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<math> \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) </math> | <math> \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) </math> | ||
| + | |||
| + | But we had taken the derivative of z(t) to get g(t) (and hence <math> g_k </math>). | ||
| + | <math> \therefore z_k = \left ( \frac{g_k}{jk\omega_o} \right ) </math> | ||
| + | |||
| + | <math> z_k = \left( \frac { \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) * (1 - (-1)^k) }{jk\pi/2} \right) </math> | ||
| + | |||
| + | <math> z_k = \frac {2}{j} \left( \frac {1}{(k\pi)^2} \sin ( \frac {k\pi}{2} ) \right) * (1 - (-1)^k) ~~\forall ~k ~\ne ~0 </math> | ||
| + | |||
| + | <math> g_o = \frac {2t_{1m}}{T_m} + \frac {2t_{1n}}{T_n} </math> | ||
| + | |||
| + | <math> \therefore g_o = 0.5 - 0.5 ~~~and \therefore z_o = 0 </math> | ||
Latest revision as of 11:45, 1 July 2008
Let $ g(t) = \left ( \frac{dz}{dt} \right ) $
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 { \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) * (1 - (-1)^k) }{jk\pi/2} \right) $
$ z_k = \frac {2}{j} \left( \frac {1}{(k\pi)^2} \sin ( \frac {k\pi}{2} ) \right) * (1 - (-1)^k) ~~\forall ~k ~\ne ~0 $
$ g_o = \frac {2t_{1m}}{T_m} + \frac {2t_{1n}}{T_n} $
$ \therefore g_o = 0.5 - 0.5 ~~~and \therefore z_o = 0 $
