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From 2, we know that the signal is bounded by 0 and 8.. this means it has an amplitude of 4 and an offset of 4. | From 2, we know that the signal is bounded by 0 and 8.. this means it has an amplitude of 4 and an offset of 4. | ||
| − | From 3, we know that every a_k not equal to 0, -1, or 1 will be equal to 0... this verifies that it's a sinewave. | + | From 3, we know that every a_k not equal to 0, -1, or 1 will be equal to 0... this verifies that it's a sinewave and real. |
From 4, we know that the period is 5, making <math>\omega=\frac{2\pi}{5}</math> | From 4, we know that the period is 5, making <math>\omega=\frac{2\pi}{5}</math> | ||
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| − | Thus this signal is <math>4sin(\frac{2\pi}{5}t)+4</math> | + | Thus this signal is.... |
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| + | <math>x(t)=4sin(\frac{2\pi}{5}t)+4</math> | ||
Latest revision as of 06:43, 25 September 2008
Hints
1. This signal is the math equivalent of a synonym for the word 'strange'.
2. This signal is never less than 0, but is always less than 8.
3. This signal's a_k is always 0 when it's greater than the absolute value of 1.
4. This signals period and the number of fingers on your hand share a common integer.
Answer
From 1, we know that the signal is odd... a sinewave is probably a safe guess.
From 2, we know that the signal is bounded by 0 and 8.. this means it has an amplitude of 4 and an offset of 4.
From 3, we know that every a_k not equal to 0, -1, or 1 will be equal to 0... this verifies that it's a sinewave and real.
From 4, we know that the period is 5, making $ \omega=\frac{2\pi}{5} $
Thus this signal is....
$ x(t)=4sin(\frac{2\pi}{5}t)+4 $
