Question

f(t) 2 -1 -2 -1 0 1 2 3 4 5 6 t 1) Write algebraic expression for time dependent function \(f(t)\) shown in the plot. 2) Find the Laplace transform \(F(s)\) of time dependent function \(f(t)\). Show all your work, do not skip steps.

          f(t)
2
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-1
0
1
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t
1) Write algebraic expression for time dependent function \(f(t)\) shown in the plot.
2) Find the Laplace transform \(F(s)\) of time dependent function \(f(t)\).
Show all your work, do not skip steps.
        
Show more…
f(t)
2
-1
-2
-1
0
1
2
3
4
5
6
t
1) Write algebraic expression for time dependent function f(t) shown in the plot.
2) Find the Laplace transform F(s) of time dependent function f(t).
Show all your work, do not skip steps.

Added by Anna F.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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3 4 1) Write algebraic expression for time dependent function f(t) shown in the plot. 2 2) Find the Laplace transform F(s) of time dependent function f(t). 1- 21 Show all your work, do not skip steps (1)f
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Transcript

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00:01 In this problem we are given the piecewise function f f of t is equal to 1 when 0 less than or equal to t less than 2 and e raised to negative t minus 2 when t is greater than or equal to 2.
00:23 The first question is to graph this piecewise function f f of t for all t greater than or equal to.
00:32 To 0.
00:33 Now this function takes the value 1 in the interval 0 2 and after that it takes the value e raise to negative of t minus 2 when t is greater than or equal to 2.
00:47 So graph this function this is the graph of the function f of t for all t greater than or equal to 0.
00:56 The next question is to use the unit step function h of t to write f of t as a single function for all t greater than or equal to zero.
01:09 Now the unit step function h of t is defined as 1 when t is greater than or equal to 0 and 0 when t is less than 0.
01:22 Also, h of t minus a is defined as 1 when t is greater than or equal to a and 0 when t is less than a.
01:37 Using the unit step function h of t minus 2, we can write f of t as f of t is equal to 1 plus h of t minus 2 times e raise to negative 1 .000.
01:53 Of t minus 2 minus 1.
01:57 Now see that when t is between 0 and 2 we have h of t minus 2 is 0 therefore f of t is equal to 1 in that interval and when t is greater than or equal to 2, h of t minus 2 is equal to 1...
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