Laplace Transform Calculation for y(t)

The problem asks us to find the Laplace transform for the given time domain function: \(y(t) = -2te^{-t} + 4e^{-t} - 4e^{-2t}\). To solve this, we will use the linearity property of the Laplace transform and standard transformation formulas for exponential and time-multiplied exponential functions.

Laplace Transform Properties Overview

The following key Laplace transform properties and formulas are essential for solving this problem:

• Linearity Property: The Laplace transform is linear. If \(a\) and \(b\) are constants, and \(f(t)\) and \(g(t)\) are functions, then: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$
• Laplace Transform of an Exponential Function: $L\{e^{at}\} = \frac{1}{s - a}$
• Laplace Transform of \(t\) multiplied by an Exponential Function: $L\{te^{at}\} = \frac{1}{(s - a)^2}$

Term-by-Term Laplace Transformation

We will apply the Laplace transform to each term of the function \(y(t)\) separately.

Term 1: \(-2te^{-t}\)

• Here, we have the form \(te^{at}\) with \(a = -1\).
• Using the formula $L\{te^{at}\} = \frac{1}{(s - a)^2}$: $L\{te^{-t}\} = \frac{1}{(s - (-1))^2} = \frac{1}{(s + 1)^2}$
• Applying the linearity property for the constant \(-2\): $L\{-2te^{-t}\} = -2 L\{te^{-t}\} = -2 \times \frac{1}{(s + 1)^2} = \frac{-2}{(s + 1)^2}$

Term 2: \(4e^{-t}\)

• Here, we have the form \(e^{at}\) with \(a = -1\).
• Using the formula $L\{e^{at}\} = \frac{1}{s - a}$: $L\{e^{-t}\} = \frac{1}{(s - (-1))} = \frac{1}{(s + 1)}$
• Applying the linearity property for the constant \(4\): $L\{4e^{-t}\} = 4 L\{e^{-t}\} = 4 \times \frac{1}{(s + 1)} = \frac{4}{(s + 1)}$

Term 3: \(-4e^{-2t}\)

• Here, we have the form \(e^{at}\) with \(a = -2\).
• Using the formula $L\{e^{at}\} = \frac{1}{s - a}$: $L\{e^{-2t}\} = \frac{1}{(s - (-2))} = \frac{1}{(s + 2)}$
• Applying the linearity property for the constant \(-4\): $L\{-4e^{-2t}\} = -4 L\{e^{-2t}\} = -4 \times \frac{1}{(s + 2)} = \frac{-4}{(s + 2)}$

Combining Transformed Expressions

Now, we sum the Laplace transforms of the individual terms to get the total Laplace transform of \(y(t)\):

$L\{y(t)\} = L\{-2te^{-t}\} + L\{4e^{-t}\} + L\{-4e^{-2t}\}$ $L\{y(t)\} = \frac{-2}{(s + 1)^2} + \frac{4}{(s + 1)} - \frac{4}{(s + 2)}$

To combine these fractions into a single expression, we find a common denominator, which is \((s + 1)^2 (s + 2)\):

$L\{y(t)\} = \frac{-2(s + 2) + 4(s + 1)(s + 2) - 4(s + 1)^2}{(s + 1)^2 (s + 2)}$

Now, let's expand and simplify the numerator:

• For the first term: $-2(s + 2) = -2s - 4$
• For the second term: $4(s + 1)(s + 2) = 4(s^2 + 2s + s + 2) = 4(s^2 + 3s + 2) = 4s^2 + 12s + 8$
• For the third term: $-4(s + 1)^2 = -4(s^2 + 2s + 1) = -4s^2 - 8s - 4$

Summing these expanded terms for the numerator:

Numerator $= (-2s - 4) + (4s^2 + 12s + 8) + (-4s^2 - 8s - 4)$ Numerator $= (4s^2 - 4s^2) + (-2s + 12s - 8s) + (-4 + 8 - 4)$ Numerator $= 0s^2 + (10s - 8s) + (4 - 4)$ Numerator $= 2s + 0$ Numerator $= 2s$

Final Laplace Transform Result

Substituting the simplified numerator back into the expression:

$L\{y(t)\} = \frac{2s}{(s + 1)^2 (s + 2)}$

This result matches one of the given options, demonstrating the application of Laplace transform properties. Mastering these transformations is crucial for solving problems in signals and systems and control theory.