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Mkey [24]
3 years ago
14

(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

se function (b) f(t) = e −t + 4e −4t + te−3t
Mathematics
1 answer:
Aloiza [94]3 years ago
3 0

Answer

(a) F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

Step-by-step explanation:

(a) f(t) = 20.5 + 10t + t^2 + δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 f(s)e^{-st} \, dt

where a = ∞

=>  F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt

where d(t) = δ(t)

=> F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt

Integrating, we have:

=> F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3}  )\left \{ {{a} \atop {0}} \right.

Inputting the boundary conditions t = a = ∞, t = 0:

F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) f(t) = e^{-t} + 4e^{-4t} + te^{-3t}

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt

F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt

F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt

Integrating, we have:

F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.

Inputting the boundary condition, t = a = ∞, t = 0:

F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

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<u>Fundamental Theorem of Calculus</u>

\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a <u>constant of integration</u>.

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\displaystyle \int x \ln x \:\: \text{d}x

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5 0
2 years ago
Given the function defined below, what is the value of the 4th term?
MakcuM [25]

Answer:

The 4 t h term is   f(4)   = 143

Step-by-step explanation:

<em>Explanation</em>:-

Given  function f(1) = -4

Given 'nth' term is  f(n) = -3f(n-1) +5

Put n =2         <em> f(2) = -3 f(2-1) +5</em>

                             = -3 f(1) +5

                             = -3 (-4) +5

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put n= 3      

                         f(n) = -3f(n-1) +5  

                      <em>   f(3) = -3 f(3-1) +5</em>

                            = -3f(2) +5

                           = -3(17) +5

                           = -51+5

                      f(3) = -46

Put n=4

                  f(n) = -3f(n-1) +5

               <em>  f(4)  = -3f(4-1) +5</em>

<em>                  f(4)  = -3f(3)+5</em>

                f(4)   = -3(-46)+5

                f(4)   = 138 +5

                f(4)   = 143

<u><em>Final answer</em></u>:-

<em>The 4 t h term is   f(4)   = 143</em>

         

6 0
3 years ago
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