# State Space Mohann

Discussion in 'General Electronics Chat' started by gusmas, Apr 8, 2013.

1. ### gusmas Thread Starter Active Member

Sep 27, 2008
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Hi

I am currently working through Power Electronics, Converters, Applications and design by Mohan. Now I am at chapter 10, where a avg state space model is derived for a dc-dc converter. Problem I am having is how he derives the state variables.

I understand how he got eq `10-63, using KVL, however I can not understand how he derived 10-64. It seems like he is using KVL again but only in the RC part of the converter??

I have a image of the page I am referring to. See attached

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The second equation is indeed obtained by the application of KVL around the right hand side loop - having regard to the currents flowing in each branch.

If you require further elaboration, I'm sure some help will be forthcoming - alternatively you might consider carefully annotating the circuit and resolving the matter yourself.

Perhaps I should at least offer a hint.

The current in the capacitor [C] + resistor [rc] branch will be related to the capacitor voltage x2 such that

$i_c=C\frac{dV_c}{dt}=C\frac{dx_2}{dt}=C\dot{x_2}$

The current in the load branch resistor [R] will then be given by

$i_R=x_1-i_c=x_1-C\dot{x_2}$

Last edited: Apr 8, 2013
3. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
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HI

I understand the maths on how they did it, I just always thought if you use KVL with the one state space variable, you have to use KCL to solve the other.. haha very stupid mistake .

4. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
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ME AGIAN

I am on the next page and got stuck again.. sigh I know..

Its how he calculates Vo...

$_{}V_o = R(x_{1}-C \dot{x_{2}})$

I have no idea what happens after that line (eq 10-70)

I can understand he is taking x1 (Indcutor current) and multiplying it with the total resistance to get the output voltage, but why is he adding the (R/(R+ rc)*x2?? Is that the voltage potential that is already inside the cap, and he is just calculating the voltage potential at the resistor using ohms law??

• ###### mohan chapt 10.jpg
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Last edited: Apr 8, 2013
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It follows from ....

$x_1=$$\frac{v_o-x_2}{r_c}$$+\frac{v_o}{R}$

hence

$v_o$\frac{1}{r_c} + \frac{1}{R}$=x_1+$$\frac{x_2}{r_c}$$$

etc....

6. ### gusmas Thread Starter Active Member

Sep 27, 2008
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The state variables are the following:

$x_1 = Inductor current$
$x_2 = Capacitor Voltage$

If I take equation 10-70, and make X1 the subject of the formula I get the following:

$x_1=C*\dot{x_2}{}+\frac{v_o}{R}$

Now because the current inside the capacitor equates to the following:
$C*\dot{x_2}$

additionally: the current through the capacitor will be same as the current through the series resistance presented by the cap (rc).

$C*\dot{x_2} = I_r_c$

thus

$I_r_c= \frac{v_r_c}{r_c}$

thus the voltage over rc is:

$v_r_c = v_o - x_2$

finally, by using ohms law the current through rc AND the capacitor is:

$I_r_c = \frac{v_o - x_2}{r_c}$

is this correct? It makes sens to me.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Your reasoning seems a bit convoluted but the outcome is correct.

I viewed the situation as follows....

The current from the inductor [x1] divides between the R branch and the rc+C branch.

The current in R branch is just Vo/R

The current in rc+C branch is [by simple deduction] (Vo-Vcap)/rc

Since Vcap=x2 then the current in rc is (Vo-x2)/rc

So x1=Vo/R + (Vo-x2)/rc

etc....

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8. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
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Ye lol I am not good with explaining things, but my way of doing it was the same as yours!! thanks a million for the help !!!