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From: Peng Liang on 17 Jun 2010 08:58
Thank you Wayne and Greg,

Yes, in this case, the imaginary part is really insignificant.
However, for the following code, I design a butter filter using butter function. and FFT convolution with the mixed sine wave. after IFFT, the imaginary part is very large. it is ever larger than real part. The code is:

%%%%write a function of customized butter filter. it is low pass filter
function Hfilter = butterfilter(order,wb,L)

[b,a] = butter(order,wb);
[H,w] = freqz(b,a,L);

N = 2*length(H);
H2 = [1:N]';
H2(1:length(H2)) = zeros;
H2(1:length(H)) = H;
Hflip = flipud(H2);
Hfilter = H2+Hflip;



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%filtering in frequency domian with customized BUTTERWORTH
%%%%filter kernel using butter function to design a digital filter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clear;
fs = 8000;
t = [0.000125:1/fs:1];
A_1 = 10;
A_2 = 4;

f_1 = 350;
s_1 = A_1*sin(2*pi*f_1*t);

f_2 = 800;
s_2 = A_2*sin(2*pi*f_2*t);

s_3 = s_1+s_2;


S_1 = fft(s_1);
S_2 = fft(s_2);
S_3 = fft(s_3);

%%%call the butter function to build the filter, which will give cutoff frequency of 480Hz
Hfilter = butterfilter(10,0.12,4000);
S_3i = S_3';
S3_f = S_3i.*Hfilter;
s3_f = ifft(S3_f);

Please check the imag(s3_f), you will find the value is quite significant. like maximum is 9.6071.

any idea why I get those large imaginary data.

Thank you so much for you guys helps.


"Wayne King" <wmkingty(a)gmail.com> wrote in message <hvbeuk$870$1(a)fred.mathworks.com>...
> "Peng Liang" <madokaaukawa(a)hotmail.com> wrote in message <hvbdjg$asl$1(a)fred.mathworks.com>...
> > TideMan, thank you for your response and helps. You might know what is going wrong in my code.
> >
> > I checked the imaginary part of data. they are sometimes very significant. I have my code below. Please let me know where I go wrong.
> >
> > clear;
> >
> > fs = 8000;
> > t = [0.000125:1/fs:1];
> > A_1 = 10;
> > A_2 = 4;
> >
> > f_1 = 350;
> > s_1 = A_1*sin(2*pi*f_1*t); % generates a sine wave with 350Hz frequency
> >
> > f_2 = 700;
> > s_2 = A_2*sin(2*pi*f_2*t); % generates a sine wave with 700Hz frequency
> >
> > s_3 = s_1+s_2; % mixed 350 and 700Hz sine wave
> >
> > S_3 = fft(s_3); % get dft data using fft algorithm, now the data is complex number
> >
> >
> > %% design a high pass filter which only have ones and zeros
> > filter = ones(size(S_1));
> > filter(1:500) = 0;
> > filter(7501:8000) = 0;
> >
> > S3_f = S_3.*filter; % filter the low frequency in frequency domain.
> >
> > s3_f = ifft(S3_f); % convert back to time domain using ifft, s3_f is time domian signal and gives significant imaginary part.
> >
> > anyone has idea why my ifft generates very large imaginary part?
> > Need helps.
> >
> > Thanks in advance.
> >
> >
> >
> > TideMan <mulgor(a)gmail.com> wrote in message <d06620a7-8ec4-480d-9458-4d6c22f1e964(a)y18g2000prn.googlegroups.com>...
> > > On Jun 16, 8:04 am, "Peng Liang" <madokaauk...(a)hotmail.com> wrote:
> > > > Hi experts,
> > > >
> > > > I am using matlab to do the FFT convolution which take the fft of the signal and filter, multiply them in frequency domain. There is a problem. the ifft gives me complex numbers which is weird. since I multiplied the fft of the input signal with a filter kernel which is after fft. both data after fft are complex number. It should be turned back to real number after ifft. but I still get the complex numbers.
> > > >
> > > > is my question clear? Please give me a hand.
> > > > thanks in advance.
> > >
> > > How large are the imaginary parts?
> > > If they are very small, you can ignore them. They are the result of
> > > floating point operations. Use real(y) to discard them.
> > > If they are not negligible, then you have screwed up your processing
> > > in the frequency domain. Without more detail of what you are doing,
> > > we cannot tell what you have screwed up.
>
> Hi, Peng, I think what Tideman told you is absolutely correct. When I run your example, the largest imaginary part I get is on the order of 10^(-15). I'm not sure why you think that is significant. Tideman explained to you why that happens. Either take his advice, or use the 'symmetric' option with ifft().
>
> Also, you should not name your variable filter. filter is a Matlab function. You will cause yourself a lot of problems naming variables the same name as Matlab functions, classes, methods, etc.
>
> Wayne
From: Wayne King on 17 Jun 2010 09:32
"Peng Liang" <madokaaukawa(a)hotmail.com> wrote in message <hvd64s$1m4$1(a)fred.mathworks.com>...
> Thank you Wayne and Greg,
>
> Yes, in this case, the imaginary part is really insignificant.
> However, for the following code, I design a butter filter using butter function. and FFT convolution with the mixed sine wave. after IFFT, the imaginary part is very large. it is ever larger than real part. The code is:
>
> %%%%write a function of customized butter filter. it is low pass filter
> function Hfilter = butterfilter(order,wb,L)
>
> [b,a] = butter(order,wb);
> [H,w] = freqz(b,a,L);
>
> N = 2*length(H);
> H2 = [1:N]';
> H2(1:length(H2)) = zeros;
> H2(1:length(H)) = H;
> Hflip = flipud(H2);
> Hfilter = H2+Hflip;
>
>
>
> %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
> %%%%filtering in frequency domian with customized BUTTERWORTH
> %%%%filter kernel using butter function to design a digital filter
> %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
> clear;
> fs = 8000;
> t = [0.000125:1/fs:1];
> A_1 = 10;
> A_2 = 4;
>
> f_1 = 350;
> s_1 = A_1*sin(2*pi*f_1*t);
>
> f_2 = 800;
> s_2 = A_2*sin(2*pi*f_2*t);
>
> s_3 = s_1+s_2;
>
>
> S_1 = fft(s_1);
> S_2 = fft(s_2);
> S_3 = fft(s_3);
>
> %%%call the butter function to build the filter, which will give cutoff frequency of 480Hz
> Hfilter = butterfilter(10,0.12,4000);
> S_3i = S_3';
> S3_f = S_3i.*Hfilter;
> s3_f = ifft(S3_f);
>
> Please check the imag(s3_f), you will find the value is quite significant. like maximum is 9.6071.
>
> any idea why I get those large imaginary data.
>
> Thank you so much for you guys helps.
>
>
> "Wayne King" <wmkingty(a)gmail.com> wrote in message <hvbeuk$870$1(a)fred.mathworks.com>...
> > "Peng Liang" <madokaaukawa(a)hotmail.com> wrote in message <hvbdjg$asl$1(a)fred.mathworks.com>...
> > > TideMan, thank you for your response and helps. You might know what is going wrong in my code.
> > >
> > > I checked the imaginary part of data. they are sometimes very significant. I have my code below. Please let me know where I go wrong.
> > >
> > > clear;
> > >
> > > fs = 8000;
> > > t = [0.000125:1/fs:1];
> > > A_1 = 10;
> > > A_2 = 4;
> > >
> > > f_1 = 350;
> > > s_1 = A_1*sin(2*pi*f_1*t); % generates a sine wave with 350Hz frequency
> > >
> > > f_2 = 700;
> > > s_2 = A_2*sin(2*pi*f_2*t); % generates a sine wave with 700Hz frequency
> > >
> > > s_3 = s_1+s_2; % mixed 350 and 700Hz sine wave
> > >
> > > S_3 = fft(s_3); % get dft data using fft algorithm, now the data is complex number
> > >
> > >
> > > %% design a high pass filter which only have ones and zeros
> > > filter = ones(size(S_1));
> > > filter(1:500) = 0;
> > > filter(7501:8000) = 0;
> > >
> > > S3_f = S_3.*filter; % filter the low frequency in frequency domain.
> > >
> > > s3_f = ifft(S3_f); % convert back to time domain using ifft, s3_f is time domian signal and gives significant imaginary part.
> > >
> > > anyone has idea why my ifft generates very large imaginary part?
> > > Need helps.
> > >
> > > Thanks in advance.
> > >
> > >
> > >
> > > TideMan <mulgor(a)gmail.com> wrote in message <d06620a7-8ec4-480d-9458-4d6c22f1e964(a)y18g2000prn.googlegroups.com>...
> > > > On Jun 16, 8:04 am, "Peng Liang" <madokaauk...(a)hotmail.com> wrote:
> > > > > Hi experts,
> > > > >
> > > > > I am using matlab to do the FFT convolution which take the fft of the signal and filter, multiply them in frequency domain. There is a problem. the ifft gives me complex numbers which is weird. since I multiplied the fft of the input signal with a filter kernel which is after fft. both data after fft are complex number. It should be turned back to real number after ifft. but I still get the complex numbers.
> > > > >
> > > > > is my question clear? Please give me a hand.
> > > > > thanks in advance.
> > > >
> > > > How large are the imaginary parts?
> > > > If they are very small, you can ignore them. They are the result of
> > > > floating point operations. Use real(y) to discard them.
> > > > If they are not negligible, then you have screwed up your processing
> > > > in the frequency domain. Without more detail of what you are doing,
> > > > we cannot tell what you have screwed up.
> >
> > Hi, Peng, I think what Tideman told you is absolutely correct. When I run your example, the largest imaginary part I get is on the order of 10^(-15). I'm not sure why you think that is significant. Tideman explained to you why that happens. Either take his advice, or use the 'symmetric' option with ifft().
> >
> > Also, you should not name your variable filter. filter is a Matlab function. You will cause yourself a lot of problems naming variables the same name as Matlab functions, classes, methods, etc.
> >
> > Wayne

Hi Peng, You should not top post. It makes it hard for people to follow the thread.

You are not conjugating the entries in the frequency response of your filter when you append the data to fill out the frequency response. Try the following.

fs = 8000;
t = [0.000125:1/fs:1];
A_1 = 10;
A_2 = 4;
f_1 = 350;
s_1 = A_1*sin(2*pi*f_1*t);
f_2 = 800;
s_2 = A_2*sin(2*pi*f_2*t);
s_3 = s_1+s_2;
S_1 = fft(s_1);
S_2 = fft(s_2);
S_3 = fft(s_3);
order = 10;
wb = 0.12;
L = 4000;
[b,a] = butter(order,wb);
[H,w] = freqz(b,a,2*L,'whole');
FiltSig = ifft(S_3.'.*H);

By the way, why are you filtering like this? Why not just

FiltSig = filter(b,a,s_3);


Wayne
From: Peng Liang on 17 Jun 2010 10:41
"Wayne King" <wmkingty(a)gmail.com> wrote in message <hvd84l$8de$1(a)fred.mathworks.com>...
> "Peng Liang" <madokaaukawa(a)hotmail.com> wrote in message <hvd64s$1m4$1(a)fred.mathworks.com>...
> > Thank you Wayne and Greg,
> >
> > Yes, in this case, the imaginary part is really insignificant.
> > However, for the following code, I design a butter filter using butter function. and FFT convolution with the mixed sine wave. after IFFT, the imaginary part is very large. it is ever larger than real part. The code is:
> >
> > %%%%write a function of customized butter filter. it is low pass filter
> > function Hfilter = butterfilter(order,wb,L)
> >
> > [b,a] = butter(order,wb);
> > [H,w] = freqz(b,a,L);
> >
> > N = 2*length(H);
> > H2 = [1:N]';
> > H2(1:length(H2)) = zeros;
> > H2(1:length(H)) = H;
> > Hflip = flipud(H2);
> > Hfilter = H2+Hflip;
> >
> >
> >
> > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
> > %%%%filtering in frequency domian with customized BUTTERWORTH
> > %%%%filter kernel using butter function to design a digital filter
> > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
> > clear;
> > fs = 8000;
> > t = [0.000125:1/fs:1];
> > A_1 = 10;
> > A_2 = 4;
> >
> > f_1 = 350;
> > s_1 = A_1*sin(2*pi*f_1*t);
> >
> > f_2 = 800;
> > s_2 = A_2*sin(2*pi*f_2*t);
> >
> > s_3 = s_1+s_2;
> >
> >
> > S_1 = fft(s_1);
> > S_2 = fft(s_2);
> > S_3 = fft(s_3);
> >
> > %%%call the butter function to build the filter, which will give cutoff frequency of 480Hz
> > Hfilter = butterfilter(10,0.12,4000);
> > S_3i = S_3';
> > S3_f = S_3i.*Hfilter;
> > s3_f = ifft(S3_f);
> >
> > Please check the imag(s3_f), you will find the value is quite significant. like maximum is 9.6071.
> >
> > any idea why I get those large imaginary data.
> >
> > Thank you so much for you guys helps.
> >
> >
> > "Wayne King" <wmkingty(a)gmail.com> wrote in message <hvbeuk$870$1(a)fred.mathworks.com>...
> > > "Peng Liang" <madokaaukawa(a)hotmail.com> wrote in message <hvbdjg$asl$1(a)fred.mathworks.com>...
> > > > TideMan, thank you for your response and helps. You might know what is going wrong in my code.
> > > >
> > > > I checked the imaginary part of data. they are sometimes very significant. I have my code below. Please let me know where I go wrong.
> > > >
> > > > clear;
> > > >
> > > > fs = 8000;
> > > > t = [0.000125:1/fs:1];
> > > > A_1 = 10;
> > > > A_2 = 4;
> > > >
> > > > f_1 = 350;
> > > > s_1 = A_1*sin(2*pi*f_1*t); % generates a sine wave with 350Hz frequency
> > > >
> > > > f_2 = 700;
> > > > s_2 = A_2*sin(2*pi*f_2*t); % generates a sine wave with 700Hz frequency
> > > >
> > > > s_3 = s_1+s_2; % mixed 350 and 700Hz sine wave
> > > >
> > > > S_3 = fft(s_3); % get dft data using fft algorithm, now the data is complex number
> > > >
> > > >
> > > > %% design a high pass filter which only have ones and zeros
> > > > filter = ones(size(S_1));
> > > > filter(1:500) = 0;
> > > > filter(7501:8000) = 0;
> > > >
> > > > S3_f = S_3.*filter; % filter the low frequency in frequency domain.
> > > >
> > > > s3_f = ifft(S3_f); % convert back to time domain using ifft, s3_f is time domian signal and gives significant imaginary part.
> > > >
> > > > anyone has idea why my ifft generates very large imaginary part?
> > > > Need helps.
> > > >
> > > > Thanks in advance.
> > > >
> > > >
> > > >
> > > > TideMan <mulgor(a)gmail.com> wrote in message <d06620a7-8ec4-480d-9458-4d6c22f1e964(a)y18g2000prn.googlegroups.com>...
> > > > > On Jun 16, 8:04 am, "Peng Liang" <madokaauk...(a)hotmail.com> wrote:
> > > > > > Hi experts,
> > > > > >
> > > > > > I am using matlab to do the FFT convolution which take the fft of the signal and filter, multiply them in frequency domain. There is a problem. the ifft gives me complex numbers which is weird. since I multiplied the fft of the input signal with a filter kernel which is after fft. both data after fft are complex number. It should be turned back to real number after ifft. but I still get the complex numbers.
> > > > > >
> > > > > > is my question clear? Please give me a hand.
> > > > > > thanks in advance.
> > > > >
> > > > > How large are the imaginary parts?
> > > > > If they are very small, you can ignore them. They are the result of
> > > > > floating point operations. Use real(y) to discard them.
> > > > > If they are not negligible, then you have screwed up your processing
> > > > > in the frequency domain. Without more detail of what you are doing,
> > > > > we cannot tell what you have screwed up.
> > >
> > > Hi, Peng, I think what Tideman told you is absolutely correct. When I run your example, the largest imaginary part I get is on the order of 10^(-15). I'm not sure why you think that is significant. Tideman explained to you why that happens. Either take his advice, or use the 'symmetric' option with ifft().
> > >
> > > Also, you should not name your variable filter. filter is a Matlab function. You will cause yourself a lot of problems naming variables the same name as Matlab functions, classes, methods, etc.
> > >
> > > Wayne
>
> Hi Peng, You should not top post. It makes it hard for people to follow the thread.
>
> You are not conjugating the entries in the frequency response of your filter when you append the data to fill out the frequency response. Try the following.
>
> fs = 8000;
> t = [0.000125:1/fs:1];
> A_1 = 10;
> A_2 = 4;
> f_1 = 350;
> s_1 = A_1*sin(2*pi*f_1*t);
> f_2 = 800;
> s_2 = A_2*sin(2*pi*f_2*t);
> s_3 = s_1+s_2;
> S_1 = fft(s_1);
> S_2 = fft(s_2);
> S_3 = fft(s_3);
> order = 10;
> wb = 0.12;
> L = 4000;
> [b,a] = butter(order,wb);
> [H,w] = freqz(b,a,2*L,'whole');
> FiltSig = ifft(S_3.'.*H);
>
> By the way, why are you filtering like this? Why not just
>
> FiltSig = filter(b,a,s_3);
>
>
> Wayne


Sorry about hard reading my thread for you guys.
Wayne, Thank you so much, I tried your code, and give Only real numbers. I am so appreciated your help.
Could you explain more details about why does your code work properly? what is meaning I am not conjugating the entries in the frequency response of filter when append the data to fill out the frequency response?
Thank you again
From: Peng Liang on 17 Jun 2010 11:07
"Wayne King" <wmkingty(a)gmail.com> wrote in message <hvd84l$8de$1(a)fred.mathworks.com>...
> "Peng Liang" <madokaaukawa(a)hotmail.com> wrote in message <hvd64s$1m4$1(a)fred.mathworks.com>...
> > Thank you Wayne and Greg,
> >
> > Yes, in this case, the imaginary part is really insignificant.
> > However, for the following code, I design a butter filter using butter function. and FFT convolution with the mixed sine wave. after IFFT, the imaginary part is very large. it is ever larger than real part. The code is:
> >
> > %%%%write a function of customized butter filter. it is low pass filter
> > function Hfilter = butterfilter(order,wb,L)
> >
> > [b,a] = butter(order,wb);
> > [H,w] = freqz(b,a,L);
> >
> > N = 2*length(H);
> > H2 = [1:N]';
> > H2(1:length(H2)) = zeros;
> > H2(1:length(H)) = H;
> > Hflip = flipud(H2);
> > Hfilter = H2+Hflip;
> >
> >
> >
> > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
> > %%%%filtering in frequency domian with customized BUTTERWORTH
> > %%%%filter kernel using butter function to design a digital filter
> > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
> > clear;
> > fs = 8000;
> > t = [0.000125:1/fs:1];
> > A_1 = 10;
> > A_2 = 4;
> >
> > f_1 = 350;
> > s_1 = A_1*sin(2*pi*f_1*t);
> >
> > f_2 = 800;
> > s_2 = A_2*sin(2*pi*f_2*t);
> >
> > s_3 = s_1+s_2;
> >
> >
> > S_1 = fft(s_1);
> > S_2 = fft(s_2);
> > S_3 = fft(s_3);
> >
> > %%%call the butter function to build the filter, which will give cutoff frequency of 480Hz
> > Hfilter = butterfilter(10,0.12,4000);
> > S_3i = S_3';
> > S3_f = S_3i.*Hfilter;
> > s3_f = ifft(S3_f);
> >
> > Please check the imag(s3_f), you will find the value is quite significant. like maximum is 9.6071.
> >
> > any idea why I get those large imaginary data.
> >
> > Thank you so much for you guys helps.
> >
> >
> > "Wayne King" <wmkingty(a)gmail.com> wrote in message <hvbeuk$870$1(a)fred.mathworks.com>...
> > > "Peng Liang" <madokaaukawa(a)hotmail.com> wrote in message <hvbdjg$asl$1(a)fred.mathworks.com>...
> > > > TideMan, thank you for your response and helps. You might know what is going wrong in my code.
> > > >
> > > > I checked the imaginary part of data. they are sometimes very significant. I have my code below. Please let me know where I go wrong.
> > > >
> > > > clear;
> > > >
> > > > fs = 8000;
> > > > t = [0.000125:1/fs:1];
> > > > A_1 = 10;
> > > > A_2 = 4;
> > > >
> > > > f_1 = 350;
> > > > s_1 = A_1*sin(2*pi*f_1*t); % generates a sine wave with 350Hz frequency
> > > >
> > > > f_2 = 700;
> > > > s_2 = A_2*sin(2*pi*f_2*t); % generates a sine wave with 700Hz frequency
> > > >
> > > > s_3 = s_1+s_2; % mixed 350 and 700Hz sine wave
> > > >
> > > > S_3 = fft(s_3); % get dft data using fft algorithm, now the data is complex number
> > > >
> > > >
> > > > %% design a high pass filter which only have ones and zeros
> > > > filter = ones(size(S_1));
> > > > filter(1:500) = 0;
> > > > filter(7501:8000) = 0;
> > > >
> > > > S3_f = S_3.*filter; % filter the low frequency in frequency domain.
> > > >
> > > > s3_f = ifft(S3_f); % convert back to time domain using ifft, s3_f is time domian signal and gives significant imaginary part.
> > > >
> > > > anyone has idea why my ifft generates very large imaginary part?
> > > > Need helps.
> > > >
> > > > Thanks in advance.
> > > >
> > > >
> > > >
> > > > TideMan <mulgor(a)gmail.com> wrote in message <d06620a7-8ec4-480d-9458-4d6c22f1e964(a)y18g2000prn.googlegroups.com>...
> > > > > On Jun 16, 8:04 am, "Peng Liang" <madokaauk...(a)hotmail.com> wrote:
> > > > > > Hi experts,
> > > > > >
> > > > > > I am using matlab to do the FFT convolution which take the fft of the signal and filter, multiply them in frequency domain. There is a problem. the ifft gives me complex numbers which is weird. since I multiplied the fft of the input signal with a filter kernel which is after fft. both data after fft are complex number. It should be turned back to real number after ifft. but I still get the complex numbers.
> > > > > >
> > > > > > is my question clear? Please give me a hand.
> > > > > > thanks in advance.
> > > > >
> > > > > How large are the imaginary parts?
> > > > > If they are very small, you can ignore them. They are the result of
> > > > > floating point operations. Use real(y) to discard them.
> > > > > If they are not negligible, then you have screwed up your processing
> > > > > in the frequency domain. Without more detail of what you are doing,
> > > > > we cannot tell what you have screwed up.
> > >
> > > Hi, Peng, I think what Tideman told you is absolutely correct. When I run your example, the largest imaginary part I get is on the order of 10^(-15). I'm not sure why you think that is significant. Tideman explained to you why that happens. Either take his advice, or use the 'symmetric' option with ifft().
> > >
> > > Also, you should not name your variable filter. filter is a Matlab function. You will cause yourself a lot of problems naming variables the same name as Matlab functions, classes, methods, etc.
> > >
> > > Wayne
>
> Hi Peng, You should not top post. It makes it hard for people to follow the thread.
>
> You are not conjugating the entries in the frequency response of your filter when you append the data to fill out the frequency response. Try the following.
>
> fs = 8000;
> t = [0.000125:1/fs:1];
> A_1 = 10;
> A_2 = 4;
> f_1 = 350;
> s_1 = A_1*sin(2*pi*f_1*t);
> f_2 = 800;
> s_2 = A_2*sin(2*pi*f_2*t);
> s_3 = s_1+s_2;
> S_1 = fft(s_1);
> S_2 = fft(s_2);
> S_3 = fft(s_3);
> order = 10;
> wb = 0.12;
> L = 4000;
> [b,a] = butter(order,wb);
> [H,w] = freqz(b,a,2*L,'whole');
> FiltSig = ifft(S_3.'.*H);
>
> By the way, why are you filtering like this? Why not just
>
> FiltSig = filter(b,a,s_3);
>
>
> Wayne

Wayne,
I don't know filter(b,a,s_3) is filtering in frequency domain or in time domain. what I want is filtering in frequency domain. that is the reason I am doing S_3.'.*H which is multiplication in frequency domain.
From: Wayne King on 17 Jun 2010 12:55
"Peng Liang" <madokaaukawa(a)hotmail.com> wrote in message <hvddmo$8pi$1(a)fred.mathworks.com>...
> "Wayne King" <wmkingty(a)gmail.com> wrote in message <hvd84l$8de$1(a)fred.mathworks.com>...
> > "Peng Liang" <madokaaukawa(a)hotmail.com> wrote in message <hvd64s$1m4$1(a)fred.mathworks.com>...
> > > Thank you Wayne and Greg,
> > >
> > > Yes, in this case, the imaginary part is really insignificant.
> > > However, for the following code, I design a butter filter using butter function. and FFT convolution with the mixed sine wave. after IFFT, the imaginary part is very large. it is ever larger than real part. The code is:
> > >
> > > %%%%write a function of customized butter filter. it is low pass filter
> > > function Hfilter = butterfilter(order,wb,L)
> > >
> > > [b,a] = butter(order,wb);
> > > [H,w] = freqz(b,a,L);
> > >
> > > N = 2*length(H);
> > > H2 = [1:N]';
> > > H2(1:length(H2)) = zeros;
> > > H2(1:length(H)) = H;
> > > Hflip = flipud(H2);
> > > Hfilter = H2+Hflip;
> > >
> > >
> > >
> > > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
> > > %%%%filtering in frequency domian with customized BUTTERWORTH
> > > %%%%filter kernel using butter function to design a digital filter
> > > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
> > > clear;
> > > fs = 8000;
> > > t = [0.000125:1/fs:1];
> > > A_1 = 10;
> > > A_2 = 4;
> > >
> > > f_1 = 350;
> > > s_1 = A_1*sin(2*pi*f_1*t);
> > >
> > > f_2 = 800;
> > > s_2 = A_2*sin(2*pi*f_2*t);
> > >
> > > s_3 = s_1+s_2;
> > >
> > >
> > > S_1 = fft(s_1);
> > > S_2 = fft(s_2);
> > > S_3 = fft(s_3);
> > >
> > > %%%call the butter function to build the filter, which will give cutoff frequency of 480Hz
> > > Hfilter = butterfilter(10,0.12,4000);
> > > S_3i = S_3';
> > > S3_f = S_3i.*Hfilter;
> > > s3_f = ifft(S3_f);
> > >
> > > Please check the imag(s3_f), you will find the value is quite significant. like maximum is 9.6071.
> > >
> > > any idea why I get those large imaginary data.
> > >
> > > Thank you so much for you guys helps.
> > >
> > >
> > > "Wayne King" <wmkingty(a)gmail.com> wrote in message <hvbeuk$870$1(a)fred.mathworks.com>...
> > > > "Peng Liang" <madokaaukawa(a)hotmail.com> wrote in message <hvbdjg$asl$1(a)fred.mathworks.com>...
> > > > > TideMan, thank you for your response and helps. You might know what is going wrong in my code.
> > > > >
> > > > > I checked the imaginary part of data. they are sometimes very significant. I have my code below. Please let me know where I go wrong.
> > > > >
> > > > > clear;
> > > > >
> > > > > fs = 8000;
> > > > > t = [0.000125:1/fs:1];
> > > > > A_1 = 10;
> > > > > A_2 = 4;
> > > > >
> > > > > f_1 = 350;
> > > > > s_1 = A_1*sin(2*pi*f_1*t); % generates a sine wave with 350Hz frequency
> > > > >
> > > > > f_2 = 700;
> > > > > s_2 = A_2*sin(2*pi*f_2*t); % generates a sine wave with 700Hz frequency
> > > > >
> > > > > s_3 = s_1+s_2; % mixed 350 and 700Hz sine wave
> > > > >
> > > > > S_3 = fft(s_3); % get dft data using fft algorithm, now the data is complex number
> > > > >
> > > > >
> > > > > %% design a high pass filter which only have ones and zeros
> > > > > filter = ones(size(S_1));
> > > > > filter(1:500) = 0;
> > > > > filter(7501:8000) = 0;
> > > > >
> > > > > S3_f = S_3.*filter; % filter the low frequency in frequency domain.
> > > > >
> > > > > s3_f = ifft(S3_f); % convert back to time domain using ifft, s3_f is time domian signal and gives significant imaginary part.
> > > > >
> > > > > anyone has idea why my ifft generates very large imaginary part?
> > > > > Need helps.
> > > > >
> > > > > Thanks in advance.
> > > > >
> > > > >
> > > > >
> > > > > TideMan <mulgor(a)gmail.com> wrote in message <d06620a7-8ec4-480d-9458-4d6c22f1e964(a)y18g2000prn.googlegroups.com>...
> > > > > > On Jun 16, 8:04 am, "Peng Liang" <madokaauk...(a)hotmail.com> wrote:
> > > > > > > Hi experts,
> > > > > > >
> > > > > > > I am using matlab to do the FFT convolution which take the fft of the signal and filter, multiply them in frequency domain. There is a problem. the ifft gives me complex numbers which is weird. since I multiplied the fft of the input signal with a filter kernel which is after fft. both data after fft are complex number. It should be turned back to real number after ifft. but I still get the complex numbers.
> > > > > > >
> > > > > > > is my question clear? Please give me a hand.
> > > > > > > thanks in advance.
> > > > > >
> > > > > > How large are the imaginary parts?
> > > > > > If they are very small, you can ignore them. They are the result of
> > > > > > floating point operations. Use real(y) to discard them.
> > > > > > If they are not negligible, then you have screwed up your processing
> > > > > > in the frequency domain. Without more detail of what you are doing,
> > > > > > we cannot tell what you have screwed up.
> > > >
> > > > Hi, Peng, I think what Tideman told you is absolutely correct. When I run your example, the largest imaginary part I get is on the order of 10^(-15). I'm not sure why you think that is significant. Tideman explained to you why that happens. Either take his advice, or use the 'symmetric' option with ifft().
> > > >
> > > > Also, you should not name your variable filter. filter is a Matlab function. You will cause yourself a lot of problems naming variables the same name as Matlab functions, classes, methods, etc.
> > > >
> > > > Wayne
> >
> > Hi Peng, You should not top post. It makes it hard for people to follow the thread.
> >
> > You are not conjugating the entries in the frequency response of your filter when you append the data to fill out the frequency response. Try the following.
> >
> > fs = 8000;
> > t = [0.000125:1/fs:1];
> > A_1 = 10;
> > A_2 = 4;
> > f_1 = 350;
> > s_1 = A_1*sin(2*pi*f_1*t);
> > f_2 = 800;
> > s_2 = A_2*sin(2*pi*f_2*t);
> > s_3 = s_1+s_2;
> > S_1 = fft(s_1);
> > S_2 = fft(s_2);
> > S_3 = fft(s_3);
> > order = 10;
> > wb = 0.12;
> > L = 4000;
> > [b,a] = butter(order,wb);
> > [H,w] = freqz(b,a,2*L,'whole');
> > FiltSig = ifft(S_3.'.*H);
> >
> > By the way, why are you filtering like this? Why not just
> >
> > FiltSig = filter(b,a,s_3);
> >
> >
> > Wayne
>
> Wayne,
> I don't know filter(b,a,s_3) is filtering in frequency domain or in time domain. what I want is filtering in frequency domain. that is the reason I am doing S_3.'.*H which is multiplication in frequency domain.

Hi Peng, Multiplying in the frequency domain is equivalent to convolution in the time domain.

filter(b,a,s_3) does what you want.

Wayne
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