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The bilateral Laplace transform of u (- t + 5) is
- a)-e-5s /s, Re(s) < />
- b)(1 - e-5s)/ s, Re(s) < />
- c)1- (-e-5s / s), Re(s) > 0
- d)-e-5s / s, Re(s) > 0
Correct answer is option 'A'. Can you explain this answer?
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The bilateral Laplace transform of u (- t + 5) isa)-e-5s /s, Re(s) b)...
Bilateral Laplace Transform of a Function
The bilateral Laplace transform of a function f(t) is defined as:
F(s) = ∫[ -∞ to +∞ ] f(t) e^(-st) dt
where F(s) is the Laplace transform of f(t) and s is the complex variable.
Given Function
In this question, we are given the function f(t) = u(-t + 5), where u(t) is the unit step function defined as u(t) = 1 for t ≥ 0 and u(t) = 0 for t < />
Bilateral Laplace Transform of the Given Function
To find the bilateral Laplace transform of f(t), we substitute the given function in the definition of the bilateral Laplace transform:
F(s) = ∫[ -∞ to +∞ ] u(-t + 5) e^(-st) dt
Since the unit step function u(-t + 5) is equal to 1 when -t + 5 ≥ 0, we can rewrite the integral limits as t ≥ 5.
F(s) = ∫[ 5 to +∞ ] e^(-st) dt
Solving the Integral
To solve the integral, we can use the property of the Laplace transform:
∫[ a to b ] e^(-st) dt = [ -e^(-st) / s ] evaluated from a to b
Applying this property to our integral, we have:
F(s) = [ -e^(-st) / s ] evaluated from 5 to +∞
F(s) = [ -e^(-s∞) / s ] - [ -e^(-s5) / s ]
Since e^(-∞) approaches 0 and e^(-s5) is a constant, the first term in the expression becomes 0. We are left with:
F(s) = [ -e^(-s5) / s ]
Final Answer
The bilateral Laplace transform of f(t) = u(-t + 5) is given by F(s) = [ -e^(-s5) / s ]. This matches with option 'A' which states -e^(-5s) / s, Re(s) > 0.
The bilateral Laplace transform of a function f(t) is defined as:
F(s) = ∫[ -∞ to +∞ ] f(t) e^(-st) dt
where F(s) is the Laplace transform of f(t) and s is the complex variable.
Given Function
In this question, we are given the function f(t) = u(-t + 5), where u(t) is the unit step function defined as u(t) = 1 for t ≥ 0 and u(t) = 0 for t < />
Bilateral Laplace Transform of the Given Function
To find the bilateral Laplace transform of f(t), we substitute the given function in the definition of the bilateral Laplace transform:
F(s) = ∫[ -∞ to +∞ ] u(-t + 5) e^(-st) dt
Since the unit step function u(-t + 5) is equal to 1 when -t + 5 ≥ 0, we can rewrite the integral limits as t ≥ 5.
F(s) = ∫[ 5 to +∞ ] e^(-st) dt
Solving the Integral
To solve the integral, we can use the property of the Laplace transform:
∫[ a to b ] e^(-st) dt = [ -e^(-st) / s ] evaluated from a to b
Applying this property to our integral, we have:
F(s) = [ -e^(-st) / s ] evaluated from 5 to +∞
F(s) = [ -e^(-s∞) / s ] - [ -e^(-s5) / s ]
Since e^(-∞) approaches 0 and e^(-s5) is a constant, the first term in the expression becomes 0. We are left with:
F(s) = [ -e^(-s5) / s ]
Final Answer
The bilateral Laplace transform of f(t) = u(-t + 5) is given by F(s) = [ -e^(-s5) / s ]. This matches with option 'A' which states -e^(-5s) / s, Re(s) > 0.
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Community Answer
The bilateral Laplace transform of u (- t + 5) isa)-e-5s /s, Re(s) b)...
X(s) = -u (- t + 5) e-st dt
= -5e-st dt
= -e-5s / s, Re(s) < />
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The bilateral Laplace transform of u (- t + 5) isa)-e-5s /s, Re(s) b)(1 - e-5s)/ s, Re(s) c)1- (-e-5s / s), Re(s) > 0d)-e-5s / s, Re(s) > 0Correct answer is option 'A'. Can you explain this answer?
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