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I am interested in knowing the step amplitude on a signal which can only be measured through a hardware RC filter. The measured signal follows this formula:

$$ V(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right) $$

Is it possible to estimate the step amplitude, $V_0$, given that we can perform two measurements on the almost linear part of the rising edge (with a known delay between them)?

Jason R
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Vandhunden
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  • Definitely. But I'd guess you'd already know that. It won't be exactly the step amplitude, but it can work for an approximate result. I remember several textbooks mentioning this method for biasing a p-n junction diode. – Naresh Mar 14 '13 at 15:51
  • Do you know the exact times $t_1$ and $t_2$ of the two measurements or just that you know that $t_1 \geq 0$ and $t_2$ is exactly $\tau$ seconds after $t_1$? The latter condition is also compatible with "known delay" since the delay $\tau$ is known exactly, but the exact locations of the sampling times on the time axis are unknown. – Dilip Sarwate Mar 14 '13 at 22:52
  • Dilip, I only know the exact time difference, not the absolute times. – Vandhunden Mar 15 '13 at 15:43

1 Answers1

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That's certainly possible, assuming you know the values of the resistance and capacitance $R$ and $C$. You can expand the signal $V(t)$ in a Taylor series about $t=0$ to arrive at an approximate linear model. First, remember the Taylor series of the exponential function:

$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \ldots $$

If you're assuming a linear model near the step input time of $t=0$, then truncate the sum to arrive at:

$$ e^x \approx 1 + x $$

Substitute this into the equation for $V(t)$ to yield:

$$ V(t) \approx V_0\left(1 - \left(1 - \frac{t}{RC}\right)\right) $$

$$ V(t) \approx V_0\frac{t}{RC} $$

Which is a pretty simple result. If you know the absolute time of the step input, then you theoretically don't even need multiple measurements; you could estimate $V_0$ directly using:

$$ V_0 \approx \frac{V(T)RC}{T} $$

where $T$ is the time at which you measure the signal. If you don't know the exact time at which the step is input, then you can take two measurements, as you suggested, at times $T_1$ and $T_2$, then form the estimate as follows:

$$ V(T_2) - V(T_1) \approx V_0\frac{T_2}{RC} - V_0\frac{T_1}{RC} $$

$$ V_0 \approx RC \frac{V(T_2) - V(T_1)}{T_2 - T_1} $$

Jason R
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