:

On comparing equation (1) & (2) we find that;

Where X(z) is z-transform of nu[n] which is:

Hence our required summation is

Concept:
The Z-transform of standard signals with their ROC are:

Application:

By applying Fourier transform,

The Discrete-Time Fourier transform of a signal of infinite duration x[n] is given by:

For a signal x(n) = aⁿ u(n), the DTFT will be:

Since u[n] is zero for n < 0 and a constant 1 for n > 0, the above summation becomes:

The geometric series states that:

Equation (1) can now be written as:

Given x[n] = 0.5n u[n], i.e. a = 0.5

The DTFT of the above given sequence using Equation (2) will be:

y(t) = x₁(t) * x₂(t)
x₁(t) = (2 cos 4t) u(t)
x₂(t) = (sin 2t) u(t)
We know convolution in time domain is multiplication in frequency domain
So, y(t) = x1(t) * x2(t)
y(t) = x₁(s) ⋅ x₂(s)

 

Put s = 0 ⇒ 0 = 4B + 16D → 1)
Put s = 1 ⇒ 4 = (A + B) (5) + (C + D) (17) → 2)
Put s = 2 ⇒ 8 = (2A + B) (8) + (2C + D) (20) → 3)
Put s = 3 ⇒ 12 = (3A + B) (13) + (3C + D) (25) → 4)
Solving 1), 2), 3) and 4) we get
A = B = 1 / 3
C = D = -1 / 3

Applying inverse Laplace transform,

y(n) = x(n) * x(n)
By taking z-transform both sides

Y(z) = X(z) ⋅ X(z)

x(n) = (0.7)ⁿ u(n)

We know that;

Frequency response of 

Input to the system is x(t) = e^-t u(t)

The output in frequency domain is given by

By applying partial fraction method,

By applying the inverse Fourier transform,
 
 
 

Concept:
The DTFT of a discrete sequence is a complex quantity, which can be defined as:

X(e^jω) = Real {X(e^jω)} + Img. {X(e^jω)}

Where the Real {X(e^jω)} is nothing but the DTFT of the even part of the signal.
Proof:
A signal can be written as the sum of the even part and odd part, is:

Taking the Fourier transform of the even part, we get:

Application:

We are required to find F^-1{Re{X(e^jω)}} at n = 0

The inverse DTFT of Re{X(e^jω)} is x_e(n)

Given:

This is an ordinary differential equation with constant co-efficient, therefore it is linear and time invariant. It is casual.

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

This is a difference equation with the constant co-efficient. Therefore, it is linear and time-invariant. It is non-casual since the output depends on future values of x.

3. y[n + 1 ] + 4y[n] = 3x [n + 1] – x[n]

Rewriting the above equation by shifting one unit.

Y[n] + 4y [n - 1] = 3x [n] – x[n - 1]

This is a difference equation with constant co-efficient. So, it is linear and time invariant. The output does not depend on future values of the input, so it is causal.

The co-efficient of y means that this is non-linear and it does not depend on t. so, it is time invariant. It is causal.