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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

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

hang on....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?

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

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

hang on....

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

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

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?

2 added 870 characters in body
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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

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

hang on....

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:

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

hang on....

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

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

hang on....

1
source | link

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:

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

hang on....