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autocorrelation of a downsampled signal

Started by Thierry July 12, 2009
On 20 July, 17:48, Verictor <stehu...@gmail.com> wrote:
> On Jul 20, 7:10&#2013266080;am, Thierry <thierrybi...@googlemail.com> wrote: > > > > > > > On 19 July, 16:09, Verictor <stehu...@gmail.com> wrote: > > > > On Jul 18, 7:15&#2013266080;am, Thierry <thierrybi...@googlemail.com> wrote: > > > > > On 18 July, 13:54, Thierry <thierrybi...@googlemail.com> wrote: > > > > > > On 14 July, 18:25, Verictor <stehu...@gmail.com> wrote: > > > > > > > On Jul 12, 5:25&#2013266080;pm, Thierry <thierrybi...@googlemail.com> wrote: > > > > > > > > Hi Guys, > > > > > > > forgive me but it has been 10 years since i took a course insignal > > > > > > > processing. i am doing some simulations and i need the expression of > > > > > > > theautocorrelation(Rx') of asignalafter downsampling using the > > > > > > >autocorrelation(Rx) of the unsampledsignal. > > > > > > > > I think it should be :Rx'(n)=(1/M).Rx(n.M). > > > > > > > However, when i use the matlab xcorr function &#2013266080;with M=2, the results > > > > > > > don't adhere to the above formula, any help please. > > > > > > > > Cheers. > > > > > > > Why doesn't Matlab match? Can you show some samples on how you tested > > > > > > it? Downsample-by-2 makes power spectrum density has lower magnitude > > > > > > and wider "shape" while FFT the power spectrum density you will get > > > > > > the downsampleautocorrelation. > > > > > > Verictor, > > > > > I don't have the file in front of me. but the program was just few > > > > > lines, mainly: > > > > > x=randn(1,1000); > > > > > y=x(1:2:1000); > > > > > > Rx=xcorr(x,x)/length(x); > > > > > Ry=xcorr(y,y)/length(y); > > > > > &#2013266080;it is almost everything > > > > > > Cheers- Hide quoted text - > > > > > > - Show quoted text - > > > > > sorry a mistake above, it is rather > > > > y=zeros(1,1000); > > > > y(1:2:1000)=x(1:2:1000);- Hide quoted text - > > > > > - Show quoted text - > > > > Well, my understanding is that this is not exactly downsample-by-2 as > > > y still has 1000 samples. This is the way for downsample-by-2 (your > > > first lines of code) > > > x = randn(1,1000); > > > y = x(1:2:1000); &#2013266080; &#2013266080; &#2013266080;% y has 500 samples now. > > > > You can also use a sine waveform to see your OP equation actually is > > > correct.- Hide quoted text - > > > > - Show quoted text - > > > You are right Verictor &#2013266080;y = x(1:2:1000); &#2013266080;the OP equation is correct > > for a sinsignalas it is &#2013266080;one frequency, but it might not remain so > > for a wide bandwidthsignal. I should consider the case further ...- Hide quoted text - > > > - Show quoted text - > > The equation is correct as long as your downsampling didn't cause > aliasing. Note that you were using randn(), downsample-by-2 will cause > aliasing. I am sure you will notice this later. You can also use > multiple frequencies too but just no aliasing.- Hide quoted text - > > - Show quoted text -
but i can;t verify this using matlab. below is my script program. OutCorr is the autocorrelation of the non-downsampled filter ouput. outcorr_s is that of the downsampled signal. the esrults of the script execution is: OutCorr(1:10)= [-0.0042,-0.0158,0.0445,0.2343,0.2637,-0.1311,-0.6974,-1.0781,-1.0527,-0.6912] OutCorr_S(1:4)=[-0.0085,0.1179,0.0663,-0.6025,-0.7993] max(OutCorr)=91.5355 max(OutCorr_S)=88.9495 any help please, i am sure i am missing something % script function [OutCorr,OutCorr_S]=corrSamp() N=512; Scale=10; In=Scale*randn(N,1); % daubechies 4 filter Flt='db2'; [La,Ha]=wfilters(Flt); % sampling ratio Out=conv(In,La); Sampling=2; Out_S=Out(1:Sampling:length(Out)); OutCorr=xcorr(Out,Out)/length(Out); OutCorr_S=xcorr(Out_S,Out_S)/length(Out_S);
On Jul 21, 8:32&#2013266080;am, Thierry <thierrybi...@googlemail.com> wrote:
> On 20 July, 17:48, Verictor <stehu...@gmail.com> wrote: > > > > > > > On Jul 20, 7:10&#2013266080;am, Thierry <thierrybi...@googlemail.com> wrote: > > > > On 19 July, 16:09, Verictor <stehu...@gmail.com> wrote: > > > > > On Jul 18, 7:15&#2013266080;am, Thierry <thierrybi...@googlemail.com> wrote: > > > > > > On 18 July, 13:54, Thierry <thierrybi...@googlemail.com> wrote: > > > > > > > On 14 July, 18:25, Verictor <stehu...@gmail.com> wrote: > > > > > > > > On Jul 12, 5:25&#2013266080;pm, Thierry <thierrybi...@googlemail.com> wrote: > > > > > > > > > Hi Guys, > > > > > > > > forgive me but it has been 10 years since i took a course insignal > > > > > > > > processing. i am doing some simulations and i need the expression of > > > > > > > > theautocorrelation(Rx') of asignalafter downsampling using the > > > > > > > >autocorrelation(Rx) of the unsampledsignal. > > > > > > > > > I think it should be :Rx'(n)=(1/M).Rx(n.M). > > > > > > > > However, when i use the matlab xcorr function &#2013266080;with M=2, the results > > > > > > > > don't adhere to the above formula, any help please. > > > > > > > > > Cheers. > > > > > > > > Why doesn't Matlab match? Can you show some samples on how you tested > > > > > > > it? Downsample-by-2 makes power spectrum density has lower magnitude > > > > > > > and wider "shape" while FFT the power spectrum density you will get > > > > > > > the downsampleautocorrelation. > > > > > > > Verictor, > > > > > > I don't have the file in front of me. but the program was just few > > > > > > lines, mainly: > > > > > > x=randn(1,1000); > > > > > > y=x(1:2:1000); > > > > > > > Rx=xcorr(x,x)/length(x); > > > > > > Ry=xcorr(y,y)/length(y); > > > > > > &#2013266080;it is almost everything > > > > > > > Cheers- Hide quoted text - > > > > > > > - Show quoted text - > > > > > > sorry a mistake above, it is rather > > > > > y=zeros(1,1000); > > > > > y(1:2:1000)=x(1:2:1000);- Hide quoted text - > > > > > > - Show quoted text - > > > > > Well, my understanding is that this is not exactly downsample-by-2 as > > > > y still has 1000 samples. This is the way for downsample-by-2 (your > > > > first lines of code) > > > > x = randn(1,1000); > > > > y = x(1:2:1000); &#2013266080; &#2013266080; &#2013266080;% y has 500 samples now. > > > > > You can also use a sine waveform to see your OP equation actually is > > > > correct.- Hide quoted text - > > > > > - Show quoted text - > > > > You are right Verictor &#2013266080;y = x(1:2:1000); &#2013266080;the OP equation is correct > > > for a sinsignalas it is &#2013266080;one frequency, but it might not remain so > > > for a wide bandwidthsignal. I should consider the case further ...- Hide quoted text - > > > > - Show quoted text - > > > The equation is correct as long as your downsampling didn't cause > > aliasing. Note that you were using randn(), downsample-by-2 will cause > > aliasing. I am sure you will notice this later. You can also use > > multiple frequencies too but just no aliasing.- Hide quoted text - > > > - Show quoted text - > > but i can;t verify this using matlab. below is my script program. > OutCorr is the autocorrelation of the non-downsampled filter ouput. > outcorr_s is that of the downsampled signal. the esrults of the script > execution is: > OutCorr(1:10)= > [-0.0042,-0.0158,0.0445,0.2343,0.2637,-0.1311,-0.6974,-1.0781,-1.0527,-0.69&#2013266093;12] > OutCorr_S(1:4)=[-0.0085,0.1179,0.0663,-0.6025,-0.7993] > max(OutCorr)=91.5355 > max(OutCorr_S)=88.9495 > any help please, i am sure i am missing something > > % script > function [OutCorr,OutCorr_S]=corrSamp() > N=512; > Scale=10; > In=Scale*randn(N,1); > % daubechies 4 filter > Flt='db2'; > [La,Ha]=wfilters(Flt); > % sampling ratio > Out=conv(In,La); > Sampling=2; > Out_S=Out(1:Sampling:length(Out)); > OutCorr=xcorr(Out,Out)/length(Out); > OutCorr_S=xcorr(Out_S,Out_S)/length(Out_S);- Hide quoted text - > > - Show quoted text -
I haven't used wavelet filter before so I can't comment on the filter you used. I don't have the toolbox either but you can check your filter response to see how good it is. Here instead, I give you a simple script to see how it works. You can then play around your filter. % simple script starts n = 512; scale = 10; in = scale * randn(n,1); % since you still prefer randn() b = fir1(128,0.2); % check how the frequency response looks like out = filter(b,1, in); % pass input through the anti- aliasing filter outcorr= xcorr(in); % in is a vector outcorr_s = xcorr(in(1:2:end)); % downsample sequency auto-correlation max(outcorr); % I got 1.7197e+004 max(outcorr_s); % I got 8.5983e+003 plot(outcorr); hold on plot(outcorr_s, 'r') % clearly you can see what happened according to the equation.
On 21 July, 16:08, Verictor <stehu...@gmail.com> wrote:
> On Jul 21, 8:32&#2013266080;am, Thierry <thierrybi...@googlemail.com> wrote: > > > > > > > On 20 July, 17:48, Verictor <stehu...@gmail.com> wrote: > > > > On Jul 20, 7:10&#2013266080;am, Thierry <thierrybi...@googlemail.com> wrote: > > > > > On 19 July, 16:09, Verictor <stehu...@gmail.com> wrote: > > > > > > On Jul 18, 7:15&#2013266080;am, Thierry <thierrybi...@googlemail.com> wrote: > > > > > > > On 18 July, 13:54, Thierry <thierrybi...@googlemail.com> wrote: > > > > > > > > On 14 July, 18:25, Verictor <stehu...@gmail.com> wrote: > > > > > > > > > On Jul 12, 5:25&#2013266080;pm, Thierry <thierrybi...@googlemail.com> wrote: > > > > > > > > > > Hi Guys, > > > > > > > > > forgive me but it has been 10 years since i took a course insignal > > > > > > > > > processing. i am doing some simulations and i need the expression of > > > > > > > > > theautocorrelation(Rx') of asignalafter downsampling using the > > > > > > > > >autocorrelation(Rx) of the unsampledsignal. > > > > > > > > > > I think it should be :Rx'(n)=(1/M).Rx(n.M). > > > > > > > > > However, when i use the matlab xcorr function &#2013266080;with M=2, the results > > > > > > > > > don't adhere to the above formula, any help please. > > > > > > > > > > Cheers. > > > > > > > > > Why doesn't Matlab match? Can you show some samples on how you tested > > > > > > > > it? Downsample-by-2 makes power spectrum density has lower magnitude > > > > > > > > and wider "shape" while FFT the power spectrum density you will get > > > > > > > > the downsampleautocorrelation. > > > > > > > > Verictor, > > > > > > > I don't have the file in front of me. but the program was just few > > > > > > > lines, mainly: > > > > > > > x=randn(1,1000); > > > > > > > y=x(1:2:1000); > > > > > > > > Rx=xcorr(x,x)/length(x); > > > > > > > Ry=xcorr(y,y)/length(y); > > > > > > > &#2013266080;it is almost everything > > > > > > > > Cheers- Hide quoted text - > > > > > > > > - Show quoted text - > > > > > > > sorry a mistake above, it is rather > > > > > > y=zeros(1,1000); > > > > > > y(1:2:1000)=x(1:2:1000);- Hide quoted text - > > > > > > > - Show quoted text - > > > > > > Well, my understanding is that this is not exactly downsample-by-2 as > > > > > y still has 1000 samples. This is the way for downsample-by-2 (your > > > > > first lines of code) > > > > > x = randn(1,1000); > > > > > y = x(1:2:1000); &#2013266080; &#2013266080; &#2013266080;% y has 500 samples now. > > > > > > You can also use a sine waveform to see your OP equation actually is > > > > > correct.- Hide quoted text - > > > > > > - Show quoted text - > > > > > You are right Verictor &#2013266080;y = x(1:2:1000); &#2013266080;the OP equation is correct > > > > for a sinsignalas it is &#2013266080;one frequency, but it might not remain so > > > > for a wide bandwidthsignal. I should consider the case further ...- Hide quoted text - > > > > > - Show quoted text - > > > > The equation is correct as long as your downsampling didn't cause > > > aliasing. Note that you were using randn(), downsample-by-2 will cause > > > aliasing. I am sure you will notice this later. You can also use > > > multiple frequencies too but just no aliasing.- Hide quoted text - > > > > - Show quoted text - > > > but i can;t verify this using matlab. below is my script program. > > OutCorr is theautocorrelationof the non-downsampledfilter ouput. > > outcorr_s is that of thedownsampledsignal. the esrults of the script > > execution is: > > OutCorr(1:10)= > > [-0.0042,-0.0158,0.0445,0.2343,0.2637,-0.1311,-0.6974,-1.0781,-1.0527,-0.69&#2013266093;&#2013266093;12] > > OutCorr_S(1:4)=[-0.0085,0.1179,0.0663,-0.6025,-0.7993] > > max(OutCorr)=91.5355 > > max(OutCorr_S)=88.9495 > > any help please, i am sure i am missing something > > > % script > > function [OutCorr,OutCorr_S]=corrSamp() > > N=512; > > Scale=10; > > In=Scale*randn(N,1); > > % daubechies 4 filter > > Flt='db2'; > > [La,Ha]=wfilters(Flt); > > % sampling ratio > > Out=conv(In,La); > > Sampling=2; > > Out_S=Out(1:Sampling:length(Out)); > > OutCorr=xcorr(Out,Out)/length(Out); > > OutCorr_S=xcorr(Out_S,Out_S)/length(Out_S);- Hide quoted text - > > > - Show quoted text - > > I haven't used wavelet filter before so I can't comment on the filter > you used. I don't have the toolbox either but you can check your > filter response to see how good it is. Here instead, I give you a > simple script to see how it works. You can then play around your > filter. > > % simple script starts > n = 512; > scale = 10; > in = scale * randn(n,1); &#2013266080; &#2013266080; &#2013266080;% since you still prefer randn() > b = fir1(128,0.2); &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; % check how the frequency response > looks like > out = filter(b,1, in); &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;% pass input through the anti- > aliasing filter > outcorr= xcorr(in); &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; % in is a vector > outcorr_s = xcorr(in(1:2:end)); % downsample sequency auto-correlation > max(outcorr); &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;% I got 1.7197e+004 > max(outcorr_s); &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;% I got 8.5983e+003 > plot(outcorr); > hold on > plot(outcorr_s, 'r') &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080; &#2013266080;% clearly you can see what happened > according to the equation.- Hide quoted text - > > - Show quoted text -
re. the autocorrelation expression above. i think we need to divide by the number of signal samples since it is an estimation