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\ x_2(t) = - \sqrt{2} cos(2\pi t) | \ x_2(t) = - \sqrt{2} cos(2\pi t) | ||
</math> | </math> | ||
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| + | as cos(t) is an even function with a Fourier Series representation that has coefficients of 0 for absolute values of k greater than one. By multiplying the cosine function by the amplitude of the square root of 2 the function will provide a | ||
Revision as of 17:29, 26 September 2008
Suppose we are given the following information about a signal x(t):
1. x(t) is real and even.
2. x(t) is periodic with period T = 4 and Fourier coefficients $ \ a_k $.
3. $ \ a_k = 0 $ for $ \left \vert k \right \vert > 1 $.
4. $ \frac{1}{2}\int_{0}^{2} \left \vert x(t) \right \vert ^2 \, dt = 1 $.
Specify two different signals that satisfy these conditions.
Two signals that would satisfy these coniditions is the input signal
$ \ x_1(t) = \sqrt{2} cos(2\pi t) $
and $ \ x_2(t) = - \sqrt{2} cos(2\pi t) $
as cos(t) is an even function with a Fourier Series representation that has coefficients of 0 for absolute values of k greater than one. By multiplying the cosine function by the amplitude of the square root of 2 the function will provide a
