The discussion centers around the proper shifting of the Heaviside step function in the context of the function f(t) defined as f(t) = { 0, 0 <= t < pi; -sin(3t), t >= pi }. The user initially expresses confusion about manipulating the function to achieve the desired form of f(t) = upi(t)sin[3(t-pi)]. The key insight provided is the transformation of the expression 3(t - pi + pi) into 3(t - pi) + 3pi, clarifying the handling of the Heaviside function and the sine term.
PREREQUISITESStudents studying calculus, particularly those focusing on piecewise functions and the Heaviside step function, as well as anyone interested in signal processing and function manipulation techniques.
What happened to the minus sign?missavvy said:Homework Statement
Hey so this is a really trivial question.
f(t) = { 0, 0<= t < pi
{ -sin3t, t>=pi
Just having trouble properly shifting this!
The Attempt at a Solution
f(t) = -upi(t)sin(3t)
= -upi(t)sin[3(t-pi+pi)]
= upi(t)sin[3(t-pi+pi)]
Hint: 3(t-pi+pi)=3(t-pi)+3piHow do I take the +pi out?
I know it should end up as:
upi(t)sin[3(t-pi)]
I just don't really understand how from my last step.
thanks for reading!