Energy of signal
in [Matlab]
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From: kk KKsingh on 28 May 2010 02:01 "Wayne King" <wmkingty(a)gmail.com> wrote in message <htmsa9$7kk$1(a)fred.mathworks.com>... > "kk KKsingh" <akikumar1983(a)gmail.com> wrote in message <htmqt9$8u9$1(a)fred.mathworks.com>... > > "kk KKsingh" <akikumar1983(a)gmail.com> wrote in message <htjsok$rl6$1(a)fred.mathworks.com>... > > > How can i calculate the Energy of the signal in Matlab > > > > > > When i try to calculate for positive half i am getting Energy at right frequency ! But some thing is wrong when i do it for whole signal i am not getting at -30 and +30... > > > N=1024; > > > dt=.001; > > > t=0:dt*dt*(N-1); > > > y=sin(2*pi*30*t); > > > e=abs(fft(y))/(N); > > > energy=e.^2; > > > k=1/dt/N*[-N/2:N/2-1] > > > plot(k,e) > > > > > > I am not able to get where i am doing wrong ! If i do it for positive half using energy=e.^2(1:N/2)..I am getting at Energy30 Hz..but if i use for full then i am not getting at -30 and +30 but getting some where near 470.. > > > > > > Thanks for your help > > > > > > kk > > > > > > Is there any difference between the energy if i calculate it in time domain and in frequency domain ! > > > > Suppose my signal is x=[1 2 3 4 5 6] > > > > So its energy will be x^2... > > > > and in Frequency domain > > > > its (fft(x))^2 > > > > As, far as i know there is no change in Energy if signal in both domains! > > Hi, it depends on how you define the Fourier transform whether it is a unitary operator or not (energy preserving). The way the DFT is defined in Matlab, it is not energy preserving: > > x = [1 2 3 4 5 6]; > xDFT = fft(x); > > Note that > norm(x,2) > is not equal to > norm(xDFT,2) > > But that > norm(xDFT,2)/sqrt(length(x)) > > equals the norm of the time domain signal. > > Wayne Thanks wayne ! Didnt get the logic of sqrt (length(x))...though its working
From: Greg Heath on 28 May 2010 05:35 % Newsgroups: comp.soft-sys.matlab % From: "kk KKsingh" <akikumar1...(a)gmail.com> % Date: Wed, 26 May 2010 19:28:20 +0000 (UTC) % Local: Wed, May 26 2010 3:28 pm % Subject: Energy of signal % % How can i calculate the Energy of the signal in Matlab % % When i try to calculate for positive half i am getting % Energy at right frequency ! But some thing is wrong % when i do it for whole signal i am not getting at -30 % and +30... % N=1024; % dt=.001; % t=0:dt*dt*(N-1); INCORRECT % y=sin(2*pi*30*t); % e=abs(fft(y))/(N); % energy=e.^2; INCORRECT % k=1/dt/N*[-N/2:N/2-1] % plot(k,e) NOT a plot of energy or power % I am not able to get where i am doing wrong ! % If i do it for positive half using energy=e.^2(1:N/2).. % I am getting at Energy30 Hz..but if i use for full then % i am not getting at -30 and +30 but getting some where % near 470.. close all, clear all, clc, k=0 N = 1024 % 1024 dt = 0.001 % 1e-3 Fs = 1/dt % 1000 fft period T = N*dt % 1.024 ifft period t = 0:dt:T-dt; f0 = 30 % 30 T0 = 1/f0 % 0.033333 signal period Nc = T/T0 % 30.72 noninteger number of cycles y = sin(2*pi*f0*t); % Power and Energy in the Time Domain p = y.^2 ; % Instantaneous Power P = sum(p) % 511.04 Total Power Pav = mean(p) % 0.49906 Average Power e = dt*p; % Instantaneous Energy Et = sum(e) % 0.51104 Total Energy Etav = mean(e) % 0.00049906 Average Energy http://groups.google.com/group/comp.soft-sys.matlab/ msg/009339c581e63467?hl=en % Power and Energy in the Freqency Domain df = Fs/N % 0.97656 = 1/T f = 0:df:Fs-df; % Unipolar freq interval fb = f - df*ceil((N-1)/2); % Bipolar freq interval Y = fft(y); % Unipolar freq spectrum Yb = fftshift(Y); % Bipolar freq spectrum % Note Y and Yb are interchangeable in most of the following absYb = abs(Yb); PSDb = absYb.^2/N; % Power Spectral Density figure plot(fb,PSDb) axis([-Fs/2 Fs/2 0 1.1*max(PSDb)]) title('Power Spectrum') % Check Parseval's Theorem (See Wikipedia) check = sum(p)-sum(PSDb) % -4.5475e-013 You can determine the appropriate expressions for the frequency domain calculations of P, Pav, Ef and Efav http://groups.google.com/group/comp.soft-sys.matlab/ msg/2222327db2ea7f51 Hope this helps. Greg
From: Wayne King on 28 May 2010 06:32 "kk KKsingh" <akikumar1983(a)gmail.com> wrote in message <htnm75$4fj$1(a)fred.mathworks.com>... > "Wayne King" <wmkingty(a)gmail.com> wrote in message <htmsa9$7kk$1(a)fred.mathworks.com>... > > "kk KKsingh" <akikumar1983(a)gmail.com> wrote in message <htmqt9$8u9$1(a)fred.mathworks.com>... > > > "kk KKsingh" <akikumar1983(a)gmail.com> wrote in message <htjsok$rl6$1(a)fred.mathworks.com>... > > > > How can i calculate the Energy of the signal in Matlab > > > > > > > > When i try to calculate for positive half i am getting Energy at right frequency ! But some thing is wrong when i do it for whole signal i am not getting at -30 and +30... > > > > N=1024; > > > > dt=.001; > > > > t=0:dt*dt*(N-1); > > > > y=sin(2*pi*30*t); > > > > e=abs(fft(y))/(N); > > > > energy=e.^2; > > > > k=1/dt/N*[-N/2:N/2-1] > > > > plot(k,e) > > > > > > > > I am not able to get where i am doing wrong ! If i do it for positive half using energy=e.^2(1:N/2)..I am getting at Energy30 Hz..but if i use for full then i am not getting at -30 and +30 but getting some where near 470.. > > > > > > > > Thanks for your help > > > > > > > > kk > > > > > > > > > Is there any difference between the energy if i calculate it in time domain and in frequency domain ! > > > > > > Suppose my signal is x=[1 2 3 4 5 6] > > > > > > So its energy will be x^2... > > > > > > and in Frequency domain > > > > > > its (fft(x))^2 > > > > > > As, far as i know there is no change in Energy if signal in both domains! > > > > Hi, it depends on how you define the Fourier transform whether it is a unitary operator or not (energy preserving). The way the DFT is defined in Matlab, it is not energy preserving: > > > > x = [1 2 3 4 5 6]; > > xDFT = fft(x); > > > > Note that > > norm(x,2) > > is not equal to > > norm(xDFT,2) > > > > But that > > norm(xDFT,2)/sqrt(length(x)) > > > > equals the norm of the time domain signal. > > > > Wayne > > Thanks wayne ! Didnt get the logic of sqrt (length(x))...though its working Hi, It's because the basis vectors for the DFT e^{j 2\pi k n/N} have unit (l2) norm when they are divided by \sqrt{N}, so that e^{j 2\pi k n/N}/\sqrt{N} k=0,1, ... N-1 form an orthonormal set of N-periodic sequences. If you wanted the DFT to be a unitary operator, you would have a 1/sqrt{N} factor both in the Fourier transform from the time domain to the frequency domain AND in the "inverse" Fourier transform from the frequency domain to the time domain. However, many people define the Fourier transform by incorporating the product of those factors 1/N into either the transform from time to frequency or from frequency to time. That is the way that MATLAB defines it. In MATLAB, the 1/N factor is incorporated in the "inverse" transform from frequency to time, therefore the l2 norm squared in time is equal to 1/N times the l2 norm squared in frequency. Wayne Wayne
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