DFT of a series of RC exponentials 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

$V(t) = \sum_{k=0}^{N-1} V_{step}(1-e^{-(t-k\Delta t_{step})/\tau})u(t-k\Delta t_{step})$

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:

$\mathfrak{F}(V(t)) = V_{step}\left(\sum_{k=0}^{N-1}e^{-j\omega k\Delta t_{step}}\right)\left(\frac{1}{j\omega} +\pi\delta(\omega) - \frac{1}{\frac{1}{\tau}+j\omega}\right)$

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.
okay, for step 1, let's get the continuous time math down right and simplified to as much as possible.  i am changing some notational convention slightly to be more consistent with Electrical Engineering practice:

$$v(t) = \sum_{k=0}^{N-1} V_\mathrm{step}(1-e^{-(t-k\Delta t_\mathrm{step})/\tau})u(t-k\Delta t_\mathrm{step})$$

$$V(j\omega) =\mathscr{F}\Big\{v(t)\Big\} = V_\mathrm{step}\left(\sum_{k=0}^{N-1}e^{-j\omega k\Delta t_\mathrm{step}}\right)\left(\frac{1}{j\omega} +\pi\delta(\omega) - \frac{1}{\frac{1}{\tau}+j\omega}\right)$$

there is a well known closed form for the summation of a geometric series:

$$ \sum_{k=0}^{N-1} x^k = \frac{x^N -1}{x-1} $$

so

$$\begin{align}
 \sum_{k=0}^{N-1} e^{-j\omega k\Delta t_\mathrm{step}} &amp;= \sum_{k=0}^{N-1} (e^{-j\omega \Delta t_\mathrm{step}})^k \\ \\
 &amp;= \frac{(e^{-j\omega \Delta t_\mathrm{step}})^N -1}{e^{-j\omega \Delta t_\mathrm{step}}-1} \\ \\
 &amp;= \frac{e^{-j\omega N \Delta t_\mathrm{step}}-1}{e^{-j\omega \Delta t_\mathrm{step}}-1} \\ \\
 &amp;= \frac{e^{-j (\omega/2) N \Delta t_\mathrm{step}}(e^{-j (\omega/2) N \Delta t_\mathrm{step}}-e^{j (\omega/2) N \Delta t_\mathrm{step}})}{e^{-j(\omega/2) \Delta t_\mathrm{step}}(e^{-j(\omega/2) \Delta t_\mathrm{step}}-e^{j(\omega/2) \Delta t_\mathrm{step}})} \\ \\
 &amp;= e^{-j (\omega/2) (N-1) \Delta t_\mathrm{step}}\frac{(e^{j (\omega/2) N \Delta t_\mathrm{step}}-e^{-j (\omega/2) N \Delta t_\mathrm{step}})/(2j)}{(e^{j(\omega/2) \Delta t_\mathrm{step}}-e^{-j(\omega/2) \Delta t_\mathrm{step}})/(2j)} \\ \\
 &amp;= e^{-j (\omega/2) (N-1) \Delta t_\mathrm{step}}\frac{\sin(\omega N \Delta t_\mathrm{step}/2)}{\sin(\omega \Delta t_\mathrm{step}/2)} \\ \\
\end{align}$$

so i think your final expression in continuous frequency of the spectrum should be

$$V(j\omega) = V_\mathrm{step}\frac{\sin(\omega N \Delta t_\mathrm{step}/2)}{\sin(\omega \Delta t_\mathrm{step}/2)} \left(\frac{1}{j\omega} +\pi\delta(\omega) - \frac{\tau}{1+j\omega \tau}\right) e^{-j (\omega/2)(N-1)\Delta t_\mathrm{step}} $$

does that look right to you?