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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
- * [[HW1.4 Ryan Scott - Period and Non-Periodic Functions_ECE301Fall2008mboutin]] * [[HW1.4 Hang Zhang - Periodic vs Non-period Functions_ECE301Fall2008mboutin]]24 KB (3,272 words) - 06:58, 1 September 2010
- The function <math>f(t)=sin(t-T)</math> is periodic, with a period of <math>T=2\pi</math>. This means that for <math>T=2n\pi</math>, n an int ...dic function does not remain self-similar for all integer multiples of its period. A decaying exponential is an example of a non-periodic function. The dis1 KB (206 words) - 07:09, 14 April 2010
- ...d be the Cosine function. Notice that it repeat itself time to time with a period of <math>2\pi</math>645 B (90 words) - 07:11, 14 April 2010
- The Function :<math>f(x) = sin(x)\,\!</math> is a periodic function with a Period :<math>T = 2*\pi\,\!</math>. This sine wave function repeats itself every period of :<math> 2*\pi,\!</math>1 KB (180 words) - 07:13, 14 April 2010
- ...xists an integer N such that x[n+N] = x[n]. The value of N is called the "Period."1 KB (212 words) - 07:15, 14 April 2010
- * We can see repeated x^2 that repeated itself every period of 4.519 B (70 words) - 07:18, 14 April 2010
- A periodic function is a function which repeats over a period of time. A good example of periodic functions are: As you can see, this function repeats itself with a period of <math>2\pi</math>1 KB (228 words) - 07:10, 14 April 2010
- 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
- ...xists an integer N such that x(n+N) = x(n). The value of N is called the "period".1 KB (265 words) - 06:12, 2 February 2011
- ...the approximation is so big that it only take approximate 1 point in each period, which is not adequate for sketching the graph.437 B (78 words) - 09:54, 6 September 2008
- ...unction is a function that repeats itself after a certain time, called the period. ...function is shown below, the tangent function. The tangent function has a period of <math>pi</math>.1,003 B (148 words) - 07:21, 14 April 2010
- Since we know that tan(t) repeats itself after every period of <math> t = \pi</math> , we know that x(t) is a periodic signal.1 KB (205 words) - 07:20, 14 April 2010
- y = sinx <-- This is very basic periodic fuction.(period time = 2<math>\pi</math>) y = cosx <-- This is also very basic periodic function.(period time = 2<math>\pi</math>)742 B (104 words) - 07:23, 14 April 2010
- ...N = [any even number] would also work. N = 2 is called the '''fundamental period''' because it is the smallest of all possible N.)2 KB (279 words) - 07:18, 14 April 2010
- A periodic function is a function that has repeating values after a certain period of time and keeps repeating as time continues.732 B (100 words) - 07:24, 14 April 2010
- ==Period Functions==20 B (2 words) - 13:51, 4 September 2008
- % Sampling period3 KB (516 words) - 16:38, 4 September 2008
- ...2\pi</math> when <math>t = \frac{\pi}{5}</math>. This is the fundamental period.879 B (140 words) - 07:25, 14 April 2010
- ...gnal. It's period is 4*k, where k is an integer. However, it's fundamental period is 4.883 B (143 words) - 07:24, 14 April 2010
- j^n has a period of 4.856 B (140 words) - 07:26, 14 April 2010
- ...mplitude of the function. When the graphs of two functions having the same period and frequency repeat at different values of the independent variable (x), t An aperiodic function (non-periodic function) is one that has no period which repeats itself (not to be confused with an antiperiodic function for2 KB (274 words) - 07:27, 14 April 2010
- ...:sinwave_ECE301Fall2008mboutin.jpg|thumb|A wavy sinusoidal wave that has a period of 2 pi]]634 B (89 words) - 07:28, 14 April 2010
- In discrete time, a function is period if there exists an integer N such that x[n+N] = x[n] ...ample of a continuous time periodic function would be x(t) = cos(t) with a period of 2*pi.796 B (137 words) - 07:18, 14 April 2010
- Periodic functions are functions that repeat over and over for a specific period. More specifically, a function is periodic if there exists some number T>0 One common example of this is the tangent function, which has a period of \pi.648 B (117 words) - 20:01, 4 September 2008
- x[n]=j^n, is periodical with fundemental period N0=4.835 B (141 words) - 07:26, 14 April 2010
- ...ere exists a number such that x(t+T) = x(t). The value of T is called the "period". ...'''''integer''''' N such that x[n+N] = x[n]. The value of N is called the "period".873 B (149 words) - 17:24, 4 September 2008
- ...has a fundamental period of two Pi. Therefore any multiple of two Pi is a period.1 KB (221 words) - 12:21, 5 September 2008
- ...strated above, the function is clearly periodic, and has 4 as the smallest period.656 B (115 words) - 06:13, 5 September 2008
- A function is period if there is a time interval <math>T > 0\!</math>, such that the value for t ...fundamental period. Every other value of <math>T\!</math> is just another period.2 KB (300 words) - 20:02, 4 September 2008
- Note: N is the period of the signal.1 KB (209 words) - 09:49, 5 September 2008
- Over an infinite period of time <br>647 B (89 words) - 21:00, 4 September 2008
- ...such that <math>\,x(t+T)=x(t)\,</math>. In this example, the fundamental period is <math>\,2\pi\,</math>, which one possible value for <math>\,T\,</math>.960 B (150 words) - 21:01, 4 September 2008
- ...> gives the original function since <math>2\pi</math> is the ''fundamental period'' of <math>x(t) = sin(t)</math> ...that <math>x[n + N] = x[n]</math> for all <math>n</math>. The fundamental period of the sine function is <math>2\pi</math>, which is not an integer. Further1 KB (189 words) - 21:21, 4 September 2008
- y = sin(3x) is a continuous function and has a period of 2pi/3, like shown on the graph below.438 B (72 words) - 21:53, 4 September 2008
- The first signal is a triangular wave which has period of 10 seconds. The second signal is a bunch of noises.656 B (87 words) - 21:36, 4 September 2008
- For all Y(t) and Y(x) Y is equal to 2. The period can be any integer greater than except 0.175 B (35 words) - 03:36, 5 September 2008
- For all Y(t) and Y(x) Y is equal to 2. The period can be any integer greater than except 0.685 B (102 words) - 07:16, 14 April 2010
- ...shift should not effect the energy or power of periodic function over one period (0 to 2<math>\pi</math> in this case).1 KB (169 words) - 18:20, 5 November 2010
- x(t) = <math>e^{n*j*w*t}</math> can be a periodic function. The period is <math>(2 * pi) / w</math>. If <math>w / (2 * pi)</math> is a rational n831 B (141 words) - 08:17, 5 September 2008
- ...r all values of t ,for some T not equal to zero,then x(t) is periodic with period T. ...(n)=x(n+N) for all values of n ,some integer N ,then x(n) is periodic with period N.960 B (171 words) - 07:13, 14 April 2010
- ...after 4 and after every 4 values. So the given function is periodic with a period of 4.566 B (79 words) - 09:16, 5 September 2008
- In CT the cosine function cos(t) is a periodic function with fundamental period T = 2π. This comes from the definition of periodic systems. A function x(563 B (104 words) - 09:23, 5 September 2008
- =Non-period Function=597 B (102 words) - 10:11, 5 September 2008
- ...ence of a single period instead of multiple periods, singly periodic) with period ...is periodic with least period <math> 2\pi </math> (often simply called the period)597 B (99 words) - 14:24, 5 September 2008
- periodic functions, all with the same period. ...eated at regular intervals. More explicitly, a function f is periodic with period P greater than zero if1 KB (253 words) - 07:04, 14 April 2010
- The power over a time period t1 to t2 is calculated by The energy of <math>y(t) = e^{t}</math> over the time period 0 to 2 seconds is found as follows:1,016 B (167 words) - 15:48, 5 September 2008
- ...odic function''' is a function that repeats its values after some definite period has been added to its independent variable. This property is called periodi ...eated at regular intervals. More explicitly, a function f is periodic with period P greater than zero if2 KB (291 words) - 07:03, 14 April 2010
- ...ence of a single period instead of multiple periods, singly periodic) with period if p is the Fundamental Period813 B (113 words) - 07:03, 14 April 2010
- ...riodic function is "a function that repeats its values after some definite period has been added to its independent variable." That is to say that x(t) = x(t688 B (113 words) - 07:12, 14 April 2010
- ...odic function''' is a function that repeats its values after some definite period has been added to its independent variable. This property is called periodi The function repeats itself in a certain period which is <math>2\pi</math> so it's a periodic function.990 B (171 words) - 18:37, 5 September 2008
- === Period Function === === Non-Period Function ===942 B (142 words) - 18:30, 5 September 2008
- ...spinding to (period + number) where "number" is less than the size of the "period" over the independent variable domain, then your function is periodic!940 B (153 words) - 18:27, 5 September 2008
- I changed the sample rate to take 25 samples per period.248 B (45 words) - 11:49, 7 September 2008
- ...3 = 0.076 and Ts = 0.07, we can see that the program only samples once per period, which is not nearly enough. The way to debug this program is to decrease T734 B (115 words) - 04:01, 9 September 2008
- ...>. Plotting this signal yields a smooth waveform that repeats itself with period <math>T=2\pi</math>. ...create a signal with an arbitrary period. Presented here is a signal with period 5.3 KB (536 words) - 11:07, 10 September 2008
- The function x(t) = cos(t) is periodic in CT, as its period is 2<math>\pi</math>. However, it is not periodic in DT. period = 5;2 KB (377 words) - 06:59, 10 September 2008
- The original signal shown in the first plot is y(t) = sin(t) with a period of <math>2\pi</math>1 KB (239 words) - 06:20, 11 September 2008
- ...Ts is too high. In the original code it only sampled enough to create two period.507 B (86 words) - 16:26, 9 September 2008
- Period is 4.<br>556 B (92 words) - 15:00, 10 September 2008
- Since the Ts value is quite large,ie .07 and the period of the sinusoidal t0 = .0769 So, we cant plot 13 ycles of 13Hz accurately a452 B (76 words) - 17:31, 9 September 2008
- ...g back my own signal, <math>y = x^2 \,</math>, i'll create a signal with a period of 5units. The graph goes back to itself with the period of 5 units1 KB (196 words) - 20:31, 10 September 2008
- ...r to fix it, we need to give points more frequently by decreasing Ts (time period).464 B (81 words) - 11:11, 10 September 2008
- What this does is repeats the function with a period of 5, and creates 4 full periods.<br>1 KB (231 words) - 06:22, 11 September 2008
- The bug in the above problem is that the step time Ts=0.07 is too big as the period for one cycle To=1/f=0.0786. Thus in one there are hardly one or two points624 B (116 words) - 14:41, 10 September 2008
- ...te, in fact it is sampling at approximately the same time intervals as the period of the signal.771 B (136 words) - 15:14, 10 September 2008
- ...'''periodic'''. For instance, let <math>k=1 (k\in\mathbb{N})</math> be the period. Then, <math>f[n+k]=f[n+1]=5 \cos(2\pi (n+1))=5 \cos(2\pi n+2\pi)=5 \cos(2 Consider the period <math>P = 2 \pi k</math>. <math>x(t+P) |_{P=2 \pi} = \sum_{k \in \mathbb{Z2 KB (312 words) - 17:53, 10 September 2008
- The bug is that the period Ts is overlarge and so it cannot display the whole wave. After trial it's ok with a period about 0.01 of it.362 B (64 words) - 16:58, 10 September 2008
- The period of the signal in the continuous time domain is <math>2\pi\!</math>. However ...d in the continuous world. In fact, any multiple or factor of the signal's period in continuous time can be used as a sampling rate that will yield a periodi1 KB (227 words) - 17:24, 10 September 2008
- T0 =1/F0; %period T0 =1/F0; %period1 KB (178 words) - 18:05, 10 September 2008
- ...he signal are at distances between the points greater than the half of the period of the waveform. As a result it does not produce the correct plot.587 B (106 words) - 12:08, 11 September 2008
- ...CT signal because it follows the rule, x(t+T)=x(t). It is periodic with a period T=<math>\barwedge</math> ...non periodic signal y=x. Now this can be made periodic by shifting it by a period of t0 and adding the sum of the shifted copies.1 KB (263 words) - 09:36, 11 September 2008
- ...the graph, the values of x[n] are periodic because they repeat after every period of <math>t = 2\pi</math>.2 KB (380 words) - 10:24, 11 September 2008
- Say I choose the sampling period <math>T_s=.1\,</math> But if I choose the sampling period to be <math>T_s=.7\,</math>1 KB (186 words) - 16:24, 11 September 2008
- ...the fact that it isn't sampling enough times to plot a decent graph. The period is is 0.7 and so is the Ts value. So therefore decrease the Ts value such461 B (80 words) - 09:55, 11 September 2008
- ...id is 1/13 = 0.077 s. With the sampling rate being roughly the same as the period it is little wonder that most of the signal was not captured.476 B (79 words) - 10:21, 11 September 2008
- ...nction was chosen at random from HW1: [[HW1.4 Hang Zhang - Periodic vs Non-period Functions_ECE301Fall2008mboutin]] If <math>x(t)</math> is sampled at <math>period=0.1</math>, the function3 KB (512 words) - 06:47, 12 September 2008
- Here I use the CT signal x=sin(2*pi*t) with a period of 1 sec:727 B (118 words) - 16:11, 12 September 2008
- However, if you sample this function with a period of anything OTHER than <math>\pi</math> then you get random dots all over t1 KB (208 words) - 14:07, 11 September 2008
- ...s that the time vector interval Ts = 0.07 sec is too small compared to the period of the signal which is 1/13 <math>\approx</math> 0.0769 sec. This bug can b573 B (96 words) - 14:13, 11 September 2008
- ...the function period, meaning that it may only mark points about once every period:526 B (91 words) - 14:20, 11 September 2008
- The bug here is that the period is 0.07 and so is the sampling rate. To fix this, we can just try a smalle260 B (47 words) - 14:20, 11 September 2008
- ...is necessary to increase the sampling frequency by decreasing the sampling period Ts.552 B (87 words) - 14:59, 11 September 2008
- %Periodic signal turned non-periodic by setting the sampling period to 0.1642 B (88 words) - 15:45, 11 September 2008
- %by sampling cosine values at whole period intervals2 KB (242 words) - 16:14, 12 September 2008
- [[Image:Period exponential_ECE301Fall2008mboutin.jpg]]774 B (126 words) - 13:28, 12 September 2008
- ...y". There reason for this is because the sampling rate is too large. The period of the wave is coded as 1/F0, which equals .0769. However, the value for T918 B (157 words) - 17:05, 11 September 2008
- which appears to be correct. However, the sampling period <math>T_{s}=0.07</math> only provides 14 points on the graph, which appears ...en. The value <math>T_{s}=0.07</math> roughly sampled one point from each period of the sinusodial wave, which was not fine enough to describe the wave suff1 KB (190 words) - 17:53, 11 September 2008
- ...y(t)=cos(t) is periodic because cos(t + T) = cos(t) where its fundamental period is 2*π1 KB (207 words) - 21:44, 11 September 2008
- period = 5259 B (39 words) - 20:45, 11 September 2008
- ...d add the signal together multiple times we can get a periodic signal with period 2.397 B (76 words) - 05:07, 12 September 2008
- ...its a 13Hz sinusoid. The samples are not frequent enough and go beyond one period. To fix this code Ts must be decreased to a reasonably small value to impro723 B (119 words) - 06:43, 12 September 2008
- Using an arbitrarily chosen period of 2, I produced the following graph1 KB (183 words) - 07:45, 12 September 2008
- ...eriod is only .076 s, so the reproduction is barely getting one sample per period373 B (63 words) - 07:58, 12 September 2008
- In, Homework 1, Problem 4 [[HW1.4 Hang Zhang - Periodic vs Non-period Functions_ECE301Fall2008mboutin]], I used the function:1 KB (217 words) - 08:58, 12 September 2008
- 1.This is a sine function of period 2. Function is sin(pi*t). Continuous Signal.642 B (86 words) - 10:23, 12 September 2008
- ...of <math>e^t</math> from the one posted under Hw 1.4 David Hartmann. I use period of 10 to create this signal to be a periodic one. The plot is shown below:310 B (57 words) - 11:05, 12 September 2008
- for count=1:period/delta:(nperiod+8)*period/delta851 B (147 words) - 15:14, 12 September 2008
- T0 is 0.07. This value is too large. It is very close to the value of the period. 0.07 vs. 0.0769. ...period. To solve this just make the value of T0 very small compared to the period.244 B (47 words) - 13:39, 12 September 2008
- ...x(t)</math> , where T is a multiple of the fundamental period, or smallest period. ...tion. Note that pi/2 is still a multiple of the inverse of the fundamental period.1 KB (215 words) - 15:10, 12 September 2008
- ...o be a factor of 2pi and since n is an integer, there is not going to be a period.486 B (101 words) - 14:44, 12 September 2008
- non-period signal with delta t = 0.3649 B (104 words) - 16:13, 12 September 2008
- To=1/13=0.076 and Ts=0.07 . Sampling rate seems to be too close to the period and based on the figure we conclude that the sampling rate is too small. On594 B (85 words) - 17:38, 12 September 2008
- ...math>\omega_o \,</math> as <math>\pi \,</math> since both functions have a period based on it.784 B (140 words) - 10:34, 20 September 2008
- 2) x(t) is periodic with period T = 2 and has Fourier coefficients <math> ak </math> 2)the signal has a period of 2615 B (95 words) - 17:51, 25 September 2008
- ...math>\omega_o \,</math> as <math>\pi \,</math> since both functions have a period based on it.1 KB (197 words) - 10:59, 16 September 2013
- ...n. The period of sin and cos is <math>2\pi</math>, therefore the combined period is also <math>2\pi</math>.2 KB (306 words) - 10:57, 16 September 2013
- ...ry even -- to find the fundamental period of the signal. In our case, the period of the overall signal is <math>2\pi</math>, so <math>\omega_0</math> will b2 KB (384 words) - 10:56, 16 September 2013
- For periodic DT signal, x[n] with fundamental period N:1 KB (230 words) - 14:22, 26 September 2008
- <math> x[4] = x[0] </math> etc, the function is periodic with period 4 Since the period is 4, N=4.2 KB (271 words) - 17:36, 25 September 2008
- ...math> because <math>\,a_0</math> is the average value of the signal over 1 period.650 B (95 words) - 07:26, 24 September 2008
- ...in(5t) </math>. The graph below proves that it is indeed periodic, with a period <math> T = 2\pi </math>. The period <math> \,\ T = 2\pi </math> so if <math> \,\ w_0 = \frac{2\pi}{T} </math>,3 KB (464 words) - 10:58, 16 September 2013
- ...th>\omega_0\!</math> = 2<math>\pi / T\!</math> (where T is the fundamental period). Therefore, the fundamental frequency is <math>1\!</math>.2 KB (384 words) - 10:55, 16 September 2013
- 2) <math>x(t)</math> is periodic with period <math>T = 5</math>369 B (58 words) - 16:46, 26 September 2008
- The fundamental period <math>T\!</math> is <math>2\pi\!</math>. Thus we use the equation <math>\o2 KB (360 words) - 10:55, 16 September 2013
- The fundamental period, denoted as <math>T\!</math>, of this signal is <math>2\pi\!</math>. The fu *Hey Virgil it's Joe.. I am not sure if its period is <math> 2\pi </math>. I would appreciate your explanation. (Jungu -Joe- C2 KB (429 words) - 10:55, 16 September 2013
- ...nt <math>a_{0}</math> is equal to the average value of the signal over one period, so <math>a_{0} = \frac{0}{2} = 0</math>.2 KB (351 words) - 20:45, 24 September 2008
- Let <math>x[n]\,</math> be a periodic DT signal with fundamental period N. Thus the fundamental period is 2.907 B (155 words) - 06:49, 25 September 2008
- 2. x(t) has period N=12.210 B (32 words) - 07:25, 25 September 2008
- 4. This signals period and the number of fingers on your hand share a common integer. From 4, we know that the period is 5, making <math>\omega=\frac{2\pi}{5}</math>801 B (140 words) - 07:43, 25 September 2008
- <math>a_0</math> is the average value of the function over one period. <math>a_0= \frac{1}{T}\int_{0}^{T} x(t)e^{0}dt\!</math>. <math>w_0</math> is 4, so the period is <math>T=\frac{2\pi}{4}=\frac{\pi}{2}</math>2 KB (363 words) - 10:56, 16 September 2013
- The astute observer will notice that, when divided by the period, this is the formula for computing <math>a_3</math>. This result along wit1 KB (203 words) - 17:19, 25 September 2008
- <br>The fundamental period is 2*pi1 KB (216 words) - 11:02, 16 September 2013
- 3. This signals period and the price of a footlong subway sandwich have the same value. Part3 tells us that the period is 5, making <math>\omega=\frac{2\pi}{5}</math>488 B (82 words) - 09:45, 25 September 2008
- ...period of the function is found from: <math>e^{j\omega_0}</math> where he period T = <math>{2\pi \over \omega_o}</math> Thus, the fundamental period = <math> {2\pi \over 3} </math>2 KB (279 words) - 10:54, 16 September 2013
- ...period of the function is found from: <math>e^{j\omega_0}</math> where he period T = <math>{2\pi \over \omega_o}</math> Thus, the fundamental period = <math> {2\pi \over 3} </math>2 KB (297 words) - 17:34, 25 September 2008
- 1. Period is equal to 21 KB (219 words) - 10:32, 25 September 2008
- 1. DT signal x[n] has a period of 2. From 1. we know the period = 2, therefore:590 B (97 words) - 10:39, 25 September 2008
- Now, each sine has its own period, and the fundamental period of the function is the greater of the separate periods. <math>N_4sin = 4\,</math>, so the overall fundamental period is3 KB (405 words) - 12:42, 25 September 2008
- 2. <math>x(t)\!</math> is periodic with period <math>T=4\!</math><br>261 B (42 words) - 16:17, 26 September 2008
- 1. x[n] is periodic and period N=6.672 B (117 words) - 13:08, 25 September 2008
- Since we know the 2 even <math>a_k\,</math>'s in the fundamental period, by property 4, we can find the 2 odd <math>a_k\,</math>'s.2 KB (328 words) - 14:01, 25 September 2008
- 2. <math>x(t)\!</math> has a fundamental period of <math>T = 9\!</math>.422 B (67 words) - 13:28, 25 September 2008
- choose <math>\,\omega_0</math> as <math>\,\pi</math>, the smallest period between the two parts.2 KB (283 words) - 10:55, 16 September 2013
- Where N is the period of the function. We will make a relatively low period, say <math>\,N=4</math>2 KB (374 words) - 14:27, 25 September 2008
- The function has a fundamental period of 4 (it can be easily shown that <math>\,x[n]=x[n+4], \forall n\in\mathbb{2 KB (375 words) - 15:10, 25 September 2008
- ...series coefficients of x[n], where x[n] is a square wave with fundamental period N = 4 with the following values:2 KB (313 words) - 14:17, 25 September 2008
- <math>K = 0,1</math> due to the fundamental period of <math>N = 2</math><br><br>770 B (140 words) - 15:12, 25 September 2008
- 1. period of the function is 31 KB (203 words) - 16:00, 25 September 2008
- 1.) The fundamental period N = 2. <math> = \frac{1}{2} (x[0+2] + x[1])</math> (since x is periodic with period 2)1 KB (186 words) - 17:47, 25 September 2008
- The fundamental frequency is <math>\frac{\pi}{6}</math> and the fundamental period is 12 The other values of <math>\ a_{k} = 0 </math> within the fundamental period.1 KB (280 words) - 15:40, 26 September 2008
- 1. x[n] is periodic with period 4 and has Fourier coefficients <math>\,a_k\,</math>. ...econd property can be used to calculate the average of the signal over one period, which is precisely what <math>\,a_0\,</math> is1 KB (250 words) - 20:40, 25 September 2008
- 2.) x(t) is periodic with period T = 43.245 B (42 words) - 20:49, 25 September 2008
- So, the fundamental period is 4.967 B (170 words) - 14:35, 26 September 2008
- 1. Period is 8.270 B (41 words) - 05:16, 26 September 2008
- The signal is DT periodic with period of 4938 B (182 words) - 07:09, 26 September 2008
- 2.x(t) is periodic with period T = 8248 B (41 words) - 07:06, 26 September 2008
- 1. DT signal is real and even with period N=12578 B (116 words) - 10:13, 26 September 2008
- fundamental period = <math>2\pi/6</math>426 B (79 words) - 10:24, 26 September 2008
- 1. x[n] is periodic with period 3504 B (78 words) - 10:58, 26 September 2008
- In order to find the period of the signal below, we need to find a value of K that will make N an integ560 B (103 words) - 12:59, 26 September 2008
- First the fundamental period has to be found. The period of the overall signal is <math>8</math>, so <math>\omega_0</math> will be <1 KB (261 words) - 13:32, 26 September 2008
- * The period of the DT signal x[n] = 2637 B (101 words) - 13:22, 26 September 2008
- 1. x[n] is periodic with period N = 7944 B (140 words) - 14:11, 26 September 2008
- The fundamental period is: ...omega_o = \frac{2\pi}{T} = \pi</math>, with <math>T = 2\,</math> being the period of the original signal.<br>1 KB (201 words) - 11:06, 16 September 2013
- 2. x[n] has period N = 10 and Fourier coefficients <math>\ a_k</math> From #2 gives the period of N=10, from that can deduce that the frequency <math>\ w = k\frac{2\pi}{1800 B (136 words) - 15:12, 26 September 2008
- The signal above x[n] is periodic with period N.1 KB (275 words) - 16:31, 26 September 2008
- 1)x[n] is periodic with period 12. Since x[n] is periodic with period 12, it can be wriiten as2 KB (426 words) - 15:21, 26 September 2008
- Let's look at: <math>x(t)=3*cos(3t)</math>, we know that the fudamental period of x(t) is877 B (147 words) - 10:53, 16 September 2013
- A) Fundamental Period = 2 ...x[0] + x[1] = 0 since <font size = '4'><math>a_0 = 0</math></font> and the period is 2.1 KB (260 words) - 16:57, 26 September 2008
- This is a signal with period <math>T = {2\pi \over w_0}</math><br><br>2 KB (311 words) - 10:57, 16 September 2013
- San (3 in Chinese). It is periodic with a period of 2.308 B (51 words) - 16:41, 26 September 2008
- 1. x[n] is periodic and period N=8.719 B (121 words) - 16:44, 26 September 2008
- 1. period N=2 1. a period of 2 gives us369 B (57 words) - 17:40, 26 September 2008
- 4) x(t) is periodic with period T = 7198 B (35 words) - 17:45, 26 September 2008
- 4. x[n] has period 2341 B (49 words) - 18:38, 26 September 2008
- *1 Lets say the period is ... 7 *Since the period is 7 ... <math> w = \frac{2\pi}{7} </math>498 B (82 words) - 17:49, 26 September 2008
- 2. x(t) is periodic with period T = 4 and Fourier coefficients <math> \ a_k </math>.992 B (159 words) - 18:33, 26 September 2008
- 1. The signal is periodic with a period of 3 seconds416 B (69 words) - 18:18, 26 September 2008
- 1. Period of x[n] is N=4.994 B (178 words) - 18:44, 26 September 2008
- 1. <math>\ x[n]</math> is periodic with period of N = 2293 B (49 words) - 19:24, 26 September 2008
- 2. x(t) is periodic with period T271 B (39 words) - 18:51, 26 September 2008
- 1. x[n] is periodic with period N=6<br><br>1 KB (186 words) - 20:38, 26 September 2008
- ...Fourier coefficients. We just need to express all the coefficients in one period. But when it is CT isn't there an infinite amount of coefficients? Can we e320 B (61 words) - 14:03, 8 October 2008
- ...fficients <math>a_{k}</math> of the Fourier series of the signal x(t) with period T = 4 defined by ...h>a_{0}</math> is simply equal to the average value of the function over 1 period.2 KB (303 words) - 19:47, 8 October 2008
- .../math> of the Fourier series of the signal <math>x(t)</math> periodic with period <math>T=4</math> defined by ...separately. However, this is simply the average of the function over one period.1 KB (195 words) - 16:16, 13 October 2008
- .../math> of the Fourier series of the signal <math>x(t)</math> periodic with period <math>T=4</math> defined by2 KB (300 words) - 07:39, 14 October 2008
- ...\!</math> of the Fourier series signal <math>x(t) \!</math> periodic with period <math>T = 4 \!</math> defined by ...ath>a_{0} \!</math>, we know that it is the average of the signal over the period. In this case,1 KB (210 words) - 16:19, 14 October 2008
- ...f the Fourier series of the signal '''<math>x\ (t)</math>''' periodic with period '''<math>T\ = 4</math>''' defined by: <math>a\ _0 = </math> average of the signal over period <math> = \frac{2}{4} = \frac{1}{2}</math>851 B (138 words) - 18:23, 14 October 2008
- .../math> of the Fourier series of the signal <math>x(t)</math> periodic with period <math>T=4</math> defined by1 KB (193 words) - 09:12, 15 October 2008
- ...ts <math>a_k</math> of the Fourier series of the signal x(t) periodic with period T = 4 defined by ...>a_0 = \frac{1}{2}</math> because it is the average of the signal over one period.1 KB (217 words) - 11:05, 15 October 2008
- ...ts <math>a_k</math> of the Fourier series of the signal x(t) periodic with period T=4 defined by764 B (143 words) - 17:52, 15 October 2008
- Let us assume that <math>T\,</math> is the sampling period of the sampling process. We can safely say that <math>\frac{2 \pi}{T} = \o2 KB (367 words) - 11:27, 7 November 2008
- ...pling function, the period, <math>T</math>, is referred to as the sampling period, and the fundamental frequency of <math>p(t)</math>,3 KB (582 words) - 06:11, 16 September 2013
- period T :sampling interval565 B (93 words) - 13:48, 9 November 2008
- With sampling period, T, samples of x(t),x(nT), can be obtained from x(t), where n = 0 +-1, +-2,500 B (101 words) - 16:21, 9 November 2008
- With sampling period T, samples of x(t),x(nT), can be obtained , where n = 0 +-1, +-2, .... Therefore, the sampling period has to be selected well.1 KB (238 words) - 16:44, 9 November 2008
- With sampling period T, samples of x(t),x(nT), can be obtained , where n = 0 +-1, +-2, .... Therefore, the sampling period has to be selected well.1 KB (238 words) - 05:31, 16 November 2008
- *<math>T</math>: Sampling period; equal to <math>\frac{2\pi}{\omega_s}</math>1 KB (178 words) - 19:31, 23 November 2008
- <math>T</math> Sampling Period2 KB (349 words) - 12:09, 10 November 2008
- Sampling at the Nyquist rate gives a signal that only shows two points of the period and a recreated signal would not be able to reconstruct the direction of th2 KB (373 words) - 14:24, 10 November 2008
- ...eries of impulses that are spaced out by a period T, known as the Sampling Period.1 KB (274 words) - 06:49, 16 September 2013
- ...an be uniquely recovered from its samples. In this case, T is the sampling period, and <math>\frac{2\pi}{T}</math> (or <math>w_s</math>) is the sampling freq663 B (118 words) - 16:28, 10 November 2008
- Since <math>x(t)</math> only has values at intervals of the sampling period <math>nT</math>, the equation can be simplified to3 KB (543 words) - 17:23, 10 November 2008
- if <math>T<\frac{1}{2}\frac{2\pi}{\omega_m}</math>, where T is the sampling period.2 KB (340 words) - 17:29, 10 November 2008
- Where <math>H(\omega)</math> is a filter with gain equal to the period of the signal and a cutoff frequency of <math>\omega_c</math>.711 B (130 words) - 19:44, 10 November 2008
- ...elta}{T} </math> where <math>\delta</math> is the average of signal over 1 period.949 B (174 words) - 16:11, 16 November 2008
- 1 KB (263 words) - 13:14, 17 November 2008
- <math>\omega_s = \frac{2\pi}{T}</math>, where T is the sampling period of <math> p(t) </math>2 KB (279 words) - 12:53, 17 November 2008
- <math>x_p(t)=x(t)\sum_{n=-\infty}^\infty \delta(t-nT)</math> where T is the period of the function Now, convert <math>x(t)</math> to <math>x(nT)</math> for period T2 KB (411 words) - 17:16, 17 November 2008
- We will first find the fourrier transform X(W) and plot out its signal over a period of frequency Wm.805 B (160 words) - 20:06, 17 November 2008
- ...=x(nT) </math> the n<sup>th</sup> sample, <math> T </math> is the sampling period, and with the substitution: <math> z = e^{sT} </math>.3 KB (537 words) - 17:27, 3 December 2008
- All DT Fourier transforms are periodic with period <math>2\pi</math>21 KB (3,312 words) - 11:58, 5 December 2008
- ===[[Chapter 3_ECE301Fall2008mboutin]]: Fourier Series Representation of Period Signals===7 KB (1,017 words) - 10:05, 11 December 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) - 14:35, 31 August 2008
- ...T)=X(t),for all values of t. The '''fundamental period''' is the smallest period of all periods of a signal (denoted by :<math>T_0</math>). A DT signal X[n] is called '''periodic''' if there exists N>0 period such that X[n+N]=X[n],for all values of n.1 KB (249 words) - 18:35, 16 March 2008
- Let x[n] be a periodic DT signal with fundamental period N, we can write...780 B (131 words) - 20:10, 22 March 2008
- 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}+j{b}_{868 B (154 words) - 17:36, 30 March 2008
- YES: <math>X(e^{j\omega})</math> is always periodic with period <math>2\pi</math>4 KB (739 words) - 20:48, 30 July 2008
- ...ies coefficients <math>a_k</math> of the signal x[n] is also periodic with period N. For the periodic signal x[n], find the values of <math>a_0,a_1,...,a_{N- 2)a)Let's consider a continuous-time periodic signal x(t) with period T = 5 whose non-zero Fourier series coefficients <math>a_k</math> are given4 KB (803 words) - 11:10, 22 July 2008
- 1(a) .... The period is not 12. Check again. 1(b) .... Even with the correct period, the answer is still wrong. Check again.124 B (20 words) - 23:09, 18 July 2008
- ...or use with the DFT. A discrete periodic signal can be used when only one period of the signal is analyzed. The DFT of a signal will be discrete and have a ...th> in the complex plane, and all <math>x_1[n],x_2[n]</math> with the same period N4 KB (695 words) - 06:46, 23 September 2011
- delta = 0.00005; %Sampling period2 KB (232 words) - 05:30, 11 July 2012
- ...A discrete signal for the amount of toilet paper and how much is used in a period of time(one day,month,year?) by one person.993 B (192 words) - 11:01, 23 June 2009
- t4 = 0:del:(1/10000); %a miniscule rest period<br>688 B (114 words) - 05:46, 25 June 2009
- Suppose that x(t) and y(t) are both periodic with Period '''T''' and that Since the product '''<math>x(t)y(t)</math>''' is also periodic with period '''T''', we can expand it in a Fourier series with Fourier series coefficie583 B (107 words) - 09:20, 8 July 2009
- The period T of a periodic signal x(t) remains unchanged when it goes through time rev539 B (102 words) - 18:39, 8 July 2009
- When a time shift is applied to a periodic signal x(t), the period T of the signal is preserved.<br>1 KB (200 words) - 03:44, 9 July 2009
- * [[Fundamental period/frequency]]1 KB (152 words) - 04:06, 23 July 2009
- A CT signal <math>x(t)</math> is periodic with period <math>T</math> if <math>x(t)=x(t+T)</math>. A DT signal <math>x(n)</math> is periodic with period <math>N</math> if <math>x(n)=x(n+N)</math>.2 KB (328 words) - 07:19, 22 July 2009
- [[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
- ...riodic. If you sample a continuous function with sample period t, the DTFT period is <math>2*\pi/t</math> radians.201 B (31 words) - 07:18, 14 November 2011
- <math>\Leftrightarrow X(\omega)</math> is periodic with period <math>2\pi</math>134 B (21 words) - 03:03, 30 July 2009
- and T = 1/Fs = sampling period k=-inf with period T, and can be expressed as a1 KB (196 words) - 04:17, 30 July 2009
- * Should you have any work to hand in during this period of time, you should proceed by uploading an electronic version of your wor2 KB (371 words) - 09:17, 10 August 2009
- * Should you have any work to hand in during this period of time, you should proceed by uploading an electronic version of your work2 KB (370 words) - 09:01, 25 August 2009
- ...n other words, each frequency component must be sampled at least twice per period. ...if it is understood that a sinsuoid must be sampled twice within a single period to determine its frequency, then one can also understand that once we break8 KB (1,452 words) - 06:49, 16 September 2013
- Now, we are going to sample <math> x(t) </math> with an impulse train with period <math> T_{s1} = 1/f_{s1} </math> and convert to a discrete time signal. Th ...This is the same process as in the previous section, but now the sampling period is <math> T_{s2} = \frac{T_{s1}}{N} </math> due to the up-sampler. This yi5 KB (840 words) - 19:08, 22 September 2009
- ...uous time signal by a train of dirac delta functions separated by the time period T. This can be mathematically represented as follows:3 KB (527 words) - 11:50, 22 September 2009
- We see that the scaled function is no longer periodic with period <math>2\pi</math>. 3. Sampling band-limited <math>x(t)</math> with period <math>T_2</math> to get <math>x_2[n]</math>4 KB (655 words) - 07:13, 23 September 2009