Use the definition of Laplace Transforms to find Lleft{f(t)right}f(t)=begin{cases}-1 & 0leq t <11 & tgeq 1end{cases}Then rewrite f(t) as a sum of step functions, u_c(t), and show that by taking Laplace transforms, this yields the same answer as your direct computation.

Braxton Pugh 2021-02-12 Answered

Use the definition of Laplace Transforms to find L{f(t)}
f(t)={10t<11t1
Then rewrite f(t) as a sum of step functions, uc(t), and show that by taking Laplace transforms, this yields the same answer as your direct computation.

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Expert Answer

Sadie Eaton
Answered 2021-02-13 Author has 104 answers

Step 1
Consider the given function:
f(t)={10t<11t1
Step 2
Now, by definition of Laplace Transform:
L{f(t)}=0estf(t)dt
=01estf(t)dt+1estf(t)dt
=01est(1)dt+1est(1)dt
=[ests]01+[ests]1
=(ess1s)+(0+ess)
L{f(t)}=2ess1s
Step 3
Now, Rewrite f(t) as the sum of step functions, uc(t):
f(t)=f(t)={10t<11t1
=(1)(u0,1(t))+(1)(u1(t))
=(1)(u0(t)u1(t))+(1)(u1(t))
=u0(t)+2u1(t)
Step 4 Now, use the following formula for Laplace transform of step function:
L{uc(t)}=ecss
Step 5
Thus, now take Laplace transform:
f(t)=u0(t)+2u1(t)
L{f(t)}=L{u0(t)}+L{2u1(t)}
=e(0)ss+2e(1)ss
L{f(t)}=1s+2ess
Step 6
Thus, from equation (1) and (2), it can be seen that the Laplace transform of f(t) that is obtained from both the methods is same that is :
L{f(t)}=2ess1s

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