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Let x1(t) = e-tu(t) and x2(t) = u(t) - u(t - 2), where u (.) denotes the unit step function.
If y(t) denotes the convolution of x1(t) and x2(t) , then = ________. (rounded off to one decimal place)
    Correct answer is '0'. Can you explain this answer?
    Most Upvoted Answer
    Let x1(t) = e-tu(t) and x2(t) = u(t) - u(t - 2), where u (.) denotes ...
    Given
    Signal x1(t) = e-tu(t)
    x2(t) = u(t) - u(t - 2)
    Method 1
    y(t ) = x1(t) Ⓧ x2(t )
    Taking Laplace transform on both sides,
    Y (s) = X1(s)X2(s); Re(s) > 0
    Applying final value theorem,
    Method 2
    x1(t) = e-tu(t)
    x2(t) = u(t) - u(t - 2)
    y(t ) = x1(t) Ⓧ x2(t )
    Hence, the correct answer is 0.
    Free Test
    Community Answer
    Let x1(t) = e-tu(t) and x2(t) = u(t) - u(t - 2), where u (.) denotes ...
    Given
    Signal x1(t) = e-tu(t)
    x2(t) = u(t) - u(t - 2)
    Method 1
    y(t ) = x1(t) Ⓧ x2(t )
    Taking Laplace transform on both sides,
    Y (s) = X1(s)X2(s); Re(s) > 0
    Applying final value theorem,
    Method 2
    x1(t) = e-tu(t)
    x2(t) = u(t) - u(t - 2)
    y(t ) = x1(t) Ⓧ x2(t )
    Hence, the correct answer is 0.
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    Let x1(t) = e-tu(t) and x2(t) = u(t) - u(t - 2), where u (.) denotes the unit step function.If y(t) denotes the convolution of x1(t) and x2(t) , then = ________. (rounded off to one decimal place)Correct answer is '0'. Can you explain this answer?
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