Unwind di Power Electronics - 2 di 5

Open Circuit Voltage at Nodes Y-Y'

SI RtNow let’s evalute the open circuit voltage at the couple of nodes Y-Y’:KatDil5L2 VegD Il iItd i iVee vo IFIiIe iavg.at avg.atVo iKat sl1VÀ e ERDIo 1aVo RThe equivalent circuit, considering the Thevenin equivalent generator becomes:Ven RZsi 1DR RCNo s1 tZThe small ripple output voltage is: RÈ thatZ slIRCsIVeg ta e 42R D2 5L 5L Dt tÈ Èsrl1 tZ R RCsi tthatVeda SI1 ERD R Kat sl1RCsI t aes'LCsl 42R D Di DIRE D5I t 5L1Kat D'Rs'LCtslD i 2DI DAt least, the control-to-output transfer function (TF) Gvd(s) is: 5L1ÈGd No Vo D'R5 s s'LCslD is Ig DIO DThis is the control-to-output TF of the CCM boost converter.In a more canonical form, this TF becomes: VoGrido Ds1 CuzGrid WzGrido D5 e525I t Woo WE No Da LeTo determine the factor Q it’s usually considered the first order term at the denominator:MR111 Le LEaae ILDI iv Q Q D D'R ieLSo: VoGrido DCuz Dira51Girds WzGrido 52 No D5I t Woo CUI leQ.IR ieLThe resonant

Gra dB5
Grado 40
dB deedi 20
dB dee0
dB UzWo 5log
Grid 5
Wo Uz 5log
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-180° while the zero introduces a phasedisplacement only of -90° (because it’s a RHP zero) for a total phase displacement of 270°. The RHP zero actslike a pole in the phase plot (because it’s a RHP zero) and this is a very big problem for the phase marginbecause an eventual compensation must be done not for -180° but for -270° and this is very difficult, in fact it’snot usually ever done. Moreover the compensation should consider the fact that the resonance frequency is notfixed but depends by the operating point.

Let’s consider the origin of this RHP zero and his implications on the time-invariant and linear CCM equivalentcircuit: diVil1 aCVg v REdè

The RHP zero comes from the current generator at the right side of the transformer, in particular from it’sorientation.Considering the 2 indipendent generators and a positive variation of the duty cycle, it happens that the 2generators fight each other, because the voltage generator

induces a current variation in this direction, while the current generator induces a current in this direction. For this reason the current generator counteracts the positive variation of the duty-cycle and induces a decreasing of the phase at the output voltage.

Let's consider now what happens in time domain when there is a RHP zero in a TF in general terms.

Let's consider a general TF like the following one:

iG Gae 0G Wzi S Go es 5 sWz 5i 5cuiiv a

This general TF has a Left Half Plane zero. The unit step response of this general TF is:

iS5I Go Go Go Go Go5 5 5 5 55Wz 5Wz5

Considering this term as yo(t), then this term corresponds to the derivative of yo(t), divided by wZ.

4 t t5 40G Gas s s a5 sWz

In time domain, the plots of the 2 functions are the following:

4 t t40 wz

To find the total response of the system, the 2 waveforms must be summed. The total plot becomes:

4 t t wz

The final waveform is quite similar to yo(t).

A very interesting case is when in the G(s) TF there is a

minus and not a plus:iG Gae 0G Wzi _S Go es 5 sWz 5i 5cuiiv aIn this case, the unit step response becomes: I1 ssI gos Go Go Go Go5 5 5 5 5 55 sWzWz Nz S5 5Considering this term as yo(t), then this term corresponds to the derivate of yo(t), divided by wZ:4 Yatt5G G jetas s s a5 sWzIn time domain, the plots of the 2 functions are the following:4 t t40 wzTo obtain the response of the system in this case, the 2 plots must be subtracted. The total plot is:4 t t wzIt’s clear that y(t) is really different in respect of the previous case. In particular, if a positive step variation isapplied at the input, the output goes in the opposite direction. So the effect of a RHP zero is to give a responsethat is opposite to the given command.Considering the RHP zero in the Gvd(s) TF, if the controller is quite slow, this opposite initial response doesn’tcreate too much problems, but if the controller requires a fast response to output variations, this opposite initialresponse is not

negligible and takes the controller unstable.

The problem of the RHP zero in Gvd(s) TF can be analysed in frequency domain considering the particular final phase of the system and in time domain considering the initial opposite response of the system.

Let's see now why a positive variation of the duty cycle gives initially a diminution of the output voltage. Let's in particular consider the CCM boost converter general circuit:

5L DtdVI e5 r

Let's suppose to drive the primary switch S with a duty cycle D and to apply a positive variation to this duty cycle. In this condition, the duty cycle of the secondary switch decreases, so the primary switch is ON for a longer time and the secondary switch is OFF for a longer time, for this reason the initial behaviour of the output voltage is to go down and not to go up, but after some time, the output voltage grows up another time and this happen because if the primary switch is closed for a longer time, much more current is injected in

The inductor. This is reflected in a bigger current that flows in the load (capacitor and resistor). From a different point of view, it can be seen that initially the capacitor discharges for a longer time on the load, for this reason the voltage decreases but after some time, the amount of current that comes from the inductor recharges the capacitor and the output voltage grows up. The quantity of time after which the voltage grows up depends by some factors like the parameters of the converter and the resonance frequency.

LESSON #13

LESSON #14

LESSON #15

Power Electronics – a.a. 2020-2021

Review of Linear, Time-Invariant Discrete-Time Systems

Luca Corradini

Power Electronics Group

University of Padova

Department of Information Engineering – DEI

Linearity and Time Invariance

Linearity:

Time Invariance:

L. Corradini, a.a. 2020-2021 3

In a SISO (Single Input Single Output) discrete-time system, with the sequence u[k] as input and the sequence y[k]

• Linearity: if 2 different inputs u1[k] and u2[k] produces 2 different outputs y1[k] and y2[k] then the system is linear if the linear combination of the inputs produces an output that is the linear combination of the outputs with the same multiplying factors;
• Time invariance: the systems produces the same output during the time. A shift in time of the input corresponds to a shift of the same time of the output;

A difference equation is the discrete-time equivalent equation of a continuous-time differential equation. The output y[k] is equal to the superposition of 2 contributions:

• The first term is a linear combination of previous N-samples of the output;
• The second term is quite similar but it's the linear combination of the present input that is u[k] and M previous input samples;

The obtained relation is fully linear because the output y[k] is the linear superposition of contributes that are all linear.

Input/Output Difference Equation

L. Corradini, a.a. 2020-2021

Let's see now what happens in the discrete-time domain, considering the following expression:

I ifi.it4 ko ko The expression is particularized in afaite fbiUko specific k0.

Let's in particular consider the case for N>=M:

ifiko.M The instantaneous N output samples. output is Present. influenced by the previous M input samples and the previous KMko ko no N output samples. M input

samples.In the opposite case, it can happen that N<=M: TheM input samples. instantaneousPresent. output isinfluenced bythe previous Minput sampleskM MKa ko ko no and the previousN outputsamples.N output samples.It’s important to note that in can happen that N=0. This means that the instantaneous output depends only by theinputs. There is no relation between the instantaneous output and the previous outputs of the system.Initial ConditionsL. Corradini, a.a. 2020−2021 6Simply looking to the definition of the output of the system and considering the case of k>=0, it&rsq