g(t)={0, 0<t<11, 1<t<36, 3<t<54, 5<t g(t)={1,3}(t)+6prod_{3,5}(t)+4u(t-5) compute the Laplace transform g(t)

Question

$g (t) = {\begin{cases} 0 & 0 < t < 11 & 1 < t < 36 & 3 < t < 54 & 5 < t \end{cases}$

$g (t) = \prod_{1, 3} (t) + 6 \prod_{3, 5} (t) + 4 u (t - 5)$
compute the Laplace transform of g(t)

Answer 1

Step 1
According to the given information it is required to compute Laplace transform of g(t), where given that g(t) is expressed as,

g (t) = \prod_{1, 3} (t) + 6 \prod_{3, 5} (t) + 4 u (t - 5)

Also from window , for

f (t) = \prod_{a, b} (t)

L (f) (s) = \frac{e^{- a s} - e^{- b s}}{s}, s > 0

and for f (t) = u (t - a)

L (f (s)) = \frac{e^{- a s}}{s}, s > 0

Step 2
Laplace transform of g(t) can be expressed as:

g (t) = L (g) (t)

= L [g_{1} (t) + 6 g_{2} (t) + 3 g_{3} (t)]

= L (g_{1}) (t) + 6 L (g_{2}) (t) + 4 L (g_{3}) (t)

since given L [f + g] = L {f} + L {g}, and L {c f} = c L {f}

Where, g_{1} (t) = \prod_{1, 3} (t)

g_{2} (t) = \prod_{3, 5} (t)

And g_{3} (t) = u (t - 5)

Step 3
Now, find

L (g_{1}) (t), L (g_{2}) (t) and L (g_{3}) (t)

As,

L (g_{1}) (t) = \frac{e^{- 1 t} - e^{- 3 t}}{t}, t > 0

(comparing by formula mentioned in previous step a=1 , b=3)

L (g_{2}) (t) = \frac{e^{- 3 t} - e^{- 5 t}}{t}, t > 0

(similarly here a=3 , b=5)

L (g_{3}) (t) = \frac{e^{- 5 t}}{t}, t > 0

(And here a=5)
Step 4
Substitute above values in formula for Laplace of g(t).

g (t) = L {(g) (t)}

= L (g_{1}) (t) + 6 L (g_{2}) (t) + 4 L (g_{3}) (t)

= \frac{e^{- t} - e^{- 3 t}}{t} + 6 (\frac{e^{- 3 t} - e^{- 5 t}}{t}) + 4 (\frac{e^{- 5 t}}{t}), t > 0

= \frac{e^{- t} - e^{- 3 t}}{t} + \frac{6 e^{- 3 t} - 6 e^{- 5 t}}{t} + \frac{4 e^{- 5 t}}{t}, t > 0

= \frac{e^{- t} - e^{- 3 t} + 6 e^{- 3 t} - 6 e^{- 5 t} + 4 e^{- 5 t}}{t}

= \frac{e^{- t} + 5 e^{- 3 t} - 2 e^{- 5 t}}{t}, t > 0

Thus , Laplace transform of g(t) is

= \frac{e^{- t} + 5 e^{- 3 t} - 2 e^{- 5 t}}{t}, t > 0

Expert Answer