m (→Solution) |
|||
| Line 17: | Line 17: | ||
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! | |
Revision as of 16:58, 25 September 2008
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!
