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Yo there! I have a RL circuit in the arrange showed in the image.

enter image description here

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 don't see where can it get the values of R and L. Thanks in advance

The graphs are double-checked, 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 don't see where to tackle the values,

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  • $\begingroup$ Yo yourself : ), I added in the image because users normally like to read without going off site. Personally, I think this is an ElectronicsSE kind of question, but the best of luck with it. $\endgroup$ – user108787 Oct 3 '16 at 16:39
  • $\begingroup$ Thanks CountTo10, It was my idea to add the picture, but the editing form says I cannot insert directly images. Thanks a lot again. $\endgroup$ – riccs_0x Oct 3 '16 at 17:35

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