Question

Express the function below using window and step functions and compute its Laplace transform. Ag(t) 10- o+ 0 2 6 8 10 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. Express g(t) using window and step functions. Choose the correct answer below. O A. g(t) = u(t - 4) + (31 - 12)114,5(t)}-( - 36 + 18)|15,6 (t) + u(t-6) O B. g(t) = (3-12)114,5(t)+(-3t+18)15,6(t) O c. g(t) = (3-4)u(t - 4)+(-3t+6)u(t-6) OD. g(t) = (3-12)114,6(t)+(-3t+18)u(-5) Comnathalanine

Answer 1

The given function is 0 throughout except between 4 and 6.

Further, between 4 and 5, the function is the straight line joining the points (4,0) and (5,3), so it's equation is given by:

${g-0\over t-4}={3-0\over 5-4}\Rightarrow g=3(t-4)=3t-12$...................(1)

Also, between 5 and 6, the function is the straight line joining the points (5,3) and (6,0), so

${g-0\over t-6}={3-0\over 5-6}\Rightarrow g=-3(t-6)=-3t+18$.......................(2)

First, let us write down the given function as a piecewise function, using the expressions (1) and (2) for representing the function in the interval (4,5) and (5,6) respectively:-

$g(t)=\begin{cases} 0 & \text{ if } 0\leq t <4 \\ 3t-12 & \text{ if } 4\leq t <5 \\ -3t+18 & \text{ if } 5\leq t <6\\ 0 & \text{ if } 6\leq t \end{cases}$

Now, we can write down this function using a single line, using the rectangular window function, which is given by:

$\Pi_{a,b}(t)=\begin{cases} 1 & \text{ if } a<t<b \\ 0 & \text{ otherwise } \end{cases}$

So, we can write down g(t) as

g(t) = (3t – 12)114,5t) +(-3t+18)115.6(t)

which is the answer to the first part (option B)

Now, we take the Laplace transform of g(t), for this, first we convert the rectangular windows function to the Heaviside step function as-

$g(t)=(3t-12)(u(t-4)-u(t-5))+(-3t+18)(u(t-5)-u(t-6))$

Simplifying and opening the brackets

$\small g(t)=u(t-4)(3t-12)-u(t-5)(3t-12)+u(t-5)(-3t+18)-u(t-6)(-3t+18)$

Further simplification gives:

$\small g(t)=u(t-4)(3t-12)+u(t-5)(-6t+30)-u(t-6)(-3t+18)$

Now, we can take the Laplace transform of this:

$\small L\{g(t)\}=L\left \{u(t-4)(3t-12) \right \}+L\left \{u(t-5)(-6t+30) \right \}-L\left \{u(t-6)(-3t+18) \right \}$

which gives us

$\small =e^{-4s}L\left \{3(t+4)-12 \right \}+e^{-5s}L\left \{-6(t+5)+30 \right \}-e^{-6s}L\left \{-3(t+6)+18 \right \}$

opening the brackets

$\small =e^{-4s}L\left \{3t \right \}+e^{-5s}L\left \{-6t \right \}-e^{-6s}L\left \{-3t \right \}$

Taking out the constants

$\small =3e^{-4s}L\left \{t \right \}-6e^{-5s}L\left \{t \right \}+3e^{-6s}L\left \{t \right \}$

Now, this comes directly from a standard laplace transform formula as

$\small =3e^{-4s}{1\over s^2}-6e^{-5s}{1\over s^2}+3e^{-6s}{1\over s^2}$

$\small L\{g(t)\}={1\over s^2}\left (3e^{-4s}-6e^{-5s}+3e^{-6s} \right )$

which is our answer