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How are you? I am working in the next circuit, so \$I_R\$,\$I_L\$ and \$I_C\$ can be calculated.


Circuit


The next data is supplied

  • \$R=?\$
  • \$L=4.3mH\$
  • \$C=220 μF\$
  • \$i(t)=15u_1(t)A\$
  • \$M_p\leq16.3\%\$
  • \$t_s\pm2\%\$ \$=10ms\$

The first thing done is calculate the model of the system, and I just used the inductor current, but it can be in terms the current in the resistance or the capacitor too, the coefficients must be the same:
$$\frac{d^2I_L}{dt^2}+\frac{1}{RC}\frac{dI_L}{dt}+\frac{1}{CL}I_L=\frac{1}{CL}i(t)$$
The next pass is about taking the parameters like the natural frecuency, damping ratio, gain; for this system:

$$\alpha=\epsilon\omega_n$$
$$2\alpha=\frac{1}{RC}$$
$$\alpha=\frac{1}{2RC}..[1]$$
$$\omega_n^2=\frac{1}{LC}$$
$$\omega_n=\frac{1}{\sqrt(LC)}..[2]$$
comparing 1 and [2] $$\epsilon=\frac{\sqrt(L)}{2R\sqrt(C)}..[3]$$
Next try to calculate \$\epsilon\$, R and the \$\[email protected]\%\$
- From the overdamping case $$\epsilon>1$$ then $$\frac{\sqrt(L)}{2R\sqrt(C)}>1$$ or $$\frac{\sqrt(L)}{2\sqrt(C)}>R$$ then \$t_s@2\%\$ \$=\frac{4}{\epsilon\omega_n}\$ (I dont remember is this numerator is 3 or 4), so $$\epsilon=\frac{4}{(10\times10^-3)(\frac{1}{\sqrt(LC)})}=0.389$$. On tho R, comparing terms $$0.389\frac{1}{\sqrt((4.3\times10^-3)(220\times10^-6))}$$$$=\frac{1}{2RC}$$
From here, R is $$\frac{1}{2(220\times10^-6)(400)}=568$$
$$\epsilon$$ of the 16.3%, ergh, say, the overpass at 16.3% is calculated from the definition $$M_p=e^{(\frac{-\epsilon\pi}{\sqrt(1-\epsilon^2)})}$$
$$Ln0.163=\frac{-\epsilon\pi}{\sqrt(1-\epsilon^2)}$$ and $$\epsilon_{16.3\%}=0.50$$

and using KCL $$I(t)=I_{R_1}+I_{L_1}+I_{C_1}$$
But isnt clear how the parameters calculated are related to the currents. Thanks in advance

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    \$\begingroup\$ What does "15u1(t)A" mean? What is "Mp"? What does "Ts@2%" mean? \$\endgroup\$ Commented Oct 6, 2016 at 21:18
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    \$\begingroup\$ \$u_1(t)\$, means the unit step function, \$M_p\$ its the value of the first peak of the system and \$T_s\pm2\%\$ is the time to reach and stay within 2% of the response \$\endgroup\$
    – riccs_0x
    Commented Oct 6, 2016 at 21:47
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    \$\begingroup\$ So current starts at 0 and goes to 15A at t=0, correct? "first peak of the system" is what, peak voltage? "But isnt clear how the parameters calculated are related to the currents" - once you know the values of all the components you should be able to calculate the currents passing through them at any time. Is that what you want to do? \$\endgroup\$ Commented Oct 6, 2016 at 22:19
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    \$\begingroup\$ Thanks for your interest Bruce, Its not peak voltage is the peak of the step response of the system like the one showed at right step response, the current is right. \$\endgroup\$
    – riccs_0x
    Commented Oct 7, 2016 at 0:22
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    \$\begingroup\$ But does the "peak of the step response" coincide with peak output voltage, or is it something else? I read further and found that 'M' represents the set of states in a dynamical system. So what does "Mp <=16%" mean? \$\endgroup\$ Commented Oct 7, 2016 at 2:55

1 Answer 1

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R=? L=4.3mH C=220μF i(t)=15u1(t)A Mp≤16.3% ts±2% =10ms

My analysis as follows;

  • Using \$Q=X(f_0)/R= \frac{1}{2\zeta}\$ at resonant \$f_0\$
  • Mp=overshoot % 16.3% and from chart below \$\zeta=0.35, \text { then } Q=1.4 \$

  • \$f_0=\frac{1}{2\pi\sqrt{LC}}=140Hz \$

  • thus \$R=Q*Z_L(140 Hz)= 1 * 2\pi 140[Hz]*4.3[mH]= 3.8 \text{ } \Omega =R \$

Initial condition \$Ic(0)=15A, I_L(0)=0A... I_R(0)=0 . . . V_R(0)=0V\$

Final condition \$Ic=0A, I_L=15A. . . I_R=0 . . .V_R=0 \$

enter image description here

  • Response can be show from normalized for Q=1
  • Ts is redundant
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