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Let's say that I have a signal in Matlab like this
x = cos(2*pi*10*t) + cos(2*pi*20*t) + cos(2*pi*50*t);
And I want to change the values between 20 and 30 hz into 0. How can I do that? I mean, those values generated from the x formula, I want to change them a little bit.
You can do it by performing FFT over x and setting to zero those values that are between 20 and 30 Hz then applying the FFT inverse on the previous values and you should get the signal without those frequencies. However, you may lose valuable information or the signal might just not look as you wish. Therefore, I recommend you to use a "Bandstop filter". The band stop filter will receive the cutoff frequencies (the limit frequencies you want to work with) and some other parameters. The bandstop filter basically removes from the signal the frequencies that you specify. And the good part is that it can be done as easy as doing what follows:
First you have to build the filter. To do so, you need to indicate the filter order which can be defined as you wish. Usually a second order works good. Also, you have to be aware of your sampling rate Fs.
d = designfilt('bandstopiir','FilterOrder',2, ...
'HalfPowerFrequency1',20,'HalfPowerFrequency2',30, ...
'SampleRate',Fs);
Now you only need to apply the filter to your desired signal.
filtered_signal_x = filtfilt(d, x)
Now, filtered_signal_x should not have the frequencies you wanted to delete. By using the bandstop you don't have to mess with the FFT and that kind of stuff and is a way faster so I think its the best option.
You can either use a filter, or you can filter it by yourself by going into Fourier space and explicitly setting the signal on the frequencies you need to zero. After that, you need to go back to the time domain. Here is a code:
t=0:0.01:0.99; % time
x = cos(2*pi*10*t) + cos(2*pi*20*t) + cos(2*pi*50*t); %signal
xf=fftshift(fft(x)); %Fourier signal
N=size(x,2); % Size of the signal
frequency=2*pi*[-N/2:N/2-1]; %frequency range
frequencyrangeplus=find(frequency/(2*pi)>=20 & frequency/(2*pi)<=30); %find positive frequencies in the required range
frequencyrangeminus=find(frequency/(2*pi)<=-20 & frequency/(2*pi)>=-30); %find negative frequencies in the required range
xf(frequencyrangeplus)=0; %set signal to zero at positive frequencies range
xf(frequencyrangeminus)=0; %set signal to zero at nagative frequencies range
xnew=ifft(ifftshift(xf)); %get the new signal in time domain
xcheck= cos(2*pi*10*t) + cos(2*pi*50*t); % to check the code
max(abs(xcheck-xnew)) % maximum difference
I have asked this question on DSP.SE before, but my question has got no attention. Maybe it was not so related to signal processing.
I needed to divide a discrete audio signal into segments to have some statistical processing and analysis on them. Therefore, segments with fixed local mean would be very helpful for my case. Length of segments are predefined, e.g. 512 samples.
I have tried several things. I do use reshape() function to divide audio signal into segments, and then calculate means of every segment as:
L = 512; % Length of segment
N = floor(length(audio(:,1))/L); % Number of segments
seg = reshape(audio(1:N*L,1), L, N); % Reshape into LxN sized matrix
x = mean(seg); % Calculate mean of each column
Subtracting x(k) from each seg(:,k) would make each local mean zero, yet it would distort audio signal a lot when segments are joined back.
So, since mean of hanning window is almost 0.5, substracting 2*x(k)*hann(L) from each seg(:,k) was the first thing I tried. But this time multiplying by 2 (to make the mean of hanning window be almost equal to 1) distorted the neighborhood of midpoints in each segments itself.
Then, I have used convolution by a smaller hanning window instead of multiplying directly, and subtracting these (as shown in figure below) from each seg(:,k).
This last step gives better results, yet it is still not very useful when segments are smaller. I have seen many amazing approaches here on this site for different problems. So I just wonder if there is any clever ways or existing methods to obtain zero local means which distorts an audio signal less. I read that, this property is useful in some decompositions such as EMD. So maybe I need such decompositions?
You can try to use a moving average filter:
x = cumsum(rand(15*512, 1)-0.5); % generate a random input signal
mean_filter = 1/512 * ones(1, 512); % generate a mean filter
mean = filtfilt(mean_filter, 1, x); % filtfilt is used instead of filter to obtain a symmetric moving average.
% plot the result
figure
subplot(2,1,1)
plot(x);
hold on
plot(mean);
subplot(2,1,2)
plot(x - mean);
You can tune the filter by changing the interval of the mean filter. Using a smaller interval, results in lower means inside each interval, but filters also more low frequencies out of your signal.
Today I have stumbled upon a strange outcome in matlab. Lets say I have a sine wave such that
f = 1;
Fs = 2*f;
t = linspace(0,1,Fs);
x = sin(2*pi*f*t);
plot(x)
and the outcome is in the figure.
when I set,
f = 100
outcome is in the figure below,
What is the exact reason of this? It is the Nyquist sampling theorem, thus it should have generated the sine properly. Of course when I take Fs >> f I get better results and a very good sine shape. My explenation to myself is that Matlab was having hardtime with floating numbers but I am not so sure if this is true at all. Anyone have any suggestions?
In the first case you only generate 2 samples (the third input of linspace is number of samples), so it's hard to see anything.
In the second case you generate 200 samples from time 0 to 1 (including those two values). So the sampling period is 1/199, and the sampling frequency is 199, which is slightly below the Nyquist rate. So there is aliasing: you see the original signal of frequency 100 plus its alias at frequency 99.
In other words: the following code reproduces your second figure:
t = linspace(0,1,200);
x = .5*sin(2*pi*99*t) -.5*sin(2*pi*100*t);
plot(x)
The .5 and -.5 above stem from the fact that a sine wave can be decomposed as the sum of two spectral deltas at positive and negative frequencies, and the coefficients of those deltas have opposite signs.
The sum of those two sinusoids is equivalent to amplitude modulation, namely a sine of frequency 99.5 modulated by a sine of frequency 1/2. Since time spans from 0 to 1, the modulator signal (whose frequency is 1/2) only completes half a period. That's what you see in your second figure.
To avoid aliasing you need to increase sample rate above the Nyquist rate. Then, to recover the original signal from its samples you can use an ideal low pass filter with cutoff frequency Fs/2. In your case, however, since you are sampling below the Nyquist rate, you would not recover the signal at frequency 100, but rather its alias at frequency 99.
Had you sampled above the Nyquist rate, for example Fs = 201, the orignal signal could ideally be recovered from the samples.† But that would require an almost ideal low pass filter, with a very sharp transition between passband and stopband. Namely, the alias would now be at frequency 101 and should be rejected, whereas the desired signal would be at frequency 100 and should be passed.
To relax the filter requirements you need can sample well above the Nyquist rate. That way the aliases are further appart from the signal and the filter has an easier job separating signal from aliases.
† That doesn't mean the graph looks like your original signal (see SergV's answer); it only means that after ideal lowpass filtering it will.
Your problem is not related to the Nyquist theorem and aliasing. It is simple problem of graphic representation. You can change your code that frequency of sine will be lower Nyquist limit, but graph will be as strange as before:
t = linspace(0,1,Fs+2);
plot(sin(2*pi*f*t));
Result:
To explain problem I modify your code:
Fs=100;
f=12; %f << Fs
t=0:1/Fs:0.5; % step =1/Fs
t1=0:1/(10*Fs):0.5; % step=1/(10*Fs) for precise graphic representation
subplot (2, 1, 1);
plot(t,sin(2*pi*f*t),"-b",t,sin(2*pi*f*t),"*r");
subplot (2, 1, 2);
plot(t1,sin(2*pi*f*t1),"g",t,sin(2*pi*f*t),"r*");
See result:
Red star - values of sin(2*pi*f) with sampling rate of Fs.
Blue line - lines which connect red stars. It is usual data representation of function plot() - line interpolation between data points
Green curve - sin(2*pi*f)
Your eyes and brain can easily understand that these graphs represent the sine
Change frequency to more high:
f=48; % 2*f < Fs !!!
See on blue lines and red stars. Your eyes and brain do not understand now that these graphs represent the same sine. But your "red stars" are actually valid value of sine. See on bottom graph.
Finally, there is the same graphics for sine with frequency f=50 (2*f = Fs):
P.S.
Nyquist-Shannon sampling theorem states for your case that if:
f < 2*Fs
You have infinite number of samples (red stars on our plots)
then you can reproduce values of function in any time (green curve on our plots). You must use sinc interpolation to do it.
copied from Matlab Help:
linspace
Generate linearly spaced vectors
Syntax
y = linspace(a,b)
y = linspace(a,b,n)
Description
The linspace function generates linearly spaced vectors. It is similar to the colon operator ":", but gives direct control over the number of points.
y = linspace(a,b) generates a row vector y of 100 points linearly spaced between and including a and b.
y = linspace(a,b,n) generates a row vector y of n points linearly spaced between and including a and b. For n < 2, linspace returns b.
Examples
Create a vector of 100 linearly spaced numbers from 1 to 500:
A = linspace(1,500);
Create a vector of 12 linearly spaced numbers from 1 to 36:
A = linspace(1,36,12);
linspace is not apparent for Nyquist interval, so you can use the common form:
t = 0:Ts:1;
or
t = 0:1/Fs:1;
and change the Fs values.
The first Figure is due to the approximation of '0': sin(0) and sin(2*pi). We can notice the range is in 10^(-16) level.
I wrote the function reconstruct_FFT that can recover critically sampled data even for short observation intervals if the input sequence of samples is periodic. It performs lowpass filtering in the frequency domain.
I want to find the peaks of the raw ecg signal so that I can calculate the beats per minute(bpm).
I Have written a code in matlab which I have attached below.In the code below I am unable to find threshold point correctly which will help me in finding the peaks and hence the bpm.
%input the signal into matlab
[x,fs]=wavread('heartbeat.wav');
subplot(2,1,1)
plot(x(1:10000),'r-')
grid on
%lowpass filter the input signal with cutoff at 100hz
h=fir1(30,0.3126); %normalized cutoff freq=0.3126
y=filter(h,1,x);
subplot(2,1,2)
plot(y(1:10000),'b-')
grid on
% peaks are seen as pulses(heart beats)
beat_count=0;
for p=2:length(y)-1
th(p)=abs(max(y(p)));
if(y(p) >y(p-1) && y(p) >y(p+1) && y(p)>th(p))
beat_count=beat_count+1;
end
end
N = length(y);
duration_seconds=N/fs;
duration_minutes=duration_seconds/60;
BPM=beat_count/duration_minutes;
bpm=ceil(BPM);
Please help me as I am new to matlab
I suggest changing this section of your code
beat_count=0;
for p=2:length(y)-1
th(p)=abs(max(y(p)));
if(y(p) >y(p-1) && y(p) >y(p+1) && y(p)>th(p))
beat_count=beat_count+1;
end
end
This is definitely flawed. I'm not sure of your logic here but what about this. We are looking for peaks, but only the high peaks, so first lets set a threshold value (you'll have to tweak this to a sensible number) and cull everything below that value to get rid of the smaller peaks:
th = max(y) * 0.9; %So here I'm considering anything less than 90% of the max as not a real peak... this bit really depends on your logic of finding peaks though which you haven't explained
Yth = zeros(length(y), 1);
Yth(y > th) = y(y > th);
OK so I suggest you now plot y and Yth to see what that code did. Now to find the peaks my logic is we are looking for local maxima i.e. points at which the first derivative of the function change from being positive to being negative. So I'm going to find a very simple numerical approximation to the first derivative by finding the difference between each consecutive point on the signal:
Ydiff = diff(Yth);
No I want to find where the signal goes from being positive to being negative. So I'm going to make all the positive values equal zero, and all the negative values equal one:
Ydiff_logical = Ydiff < 0;
finally I want to find where this signal changes from a zero to a one (but not the other way around)
Ypeaks = diff(Ydiff_logical) == 1;
Now count the peaks:
sum(Ypeaks)
note that for plotting purpouse because of the use of diff we should pad a false to either side of Ypeaks so
Ypeaks = [false; Ypeaks; false];
OK so there is quite a lot of matlab there, I suggest you run each line, one by one and inspect the variable by both plotting the result of each line and also by double clicking the variable in the matlab workspace to understand what is happening at each step.
Example: (signal PeakSig taken from http://www.mathworks.com/help/signal/ref/findpeaks.html) and plotting with:
plot(x(Ypeaks),PeakSig(Ypeaks),'k^','markerfacecolor',[1 0 0]);
What do you think about the built-in
findpeaks(data,'Name',value)
function? You can choose among different logics for peak detection:
'MINPEAKHEIGHT'
'MINPEAKDISTANCE'
'THRESHOLD'
'NPEAKS'
'SORTSTR'
I hope this helps.
You know, the QRS complex does not always have the maximum amplitude, for pathologic ECG it can be present as several minor oscillations instead of one high-amplitude peak.
Thus, you can try one good algothythm, tested by me: the detection criterion is assumed to be high absolute rate of change in the signal, averaged within the given interval.
Algorithm:
- 50/60 Hz filter (e.g. for 50 Hz sliding window of 20 msec will be fine)
- adaptive hipass filter (for baseline drift)
- find signal's first derivate x'
- fing squared derivate (x')^2
- apply sliding average window with the width of QRS complex - approx 100-150 msec (you will get some signal with 'rectangles', which have width of QRS)
- use simple threshold (e.g. 1/3 of maximum of the first 3 seconds) to determine approximate positions or R
- in the source ECG find local maximum within +-100 msec of that R position.
However, you still have to eliminate artifacts and outliers (e.g. surges, when the electrod connection fails).
Also, you can find a lot of helpful information from this book: "R.M. Rangayyan - Biomedical Signal Analysis"
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 ?