Math Problem Statement

Using Convolution theorem solve: 𝐿 βˆ’1 ( 𝑠 (𝑠 2+1)(𝑠 2+16) )

Solution

To solve the given problem using the Convolution theorem, we will first need to find the inverse Laplace transforms of the individual factors and then convolve the resulting time-domain functions.

Given:

Lβˆ’1(s(s2+1)(s2+16))L^{-1} \left( \frac{s}{(s^2+1)(s^2+16)} \right)

Let's denote:

F(s)=s(s2+1)(s2+16)F(s) = \frac{s}{(s^2+1)(s^2+16)}

Step 1: Partial Fraction Decomposition

We need to decompose F(s)F(s) into simpler fractions:

s(s2+1)(s2+16)=As+Bs2+1+Cs+Ds2+16\frac{s}{(s^2+1)(s^2+16)} = \frac{A s + B}{s^2+1} + \frac{C s + D}{s^2+16}

Multiplying both sides by (s2+1)(s2+16)(s^2+1)(s^2+16) to clear the denominators:

s=(As+B)(s2+16)+(Cs+D)(s2+1)s = (A s + B)(s^2 + 16) + (C s + D)(s^2 + 1)

Expanding and equating coefficients:

s=As3+16As+Bs2+16B+Cs3+Cs+Ds2+Ds = A s^3 + 16 A s + B s^2 + 16 B + C s^3 + C s + D s^2 + D

Grouping the terms:

s=(A+C)s3+(B+D)s2+(16A+C)s+(16B+D)s = (A+C)s^3 + (B+D)s^2 + (16A+C)s + (16B+D)

Now, we equate the coefficients of corresponding powers of ss:

For s3s^3: A+C=0A + C = 0

For s2s^2: B+D=0B + D = 0

For s1s^1: 16A+C=116A + C = 1

For s0s^0: 16B+D=016B + D = 0

From A+C=0A + C = 0, we have C=βˆ’AC = -A.

From B+D=0B + D = 0, we have D=βˆ’BD = -B.

Substitute C=βˆ’AC = -A into 16A+C=116A + C = 1:

16Aβˆ’A=116A - A = 1 15A=115A = 1 A=115A = \frac{1}{15}

Then C=βˆ’115C = -\frac{1}{15}.

Now substitute D=βˆ’BD = -B into 16B+D=016B + D = 0:

16Bβˆ’B=016B - B = 0 15B=015B = 0 B=0B = 0

Thus, D=0D = 0.

So the partial fraction decomposition is:

s(s2+1)(s2+16)=115ss2+1+βˆ’115ss2+16\frac{s}{(s^2+1)(s^2+16)} = \frac{\frac{1}{15}s}{s^2+1} + \frac{-\frac{1}{15}s}{s^2+16}

Step 2: Inverse Laplace Transform

Now, we find the inverse Laplace transform of each term separately.

  1. For 115ss2+1\frac{\frac{1}{15}s}{s^2+1}:

Lβˆ’1(115ss2+1)=115cos⁑(t)L^{-1}\left(\frac{\frac{1}{15}s}{s^2+1}\right) = \frac{1}{15} \cos(t)

  1. For βˆ’115ss2+16\frac{-\frac{1}{15}s}{s^2+16}:

Lβˆ’1(βˆ’115ss2+16)=βˆ’115cos⁑(4t)L^{-1}\left(\frac{-\frac{1}{15}s}{s^2+16}\right) = -\frac{1}{15} \cos(4t)

Step 3: Combining Results

The inverse Laplace transform of the original function is the sum of the individual inverse Laplace transforms:

Lβˆ’1(s(s2+1)(s2+16))=115cos⁑(t)βˆ’115cos⁑(4t)L^{-1} \left( \frac{s}{(s^2+1)(s^2+16)} \right) = \frac{1}{15} \cos(t) - \frac{1}{15} \cos(4t)

Thus, the solution is:

f(t)=115cos⁑(t)βˆ’115cos⁑(4t)f(t) = \frac{1}{15} \cos(t) - \frac{1}{15} \cos(4t)

Do you want more details or have any questions?

Here are 5 more questions you might ask:

  1. How do you perform partial fraction decomposition for different types of functions?
  2. Can you explain the steps to find the inverse Laplace transform in more detail?
  3. What are some applications of the convolution theorem in solving differential equations?
  4. How do you find the inverse Laplace transform of more complex functions?
  5. Can you explain the properties and uses of the Laplace transform in engineering?

Tip: When performing partial fraction decomposition, ensure all coefficients are correctly matched with their corresponding terms to avoid calculation errors.

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Math Problem Analysis

Mathematical Concepts

Laplace Transform
Inverse Laplace Transform
Partial Fraction Decomposition

Formulas

Laplace Transform of derivatives
Convolution Theorem

Theorems

Convolution Theorem

Suitable Grade Level

Advanced Undergraduate