Stating the problem:
A capacitance is loaded through a resistor according to the formula $$U(t) = U_{m}(1-e^{-\alpha t})\tag{1}\label{eq1}$$
At time $T$ the voltage $U(T)$ reaches the critical value $U_{0}<U_{m}$ and with the help of an active element the voltage is reset to $0$ and the charging starts again. This leads to a periodic signal with period $T$. The values of $U_{m}$ and $T$ are fixed. we may choose the values of $U_{0}$ and $\alpha$.
a) The period $T$ should not be sensitive to small variations of $U_{0}$ e.g. caused by temperature variations. Verify that this condition is satisfied if the slope of the voltage $U(t)$ at time $t = T$ is maximal. The answer is that
$$U_{0} \approx \frac{\partial U}{\partial T} \Delta T$$
For $\Delta T$ to be minimal for a given value of $\Delta U_{0}$ we need thus a large derivative $\frac{\partial U}{\partial T}$
b) Find the optimal value for $\alpha$ and $U_{0}$
Answer: the slope of the curve is determined by $$U'(t) = U_{m}\alpha e^{-\alpha t}\tag{2}\label{eq2}$$ Now this new expression has to be as large as possible, as function of the parameter $\alpha$.
$$\text{Thus we need the zeros of the derivative with respect to $\alpha$}\tag{3}\label{eq3}$$
$$\frac{\partial}{\partial \alpha}U'(T) = U_{m}e^{-\alpha T} (1-\alpha T)\tag{4}\label{eq4}$$
This derivative vanishes for $\alpha = \frac{1}{T}$ and thus the optimal choice is
$$U_{0} = U_{m}(1-e^{-\alpha T}) = U_{m}(1-\frac{1}{e})\tag{5}\label{eq5}$$
Here is what I don't understand:
- why do we need the zero of the derivative of (2) with respect $\alpha$ to find that the expression (2) be as large as possible
- Which derivative rule is used to get from (2) to (4)?
- I also don't understand why $\alpha = \frac{1}{T}$ is the optimal choice
Optimisation: when the derivative is zero there is a stationary point, indicating maxima/minima which can be determined using second derivatives.
Product rule: $(fg)'=f'g+fg'$ with respect to the variable you are trying to differentiate
Set $U_{m}e^{-\alpha T} (1-\alpha T)=0$. What are the options?