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The answer is <math>{1 \over 2}\sin({\pi \over 4}t)</math>
 
The answer is <math>{1 \over 2}\sin({\pi \over 4}t)</math>
  
  Surprise!
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Surprise!

Revision as of 17:58, 25 September 2008

<< Back to Homework 4

Homework 4 Ben Horst: 4.1 :: 4.3 :: 4.5


Problem

1. The signal has only two non-zero Fourier coefficients (-1 and 1).

2. The fundamental period of the signal is 8.

3. The function is sinusoidal.

4. Both Fourier coefficients are non-real.

5. The signal has a maximum value of 0.5 and a minimum of -0.5.

Solution

The answer is $ {1 \over 2}\sin({\pi \over 4}t) $

Surprise!

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett