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- Thus the digital camera '''''samples''''' the continuous signal, with a period <math>T</math> (shutter speed) and "on" for length <math>tau</math> (relate6 KB (914 words) - 12:07, 22 October 2009
- Definition: let x[n] be a DT signal with Period N.3 KB (640 words) - 13:17, 2 December 2011
- ...s simply multiplied by a low-pass filter with height equal to the sampling period T, <math>{H_r}(f)</math>, to isolate the original signal. ...>X_s(f)</math> in the time domain is the sum of a train of deltas at every period of sampling, T2 KB (436 words) - 19:51, 22 September 2009
- sampling period of 4T would yield sampling period of 2T yields:2 KB (279 words) - 18:30, 22 September 2009
- Let X[n] be a DT signal with period N2 KB (491 words) - 23:03, 22 September 2009
- A rep function periodically repeats another function with some specified period T ( basically sampling time). Mathematically a rep operator is a function x ...th period T. To keep x(t) repeating without aliasing, the minimum sampling period should be <math> \geqq 2 * T </math>2 KB (365 words) - 07:38, 25 August 2010
- Since the voiced section is approximately period, it can be modeled as a pulse train. Similarly, the unvoiced section can b ...{2 \pi T} </math> , where F is the formant frequency and T is the sampling period.5 KB (841 words) - 15:26, 10 April 2013
- the sampling period. (same thing as wT = theta - wideband uses window length = one period2 KB (390 words) - 07:46, 14 November 2011
- the sampling period. (same thing as wT = theta - wideband uses window length = one period2 KB (387 words) - 07:47, 14 November 2011
- | Let x[n] be a periodic DT signal, with period N. | <math> x[n] \ \text{ (period } N) </math>4 KB (633 words) - 15:28, 23 April 2013
- is periodic with period $|N|$. is periodic with period 4, because it is the repetition of the <math>\frac{sin(t)}{t}</math> part o4 KB (736 words) - 17:25, 23 April 2013
- ...that the function <span class="texhtml">''X''(ω)</span> is periodic with period <span class="texhtml">2π</span>. <br>Turns out we can clearly see where the major variations occurred over the period of two semesters.We can account for each those big bumps( or dips) by seein13 KB (2,348 words) - 13:25, 2 December 2011
- As president of a major U.S. university during a period of deep my wife and I did it. When tuition doubles over a time period4 KB (663 words) - 19:53, 20 July 2010
- ...did in class) and obtain its CTFT <math>X(f)</math>. Then pick a sampling period <math>T_1</math> for which no aliasing occurs and obtain the DTFT of the sa2 KB (408 words) - 04:59, 7 September 2010
- ...ncy axis. First, one must not forget that the DTFT is ALWAYS periodic with period <math>2\pi</math>. Second, the amplitude of the FT changes when sampling. T1 KB (222 words) - 08:07, 7 September 2010
- ...ath>X_1(\omega)</math> and <math>X_2(\omega)</math> are both periodic with period <math>2 \pi</math>, and that they both sound like a pure, but different, fr4 KB (567 words) - 12:09, 13 September 2010
- *Fourier series of a continuous-time signal x(t) periodic with period T *Fourier series coefficients of a continuous-time signal x(t) periodic with period T5 KB (797 words) - 09:43, 29 December 2010
- *Fourier series of a discrete-time signal x[n] periodic with period N *Fourier series coefficients of a continuous-time signal x(t) periodic with period T2 KB (355 words) - 09:44, 29 December 2010
- x(t) periodic with period two. &= \text{ average of } x(t) \text{ over one period} \\2 KB (324 words) - 08:08, 15 February 2011
- | <math>a.\ Assign\ sampling\ period\ T_1=\frac{1}{1000}</math> ...nterval of <span class="texhtml">[ − π,π]</span>, which represents one period, the frequcy spectrum remains the same as Fig a-1.3 KB (408 words) - 12:27, 15 September 2010
- The first step is to figure out the period N of x[n].982 B (156 words) - 16:41, 30 November 2010
- ...text{ when } 1\leq t <2\end{array} \right. \text{ and is periodic with the period of two.}</math>2 KB (315 words) - 10:39, 11 November 2011
- Since <math>X(e^{jw})</math> has a period of <math>2\pi</math>, <math>Z(e^{jw}) \text{ will have a period of } \frac{2\pi}{L}. \,\!</math>3 KB (467 words) - 19:52, 20 September 2010
- <math>\text{ Sample a continuous signal x(t)=sin(}\omega t)\text{ with period of T, }</math>2 KB (373 words) - 10:41, 11 November 2011
- Then <math>X(e^{j\omega})</math> has a period of <math>2\pi</math>.666 B (121 words) - 12:21, 29 September 2010
- <math>\;\;\;x(t)=|\text{sin}(\pi t)|, \;\; \text{with the period }T=1</math> ::<span style="color:green"> Note: It is not necessary to state the period in the question, as one can figure it out from the signal itself. </span> -6 KB (999 words) - 13:00, 16 September 2013
- *WARNING: The function you get MUST be periodic with period <span class="texhtml">2π</span>. This one is not. --[[User:Mboutin|Mboutin **The DTFT is always periodic with a period <math>2\pi</math>. --[[User:Mboutin|Mboutin]] 13:11, 1 October 2010 (UTC)5 KB (778 words) - 09:11, 1 October 2010
- ...tput of a filter when we input the impulse train of <math>x(t)</math> with period <math>T</math>.4 KB (751 words) - 04:56, 2 October 2011
- *Fourier series of a continuous-time signal x(t) periodic with period T *Fourier series coefficients of a continuous-time signal x(t) periodic with period T2 KB (292 words) - 17:13, 30 September 2010
- *Fourier series of a continuous-time signal x(t) periodic with period T *Fourier series coefficients of a continuous-time signal x(t) periodic with period T1 KB (241 words) - 06:50, 30 September 2010
- Definition: let x[n] be a discrete-time signal with Period N. Then the Discrete Fourier Transform X[k] of x[n] is the discrete-time si521 B (90 words) - 15:35, 8 October 2010
- Definition: let x[n] be a discrete-time signal with Period N. Then the Discrete Fourier Transform X[k] of x[n] is the discrete-time si1 KB (219 words) - 13:25, 2 December 2011
- ...n. The formula was observed to hold whenever the periodic repetition has a period that is at least as long as the signal duration. We finished the lecture by1 KB (158 words) - 15:59, 8 October 2010
- ...o to be correct, one should write "the DFT of the periodic repetition with period N of a signal with finite duration N"). If you found my diagrams hard to re1,003 B (148 words) - 16:16, 8 October 2010
- ...iscrete nonperiodic function. Nonperiodic function is also a function with period <math>\infty</math>. Therefore DTFT of this function can be DFT with <math>3 KB (456 words) - 13:44, 30 April 2015
- ...x_p[n]</math> is the periodical extension of <math>\bar x[n]</math> with a period P10 KB (1,690 words) - 17:44, 23 October 2011
- ...the DTFT is a transform to a "finite" frequency-domain (the length of one period), rather than to the entire real line. <math>X(\omega)</math> is periodic with period <math>2\pi</math>4 KB (746 words) - 08:47, 11 November 2013
- ...idea to check that <math class="inline">X(\omega)</math> is periodic with period <math class="inline">2\pi</math> We have So yes, the expression defines a function that is periodic with period <math>2\pi</math>.7 KB (1,273 words) - 12:42, 26 November 2013
- ...: the picture of filter I drew should have been repeated periodically with period <math>2 \pi</math>. It is important to remember this for any filter in disc2 KB (260 words) - 12:42, 22 October 2010
- ...be to insure that the result is the same as the periodic repetition (with period N) of the usual convolution between x[n] and h[n]?2 KB (366 words) - 12:16, 27 October 2010
- X[k] is defined for <math>0 <= k <= N - 1</math> and periodic with period N X[n] is defined for <math>0 <= n <= N - 1</math> and also periodic with period N1 KB (166 words) - 08:50, 11 November 2013
- Definition: let x[n] be a DT signal with Period N. Then, ...ous periodic version of x[n]. <br> The DFT of the N samples comprising one period of x[n] equals N times the Fourier series coefficients.19 KB (3,208 words) - 11:23, 30 October 2011
- In order for the periodic repetition (with period N) of the usual convolution between x[n] and h[n] to be the same with the N6 KB (1,128 words) - 16:16, 2 December 2010
- ...of rate 15 inquiries per minute. Find the probability that in a 1-minute period, 3 inquiries arrive during the first 10 seconds and 2 inquiries arrive duri7 KB (1,192 words) - 08:22, 27 June 2012
- [-1,1] as a function with period 2.6 KB (1,054 words) - 09:24, 1 December 2010
- # It can track the growth of Rhea. In other words, it will limit the period that the analysis applies.2 KB (237 words) - 06:22, 6 April 2011
- %Each note as a singular array lasting the appropriate time period<br> %Each note as a singular array lasting the appropriate time period<br>3 KB (541 words) - 08:10, 19 January 2011
- Find the fundamental period of the following signals.3 KB (478 words) - 05:17, 25 January 2011
- ...e periodic, we described a simple method for determining their fundamental period.1 KB (178 words) - 13:24, 31 January 2011
- ...cular, you do not need the peak value of that functions. Try to guess the period directly by looking at the sum. If you have no idea how to do this, read th2 KB (445 words) - 15:40, 2 February 2011
