Why Did They Add 2π When Shifting AC Circuits?

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SUMMARY

The discussion centers on the mathematical manipulation of the function v=40 cos(100πt + π/3) to determine the necessary leftward shift in time. The key insight is that adding 2π to the argument of the cosine function allows for a positive time shift, which is essential for finding the minimum time required. The initial approach led to a negative time shift, which was incorrect given the problem's constraints. The addition of 2π ensures that the solution remains within the periodic nature of the cosine function.

PREREQUISITES
  • Understanding of trigonometric functions, specifically cosine.
  • Familiarity with phase shifts in periodic functions.
  • Basic knowledge of angular frequency and its representation in equations.
  • Ability to manipulate algebraic expressions involving trigonometric identities.
NEXT STEPS
  • Study the concept of phase shifts in trigonometric functions.
  • Learn about the periodic properties of the cosine function and how they affect shifts.
  • Explore the implications of angular frequency in AC circuit analysis.
  • Review the use of radians in trigonometric equations and their significance in shifts.
USEFUL FOR

Students studying electrical engineering, particularly those focusing on AC circuit analysis, as well as anyone interested in the mathematical principles behind trigonometric functions and their applications in real-world scenarios.

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Hi everyone,
I ran into the following little problem:

Homework Statement


Given v=40 cos(100*pi*t + pi/3), the question is:
what is the minimum number of millisec that the function must be shifted to the left if the expression for v is 40 cos(100*pi*t)


Homework Equations





The Attempt at a Solution


ok, I started with:
40 cos(100*pi*(t+t0) + pi/3) = 40 cos(100*pi*t)
hence 100*pi*(t+t0) + pi/3 = 100*pi*t
the 100*pi*t cancels out on each side and we can solve for t0 but t0 < 0

BUT the book starts with:
40 cos(100*pi*(t+t0) + pi/3) = 40 cos(100*pi*t + 2pi)
and they solved for t0... and t0 is now > 0

My question why they added the 2pi! Although the more I think about it the more it makes sense to me but I'd like to be sure... Is it because we look for the min time but still positive?
If that's the case, what bothers me is: how are we supposed to know ahead of time to add this '2pi'?

Thanks for your answer(s) o:)
 
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I think the answer is that they said "shift the function to the left" implying a positive time shift was being looked for.
 
Dick said:
I think the answer is that they said "shift the function to the left" implying a positive time shift was being looked for.

Oh my god... you are right, I can't believe I didn't think of it this way... That totally makes sense !:biggrin:

Thx
 

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