Mixing
DFT of a series of RC exponentials
$begingroup$
Context: I'm trying use matlab to apply a single-pole filter to a time-domain ramp waveform that is generated by a sequence of time-shifted "RC steps" that are added together.
The time domain voltage waveform is
$V(t) = sum_{k=0}^{N-1} V_{step}(1-e^{-(t-kDelta t_{step})/tau})u(t-kDelta t_{step})$
where $V_{step}$ and $Delta t_{step}$ are constants
Using the fft of a one-sided exponential decay, the unit-step, and the time-shift property of the fft I get a frequency domain representation of:
$mathfrak{F}(V(t)) = V_{step}left(sum_{k=0}^{N-1}e^{-jomega kDelta t_{step}}right)left(frac{1}{jomega} +pidelta(omega) - frac{1}{frac{1}{tau}+jomega}right)$
So now I want to generate this complex-valued fft manually in matlab, multiply by a filter response, and inverse fft. To start, I'm checking the frequency response of the input and its ifft to make sure it looks right:
N=32;
Vstep=25e-3;
tstep=10e-12;
tau=5e-12;
ts=0.1e-12;
Nfft = 2^nextpow2(max([N*tstep/ts N*tau*5/ts])); %Get enough points for the whole ramp with at least 5 tau's per exponenetial
w=(0:Nfft-1)*2*pi/Nff;t
f=w/2/pi/ts;
timeshifts=sum(exp(-1i*(0:N-1)'*w*tstep));
step=[pi 1./(1i*w(2:end))];
expdecay=1./(1/tau+1i*w);
Vf=Vstep*timeshifts.*(step-expdecay);
vt=ifft(V);
So the frequency domain Vf looks reasonable in amplitude. However, when I take the inverse and plot vt, it is definitely not correct.
Where am I going wrong? I suspect there's something I'm missing with the fact that this is a DFT not fourier transform. Also I know there's some subtlety to the FT/DFT of the heaviside, particularly at w=0, and I know that the fft at 0 will just be the average (times N) and I'm not sure my script accomplishes this.
I do know that I can just start with the time domain and then fft it, but I'm rather curious now.
fourier-analysis matlab fast-fourier-transform
asked Feb 1 at 23:20
KaysonKayson
613
$endgroup$
add a comment |
$begingroup$
Context: I'm trying use matlab to apply a single-pole filter to a time-domain ramp waveform that is generated by a sequence of time-shifted "RC steps" that are added together.
The time domain voltage waveform is
$V(t) = sum_{k=0}^{N-1} V_{step}(1-e^{-(t-kDelta t_{step})/tau})u(t-kDelta t_{step})$
where $V_{step}$ and $Delta t_{step}$ are constants
Using the fft of a one-sided exponential decay, the unit-step, and the time-shift property of the fft I get a frequency domain representation of:
$mathfrak{F}(V(t)) = V_{step}left(sum_{k=0}^{N-1}e^{-jomega kDelta t_{step}}right)left(frac{1}{jomega} +pidelta(omega) - frac{1}{frac{1}{tau}+jomega}right)$
So now I want to generate this complex-valued fft manually in matlab, multiply by a filter response, and inverse fft. To start, I'm checking the frequency response of the input and its ifft to make sure it looks right:
N=32;
Vstep=25e-3;
tstep=10e-12;
tau=5e-12;
ts=0.1e-12;
Nfft = 2^nextpow2(max([N*tstep/ts N*tau*5/ts])); %Get enough points for the whole ramp with at least 5 tau's per exponenetial
w=(0:Nfft-1)*2*pi/Nff;t
f=w/2/pi/ts;
timeshifts=sum(exp(-1i*(0:N-1)'*w*tstep));
step=[pi 1./(1i*w(2:end))];
expdecay=1./(1/tau+1i*w);
Vf=Vstep*timeshifts.*(step-expdecay);
vt=ifft(V);
So the frequency domain Vf looks reasonable in amplitude. However, when I take the inverse and plot vt, it is definitely not correct.
Where am I going wrong? I suspect there's something I'm missing with the fact that this is a DFT not fourier transform. Also I know there's some subtlety to the FT/DFT of the heaviside, particularly at w=0, and I know that the fft at 0 will just be the average (times N) and I'm not sure my script accomplishes this.
I do know that I can just start with the time domain and then fft it, but I'm rather curious now.
fourier-analysis matlab fast-fourier-transform
asked Feb 1 at 23:20
KaysonKayson
613
$endgroup$
add a comment |
$begingroup$
Context: I'm trying use matlab to apply a single-pole filter to a time-domain ramp waveform that is generated by a sequence of time-shifted "RC steps" that are added together.
The time domain voltage waveform is
$V(t) = sum_{k=0}^{N-1} V_{step}(1-e^{-(t-kDelta t_{step})/tau})u(t-kDelta t_{step})$
where $V_{step}$ and $Delta t_{step}$ are constants
Using the fft of a one-sided exponential decay, the unit-step, and the time-shift property of the fft I get a frequency domain representation of:
$mathfrak{F}(V(t)) = V_{step}left(sum_{k=0}^{N-1}e^{-jomega kDelta t_{step}}right)left(frac{1}{jomega} +pidelta(omega) - frac{1}{frac{1}{tau}+jomega}right)$
So now I want to generate this complex-valued fft manually in matlab, multiply by a filter response, and inverse fft. To start, I'm checking the frequency response of the input and its ifft to make sure it looks right:
N=32;
Vstep=25e-3;
tstep=10e-12;
tau=5e-12;
ts=0.1e-12;
Nfft = 2^nextpow2(max([N*tstep/ts N*tau*5/ts])); %Get enough points for the whole ramp with at least 5 tau's per exponenetial
w=(0:Nfft-1)*2*pi/Nff;t
f=w/2/pi/ts;
timeshifts=sum(exp(-1i*(0:N-1)'*w*tstep));
step=[pi 1./(1i*w(2:end))];
expdecay=1./(1/tau+1i*w);
Vf=Vstep*timeshifts.*(step-expdecay);
vt=ifft(V);
So the frequency domain Vf looks reasonable in amplitude. However, when I take the inverse and plot vt, it is definitely not correct.
Where am I going wrong? I suspect there's something I'm missing with the fact that this is a DFT not fourier transform. Also I know there's some subtlety to the FT/DFT of the heaviside, particularly at w=0, and I know that the fft at 0 will just be the average (times N) and I'm not sure my script accomplishes this.
I do know that I can just start with the time domain and then fft it, but I'm rather curious now.
fourier-analysis matlab fast-fourier-transform
asked Feb 1 at 23:20
KaysonKayson
613
$endgroup$
Context: I'm trying use matlab to apply a single-pole filter to a time-domain ramp waveform that is generated by a sequence of time-shifted "RC steps" that are added together.
The time domain voltage waveform is
$V(t) = sum_{k=0}^{N-1} V_{step}(1-e^{-(t-kDelta t_{step})/tau})u(t-kDelta t_{step})$
where $V_{step}$ and $Delta t_{step}$ are constants
Using the fft of a one-sided exponential decay, the unit-step, and the time-shift property of the fft I get a frequency domain representation of:
$mathfrak{F}(V(t)) = V_{step}left(sum_{k=0}^{N-1}e^{-jomega kDelta t_{step}}right)left(frac{1}{jomega} +pidelta(omega) - frac{1}{frac{1}{tau}+jomega}right)$
So now I want to generate this complex-valued fft manually in matlab, multiply by a filter response, and inverse fft. To start, I'm checking the frequency response of the input and its ifft to make sure it looks right:
N=32;
Vstep=25e-3;
tstep=10e-12;
tau=5e-12;
ts=0.1e-12;
Nfft = 2^nextpow2(max([N*tstep/ts N*tau*5/ts])); %Get enough points for the whole ramp with at least 5 tau's per exponenetial
w=(0:Nfft-1)*2*pi/Nff;t
f=w/2/pi/ts;
timeshifts=sum(exp(-1i*(0:N-1)'*w*tstep));
step=[pi 1./(1i*w(2:end))];
expdecay=1./(1/tau+1i*w);
Vf=Vstep*timeshifts.*(step-expdecay);
vt=ifft(V);
So the frequency domain Vf looks reasonable in amplitude. However, when I take the inverse and plot vt, it is definitely not correct.
Where am I going wrong? I suspect there's something I'm missing with the fact that this is a DFT not fourier transform. Also I know there's some subtlety to the FT/DFT of the heaviside, particularly at w=0, and I know that the fft at 0 will just be the average (times N) and I'm not sure my script accomplishes this.
I do know that I can just start with the time domain and then fft it, but I'm rather curious now.
fourier-analysis matlab fast-fourier-transform
fourier-analysis matlab fast-fourier-transform
asked Feb 1 at 23:20
KaysonKayson
613
asked Feb 1 at 23:20
KaysonKayson
613
asked Feb 1 at 23:20
KaysonKayson
613
asked Feb 1 at 23:20
KaysonKayson
613
asked Feb 1 at 23:20
KaysonKayson
613
613
add a comment |
add a comment |
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