All questions of Signals & Systems for Electronics and Communication Engineering (ECE) Exam

Unilateral Laplace Transform for Constant Coefficient Differential Equations with Non-zero Initial Condition

Unilateral Laplace Transform is a mathematical tool used to solve linear constant coefficient differential equations. It is particularly useful in solving differential equations with non-zero initial conditions. Let us discuss in detail the application of Unilateral Laplace Transform for constant coefficient differential equations with a non-zero initial condition.

Constant Coefficient Differential Equations

A constant coefficient differential equation is a differential equation in which the coefficients are constant. These types of differential equations are commonly found in physics and engineering. Examples of constant coefficient differential equations include the following:

- d^2y/dt^2 + 3dy/dt + 2y = 0
- d^2y/dt^2 + 2dy/dt + 2y = f(t)

Unilateral Laplace Transform

Unilateral Laplace Transform is a mathematical tool used to solve linear constant coefficient differential equations. It is defined as follows:

L{f(t)} = F(s) = ∫0∞ e^-st f(t) dt

where L denotes the Laplace transform, f(t) is the function to be transformed, s is the complex variable, and F(s) is the Laplace transform of f(t). The Laplace transform is a powerful tool that can be used to solve differential equations.

Application of Unilateral Laplace Transform for Constant Coefficient Differential Equations with Non-zero Initial Condition

The Unilateral Laplace Transform can be used to solve constant coefficient differential equations with non-zero initial conditions. The general process involves the following steps:

- Take the Laplace transform of both sides of the differential equation.
- Solve for the Laplace transform of the dependent variable.
- Take the inverse Laplace transform to find the solution to the original differential equation.

For example, consider the following differential equation:

d^2y/dt^2 + 2dy/dt + 2y = 0, y(0) = 1, dy/dt(0) = 0

Taking the Laplace transform of both sides of the differential equation, we get:

s^2Y(s) - s + 2sY(s) + 2Y(s) = Y(0) + sdy/dt(0) + 2y(0)

Simplifying this equation, we get:

Y(s) = (s+1)/(s^2 + 2s + 2)

Taking the inverse Laplace transform, we get:

y(t) = e^-t(cos(t) - sin(t))

Thus, we have solved the differential equation with non-zero initial conditions using the Unilateral Laplace Transform.

Conclusion

In conclusion, the Unilateral Laplace Transform is a powerful tool that can be used to solve constant coefficient differential equations with non-zero initial conditions. By taking the Laplace transform of the differential equation, solving for the Laplace transform of the dependent variable, and taking the inverse Laplace transform, we can find the solution to the original differential equation.

Understanding the Problem
The given periodic signal is defined as y[n] = x[n] - x[n + N/2], where x[n] is a periodic signal with period N. Here, N is even, indicating that the signal repeats every N samples.
Fourier Series Coefficients
The Fourier Series (FS) coefficients X[k] represent the frequency components of the periodic signal x[n]. To find the FS coefficients Y[k] of the modified signal y[n], we need to analyze the impact of the subtraction operation in y[n]:
- The term x[n + N/2] represents the signal shifted by half its period.
- Since N is even, x[n + N/2] will have a specific relationship with x[n].
Calculating the Coefficients
Using the properties of Fourier series:
- The FS coefficients of x[n + N/2] can be derived from X[k], specifically:
- X[k + N/2] if k is shifted by N/2.
Thus, for y[n]:
Y[k] = X[k] - X[k + N/2]
Now, we evaluate how this affects the coefficients:
- Since X[k + N/2] is the FS representation of x[n + N/2], and because x[n] is periodic, we can deduce that this will produce a specific pattern in the coefficients.
Identifying the Correct Option
After analysis, we can conclude:
- When analyzing the subtraction, we see that:
- The coefficients will cancel out in certain frequencies depending on the periodic nature.
- The correct expression that represents this relationship in the question is:
Y[k] = (1 - (-1)^k)X[k]
This indicates that Y[k] takes into account the effect of the periodic nature of x[n] and its shifted version. Each term contributes to Y[k] based on the evenness of the index k.
Final Conclusion
Thus, the correct answer is option 'B':
Y[k] = (1 - (-1)^k)X[k].
This option accurately reflects the properties of the periodic signal and its Fourier Series representation.