fft matlab, scaling amplitude problem

2013 5 30
2 个回答
10 次查看（30 天）

0 个投票

Hi I've faced a problem in my homework coding. please help me to solve it.
I wrote a code for a problem, which i have its results. but after taking fft function from input, the amplitude of output is twice more than expected result.
what are the reasons that might be caused this?
I really appreciate it.

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Bob GH
Bob GH 2013-5-30
I wrote a part of my code related to fft :
clc; clear all;
fm=40; T=10; a=0; N=32768; %2^15 Cm=1.5; h=(T-a)/N; omega=2*pi*fm/T; hh=1/h; Iamp=2.25e3; V(1)=-65;
mhf(1)=mhfinf(V(1));
mhs(1)=mhsinf(V(1));
mNap(1)=mNapinf(V(1));
EL=leak(V(1),mhf(1),mhs(1),mNap(1));
t(1)=a;
Iinp(1)=a;
for j=1:N-1
Iinp(j+1)=Iamp*sin(omega*(t(j)^2));
%%%%%%%%% 4 step method of Runge-Kutta
K1=h*fftNapH(t(j),[V(j); mhf(j); mhs(j); mNap(j)],EL,Iinp(j+1)); deffv1=K1(1,1);mhf1=K1(2,1);mhs1=K1(3,1);mNap1=K1(4,1);
K2=h*fftNapH(t(j)+(h/2),[V(j)+(deffv1/2);mhf(j)+(mhf1/2);mhs(j)+(mhs1/2);mNap(j)+(mNap1/2)],EL,Iinp(j+1)); deffv2=K2(1,1);mhf2=K2(2,1);mhs2=K2(3,1);mNap2=K2(4,1);
K3=h*fftNapH(t(j)+(h/2),[V(j)+(deffv2/2);mhf(j)+(mhf2/2);mhs(j)+(mhs2/2);mNap(j)+(mNap2/2)],EL,Iinp(j+1)); deffv3=K3(1,1);mhf3=K3(2,1);mhs3=K3(3,1);mNap3=K3(4,1);
K4=h*fftNapH(t(j)+h,[V(j)+deffv3;mhf(j)+mhf3;mhs(j)+mhs3;mNap(j)+mNap3],EL,Iinp(j+1)); deffv4=K4(1,1);mhf4=K4(2,1);mhs4=K4(3,1);mNap4=K4(4,1);
V(j+1)=V(j)+1/6*(deffv1+(2*deffv2)+(2*deffv3)+deffv4); mhf(j+1)=mhf(j)+1/6*(mhf1+2*mhf2+2*mhf3+mhf4); mhs(j+1)=mhs(j)+1/6*(mhs1+2*mhs2+2*mhs3+mhs4); mNap(j+1)=mNap(j)+1/6*(mNap1+2*mNap2+2*mNap3+mNap4);
t(j+1)=t(j)+h;
end
V2=fft(V1,N); V3=V2(1:N/2); amp=abs(V3); ampb=amp(2:length(amp)); % remove DC part f= (0:N/2-1)*hh/N;
figure(2),plot(f,ampb) xlabel('Freq') ylabel('Amplitude') axis([0 50 0 2000])

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回答（2 个）

Wayne King
Wayne King 2013-5-30

1 个投票

You should always show your code:
fs = 1000;
t = 0:1/fs:1-1/fs;
L = length(x);
xdft = fft(x)/L;
plot(abs(xdft))
Exactly as I expect two peaks with amplitude 0.5
Or
xdft = 2*fft(x)/L;
xdft = xdft(1:length(x)/2+1);
plot(abs(xdft))

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Bob GH
Bob GH 2013-5-30
Thanks for your reply. i tried it, if i apply (/L) the result is absolutely different with expected result, but if i remove (/L) the result is same as mine.
Wayne King
Wayne King 2013-5-30
Again, you're not showing your code (not sure why). If you are using the fft() for amplitude estimation, you want to divide by the length of the input signal.
Bob GH
Bob GH 2013-5-30
thanks for your reply Wayne King. i wrote a relevant part of my code
Bob GH
Bob GH 2013-5-30
I wrote a part of my code related to fft :
clc; clear all;
fm=40; T=10; a=0; N=32768; %2^15 Tfft=10; Cm=1.5; h=(T-a)/N; omega=2*pi*fm/T; hh=1/h; Iamp=2.25e3; V(1)=-65;
mhf(1)=mhfinf(V(1));
mhs(1)=mhsinf(V(1));
mNap(1)=mNapinf(V(1));
EL=leak(V(1),mhf(1),mhs(1),mNap(1));
t(1)=a;
Iinp(1)=a;
for j=1:N-1
Iinp(j+1)=Iamp*sin(omega*(t(j)^2));
%%%%%%%%% 4 step method of Runge-Kutta
K1=h*fftNapH(t(j),[V(j); mhf(j); mhs(j); mNap(j)],EL,Iinp(j+1)); deffv1=K1(1,1);mhf1=K1(2,1);mhs1=K1(3,1);mNap1=K1(4,1);
K2=h*fftNapH(t(j)+(h/2),[V(j)+(deffv1/2);mhf(j)+(mhf1/2);mhs(j)+(mhs1/2);mNap(j)+(mNap1/2)],EL,Iinp(j+1)); deffv2=K2(1,1);mhf2=K2(2,1);mhs2=K2(3,1);mNap2=K2(4,1);
K3=h*fftNapH(t(j)+(h/2),[V(j)+(deffv2/2);mhf(j)+(mhf2/2);mhs(j)+(mhs2/2);mNap(j)+(mNap2/2)],EL,Iinp(j+1)); deffv3=K3(1,1);mhf3=K3(2,1);mhs3=K3(3,1);mNap3=K3(4,1);
K4=h*fftNapH(t(j)+h,[V(j)+deffv3;mhf(j)+mhf3;mhs(j)+mhs3;mNap(j)+mNap3],EL,Iinp(j+1)); deffv4=K4(1,1);mhf4=K4(2,1);mhs4=K4(3,1);mNap4=K4(4,1);
V(j+1)=V(j)+1/6*(deffv1+(2*deffv2)+(2*deffv3)+deffv4); mhf(j+1)=mhf(j)+1/6*(mhf1+2*mhf2+2*mhf3+mhf4); mhs(j+1)=mhs(j)+1/6*(mhs1+2*mhs2+2*mhs3+mhs4); mNap(j+1)=mNap(j)+1/6*(mNap1+2*mNap2+2*mNap3+mNap4);
t(j+1)=t(j)+h;
end
V2=fft(V1,N); V3=V2(1:N/2); amp=abs(V3); ampb=amp(2:length(amp)); % remove DC part f= (0:N/2-1)*hh/N;
figure(2),plot(f,ampb) xlabel('Freq') ylabel('Amplitude') axis([0 50 0 2000])

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Azzi Abdelmalek
Azzi Abdelmalek 2013-5-30

0 个投票

That means that, before calculating the fft, you've made some errors, which we can't find, because you have not posted the code.

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Bob GH
Bob GH 2013-5-30
thanks for your reply. actually, It has 10++ line of code and i do not know which part do you need? is there any distinguished reason for my problem? because, my result is exactly twice of expected result.
Wayne King
Wayne King 2013-5-30
编辑：Wayne King 2013-5-30
Does 10++ mean 12 lines? If it is a reasonable number, please post it all. Did you look at the code I posted below?

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Bob GH
Bob GH
2013-5-30

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