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- 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