As above lets Fs is sampling frequency, L is signal's length and t is time range.
As using mdwtdec in Matlab in order to decompose multi-raw signal into specific frequency band, I just notice that decomposed signal's length at 1st level is split into half, and keep slit into half of 1st level signal at 2nd level.
Raw signal's time range calculation: t = 0 --> (L/Fs)
My question is in every decomposition level the Sampling frequency Fs is still the same? and at every decomposition level how I can calculate the time range of each Detail and Approximation coefficient.
Also as verify the frequency band of Discrete Wavelet Transform I applied FFT at each level following this post: https://jp.mathworks.com/help/matlab/ref/fft.html?lang=en
According to this post my first question need to be answered.
Thank you very much.
I'm pretty positive that in discrete wavelet transform, the time series data or signals, if we wish, would be downsampled by a factor of 2, which means that if we would be having 2^10 or 1024 datapoints in our original time series data, in the first level, it would be divided into 2 and our level one sampling frequency would be 2^9 or 512, in second level would decrease to 256, and so on.
However, in continuous wavelet transform, it would most likely remain the same.
Based on the references, I copy some codes here that you might want to test and see, you might want to reduce the number of levels and Fs here and you can define your own x if you wish:
Fs = 1e6;
t = 0:1/Fs:1-1/Fs;
x = cos(2*pi*50*t);
[C,L] = wavedec(x,15,'db4');
details = detcoef(C,L,'cells');
d14recon = wrcoef('d',C,L,'db4',14);
plot(d14recon,'k');
d13recon = wrcoef('d',C,L,'db4',13);
hold on;
plot(d13recon,'r'); %look how small the amplitude is
a13recon = wrcoef('a',C,L,'db4',13);
plot(a13recon,'b');
Useful Links:
I'm not expert about it, you can likely read more about it and find out your desired information. There are also lots of YouTube videos about it.
Discrete wavelet transform relation to sampling frequency of the signal
Single-level discrete 2-D wavelet transform
Related
I am trying to apply a high-pass filter to a signal (column or row vector) consisting of 1-pixel-wide lines taken from a black-and-white image. I know the resolution of the image (res in the code below, given in mm/pixel). How can I filter these line data in MATLAB to discard certain low frequencies (waviness) or large wavelengths, say >10 mm, using a Butterworth filter or any other?
Line data are not centered at zero.
Fs = 1; % I do not know if this assumption is correct for the image.
Fn = Fs/2; % Nyquist frequency.
lambda = 10; % Cut-off wavelength in mm, given.
samples_in_lambda = lambda/res; % divide by resolution to get samples.
fc = 1/samples_in_lambda; % Cut-off frequency from lambda.
I tried : [z, p, k] = butter(9, fc/fn, 'high'); % I see the filter is high pass on plotting.
Can I filter the line data using the above given and assumed values? If not, is there a way that I can filter the data using a cut-off wavelength?
The highest linear spatial frequency you can represent without aliasing is 1 wave cycle per 2 pixels. This means a spatial Nyquist frequency of 1 wave cycle per 2*(res*1e-3) meters, or 1000/(res*2) reciprocal meters. (Confront this with temporal frequencies, which are measured in reciprocal seconds a.k.a. hertz).
In terms of wavelengths: the shortest wave you can represent without aliasing is 2 pixels long per wave cycle. This means a spatial "Nyquist wavelength" of res*2e-3 meters. (Confront this with temporal "wavelengths" a.k.a. periods, which are measured in seconds.)
If you want to set a cutoff wavelength of 10 mm, that corresponds to a spatial frequency of 100 reciprocal meters. Since the butter() function takes as its second input argument (Wn, the cutoff frequency) an arbitrary fraction of the (spatial) Nyquist frequency (the MATLAB documentation calls it "half the sampling rate"), you merely need to set Wn=100/(1000/(res*2)), i.e. Wn=res/5.
Even though your definition of the spatial sampling frequency is not quite correct (unless you are intentionally measuring it in reciprocal pixels), your final result ended up being equivalent to Wn=res/5, so you should be fine using the call to butter() that you indicated.
This question relates to SciPy's Short-time Fourier Transform function for signal processing.
For some reason I don't understand, the size of the output 'array of sample frequencies' is exactly equal to the hop size. From the documentation:
nperseg : int, optional
Length of each segment. Defaults to 256.
noverlap : int, optional
Number of points to overlap between segments. If None, noverlap = nperseg // 2. Defaults to None. When specified, the COLA constraint must be met (see Notes below).
f : ndarray
Array of sample frequencies.
hop size H = nperseg - noverlap
I'm new to signal processing and Fourier transforms, but as far as I understand a STFT is just chopping an audio file into segments ('time frames') on which you perform a Fourier transform. So if I want to do a STFT on 100 time frames, I'd expect the output to be a matrix of size 100 x F, where F is an array of measured frequencies ('measured' probably isn't the right word here but you know what I mean).
This is kinda what SciPy's implementation does, but the size of f here is what bothers me. It's supposed to be an array describing the different frequencies, like [0Hz 500Hz 1000Hz], and it does, but for some reasons its size exactly the same as the hop size. If the hop size is 700, the number of measured frequencies is 700.
The hop size is the number of samples (i.e. time) between each time frame, and is correctly calculated as H = nperseg - noverlap, but what does this have to do with the frequency array?
Edit: Related to this question
An FFT is an square matrix transform from one orthogonal basis to another of the same dimension. This is because N is the exact number of orthogonal (e.g. that don't interfere with one another) complex sinusoids that fit in a time domain vector of length N.
A longer time vector can contain more frequency information (e.g. it's hard to tell 2 frequencies apart using just 3 sample points, but much easier with 3000 samples, etc.)
You can zero-pad your short time vector of length N to use a longer FFT, but that is identical to interpolating a nice curve between N frequency points, which makes all the FFT results interdependent.
For many purposes (visualization, etc.) an STFT is overlapped, where the adjacent segments share some overlapped data instead of just being end-to-end. This gives better time locality (e.g. the segments can be spaced closer but still be long enough so that each one can provide the frequency resolution required).
I am trying to implement a frequency domain phase shift but there are few points on which I am not sure.
1- I am able to get a perfect reconstruction from a sine or sweep signal using a hanning window with a hop size of 50%. Nevertheless, how should I normalise my result when using a hop size > 50%?
2- When shifting the phase of low frequency signals (f<100, window size<1024, fs=44100) I can clearly see some non-linearity in my result. Is this because of the window size being to short for low frequencies?
Thank you very much for your help.
clear
freq=500;
fs=44100;
endTime=0.02;
t = 1/fs:1/fs:(endTime);
f1=linspace(freq,freq,fs*endTime);
x = sin(2*pi*f1.*t);
targetLength=numel(x);
L=1024;
w=hanning(L);
H=L*.50;% Hopsize of 50%
N=1024;
%match input length with window length
x=[zeros(L,1);x';zeros(L+mod(length(x),H),1)];
pend=length(x)- L ;
pin=0;
count=1;
X=zeros(N,1);
buffer0pad= zeros(N,1);
outBuffer0pad= zeros(L,1);
y=zeros(length(x),1);
delay=-.00001;
df = fs/N;
f= -fs/2:df:fs/2 - df;
while pin<pend
buffer = x(pin+1:pin+L).*w;
%append zero padding in the middle
buffer0pad(1:(L)/2)=buffer((L)/2+1: L);
buffer0pad(N-(L)/2+1:N)=buffer(1:(L)/2);
X = fft(buffer0pad,N);
% Phase modification
X = abs(X).*exp(1i*(angle(X))-(1i*2*pi*f'*delay));
outBuffer=real(ifft(X,N));
% undo zero padding----------------------
outBuffer0pad(1:L/2)=outBuffer(N-(L/2-1): N);
outBuffer0pad(L/2+1:L)=outBuffer(1:(L)/2);
%Overlap-add
y(pin+1:pin+L) = y(pin+1:pin+L) + outBuffer0pad;
pin=pin+H;
count=count+1;
end
%match output length with original input length
output=y(L+1:numel(y)-(L+mod(targetLength,H)));
figure(2)
plot(t,x(L+1:numel(x)-(L+mod(targetLength,H))))
hold on
plot(t,output)
hold off
Anything below 100 Hz has less than two cycles in your FFT window. Note that a DFT or FFT represents any waveform, including a single non-integer-periodic sinusoid, by possibly summing up of a whole bunch of sinusoids of very different frequencies. e.g. a lot more than just one. That's just how the math works.
For a von Hann window containing less than 2 cycles, these are often a bunch of mostly completely different frequencies (possibly very far away in terms of percentage from your low frequency). Shifting the phase of all those completely different frequencies may or may not shift your windowed low frequency sinusoid by the desired amount (depending on how different in frequency your signal is from being integer-periodic).
Also for low frequencies, the complex conjugate mirror needs to be shifted in the opposite direction in phase in order to still represent a completely real result. So you end up mixing 2 overlapped and opposite phase changes, which again is mostly a problem if the low frequency signal is far from being integer periodic in the DFT aperture.
Using a longer window in time and samples allows more cycles of a given frequency to fit inside it (thus possibly needing a lesser power of very different frequency sinusoids to be summed up in order to compose, make up or synthesize your low frequency sinusoid); and the complex conjugate is farther away in terms of FFT result bin index, thus reducing interference.
A sequence using any hop of a von Hann window that in 50% / (some-integer) in length is non-lossy (except for the very first or last window). All other hop sizes modulate or destroy information, and thus can't be normalized by a constant for reconstruction.
I am confused about the terminology used in scipy.signal.periodogram, namely:
scaling : { 'density', 'spectrum' }, optional
Selects between computing the power spectral density ('density')
where Pxx has units of V*2/Hz if x is measured in V and computing
the power spectrum ('spectrum') where Pxx has units of V*2 if x is
measured in V. Defaults to 'density'
(see: http://docs.scipy.org/doc/scipy/reference/generated/scipy.signal.periodogram.html)
1) a few tests show that result for option 'density' is dependent on signal and window length and sampling frequency (grows when signal length increases). How come? I would say that it is exactly density that should be not dependent on these things. If I take a longer signal I should just get more accurate estimation, not different result. Not to mention that dependence on window length is also very surprising.
Result diverges in the limit of infinite signal, which could be a feature of energy, but not power. Shouldn't the periodogram converge to real theoretical PSD when length increases? If, so, am I supposed to perform another normalisation outside of the signal.periodogram method?
2) to the contrary I see that alternative option 'spectrum' gives what I would previously call Power Spectrum Density, that is, it gives a resuls independent on window segment and window length and consistent with theoretical calculation. For instance for Asin(2PIft) a two sided solution yields two peaks at -f and f, each of height 0.25*A^2.
There is a lot of literature on this subject, but I get an impression that also there is a lot of incompatibile terminology, so I will be thankful for any clarification. The straightforward question is how to interpret these options and their units. (I am used to seeing V^2/Hz which are labeled "Power Spectrum Density").
Let's take a real array called data, of length N, and with sampling frequency fs. Let's call the time bin dt=1/fs, and T = N * dt. In frequency domain, the frequency bin df = 1/T = fs/N.
The power spectrum PS (scaling='spectrum' in scipy.periodogram) is calculated as follow:
import numpy as np
import scipy.fft as fft
dft = fft.fft(data)
PS = np.abs(dft)**2 / N ** 2
It has the units of V^2. It can be understood as follow. By analogy to the continuous Fourier transform, the energy E of the signal is:
E := np.sum(data**2) * dt = 1/N * np.sum(np.abs(dft)**2) * dt
(by Parseval's theorem). The power P of the signal is the total energy E divided by the duration of the signal T:
P := E/T = 1/N**2 * np.sum(np.abs(dft)**2)
The power P only depends on the Discrete Fourier Transform (DFT) and the number of samples N. Not directly on the sampling frequency fs or signal duration T. And the power per frequency channel, i.e., power spectrum SP, is thus given by the formula above:
PS = np.abs(dft)**2 / N ** 2
For the power spectrum density PSD (scaling='density' in scipy.periodogram), one needs to divide the PS by the frequency bin of the DFT, df:
PSD := PS/df = PS * N * dt = PS * N / fs
and thus:
PSD = np.abs(dft)**2 / N * dt
This has the units of V^2/Hz = V^2 * s, and now depends on the sampling frequency. That way, integrating the PSD over the frequency range gives the same result as summing the individual values of the PS.
This should explain the relations that you see when changing the window, sampling frequency, duration.
scipy.signal.peridogram uses the scipy.signal.welch function with 0 overlap. Therefore, the scaling is similar to the one provided by the welch function, density or spectrum.
In case of the density scaling, the amplitude will vary with window length, as the longer the window the higher the frequency resolution e.g. the \Delta_f is smaller. Since the estimated density is the average one, the smaller the \Delta_f the less zero energy is considered in the averaging.
As you have mentioned spectrum scaling is an integration of the energy density over the spectrum to produce the energy. Therefore, the integration over zero values does not affect the final value.
Fourier transform actually requires finite energy in an infinite duration of time series (like a decay). So, If you just make your time series sample longer by "duplicating", the energy will be infinite with an infinite duration.
My main confusion was on the "spectrum" option for scipy.signal.periodogram, which seems to create a constant energy spectrum even when the time series become longer.
Normally, 0.5*A^2=S(f)*delta_f, where S(f) is the power density spectrum. S(f)*delta_f, representing energy is constant if A is constant. But when using a longer duration of time series, delta_f (i.e. incremental frequency) is reduced accordingly, based on FFT procedure. For example, 100s time series will lead to a delta_f=0.01Hz, while 1000s time series will have a delta_f=0.001Hz. S(f) representing density will accordingly change.
I've been playing around a little with the Exocortex implementation of the FFT, but I'm having some problems.
Whenever I modify the amplitudes of the frequency bins before calling the iFFT the resulting signal contains some clicks and pops, especially when low frequencies are present in the signal (like drums or basses). However, this does not happen if I attenuate all the bins by the same factor.
Let me put an example of the output buffer of a 4-sample FFT:
// Bin 0 (DC)
FFTOut[0] = 0.0000610351563
FFTOut[1] = 0.0
// Bin 1
FFTOut[2] = 0.000331878662
FFTOut[3] = 0.000629425049
// Bin 2
FFTOut[4] = -0.0000381469727
FFTOut[5] = 0.0
// Bin 3, this is the first and only negative frequency bin.
FFTOut[6] = 0.000331878662
FFTOut[7] = -0.000629425049
The output is composed of pairs of floats, each representing the real and imaginay parts of a single bin. So, bin 0 (array indexes 0, 1) would represent the real and imaginary parts of the DC frequency. As you can see, bins 1 and 3 both have the same values, (except for the sign of the Im part), so I guess bin 3 is the first negative frequency, and finally indexes (4, 5) would be the last positive frequency bin.
Then to attenuate the frequency bin 1 this is what I do:
// Attenuate the 'positive' bin
FFTOut[2] *= 0.5;
FFTOut[3] *= 0.5;
// Attenuate its corresponding negative bin.
FFTOut[6] *= 0.5;
FFTOut[7] *= 0.5;
For the actual tests I'm using a 1024-length FFT and I always provide all the samples so no 0-padding is needed.
// Attenuate
var halfSize = fftWindowLength / 2;
float leftFreq = 0f;
float rightFreq = 22050f;
for( var c = 1; c < halfSize; c++ )
{
var freq = c * (44100d / halfSize);
// Calc. positive and negative frequency indexes.
var k = c * 2;
var nk = (fftWindowLength - c) * 2;
// This kind of attenuation corresponds to a high-pass filter.
// The attenuation at the transition band is linearly applied, could
// this be the cause of the distortion of low frequencies?
var attn = (freq < leftFreq) ?
0 :
(freq < rightFreq) ?
((freq - leftFreq) / (rightFreq - leftFreq)) :
1;
// Attenuate positive and negative bins.
mFFTOut[ k ] *= (float)attn;
mFFTOut[ k + 1 ] *= (float)attn;
mFFTOut[ nk ] *= (float)attn;
mFFTOut[ nk + 1 ] *= (float)attn;
}
Obviously I'm doing something wrong but can't figure out what.
I don't want to use the FFT output as a means to generate a set of FIR coefficients since I'm trying to implement a very basic dynamic equalizer.
What's the correct way to filter in the frequency domain? what I'm missing?
Also, is it really needed to attenuate negative frequencies as well? I've seen an FFT implementation where neg. frequency values are zeroed before synthesis.
Thanks in advance.
There are two issues: the way you use the FFT, and the particular filter.
Filtering is traditionally implemented as convolution in the time domain. You're right that multiplying the spectra of the input and filter signals is equivalent. However, when you use the Discrete Fourier Transform (DFT) (implemented with a Fast Fourier Transform algorithm for speed), you actually calculate a sampled version of the true spectrum. This has lots of implications, but the one most relevant to filtering is the implication that the time domain signal is periodic.
Here's an example. Consider a sinusoidal input signal x with 1.5 cycles in the period, and a simple low pass filter h. In Matlab/Octave syntax:
N = 1024;
n = (1:N)'-1; %'# define the time index
x = sin(2*pi*1.5*n/N); %# input with 1.5 cycles per 1024 points
h = hanning(129) .* sinc(0.25*(-64:1:64)'); %'# windowed sinc LPF, Fc = pi/4
h = [h./sum(h)]; %# normalize DC gain
y = ifft(fft(x) .* fft(h,N)); %# inverse FT of product of sampled spectra
y = real(y); %# due to numerical error, y has a tiny imaginary part
%# Depending on your FT/IFT implementation, might have to scale by N or 1/N here
plot(y);
And here's the graph:
The glitch at the beginning of the block is not what we expect at all. But if you consider fft(x), it makes sense. The Discrete Fourier Transform assumes the signal is periodic within the transform block. As far as the DFT knows, we asked for the transform of one period of this:
This leads to the first important consideration when filtering with DFTs: you are actually implementing circular convolution, not linear convolution. So the "glitch" in the first graph is not really a glitch when you consider the math. So then the question becomes: is there a way to work around the periodicity? The answer is yes: use overlap-save processing. Essentially, you calculate N-long products as above, but only keep N/2 points.
Nproc = 512;
xproc = zeros(2*Nproc,1); %# initialize temp buffer
idx = 1:Nproc; %# initialize half-buffer index
ycorrect = zeros(2*Nproc,1); %# initialize destination
for ctr = 1:(length(x)/Nproc) %# iterate over x 512 points at a time
xproc(1:Nproc) = xproc((Nproc+1):end); %# shift 2nd half of last iteration to 1st half of this iteration
xproc((Nproc+1):end) = x(idx); %# fill 2nd half of this iteration with new data
yproc = ifft(fft(xproc) .* fft(h,2*Nproc)); %# calculate new buffer
ycorrect(idx) = real(yproc((Nproc+1):end)); %# keep 2nd half of new buffer
idx = idx + Nproc; %# step half-buffer index
end
And here's the graph of ycorrect:
This picture makes sense - we expect a startup transient from the filter, then the result settles into the steady state sinusoidal response. Note that now x can be arbitrarily long. The limitation is Nproc > 2*min(length(x),length(h)).
Now onto the second issue: the particular filter. In your loop, you create a filter who's spectrum is essentially H = [0 (1:511)/512 1 (511:-1:1)/512]'; If you do hraw = real(ifft(H)); plot(hraw), you get:
It's hard to see, but there are a bunch of non-zero points at the far left edge of the graph, and then a bunch more at the far right edge. Using Octave's built-in freqz function to look at the frequency response we see (by doing freqz(hraw)):
The magnitude response has a lot of ripples from the high-pass envelope down to zero. Again, the periodicity inherent in the DFT is at work. As far as the DFT is concerned, hraw repeats over and over again. But if you take one period of hraw, as freqz does, its spectrum is quite different from the periodic version's.
So let's define a new signal: hrot = [hraw(513:end) ; hraw(1:512)]; We simply rotate the raw DFT output to make it continuous within the block. Now let's look at the frequency response using freqz(hrot):
Much better. The desired envelope is there, without all the ripples. Of course, the implementation isn't so simple now, you have to do a full complex multiply by fft(hrot) rather than just scaling each complex bin, but at least you'll get the right answer.
Note that for speed, you'd usually pre-calculate the DFT of the padded h, I left it alone in the loop to more easily compare with the original.
Your primary issue is that frequencies aren't well defined over short time intervals. This is particularly true for low frequencies, which is why you notice the problem most there.
Therefore, when you take really short segments out of the sound train, and then you filter these, the filtered segments wont filter in a way that produces a continuous waveform, and you hear the jumps between segments and this is what generates the clicks you here.
For example, taking some reasonable numbers: I start with a waveform at 27.5 Hz (A0 on a piano), digitized at 44100 Hz, it will look like this (where the red part is 1024 samples long):
So first we'll start with a low pass of 40Hz. So since the original frequency is less than 40Hz, a low-pass filter with a 40Hz cut-off shouldn't really have any effect, and we will get an output that almost exactly matches the input. Right? Wrong, wrong, wrong – and this is basically the core of your problem. The problem is that for the short sections the idea of 27.5 Hz isn't clearly defined, and can't be represented well in the DFT.
That 27.5 Hz isn't particularly meaningful in the short segment can be seen by looking at the DFT in the figure below. Note that although the longer segment's DFT (black dots) shows a peak at 27.5 Hz, the short one (red dots) doesn't.
Clearly, then filtering below 40Hz, will just capture the DC offset, and the result of the 40Hz low-pass filter is shown in green below.
The blue curve (taken with a 200 Hz cut-off) is starting to match up much better. But note that it's not the low frequencies that are making it match up well, but the inclusion of high frequencies. It's not until we include every frequency possible in the short segment, up to 22KHz that we finally get a good representation of the original sine wave.
The reason for all of this is that a small segment of a 27.5 Hz sine wave is not a 27.5 Hz sine wave, and it's DFT doesn't have much to do with 27.5 Hz.
Are you attenuating the value of the DC frequency sample to zero? It appears that you are not attenuating it at all in your example. Since you are implementing a high pass filter, you need to set the DC value to zero as well.
This would explain low frequency distortion. You would have a lot of ripple in the frequency response at low frequencies if that DC value is non-zero because of the large transition.
Here is an example in MATLAB/Octave to demonstrate what might be happening:
N = 32;
os = 4;
Fs = 1000;
X = [ones(1,4) linspace(1,0,8) zeros(1,3) 1 zeros(1,4) linspace(0,1,8) ones(1,4)];
x = ifftshift(ifft(X));
Xos = fft(x, N*os);
f1 = linspace(-Fs/2, Fs/2-Fs/N, N);
f2 = linspace(-Fs/2, Fs/2-Fs/(N*os), N*os);
hold off;
plot(f2, abs(Xos), '-o');
hold on;
grid on;
plot(f1, abs(X), '-ro');
hold off;
xlabel('Frequency (Hz)');
ylabel('Magnitude');
Notice that in my code, I am creating an example of the DC value being non-zero, followed by an abrupt change to zero, and then a ramp up. I then take the IFFT to transform into the time domain. Then I perform a zero-padded fft (which is done automatically by MATLAB when you pass in an fft size bigger than the input signal) on that time-domain signal. The zero-padding in the time-domain results in interpolation in the frequency domain. Using this, we can see how the filter will respond between filter samples.
One of the most important things to remember is that even though you are setting filter response values at given frequencies by attenuating the outputs of the DFT, this guarantees nothing for frequencies occurring between sample points. This means the more abrupt your changes, the more overshoot and oscillation between samples will occur.
Now to answer your question on how this filtering should be done. There are a number of ways, but one of the easiest to implement and understand is the window design method. The problem with your current design is that the transition width is huge. Most of the time, you will want as quick of transitions as possible, with as little ripple as possible.
In the next code, I will create an ideal filter and display the response:
N = 32;
os = 4;
Fs = 1000;
X = [ones(1,8) zeros(1,16) ones(1,8)];
x = ifftshift(ifft(X));
Xos = fft(x, N*os);
f1 = linspace(-Fs/2, Fs/2-Fs/N, N);
f2 = linspace(-Fs/2, Fs/2-Fs/(N*os), N*os);
hold off;
plot(f2, abs(Xos), '-o');
hold on;
grid on;
plot(f1, abs(X), '-ro');
hold off;
xlabel('Frequency (Hz)');
ylabel('Magnitude');
Notice that there is a lot of oscillation caused by the abrupt changes.
The FFT or Discrete Fourier Transform is a sampled version of the Fourier Transform. The Fourier Transform is applied to a signal over the continuous range -infinity to infinity while the DFT is applied over a finite number of samples. This in effect results in a square windowing (truncation) in the time domain when using the DFT since we are only dealing with a finite number of samples. Unfortunately, the DFT of a square wave is a sinc type function (sin(x)/x).
The problem with having sharp transitions in your filter (quick jump from 0 to 1 in one sample) is that this has a very long response in the time domain, which is being truncated by a square window. So to help minimize this problem, we can multiply the time-domain signal by a more gradual window. If we multiply a hanning window by adding the line:
x = x .* hanning(1,N).';
after taking the IFFT, we get this response:
So I would recommend trying to implement the window design method since it is fairly simple (there are better ways, but they get more complicated). Since you are implementing an equalizer, I assume you want to be able to change the attenuations on the fly, so I would suggest calculating and storing the filter in the frequency domain whenever there is a change in parameters, and then you can just apply it to each input audio buffer by taking the fft of the input buffer, multiplying by your frequency domain filter samples, and then performing the ifft to get back to the time domain. This will be a lot more efficient than all of the branching you are doing for each sample.
First, about the normalization: that is a known (non) issue. The DFT/IDFT would require a factor 1/sqrt(N) (apart from the standard cosine/sine factors) in each one (direct an inverse) to make them simmetric and truly invertible. Another possibility is to divide one of them (the direct or the inverse) by N, this is a matter of convenience and taste. Often the FFT routines do not perform this normalization, the user is expected to be aware of it and normalize as he prefers. See
Second: in a (say) 16 point DFT, what you call the bin 0 would correspond to the zero frequency (DC), bin 1 low freq... bin 4 medium freq, bin 8 to the highest frequency and bins 9...15 to the "negative frequencies". In you example, then, bin 1 is actually both the low frequency and medium frequency. Apart from this consideration, there is nothing conceptually wrong in your "equalization". I don't understand what you mean by "the signal gets distorted at low frequencies". How do you observe that ?