The Laplace transform of u(t−2) is (a) 1s+2 (b) 1s−2 (c) e2ss(d)e−2ss

Name: The Laplace transform of u(t−2) is (a) 1s+2 (b) 1s−2 (c) e2ss(d)e−2ss
Uploaded: 2022-02-23T17:22:57+00:00
Description: The Laplace transform of u(t−2) is (a) 1s+2 (b) 1s−2 (c) e2ss(d)e−2ss

Question

The Laplace transform of u(t−2) is   (a) 1s+2  (b) 1s−2  (c) e2ss(d)e−2ss

Bentley Leach · Accepted Answer

We know that:  L(u(t))=1s  By t shifting theorem  L(u(t−2))=e−2sL(u(t))=e−2ss  Answer would be  (d)e−2ss

Andre BalkonE · Accepted Answer

To solve the problem, we&#039;ll start by recalling the definition of the unit step function. The unit step function, denoted as u(t), is defined as follows:u(t)={0if&amp;nbsp;t&amp;lt;01if&amp;nbsp;t≥0Now, we are given the Laplace transform of u(t−2) and we need to determine its value. We&#039;ll denote the Laplace transform operator as ℒ. The Laplace transform of a function f(t) is given by:ℒ{f(t)}=F(s)=∫0∞f(t)e−stdtApplying the Laplace transform to u(t−2), we have:ℒ{u(t−2)}=U(s)=∫0∞u(t−2)e−stdtSince u(t−2) is a shifted version of the unit step function, it will be zero for t&amp;lt;2 and one for t≥2. Therefore, we can rewrite the integral limits as follows:U(s)=∫2∞e−stdtTo evaluate this integral, we&#039;ll perform a change of variables. Let u=st, which implies du=sdt. Adjusting the integral limits accordingly, we get:U(s)=1s∫2s∞e−uduNow, we can solve this integral:U(s)=1s[−e−u]2s∞Substituting the limits:U(s)=1s(−e−∞+e−2s)Since e−∞ approaches zero, we can simplify further:U(s)=1s·e−2sTherefore, the Laplace transform of u(t−2) is:ℒ{u(t−2)}=e−2ssSo, the correct option is (d) e−2ss.

Nick Camelot · Accepted Answer

The Laplace transform of u(t−2) can be solved as follows:Using the definition of the unit step function, u(t)=1 for t≥0 and u(t)=0 for t&amp;lt;0.Therefore, u(t−2)=1 for t≥2 and u(t−2)=0 for t&amp;lt;2.The Laplace transform of u(t−2) can be computed as:ℒ{u(t−2)}=∫0∞u(t−2)e−stdtSince u(t−2)=1 for t≥2, we can rewrite the integral as:ℒ{u(t−2)}=∫2∞e−stdtIntegrating the expression, we get:ℒ{u(t−2)}=[−e−sts]2∞Substituting the limits of integration, we have:ℒ{u(t−2)}=(−e−s∞s)−(−e−2ss)Since e−s∞=0 (as e−∞=0), the expression simplifies to:ℒ{u(t−2)}=−(−e−2ss)Therefore, the Laplace transform of u(t−2) is e−2ss.Hence, the correct option is (d)&amp;nbsp;e−2ss.

The Laplace transform of u(t - 2) is a) 1/s + 2 b) 1/s - 2

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