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Fourier of Half-Wave Rectified

  • Thread starter izen
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1. Homework Statement

half_wave.jpg


How did [itex]\frac{1}{2}x[/itex] come from at k=1?

2. Homework Equations


3. The Attempt at a Solution

because k=1 will make the first term at denominator 2(k-1) = [itex]\frac{0}{0}[/itex]
 

SammyS

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1. Homework Statement

half_wave.jpg


How did [itex]\frac{1}{2}x[/itex] come from at k=1?

2. Homework Equations

3. The Attempt at a Solution

because k=1 will make the first term at denominator 2(k-1) = [itex]\frac{0}{0}[/itex]
Yes. The [itex]\ \frac{1}{2}x\ [/itex] comes from the fact that the first term has the form 0/0 as k → 1.

What is [itex]\displaystyle \ \lim_{t\to0}\frac{\sin(t)}{t}\ ?[/itex]
 
51
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Yes. The [itex]\ \frac{1}{2}x\ [/itex] comes from the fact that the first term has the form 0/0 as k → 1.

What is [itex]\displaystyle \ \lim_{t\to0}\frac{\sin(t)}{t}\ ?[/itex]
Use L' Hopital's Rule --> [itex]\displaystyle \ \lim_{t\to0}cos(t) = 1 [/itex]

thank you SammyS
 

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