(→Problem 7 Part b) |
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==Problem 7 Part b== | ==Problem 7 Part b== | ||
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Let <math> g(t) = \left ( \frac{dz}{dt} \right ) </math> | Let <math> g(t) = \left ( \frac{dz}{dt} \right ) </math> | ||
[[Image:ECE301Summer2008_San_Exam_7b1.jpg|400px|]] | [[Image:ECE301Summer2008_San_Exam_7b1.jpg|400px|]] | ||
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Therefore, <math> 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)</math> | Therefore, <math> 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)</math> | ||
Revision as of 10:47, 16 November 2008
Problem 7 Part b
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 $
