Revision as of 11:25, 1 July 2008 by Rvalluri (Talk)

7b OldKiwi.jpg

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

7b1 OldKiwi.jpg 7b2 OldKiwi.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 ) $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva