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- A periodic function is one that repeats over a certain time interval. The period of the following, <math>y=sin(x)</math>, is <math>2\pi </math>.574 B (85 words) - 07:19, 14 April 2010
- =Non-period Function=597 B (102 words) - 10:11, 5 September 2008
- [[Fundamental period/frequency -- Yicheng Guo]]47 B (6 words) - 16:55, 22 July 2009
- For example, x(t)= cosw1t + cosw2t, determine the fundamental period and frequency of the signal. period T of x(t) must be s.t. T*w1 = N*2<math>\pi</math>465 B (96 words) - 17:08, 22 July 2009
- x(t) periodic with period two. &= \text{ average of } x(t) \text{ over one period} \\2 KB (324 words) - 08:08, 15 February 2011
Page text matches
- YES: <math>X(e^{j\omega})</math> is always periodic with period <math>2\pi</math>4 KB (777 words) - 11:49, 21 November 2008
- *1(a) .... The period is not 12. Check again. *1(b) .... Even with the correct period, the answer is still wrong. Check again.4 KB (815 words) - 10:57, 21 November 2008
- ...x(t)\ </math>, for all values of t. The fundamental period is the smallest period of all periods of a signal (denoted by <math> T_0\ </math>). ...x[n]\ </math>, for all values of n. the fundamental period is the smallest period of all periods of a signal (denoted by <math> N_0\ </math>).1 KB (206 words) - 16:58, 23 April 2013
- ...k, make sure you take it over one period. Since with voice recordings each period wont be exactly the same, it's a good idea to just do it over one.5 KB (834 words) - 17:26, 23 April 2013
- ...at an aperiodic signal can be viewed as a periodic singal with an infinite period." An example of choosing the decision (this is a link)!!##!!3 KB (431 words) - 17:29, 23 April 2013
- Let <math> x[n]\ </math> be a real periodic sequence with fundamental period <math> N_0\ </math> and Fourier coefficients <math> c_k = a_k+jb_k\ </math>936 B (157 words) - 12:11, 12 December 2008
- ...er than 2*B (called the Nyquist rate), in other words, more than twice per period of the highest frequency component. : To see why, look at the figure below, which is sampled at exactly twice per period, or the Nyquist rate for the signal:3 KB (591 words) - 17:24, 23 April 2013
- ...al <math> x(t)\ </math>, the signal is sampled with shifted deltas by some period T. [<math> x(t)\ </math> multiplied by <math> \sum_{-\infty}^{\infty} \delt4 KB (689 words) - 12:48, 12 December 2008
- ...ange is in the expression for sum, the last number (in this case 6) is the period (T). Note: This is what I got for problem 3.22(e).808 B (131 words) - 13:04, 18 December 2008
- ...(t+T) = x(t),<math>\forall t</math> then we say that x(t) is periodic with period T</li></ul>3 KB (532 words) - 06:43, 16 September 2013
- *<math>T</math>: Sampling period; equal to <math>\frac{2\pi}{\omega_s}</math>2 KB (406 words) - 11:08, 12 November 2010
- ...ive with period <math>2\pi</math>. So drawing phase and magnitude for one period is sufficient.1 KB (173 words) - 17:22, 2 February 2009
- ...th constant amplitude one. Once filtered (for part a), the signal is once period of the original signal that starts at -1e-4 and goes to 1e-4. The ideal sa ...m not sure what to do. I feel as if the amplitude should be adjusted by 1/period, but I am not sure. Currently, we are in the time domain, but the digital844 B (152 words) - 18:26, 11 February 2009
- Though we already know that it's just some shift/scale version with period 2*pi, here is the math behind it.2 KB (374 words) - 12:35, 17 February 2009
- ..., personally, I usually find myself can't focus on studying within a break period. For those reasons, I suggest to put the exam as where it was. BTW, just fo12 KB (2,099 words) - 07:41, 21 March 2009
- <p>Recall DFT of x(n) periodic w\ period N<br/>2 KB (376 words) - 06:44, 16 September 2013
- ...12 straitions between them in the middle of the segment, making the pitch period <math>\frac{0.1}{12} = 8.333</math>msec. I could be way off though. --[[U746 B (132 words) - 07:54, 20 April 2009
- a.pitch period is 50/10000 = 2ms. c. using the pitch period and formant frequency, draw the graph.926 B (147 words) - 07:56, 20 April 2009
- ...over period 'p' and time 't' in years. The investment 'A' is added every period. I originally got <math> A \sum_{t=0}^{n} \frac{r^t}{p^{t}} dt </math>, bu ...ct. i equals r/p. I forgot to add the investment again and again at each period. Otherwise, I should have got a simple exponential I guess. --[[User:Gbri1 KB (270 words) - 09:43, 7 October 2008
- 2. every planet having their orbit sweep out same areas while equal period. 3. if we set P as a period of revolution and set A as a radius of orbit from a foci then,397 B (73 words) - 14:30, 31 August 2008