Revision as of 11:29, 1 July 2008 by Rvalluri (Talk)

7b Old Kiwi.jpg

Let $ g(t) = \left ( \frac{dz}{dt} \right ) $

7b1 Old Kiwi.jpg 7b2 Old Kiwi.jpg

Therefore, $ m_k = \left ( \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) \right) , n_k = \left( \frac {-1}{k\pi} \sin ( \frac {k\pi}{2} ) e^\frac{-j2k\pi2}{4} \right) $

But $ g_k = m_k + n_k = \left ( \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) \right) + \left( \frac {-1}{k\pi} \sin ( \frac {k\pi}{2} ) e^\frac{-j2k\pi2}{4} \right) $

$ \therefore g_k = \left ( \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) \right) + \left( \frac {-1}{k\pi} \sin ( \frac {k\pi}{2} ) (-1)^k \right) $

But we had taken the derivative of z(t) to get g(t) (and hence $ g_k $). $ \therefore z_k = \left ( \frac{g_k}{jk\omega_o} \right ) $

$ z_k = \left( \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) * (1 - (-1)^k) \right) $

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett