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I am struggling with the derivation of average value of input current as below. The image is from the lecture here (page 26).

Here is what I got so far. How did the lecture derive the final step here?

$$ \langle i_g(t)\rangle_{T_s}=\frac{1}{T_s} \int_{0}^{T_s} i_g(t) dt =\frac{1}{T_s} \int_{0}^{dT_s} \langle i(t)\rangle_{T_s} dt = d(t)\langle i(t)\rangle_{T_s} $$

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$$ \langle i_g(t)\rangle_{T_s}=\frac{1}{T_s} \int_{0}^{T_s} i_g(t) dt =\frac{1}{T_s} \int_{0}^{dT_s} \langle i(t)\rangle_{T_s} dt = d(t)\langle i(t)\rangle_{T_s} $$

Lets consider i(t). Taking the integral of i(t) with respect to t will result in ti(t). I am assuming also Ts is a subscript in this derivation. The steps missing would be the following:

$$ \frac{dT_s}{T_s}i(t)_{Ts} - \frac{0}{T_s}i(t)_{Ts} = d(t)i(t)_{Ts} $$

d(t) needs to be taken with respect to t which is why it appears in the final derivation. The computation of the integral is really the only step missing in this case.

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  • \$\begingroup\$ Thanks. You are right. But I think there is a problem with this. The average value <i(t)> is not a constant, right? \$\endgroup\$ – anhnha Dec 6 '16 at 21:53
  • \$\begingroup\$ Replace I(t) by y. It will make more sense to your eyes I believe. You are not taking the integral in function of dy. Therefore, I(t) should be treated as a constant I believe. \$\endgroup\$ – 12Lappie Dec 7 '16 at 21:47

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