Can someone help me solve this Laplace Transform exercise?It is given that:&nbsp;X(s)=s+2s2+4s+5&nbsp;and i need to calculate the L.T. of:&nbsp;y(t)=x(2t-1)u(2t-1)&nbsp;(without calculating the ILT)&nbsp;Please help, im stuck for days.&nbsp;Thank you!

We have X(s)=s+2s2+4s+5, and we want to find the Laplace transform of y(t)=x(2t−1)u(2t−1), where u(t) is the unit step function.First, we can rewrite y(t) as:y(t)=x(2t−1)u(2t−1)={0,t&lt;12x(2t−1),t≥12Now we can apply the time-shifting property of the Laplace transform, which states that:ℒ{f(t−a)u(t−a)}(s)=e−asℒ{f(t)}(s)Using this property with f(t)=x(t) and a=12, we get:ℒ{y(t)}(s)=e−s2ℒ{x(2t−1)}(s)=e−s2ℒ{x(t−12)}(s)=e−s2X(s)e−s2=s+2(s2+4s+5)(s+5)=As+5+Bs+Cs2+4s+5To solve for A, B, and C, we can use partial fractions. Multiplying both sides by the denominator and simplifying, we get:s+2=A(s2+4s+5)+(Bs+C)(s+5)Setting s=−5, we get A=−325. Setting s=−12, we get C=−125. Finally, setting s=0, we get B=11105.Therefore, we have:ℒ{y(t)}(s)=−325s+5+11105s−125s2+4s+5ℒ{y(t)}(s)=11s−25105(s2+4s+5)(s+5)

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High School
Calculus and Analysis

Answered question

2022-05-04

Can someone help me solve this Laplace Transform exercise?

It is given that:

$X (s) = \frac{s + 2}{s^{2} + 4 s + 5}$

and i need to calculate the L.T. of:

$y (t) = x (2 t - 1) u (2 t - 1)$

(without calculating the ILT)

Please help, im stuck for days.

Thank you!

Answer & Explanation

xleb123

Skilled2023-05-04Added 181 answers

We have

X (s) = \frac{s + 2}{s^{2} + 4 s + 5}

, and we want to find the Laplace transform of

y (t) = x (2 t - 1) u (2 t - 1)

, where

u (t)

is the unit step function.
First, we can rewrite

y (t)

as:

y (t) = x (2 t - 1) u (2 t - 1) = {\begin{matrix} 0, & t < \frac{1}{2} \\ x (2 t - 1), & t \geq \frac{1}{2} \end{matrix}

Now we can apply the time-shifting property of the Laplace transform, which states that:

ℒ {f (t - a) u (t - a)} (s) = e^{- a s} ℒ {f (t)} (s)

Using this property with

f (t) = x (t)

and

a = \frac{1}{2}

, we get:

ℒ {y (t)} (s) = e^{- \frac{s}{2}} ℒ {x (2 t - 1)} (s)

= e^{- \frac{s}{2}} ℒ {x (t - \frac{1}{2})} (s)

= e^{- \frac{s}{2}} X (s) e^{- \frac{s}{2}}

= \frac{s + 2}{(s^{2} + 4 s + 5) (s + \sqrt{5})}

= \frac{A}{s + \sqrt{5}} + \frac{B s + C}{s^{2} + 4 s + 5}

To solve for

A

B

, and

C

, we can use partial fractions. Multiplying both sides by the denominator and simplifying, we get:

s + 2 = A (s^{2} + 4 s + 5) + (B s + C) (s + \sqrt{5})

Setting

s = - \sqrt{5}

, we get

A = - \frac{3}{2 \sqrt{5}}

. Setting

s = - \frac{1}{2}

, we get

C = - \frac{1}{2 \sqrt{5}}

. Finally, setting

s = 0

, we get

B = \frac{11}{10 \sqrt{5}}

.
Therefore, we have:

ℒ {y (t)} (s) = \frac{- \frac{3}{2 \sqrt{5}}}{s + \sqrt{5}} + \frac{\frac{11}{10 \sqrt{5}} s - \frac{1}{2 \sqrt{5}}}{s^{2} + 4 s + 5}

ℒ {y (t)} (s) = \frac{11 s - 2 \sqrt{5}}{10 \sqrt{5} (s^{2} + 4 s + 5) (s + \sqrt{5})}

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