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I have got this problem that I can't solve :

Let $y_1(t)=K(1-e^{-\frac{t}{\tau}})$

Find a way to determine K and $\tau$ from a graph

I answered $$K= \lim_{t \to \infty} y_1(t)$$ and with $$ y(t=\tau)=(1-e^{-1})K\approx 0.63 \tau$$ but I also have to find K and $\tau$ in $y_2(t)=K(t-\tau+\tau e^{-\frac{t}{\tau}})$

These functions come from the study of a first order LTIS ($y_1$) followed by an integrator ($y_2$)

I can use tangents for this problem.

Is my proof for $y_1$ correct ? how can i do for $y_2$ ?

thank you, T.D

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For $y_1$ an alternative solution is to take the slope of the curve at $t=0$. The derivative is $$y'_1(t)=\frac{K}{\tau}e^{-\frac{t}{\tau}}$$ so $$y'_1(0)=\frac{K}{\tau}$$

For $y_2$, you might notice that for large $t$, $y_2(t)\approx Kt-K\tau$. So taking the slope of the asymptote at large $t$ will give you $K$. If you plot the asymptote, the value at t=0 is $K\tau$.

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