Laplace transform of f(t-a)u(t-a), the shifted unit step function 

Sue Kaur you're welcome!

NIce pun

@jesslyn baolin f**

You’re welcome!

V is a dummy variable, hence the result of the definite integral is the same in both cases and you don't need to change back

Yup!

too much clout

Depends on what you mean by elementary functions. If you include the sign function, then the answer is yes. u(t) = 1/2 + sgn(t)/2

The graph of e^(-st)f(t-a) won’t be affected at x>a, where a could be any real number.

@@williamadams137 I mean whats the laplace transform of u(t+1)? is it 1/s ? and is L{(t+1)u(t+1)} = L{t+1} = 1/s^2 + 1/s. since u(t+1) = 1 for all t>0? (where L is the laplace transform)

You're welcome!

: )

you're welcome!

Samuel Andrade Unit Step Function

Heaviside functions, just look it up in your text book

thank you man

Jossy Beth definition of the step function?

thanks!

Part 1: Given: L{u(t^2 - 5*t + 6) * u(t)} Find the roots of t^2 - 5*t + 6. Where it is positive, u(this) will equal 1. Where it is negative, u(this) will equal zero. t^2 - 5*x + 6 = (t - 3)*(t - 2) Thus its roots are t=2 and t=3. Because this expression of t is positive for the t^2 coefficient, this means it is concave-up, and is thus positive when x3, and negative in between these values. Thus, our expression initially jumps to 1 at t=0 for the remaining u(t) function. It then drops down to 0 at t=2, and then jumps back up to 1 at t=3. Construct a linear combination of the unit step function to reflect this: u(t) - u(t - 2) + u(t - 3) Take the Laplace transform of each term, to find our result: 1/s - e^(-2*s)/s + e^(-3*s)/s Part 2: Given: L{u(t^3 - t^2 - 5*t + 5)*u(t - 0.5)*u(t)} Find roots of the cubic. It has a rational root at t=1. Use polynomial division to find quadratic factor: (t - 1)*(t^2 - 5) Thus, its roots are at t=-sqrt(5), t=1, and t=sqrt(5). We don't care about the negative root, since the remaining u(t) is off at that point. Since the cubic is increasing after its final root, this means it starts positive at t=0, becomes negative at t=1, and switches back to positive at t=sqrt(5). Multiplying with u(0.5) this means the function doesn't start until t=0.5. So it starts equaling 1 at t=0.5, switches to equaling zero at t=1, and returns to equaling 1 at x=sqrt(5). Construct a linear combination of unit steps to match this behavior. u(t - 0.5) - u(t - 1) + u(t - sqrt(5)) Take Laplace transform of each term for our result: e^(-0.5*s)/s - e^(-s)/s + e^(-sqrt(5)*s)/s

@Osman Navdar midterm2=31