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Try looking at the frequency content of a few other signals. Note that the fall signal happens to have an even length, so N/2 is an integer. If the length is odd, you may have indexing problems, so it is easiest to just omit the last sample, as in x=x(1:length(x)-1); .

After you make modifications of a signal in the frequency domain, you typically want to get back to the time domain. The MATLAB command ifft will accomplish this task. >>xnew = real(ifft(X)); You need the real command because the inverse Fourier transform returns a vector that is complex-valued, since some changes that you make in the frequence domain could result in that. If your changes maintain complex symmetry in the frequency domain, then the imaginary components should be zero (or very close), but you still need to get rid of them if you want to use the sound command to listen to your signal.

Low-pass filtering

An ideal low-pass filter eliminates high frequency components entirely, as in: H L i d e a l ( ω ) = { 1 | ω | B 0 | ω | > B } MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIeadaqhaaWcbaGaamitaaqaaiaadMgacaWGKbGaamyzaiaadggacaWGSbaaaOGaaiikaiabeM8a3jaacMcacqGH9aqpdaGadaqaauaabeqaciaaaeaacaaIXaaabaWaaqWaaeaacqaHjpWDaiaawEa7caGLiWoacqGHKjYOcaWGcbaabaGaaGimaaqaamaaemaabaGaeqyYdChacaGLhWUaayjcSdGaeyOpa4JaamOqaaaaaiaawUhacaGL9baaaaa@525A@ A real low-pass filter typically has low but non-zero values for | H L ( ω ) | MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaamisamaaBaaaleaacaWGmbaabeaakiaacIcacqaHjpWDcaGGPaaacaGLhWUaayjcSdaaaa@3DFE@ at high frequencies, and a gradual (rather than an immediate) drop in magnitude as frequency increases. The simplest (and least effective) low-pass filter is given by (e.g. using an RC circuit): H L ( ω ) = α α + j ω , α = cutoff frequency . MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIeadaWgaaWcbaGaamitaaqabaGccaGGOaGaeqyYdCNaaiykaiabg2da9maalaaabaGaeqySdegabaGaeqySdeMaey4kaSIaamOAaiabeM8a3baacaGGSaGaaeiiaiabeg7aHjabg2da9iaabogacaqG1bGaaeiDaiaab+gacaqGMbGaaeOzaiaabccacaqGMbGaaeOCaiaabwgacaqGXbGaaeyDaiaabwgacaqGUbGaae4yaiaabMhacaqGUaaaaa@5620@

This low-pass filter can be implemented in MATLAB using what we know about the Fourier transform. Remember that multiplication in the Frequency domain equals convolution in the time domain. If our signal and filter are both in the frequency domain, we can simply multiply them to produce the result of the system. y ( t ) = x ( t ) h ( t ) Y ( ω ) = X ( ω ) H ( ω ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamyEaiaacIcacaWG0bGaaiykaiabg2da9iaadIhacaGGOaGaamiDaiaacMcacqGHxiIkcaWGObGaaiikaiaadshacaGGPaaabaGaamywaiaacIcacqaHjpWDcaGGPaGaeyypa0JaamiwaiaacIcacqaHjpWDcaGGPaGaamisaiaacIcacqaHjpWDcaGGPaaaaaa@4EBC@ Below is an example of using MATLAB to perform low-pass filtering on the input signal x with the FFT and the filter definition above.

The cutoff of the low-pass filter is defined by the constant a . The low-pass filter equation above defines the filter H in the frequency domain. Because the definition assumes the filter is centered around w = 0, the vector w is defined as such. >>load fall %load in the signal>>x = fall;>>X = fft(x); % get the Fourier transform (uncentered)>>N = length(X);>>a = 100*2*pi;>>w = (-N/2+1:(N/2))*Fs/N*2*pi; % centered frequency vector (rad/s)>>H = a ./ (a + i*w); % generate centered sampling of H>>plot(w/(2*pi),abs(H)) % w converted back to Hz for plotting The plot will show the form of the frequency response of a system that we are used to looking at, but we need to shift it to match the form that the fft gave us for x. >>Hshift = fftshift(H); % uncentered version of H>>Y = X .* Hshift'; % filter the signal

If you are having problems multiplying vectors together, make sure that the vectors are the exact same size. Also, even if two vectors are the same length, they may not be the same size. For example, a row vector and column vector of the same length cannot be multiplied element-wise unless one of the vectors is transposed. The ' operator transposes vectors/matrices in MATLAB.
Now that we have the output of the system in the frequency domain, it must be transformed back to the time domain using the inverse FFT. Play the original and modified sound to see if you can hear a difference. Remember to use the sampling frequency Fs. >>y = real(ifft(Y));>>sound(x, Fs) % original sound>>sound(y, Fs) % low-pass-filtered sound The filter reduced the signal amplitude, which you can hear when you use the sound command but not with the soundsc which does automatic scaling. Replay the sounds with the soundsc and see what other differences there are in the filtered vs. original signals. What changes could you make to the filter to make a greater difference?
Sometimes, you may want to amplify the signal so that it has the same height as the original, e.g., for plotting purposes. >>y = y * (max(abs(x))/max(abs(y)))

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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Abhi
how do they get the third part x = (32)5/4
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
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I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
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I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
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is it a question of log
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Commplementary angles
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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Differences Between Laspeyres and Paasche Indices
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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what is system testing?
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preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
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Azam
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silver nanoparticles could handle the job?
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I'm interested in Nanotube
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Continuous time linear systems laboratory (ee 235). OpenStax CNX. Sep 28, 2007 Download for free at http://cnx.org/content/col10374/1.8
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