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To perform the partial fraction expansion on the given transfer function \( Y(s) \), let's break it down.

The transfer function is:
\[
Y(s) = \frac{4(s + 50)}{s^2 + 30s + 200}
\]

### Step 1: Factor the denominator
First, we need to factor the denominator \( s^2 + 30s + 200 \). The roots of this quadratic equation are given by:
\[
s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \( a = 1 \), \( b = 30 \), and \( c = 200 \).

Let's calculate these roots.

The roots of the quadratic equation \( s^2 + 30s + 200 = 0 \) are \( s = -10 \) and \( s = -20 \). Therefore, the denominator can be factored as:
\[
s^2 + 30s + 200 = (s + 10)(s + 20)
\]

### Step 2: Express the function as a sum of partial fractions
Now, the partial fraction expansion of \( Y(s) \) will have the form:
\[
\frac{4(s + 50)}{(s + 10)(s + 20)} = \frac{A}{s + 10} + \frac{B}{s + 20}
\]

### Step 3: Solve for coefficients \( A \) and \( B \)
Multiplying both sides by \( (s + 10)(s + 20) \) and equating coefficients, we get:
\[
4(s + 50) = A(s + 20) + B(s + 10)
\]

Let's solve for \( A \) and \( B \).It seems there was an issue while solving for the coefficients \( A \) and \( B \). I'll correct the approach and solve it step by step to find these coefficients.

The equation we have is:
\[
4(s + 50) = A(s + 20) + B(s + 10)
\]

To determine \( A \) and \( B \), we can substitute specific values for \( s \) to simplify the calculations.

- Let \( s = -10 \) to eliminate \( B \).
- Let \( s = -20 \) to eliminate \( A \).

Let's calculate these values now.The coefficients for the partial fraction expansion are:

\[
A = 16 \quad \text{and} \quad B = -12
\]

### Final Partial Fraction Expansion
The partial fraction expansion of the given function is:
\[
\frac{4(s + 50)}{(s + 10)(s + 20)} = \frac{16}{s + 10} - \frac{12}{s + 20}
\]

Would you like a more detailed explanation, or do you have any questions?

### Follow-up Questions:
1. How would you convert the partial fractions back to the original form?
2. What is the significance of partial fraction expansion in control systems?
3. Can you solve the time-domain response from this Laplace transform?
4. What happens if the denominator cannot be factored with real roots?
5. How does the choice of roots affect the coefficients?

### Tip:
Partial fraction expansion is a powerful tool in inverse Laplace transforms, often used in solving differential equations in engineering.

Explore the detailed step-by-step solution for performing partial fraction expansion on the transfer function Y(s) = 4(s + 50) / (s^2 + 30s + 200). Learn how to factor quadratic equations and solve for coefficients A and B. Suitable for college-level students and engineering enthusiasts.

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