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In the book 'Fast analytical techniques for electrical and electronic circuits' by Vatché Vorpérian, the DC and small-signal equivalent circuit model of the buck-boost converter is obtained by replacing the PWM switch with its equivalent circuit model as shown:

Small signal DC and AC

The book states that all of the small-signal transfer functions have the same denominator which can be determined by setting all the independent excitations in the above circuit to obtain the following circuit

No excitation

The 2-EET can is now applied by taking out the inductor and the capacitor as shown in the reference circuit below:

Reference circuit

From which the denominator is given by the equation:

Denominator equation

I'm having trouble understanding how the author got:

enter image description here

From the circuit below:

Resistor circuit to be analyzed

I would appreciate if someone walks me through what he did. Here is my attempt:

my attempt

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    \$\begingroup\$ To determine an impedance or a resistance (as here), the excitation signal is a current source \$I_T\$ while the response is the voltage obtained across its connecting terminal. Inject \$I_T\$ in port (1), determine \$V_T\$ and you have \$R=\frac{V_T}{I_T}\$. I used these FACTs, together with multiple variations of the PWM switch model in my 2021 book on transfer functions. \$\endgroup\$ Commented Sep 25 at 18:52

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If you want to determine the resistance offered by a port, you should use a current source \$I_T\$ to excite its connecting terminals. This current source - the stimulus - will generate a voltage \$V_T\$ - the response - and the ratio of the response over the stimulus is the resistance you want: \$R=\frac{V_T}{I_T}\$.

With the PWM switch model, you can use the transformer with a 1:D ratio, but I prefer resorting to its equivalent model featuring current and voltage sources. It is easier to write the equations (in my opinion) and you can quickly run a simulation to verify your results. The trick is to rearrange these sources to reveal a simple sketch to look at:

enter image description here

You can write a few equations describing the voltages at node p (please note that since node a is 0 V, the Vap source is reversed to keep Vp alone) and if you consider current Ic as leaving terminal c, it is negative in this case and equal to \$-I_T\$. The final result is verified by SPICE and is correct.

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  • \$\begingroup\$ Appreciate the new insight. Also wanted to add that as the book also states, the r can also be modeled as a dependent voltage source with current as the controlling source in which voltage is given as v = Ir, whereby the current direction is into the voltage source. using NEET, the inverse gain can be obtained by replacing the voltage source with a test voltage source of inverted polarity to the initial one, and reciprocal of resistance R. I redid the calculations and got the expected answer D'^2R. My mistake in the initial attempt was in writing the KVL equations. \$\endgroup\$ Commented Sep 27 at 12:15

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