How to know the values of R and L?
Yo there! I have a RL circuit in the arrange showed in the image . circuit, step and impulse response
Ok, the trick is how can determine the values of the components based on the graphics of a step and impulse response, I detail the process and the graps in the images; basically what is done is to determine the model, calculate the unit and step response, got a 3 equations systems, but at the end, it is only a quotient, but aside that I dont see where can it get the values of R and L. Thanks in advance
The graphs are double-cheked, it seems odd but are ok. 1st , it gets the math model of the system
$$ i(t){\displaystyle \frac{R_{1}}{L}={\displaystyle \frac{dI_{L}}{dt}+({\displaystyle \frac{R_{2}+R_{1}}{L})I_{L}}}} $$
Calculating unit step and impulse response applying the input $i(t)=10_{u}(t)[A]$
$$ I_{L}(t)={\displaystyle \frac{10R_{1}}{R_{1}+R_{2}}(1)-{\displaystyle \frac{10R_{1}}{R_{1}+R_{2}}e{\displaystyle \frac{-R_{1}+R_{2}}{L}t}}} $$ and $$ h(t)={\displaystyle \frac{10R_{1}}{L}e{\displaystyle \frac{-R_{1}+R_{2}}{L}t}} $$ In the step graph it says $I(t)$ but there is trouble trying to fit the data, perhaps it should say $I_{L}(t)$, taking the graph this way: From the step graph $$ {\displaystyle \frac{10R_{1}}{R_{1}+R_{2}}=5...[1a]} $$ and $$ {\displaystyle \frac{10R_{1}}{R_{1}+R_{2}}(1-e^{-{\displaystyle \frac{R_{1}+R_{2}}{L}(5\mu s)}})=4.325...[2a]} $$ 1a in 2a $$ 4.325=5(1-e^{-{\displaystyle \frac{R_{1}+R_{2}}{L}(5\mu s)}})...[3a] $$ from the impulse $$ 68000={\displaystyle \frac{5R_{1}}{L}e^{{\displaystyle \frac{-R_{1}+R_{2}}{L}(2.5\mu s)}}...[4a]} $$ from [3a] $$ {\displaystyle \frac{4.325}{5}=(1-e^{-{\displaystyle \frac{R_{1}+R_{2}}{L}(5\mu s)}})...[5a]} $$ Grouping terms and aplying the natural log $$ ln(e^{-{\displaystyle \frac{R_{1}+R_{2}}{L}(5\mu s)}})=ln(1-{\displaystyle \frac{4.325}{5})...[5b]} $$
$$ {\displaystyle \frac{R_{1}+R_{2}}{L}={\displaystyle \frac{-2.0024}{-5\mu}=400,490...[5c]}} $$ From 1a $$ 10R_{1}=5(R_{1}+R_{2})...[6a]5R_{1}=5R_{2}...[6b]R_{1}=R_{2}...[6c] $$ from 6c in 5c $$ {\displaystyle \frac{2R}{L}=400,490} $$ $$ {\displaystyle \frac{R}{L}=200,245...[7a]} $$
This one is where it gets stalled because I dont see where to tackle the values, thanks in advance!! 374