Question

A system has an output y(t) related to the input x(1) by the differential equation given below. What will be the output from the system (in 's' domain) when it is subject to input (20 sin 20t)? Initially both the output and input are zero. d2y/dt2 + 5 dy/dt + 6y = x

Answer 1

The output y(t) in the time domain when the input is x(t) = 20sin(20t) is:

y(t) = -20e^(2t) + 20e^(3t)

To find the output Y(s) in the Laplace domain (s-domain) when the system is subject to the input X(s) = 20sin(20t), we will apply the Laplace transform to the given differential equation and solve for Y(s).

The given differential equation is:

`d^2y/dt^2 + 5dy/dt + 6y = x`

Where:

- y(t) is the output.

- x(t) = 20sin(20t) is the input.

Step 1: Take the Laplace Transform of the Differential Equation:

Apply the Laplace transform to both sides of the differential equation. We'll use the following Laplace transform pairs:

- Laplace
`{d^2y/dt^2} = s^2Y(s) - sy(0) - y'(0)`

- Laplace
`{dy/dt} = sY(s) - y(0)`

The Laplace transform of the given equation becomes:

`s^2Y(s) - sy(0) - y'(0) + 5(sY(s) - y(0)) + 6Y(s) = 20sin(20t)`

Step 2: Initial Conditions:

You mentioned that initially both the output and input are zero. This means that y(0) = 0 and y'(0) = 0.

Step 3: Solve for Y(s):

Substitute the initial conditions into the equation:

`s^2Y(s) - 5sY(s) + 6Y(s) = 20sin(20t)`

Factor Y(s) out:

`Y(s) * (s^2 - 5s + 6) = 20sin(20t)`

Now, solve for Y(s):

`Y(s) = (20sin(20t)) / (s^2 - 5s + 6)`

Step 4: Partial Fraction Decomposition:

To find the inverse Laplace transform, we'll need to decompose the expression on the right-hand side into partial fractions. Factoring the denominator:

`s^2 - 5s + 6 = (s - 2)(s - 3)`

Now, we can express Y(s) as a sum of partial fractions:

Y(s) = A/(s - 2) + B/(s - 3)

Step 5: Find A and B:

To find the values of A and B, we'll clear the denominators and equate coefficients:

Y(s) = (A(s - 3) + B(s - 2)) / ((s - 2)(s - 3))

Now, equate numerators:

20sin(20t) = A(s - 3) + B(s - 2)

Step 6: Solve for A and B:

Let's first find A:

When s = 3, we get:

20sin(20t) = A(3 - 3) + B(3 - 2)

20sin(20t) = B

Now, let's find B:

When s = 2, we get:

20sin(20t) = A(2 - 3) + B(2 - 2)

20sin(20t) = -A

So, A = -20 and B = 20.

Step 7: Express Y(s) with A and B:

Now that we have A and B, we can express Y(s):

Y(s) = (-20/(s - 2)) + (20/(s - 3))

Step 8: Take the Inverse Laplace Transform

Now, we take the inverse Laplace transform of Y(s) to find the output y(t):

`y(t) = -20e^(2t) + 20e^(3t)`