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| − | Quiz Pool Question1 Solution: | + | == Quiz Pool Question1 Solution: == |
Recall: <math>\omega =2\pi f</math> | Recall: <math>\omega =2\pi f</math> | ||
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According to the Sampling Theorem, sample rate <math>f_s>=2f</math> | According to the Sampling Theorem, sample rate <math>f_s>=2f</math> | ||
| − | So the maximum T is <math>\frac{ | + | So the maximum T is <math>\frac{\pi}{\omega}</math>. |
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| − | [[ECE438_Week6_Quiz|Back to Pool]] | + | [[ECE438_Week6_Quiz|Back to Week6 Question Pool]] |
Latest revision as of 12:20, 29 September 2010
Quiz Pool Question1 Solution:
Recall: $ \omega =2\pi f $
The highest frequency of the continuous signal is $ f=\frac{\omega}{2\pi} $
According to the Sampling Theorem, sample rate $ f_s>=2f $
So the maximum T is $ \frac{\pi}{\omega} $.
