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- If we imagine that our slit aperture can be thought of as a spatial filter that only passes light through the slit and blocks light everywhere else, t2 KB (286 words) - 10:19, 23 September 2009
- *To filter this out, we can apply a low-pass filter with a cutoff frequency of around 2000 Hz. *So adding a high pass filter should remove the bear's voice completely.5 KB (822 words) - 11:54, 21 September 2012
- Two basic filters used are 1> '''Average Filter''' == '''Average Filter''' ==3 KB (426 words) - 06:03, 14 October 2009
- title('Image using Average Filter');782 B (107 words) - 06:01, 14 October 2009
- title('Image using Average Filter');787 B (108 words) - 06:06, 14 October 2009
- ...er:mboutin|Prof. Boutin]]: graph of the magnitude of the DFT of a windowed filter= Consider the ideal low-pass filter1 KB (212 words) - 11:50, 24 October 2011
- The end result after modeling is a transfer function that is an all-pole filter with a gain and a time delay. As noted above, the transfer function is usually an all-pole filter. We can observe what the resonances, or formants, are just by looking at t5 KB (841 words) - 15:26, 10 April 2013
- periodic filter phoneme - Generally, the vocal tract transfer function is an all-pole filter2 KB (390 words) - 07:46, 14 November 2011
- ...the series for tons of translations and dilations, uses a faster approach: filter banks. ==Multi-Resoltion Analysis using Filter Banks==10 KB (1,646 words) - 11:26, 18 March 2013
- The following pictures show the original image (Lena), the image of the filter, and the filtered image (done with conv2). Note that the FFT2 are plotted a ...e filter = 1/16*[1 2 1; 2 4 2; 1 2 1], commonly referred to as an "average filter":'''</u>8 KB (1,397 words) - 11:23, 18 March 2013
- periodic filter phoneme - Generally, the vocal tract transfer function is an all-pole filter2 KB (387 words) - 07:47, 14 November 2011
- Plot of the frequency response of the average filter: Plot of the frequency response of the filter:1 KB (163 words) - 12:50, 26 November 2014
- ...g how to handle the boundaries is as important as knowing how to apply the filter to the image (at least in the context of this course). -[[User:crtaylor|Ry945 B (155 words) - 17:55, 8 December 2009
- ...of some non-linear filters for face-enhancement, such as the Perona-Malik filter that preserve sharpness and details, whilst removing noise at the same time235 B (33 words) - 22:21, 6 December 2009
- ...though I got good grade on ECE301. And the second part is basically about filter. Before you take ECE438, I also recommend that you should figure out how to17 KB (3,004 words) - 08:11, 15 December 2011
- h(x,y) → filter FREE → filter image based only on pixels: {H, K, P}5 KB (811 words) - 16:19, 19 December 2009
- ...ion evolves into a great estimation method. The more I know about particle filter for object tracking, the more I get impressions about power of random sampl6 KB (884 words) - 16:26, 9 May 2010
- ...al ( this is of course easier said than done because we then have to use a filter of the appropriate length and so on, which means another two pages of math) ...h cut off frequencies that represented those weekly variations. A low pass filter is one that passes low frequency signals but attenuates signals that have f13 KB (2,348 words) - 13:25, 2 December 2011
- **[[Practice_Question_5_ECE438F10|Practice Question 5 (filter design)]]9 KB (1,221 words) - 11:00, 22 December 2014
- *Week 7-8: Filtering (Systems defined by Difference equations, Filter Design, DFT view of Filtering) ***Prof. Bouman's lecture notes on digital Filter design: [https://engineering.purdue.edu/~bouman/ece438/lecture/module_1/1.79 KB (1,331 words) - 07:15, 29 December 2010
- ...system is an all pole filter cascaded with a time delay. The poles of this filter determine the location of the local maxima of the voiced phonemes we pronou1 KB (151 words) - 12:53, 8 November 2010
- ...o get rid of aliases, what is the cutoff frequency of digital LPF(Low-Pass Filter)?}\,\!</math>2 KB (315 words) - 10:39, 11 November 2011
- ...ore, the cut-off frequency, <math>f_c</math> of the discrete-time low-pass filter (LPF) is <math>f_c=\frac{\pi}{L}</math>, in general.3 KB (467 words) - 19:52, 20 September 2010
- ...s relationship, we concluded that, under certain circumstances, a low-pass filter could be applied to this upsampling so to obtain the signal1 KB (220 words) - 16:07, 22 September 2010
- ...rder to get rid of aliasing, what is the cut-off frequency of the low pass filter? Explain your answer. Assume that the input signal X(f) and the continuous time filter H(f) are both band limited to 1/(2T).2 KB (373 words) - 10:41, 11 November 2011
- Thus, the cut-off frequency of the LP filter is <math>\frac{\pi}{L}</math>.666 B (121 words) - 12:21, 29 September 2010
- ...it affects the reasoning here). And finally sending it through a low pass filter, the "extra" rects get filtered out so when you end up with non-zero freque5 KB (778 words) - 09:11, 1 October 2010
- ...plies that the reconstructed signal <math>x_r(t)</math> is the output of a filter when we input the impulse train of <math>x(t)</math> with period <math>T</m ...math>\text{sinc}(t/T)</math>, whose frequency response is a ideal low-pass filter with the cut-off frequency of <math>1/(2T)</math>.4 KB (751 words) - 04:56, 2 October 2011
- Q2. Suppose that the LTI filter <math>h_1</math> satifies the following difference equation between input < Then, find the inverse LTI filter <math>h_2</math> of <math>h_1</math>, which satisfies the following relatio3 KB (462 words) - 10:42, 11 November 2011
- ...aracteristics (Filter A) and another filer with band-pass characteristics (Filter B). The behavior of these two filters will be further studied when we retur867 B (122 words) - 16:21, 8 October 2010
- ...at a specific low-pass filter (filter A) and a specific band-pass filter (filter B). We noticed the two different ways of writing the transfer function (as2 KB (393 words) - 07:25, 25 October 2010
- equation (moving average filter) :a. Find the impulse response h[n] for this filter. Is it of finite or infinite duration?4 KB (661 words) - 11:22, 30 October 2011
- Q1. Consider the following second order FIR filter with the two zeros on the unit circle as shown below. The transfer function for this filter is given by <math> H(z) = (1-e^{j\theta}z^{-1})(1-e^{-j\theta}z^{-1})=1-2\c3 KB (480 words) - 10:42, 11 November 2011
- d. If we further look at the frequency response of this filter, :therefore, when <math>\theta=\pi/2</math>, it is a bandstop filter.2 KB (437 words) - 12:00, 19 October 2010
- Today in the lecture, we continued talking about filters and filter design using the transfer function. It seems like many students find it dif1 KB (174 words) - 03:53, 21 October 2010
- ...y with period <math>2 \pi</math>. It is important to remember this for any filter in discrete-time. In the last part of the lecture, we saw how the heat equa *[[ECE_301_Fall_2007_mboutin_Filter_Types|Ideal filter types in continuous-time]]: do not forget to repeat periodically every <ma2 KB (260 words) - 12:42, 22 October 2010
- Topic: Filter Design Define a two-pole band-pass filter such that2 KB (322 words) - 13:00, 26 November 2013
- d. Determineif the filter represented by the difference equation is FIR or IIR. Give reasons for your3 KB (479 words) - 10:42, 11 November 2011
- d. Filter represented by this difference equation is IIR. Because the transfer functi2 KB (441 words) - 05:42, 28 October 2010
- Q2. Consider a causal FIR filter of length M = 2 with impulse response Q5. Define a two-zero band-stop filter such that3 KB (462 words) - 10:42, 11 November 2011
- Suppose the transfer function of the filter has the form Where <math>z_1,z_2</math> are zeros of the filter.2 KB (279 words) - 17:23, 3 November 2010
- Q1. Consider a causal FIR filter of length M = 2 with impulse response3 KB (561 words) - 10:43, 11 November 2011
- Q4. Consider a 3X3 FIR filter with coefficients h[m,n] <br/> a. Find a difference equation that can be used to implement this filter.<br/>3 KB (398 words) - 10:43, 11 November 2011
- ...that <math class="inline">h\left(t\right)</math> acts as a crude low-pass filter that attenuates high-frequency power.3 KB (498 words) - 07:16, 1 December 2010
- d. Describe how the filter behaves when <math>\lambda</math> is positive and large. <br/> e. Describe how the filter behaves when <math>\lambda</math> is negative and bigger than -1. <br/>3 KB (515 words) - 10:43, 11 November 2011
- d. For large values of <math>\lambda</math>, the filter performs sharpening.<br/> e. For -1 < <math>\lambda</math> < 0, the filter performs blurring.<br/>2 KB (275 words) - 13:34, 28 November 2010
- ...quency domain perspective. We looked an an example in detail (the low-pass filter illlustrated on top of [[ECE_438_Fall_2009_mboutin_plotCSFTofbasicfilters|t808 B (107 words) - 10:56, 29 November 2010
- ...ral input signal <math>x[m,n]</math> we get the difference equation of the filter. b. Place the center of filter (i.e. where m=0,n=0) upon the pixel of image. Multiply h[m,n] with x[m,n] o2 KB (391 words) - 07:16, 30 November 2010
- ...lting in huge disturbance to the game, I feel strong obligated to design a filter to remove the noise. Each filter is used to remove corresponding frequency component.3 KB (409 words) - 08:53, 11 November 2013
- ...ms]] from [[2011 Spring ECE 301 Boutin|ECE301]]!) are utilized to convert, filter, and combine these signals and produce the images used in diagnostics.17 KB (2,368 words) - 10:53, 6 May 2012
- ...y that the Fourier transform is a non-zero constant multiple of a low-pass filter with gain 1 and cutoff <math>3 \pi</math> and conclude from there; you woul3 KB (431 words) - 10:28, 11 November 2011
- ...'c''</sub>''t'').</span> Then feed the resulting signal through a low pass filter with a gain of 2 and a cutoff frequency of <span class="texhtml">ω<sub>''c b) Multiply by cos(w<sub>c</sub>t) then pass it through a Low Pass Filter with a gain of 2 and a cutoff f of w<sub>c</sub>2 KB (400 words) - 10:31, 11 November 2011
- ...ignal can be recovered by filtering the sampled signal using the following filter: ...DT, one must first convert the samples to a pulse-train, and then low-pass filter. -pm </font>9 KB (1,462 words) - 07:01, 22 April 2011
- ...</sub> = 1000π</span> and gain 2. The frequency response of this low pass filter is: Note that the cut-off frequency of the low pass filter can actually be anywhere between <span class="texhtml">ω<sub>''M''</sub></12 KB (2,109 words) - 05:58, 22 April 2011
- **[[Practice_Question_5_ECE438F10|Practice Question 5 (filter design)]] ....m.zip zpgui3.m] A MATLAB GUI showing the effect of poles and zeros during filter design.10 KB (1,359 words) - 03:50, 31 August 2013
- *[[ECE438_Week9_Quiz|LTI system and filter design]] <br/>900 B (121 words) - 10:39, 11 November 2011
- *Week (7)-8: Filtering (Systems defined by Difference equations, Filter Design, DFT view of Filtering) ***Prof. Bouman's lecture notes on digital Filter design: [https://engineering.purdue.edu/~bouman/ece438/lecture/module_1/1.79 KB (1,341 words) - 03:52, 31 August 2013
- =Image processing on an Android phone - Lowpass filter an Image - C++ code= ...ore used to. The filtering itself is not really complicated: We just use a filter in form of a matrix and perform a convolution. As this is not really an iss3 KB (486 words) - 11:58, 20 April 2012
- ...revious lecture, we observed that, under certain circumstances, a low-pass filter could be applied to this upsampling so to obtain the signal1 KB (213 words) - 06:24, 11 September 2013
- ...this by computing the frequency response and the transfer function of that filter. In particular, we noted how the location of the poles and the zeros of the [[Category:Filter]]998 B (143 words) - 06:27, 11 September 2013
- **[[Practice_Question_5_ECE438F10|Practice Question 5 (filter design)]]9 KB (1,273 words) - 20:52, 15 October 2011
- ...n the location of the poles and the zeros of the transfer function of this filter and the amplitude of its frequency response. [[Category:Filter]]953 B (132 words) - 06:27, 11 September 2013
- equation (moving average filter) :a. Find the impulse response h[n] for this filter. Is it of finite or infinite duration?5 KB (916 words) - 03:56, 31 August 2013
- Today we discussed the relevance of "filter design" in today's world, including some open problems for which research i [[Category:Filter]]1 KB (221 words) - 06:28, 11 September 2013
- ...e unit impulse response of an ideal filter in order to obtain a causal FIR filter. A MATLAB plot of the example presented in class can be viewed on [[ECE_4381 KB (164 words) - 06:30, 11 September 2013
- ::#low pass filter the repeated part7 KB (1,108 words) - 06:02, 23 September 2014
- A LPF (low-pass-filter) will usually be used before down-sampling to reduce aliasing. In this  ...f 4, 8, and 16. In this project we are using FIR (finite impulse response) filter.<br>The audio signal we use is part of Waving Flag, the theme song of 201010 KB (1,707 words) - 10:44, 6 May 2012
- *Filter design **[[Practice_Question_5_ECE438F10|Practice question on filter design]]6 KB (801 words) - 22:04, 19 April 2015
- ...n]. We then proceed to demonstrate how to use the formula using an average filter and a 6x6 digital image. The issue of the boundary conditions was discussed2 KB (301 words) - 06:32, 11 September 2013
- ...sample by factor of 5 then down sample by factor 3. To avoid aliasing, the filter was build by MATLAB embedded function "fir1" with order= 20, cut-of frequen2 KB (389 words) - 06:37, 25 September 2013
- ...arate it. We then considered another filter (edge detector). Although that filter is not separable, we were able to write it as a sum of two separable filter2 KB (213 words) - 06:32, 11 September 2013
- Consider the following FIR filter: a) Write a difference equation that can be used to implement this filter.2 KB (270 words) - 03:59, 31 August 2013
- ...other type of filter introduced by Perona and Malik. We observed that this filter is not linear and that it allows one to smooth out an image without blurrin1 KB (157 words) - 06:33, 11 September 2013
- Therefore the filter can be separate into two 1-D filters.3 KB (355 words) - 13:42, 4 December 2011
- =Image processing on an Android phone - Lowpass filter an Image - Java code= ...roject was to lowpass filter an image on an Android phone using a Gaussian filter. Therefore it should be possible to either take a picture with the integrat7 KB (1,278 words) - 11:57, 20 April 2012
- ...tion, signal mixing, encoding and decoding of audio signals, and real-time filter implementation. Additionally, I was involved with the development of a filt4 KB (676 words) - 12:21, 9 February 2012
- ...ng a proof of the visit to a doctor with a date and time on it, so it will filter out such students who use this excuse frequently and know all the details h6 KB (1,023 words) - 09:24, 16 March 2012
- ...cy filtering removes noise, but also blurs images as a result. The unsharp filter accentuates the edges of images, in an emboss like feature. The parts of th1 KB (196 words) - 17:45, 21 April 2013
- so filter out them. /* Filter in daily returns on earning reports dates */11 KB (1,577 words) - 08:35, 23 April 2012
- 3. \text{ Multiply step 2 by the filter } H(\rho) = |\rho| = f_c \left [ rect(\frac{f}{2f_c}) - \Lambda(\frac{f}{f_ 2. \text{ Filter the projections } \rho_{\theta}(r) \text{ with } h(r) \text{, where } H(\rh17 KB (2,783 words) - 01:51, 31 March 2015
- ...t from probabilistic noise-reducing filters such as the Bayesian or Kalman filter.8 KB (1,176 words) - 15:15, 1 May 2016
- ** Low Pass Filter: Smoothing (less sharp edges or details but reduces some static noise) ** High Pass Filter: Sharping (clear edge and enhance details but also emphasize noise)3 KB (555 words) - 08:09, 9 April 2013
- b) Create Gaussian filter of size 5x5 with mean 0 and standard deviation 3. c) Plot Fourier Transform of filter’s impulse response in 3D.4 KB (573 words) - 10:15, 15 May 2013
- ...l has probability of {1/10,2/10,4/10,2/10,1/10}. The signal goes through a filter, Z=2X^2+1. Z: output after the filter<br>2 KB (299 words) - 18:13, 27 February 2013
- filter <math>h(m,n)</math> is a <math>(2N+1)\times(2N+1)</math> filter, and for each location we need 2 multiplies, so in total, we need <math>2(2 ...ated offline, if we consider that <math> a_j b_i </math> are merged in the filter <math> h(m,n)</math>, then will need <math> (2N+1)^2 </math> multiplies to4 KB (739 words) - 10:07, 13 September 2013
- ''b) Create Gaussian filter of size 5x5 with mean 0 and standard deviation 3.'' ''c) Plot Fourier Transform of filter’s impulse response in 3D.''2 KB (348 words) - 10:50, 11 March 2013
- The upper is the Gaussian filter, while bottom is the unsharp.1 KB (174 words) - 11:34, 11 March 2013
- ...eed a 2D convolution function. So you can take the 2d fft multiply by your filter and invert. However this seems like allot of work, and I have never implime ...hat way. The example file gives you an idea of how to do this for a simple filter. Good luck!10 KB (1,756 words) - 08:05, 9 April 2013
- ...this process. These artifacts can be reduced by filtering with a high pass filter prior to back-projection. For this reason, the process is also commonly ref # Filter the projections to obtain <math>g_{\theta}(r) = h(r)*p_{\theta}(r)</math>.9 KB (1,486 words) - 07:25, 26 February 2014
- ...function is basically the frequency response of an ideal digital low pass filter. So if you were to build a low pass filter, its impulse response would be a sampled sinc in time as shown in figure 6.10 KB (1,726 words) - 07:26, 26 February 2014
- ....m.zip zpgui3.m] A MATLAB GUI showing the effect of poles and zeros during filter design.8 KB (1,096 words) - 06:44, 14 December 2013
- *Week (7)-8: Filtering (Systems defined by Difference equations, Filter Design, DFT view of Filtering) ***Prof. Bouman's lecture notes on digital Filter design: [https://engineering.purdue.edu/~bouman/ece438/lecture/module_1/1.79 KB (1,353 words) - 09:04, 11 November 2013
- ...nd last questions in part 2 of this lab, does anybody know they the RLC BP filter isn't used for the two-tone test, or why the '''in-band''' third-order inte ...in frequency) to the input signal. Thus, it would be difficult to design a filter to separate the input frequencies from these in-band distortion terms.1 KB (204 words) - 10:09, 15 January 2014
- ...gnal in the frequency domain first. (Recall that you just need to low-pass-filter the ideal sampling.) Then invert the Fourier transform to get the reconstru2 KB (362 words) - 13:59, 26 September 2013
- ...(t) was band-limited with <math>f_{max}<\pi /D</math> and apply a low-pass filter with gain D and cut-off <math>\pi/D</math> to <math>x_2[n]</math>.2 KB (335 words) - 05:55, 27 September 2013
- 1. Use the signal generators and filters in the lab to generate and filter noise and various types of periodic signals. ...rfectly band-limited to 0-10MHz, is passed through a perfectly rectangular filter of bandwidth 18 kHz, gain 3 dB, and center frequency 455 kHz. If the RMS vo14 KB (2,228 words) - 12:03, 15 January 2014
- ...r. Be sure to show how you calculated the cutoff frequency for the digital filter. ...is. Be sure to show how you calculated the cutoff frequency of the digital filter.3 KB (480 words) - 09:13, 27 September 2013
- We need a high pass filter that filters our everything below 60 Hz. ...equency component. In order to remove the annual cycle, we need a low pass filter.6 KB (1,018 words) - 12:18, 30 September 2013
- ...izing a few important facts about LTI systems, we defined a first filter, "Filter A", which we found had low-pass characteristics.2 KB (294 words) - 05:58, 14 October 2013
- ...ssion on the topic of filtering, we defined another simple filter, called "filter B", and analysed its properties using the concept of frequency response and2 KB (300 words) - 05:47, 16 October 2013
