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==Problem== | ==Problem== | ||
| − | 1. The signal has only two Fourier coefficients ( | + | 1. The signal has only two non-zero Fourier coefficients (-1 and 1). |
| − | 2. The fundamental period of the signal is 8 | + | 2. The fundamental period of the signal is 8. |
| − | 3. | + | 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== | ==Solution== | ||
| − | 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 17:57, 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!
