Math Problem Statement

solve problem 2

Solution

To solve problem 2, where we are asked to find the impulse response of the transfer function, we proceed as follows:

Given Transfer Function:

Y(s)R(s)=4(s+2)s2+8s+20\frac{Y(s)}{R(s)} = \frac{4(s+2)}{s^2 + 8s + 20}

Impulse Response:

For an impulse input r(t)=δ(t)r(t) = \delta(t), the Laplace transform R(s)=1R(s) = 1. Therefore, the output in the Laplace domain is directly the transfer function:

Y(s)=4(s+2)s2+8s+20Y(s) = \frac{4(s+2)}{s^2 + 8s + 20}

Step 1: Factor the Denominator

First, factor the denominator s2+8s+20s^2 + 8s + 20:

s2+8s+20=(s+4)2+4s^2 + 8s + 20 = (s + 4)^2 + 4

So, the transfer function becomes:

Y(s)=4(s+2)(s+4)2+4Y(s) = \frac{4(s+2)}{(s+4)^2 + 4}

Step 2: Decompose the Transfer Function

The transfer function can be decomposed into simpler fractions. Notice that the denominator is in the form of a standard second-order system:

Y(s)=4(s+2)(s+4)2+4Y(s) = \frac{4(s+2)}{(s+4)^2 + 4}

We can write this in the form:

Y(s)=4s+42(s+4)2+4Y(s) = 4 \cdot \frac{s+4 - 2}{(s+4)^2 + 4}

This separates into:

Y(s)=4(s+4(s+4)2+42(s+4)2+4)Y(s) = 4 \cdot \left(\frac{s+4}{(s+4)^2 + 4} - \frac{2}{(s+4)^2 + 4}\right)

Step 3: Identify Standard Laplace Transforms

The expression can now be recognized as the sum of standard Laplace transforms:

  1. The first term s+4(s+4)2+4\frac{s+4}{(s+4)^2 + 4} corresponds to the Laplace transform of e4tcos(2t)e^{-4t} \cos(2t).
  2. The second term 2(s+4)2+4\frac{2}{(s+4)^2 + 4} corresponds to the Laplace transform of e4tsin(2t)e^{-4t} \sin(2t).

Step 4: Inverse Laplace Transform

Taking the inverse Laplace transform of each term:

  1. L1{s+4(s+4)2+4}=e4tcos(2t)\mathcal{L}^{-1} \left\{ \frac{s+4}{(s+4)^2 + 4} \right\} = e^{-4t} \cos(2t)
  2. L1{2(s+4)2+4}=e4tsin(2t)\mathcal{L}^{-1} \left\{ \frac{2}{(s+4)^2 + 4} \right\} = e^{-4t} \sin(2t)

So, the impulse response y(t)y(t) is:

y(t)=4[e4tcos(2t)e4tsin(2t)]y(t) = 4 \left[ e^{-4t} \cos(2t) - e^{-4t} \sin(2t) \right]

This simplifies to:

y(t)=4e4t[cos(2t)sin(2t)]y(t) = 4e^{-4t} \left[ \cos(2t) - \sin(2t) \right]

Final Answer:

The impulse response of the system is:

y(t)=4e4t[cos(2t)sin(2t)]y(t) = 4e^{-4t} \left[ \cos(2t) - \sin(2t) \right]

This represents the time-domain output of the system when subjected to a unit impulse input.


Would you like further explanation or details on how these steps were derived? Here are some related questions you might find interesting:

  1. What is the significance of the damping factor in a second-order system?
  2. How does the frequency of oscillation relate to the coefficients in the transfer function?
  3. How can you determine the stability of a system by examining its impulse response?
  4. How does the response change if the damping ratio is altered?
  5. What are the conditions under which the impulse response would be purely oscillatory?

Tip: Understanding the forms of standard Laplace transforms is crucial when working with transfer functions. It can simplify the process of finding time-domain responses significantly.

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Math Problem Analysis

Mathematical Concepts

Laplace Transform
Transfer Function
Impulse Response

Formulas

Laplace transform of sinusoidal functions

Theorems

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Suitable Grade Level

Advanced Undergraduate