Generalized Averaged Model

Abstract

This chapter approaches methodologies of deriving averaged models able to also represent behavior of converters containing AC stages. This time, modeling is not restricted to DC variables and the resultant models – called generalized averaged models (GAM) – can handle averages of higher-order harmonics.

Appendices

Problems

Problem 5.1

The diagram in Fig. 5.30 shows a circuit used to convey the energy between a variable DC source (e.g., a PV array) and a single-phase strong AC grid. There are two power stages, a DC-DC boost converter and a voltage-source inverter (VSI) connected by means of a DC-bus. Therefore, the system has two control inputs – u ₁ acting on the DC-DC converter and u ₂ acting on the VSI – and the disturbance input E, due to changes in the primary resource.

Fig. 5.30

Grid interface circuit used as power conveyor between variable DC source and AC power grid

Modeling assumptions limit the variables spectra at the DC component for the boost power stage and at the first harmonic for the voltage-source inverter. The inverter provides only active power to the grid. The purpose of control is to ensure the power flux between the primary DC source and the AC grid, while preserving the rated operating conditions (in terms of voltage and current).

Using the information in Chaps. 3, 4, and 5, perform the following actions:

• (a) obtain the large-signal system model using the generalized average modeling technique;

• (b) obtain the small-signal model by perturbation and linearization around a typical operating point. Give the transfer function having 〈u ₂〉₁ as input and 〈v ₀〉₀ as output.

Solution

(a) State-space modeling begins with describing the power imbalance in the energy accumulations (inductors and capacitors), and finally giving the switched model (by combining Eqs. (3.11) and (5.39)):

$$ \left\{\begin{array}{l}{L}_1\cdot \overset{\cdotp }{i_{L1}}=-\left(1-{u}_1\right)\cdot {v}_0+E\hfill \\ {}{L}_2\cdot \overset{\cdotp }{i_{L2}}=e-{v}_0\cdot {u}_2-{i}_{L2}\cdot r\hfill \\ {}C\overset{\cdotp }{\cdot {v}_0}=\left(1-{u}_1\right)\cdot {i}_{L1}-{v}_0/R-{i}_L\cdot {u}_2.\hfill \end{array}\right. $$

(5.82)

Averaging of the previous model gives after some algebra (the reactive power component being neglected):

$$ \left\{\begin{array}{l}\frac{d{\left\langle {i}_{L1}\right\rangle}_0}{ dt}=\frac{1}{L_1}\left[E-{v}_{DC}\left(1-\upalpha \right)\right]\hfill \\ {}\frac{d{i}_d}{ dt}=\frac{V}{L_2}-\frac{v_{DC}}{L_2}\cdot {\upbeta}_d-\frac{i_d\cdot r}{L_2}\hfill \\ {}\frac{d{v}_{DC}}{ dt}=\frac{1}{C}{\left\langle {i}_{L1}\right\rangle}_0\left(1-\upalpha \right)-\frac{1}{ RC}{v}_{DC}-\frac{i_d}{2C}\cdot {\upbeta}_d,\hfill \end{array}\right. $$

(5.83)

where v _DC  = 〈v ₀〉₀, i _d  = 〈i _L2〉₁ represents the magnitude of the grid inductor current and β _d  = 〈u ₂〉₁ is the inverter’s single control input (the active component on the d-axis). The whole system’s state vector is $\mathbf{x}={\left[\begin{array}{ccc}\hfill {\left\langle {i}_{L1}\right\rangle}_0\hfill & \hfill {i}_d\hfill & \hfill {v}_{DC}\hfill \end{array}\right]}^T$ and its input vector is $\mathbf{u}={\left[\begin{array}{ccc}\hfill \upalpha \hfill & \hfill {\upbeta}_d\hfill & \hfill E\hfill \end{array}\right]}^T$, which is composed of the control input vector ${\left[\begin{array}{cc}\hfill \upalpha \hfill & \hfill {\upbeta}_d\hfill \end{array}\right]}^T$ and the disturbance input E.

(b) Let us consider a steady-state operating point determined by the input ${\left[\begin{array}{ccc}\hfill {\upalpha}_e\hfill & \hfill {\upbeta}_{de}\hfill & \hfill {E}_e\hfill \end{array}\right]}^T$ and the state ${\left[\begin{array}{ccc}\hfill {i}_{L1e}\hfill & \hfill {i}_{de}\hfill & \hfill {v}_{DCe}\hfill \end{array}\right]}^T$. By zeroing the derivatives in Eq. (5.83), one can write

$$ \left\{\begin{array}{l}{E}_e-\left(1-{\upalpha}_e\right){v}_{DCe}=0\hfill \\ {}V-{v}_{DCe}{\upbeta}_{de}-{i}_{de}r=0\hfill \\ {}{i}_{L1}\left(1-{\upalpha}_e\right)-{v}_{DCe}/R-{i}_{de}{\upbeta}_{de}/2=0.\hfill \end{array}\right. $$

(5.84)

Let us consider small variations around this operating point: $\upalpha =\tilde{\upalpha}+{\upalpha}_e$, ${\upbeta}_d=\tilde{\upbeta_d}+{\upbeta}_{de}$, $E=\tilde{E}+{E}_e$ and ${i}_{L1}=\tilde{i_{L1}}+{i}_{L1e}$, ${i}_d=\tilde{i_d}+{i}_{de}$, ${v}_{DC}=\tilde{v_{DC}}+{v}_{DC e}$. Perturbing the state-space equations from (5.83) around the previously defined steady-state operating point gives the small-signal state-space equations in the form

$$ \left\{\begin{array}{c}\kern-8.90em {L}_1\cdot \overset{\cdotp }{\tilde{i_{L1}}}=-\left(1-{\upalpha}_e\right)\cdot \tilde{v_{DC}}+{v}_{DC e}\cdot \tilde{\upalpha}+\tilde{E}\\ {}\kern-9.90em {L}_2\cdot \overset{\cdotp }{\tilde{i_d}}=-r\cdot \tilde{i_d}-{\upbeta}_{de}\cdot \tilde{v_{DC}}-{v}_{DC e}\cdot \tilde{\upbeta_d}\\ {}C\cdot \overset{\cdotp }{\tilde{v_{DC}}}=\left(1-{\upalpha}_e\right)\cdot \tilde{i_{L1}}-\frac{\upbeta_{de}}{2}\cdot \tilde{i_d}-\frac{1}{R}\cdot \tilde{v_{DC}}-{i}_{L1e}\cdot \tilde{\upalpha}-\frac{i_{de}}{2}\cdot \tilde{\upbeta_d}.\end{array}\right. $$

With $\tilde{\mathbf{x}}={\left[\begin{array}{ccc}\hfill \tilde{i_{L1}}\hfill & \hfill \tilde{i_d}\hfill & \hfill \tilde{v_{DC}}\hfill \end{array}\right]}^T$ as the state vector and by $\tilde{\mathbf{u}}={\left[\begin{array}{ccc}\hfill \tilde{\upalpha}\hfill & \hfill \tilde{\upbeta_d}\hfill & \hfill \tilde{E}\hfill \end{array}\right]}^T$ the input vector of the small-signal model, one can write the small-signal state-space matrix representation, as

$$ \overset{\cdotp }{\tilde{\mathbf{x}}}=\underset{\mathbf{A}}{\underbrace{\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill -\frac{1-{\upalpha}_e}{L_1}\hfill \\ {}\hfill 0\hfill & \hfill -\frac{r}{L_2}\hfill & \hfill -\frac{\upbeta_{de}}{L_2}\hfill \\ {}\hfill \frac{1-{\upalpha}_e}{C}\hfill & \hfill -\frac{\upbeta_{de}}{2C}\hfill & \hfill -\frac{1}{2C}\hfill \end{array}\right]}}\cdot \tilde{\mathbf{x}}+\underset{\mathbf{B}}{\underbrace{\left[\begin{array}{ccc}\hfill \frac{v_{DCe}}{L_1}\hfill & \hfill 0\hfill & \hfill \frac{1}{L_1}\hfill \\ {}\hfill 0\hfill & \hfill -\frac{v_{DCe}}{L_2}\hfill & \hfill 0\hfill \\ {}\hfill -\frac{i_{Le}}{C}\hfill & \hfill -\frac{i_{de}}{2C}\hfill & \hfill 0\hfill \end{array}\right]}}\cdot \tilde{\mathbf{u}}, $$

(5.85)

which makes apparent the state matrix A and the output matrix B. One is interested in considering $\tilde{v_{DC}}$ as the output variable; consequently, to the state-space equation in (5.85) the output equation is added, which introduces the output matrix C:

$$ \mathbf{y}=\tilde{v_{DC}}=\underset{\mathbf{C}}{\underbrace{\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]}}\cdot \tilde{\mathbf{x}}. $$

(5.86)

Note that the 1 × 3 transfer matrix

$$ \mathbf{H}(s)=\mathbf{C}\cdot {\left(s\mathbf{I}-\mathbf{A}\right)}^{-1}\cdot \mathbf{B}=\left[\begin{array}{ccc}\hfill {H}_{11}(s)\hfill & \hfill {H}_{12}(s)\hfill & \hfill {H}_{13}(s)\hfill \end{array}\right] $$

contains the transfer functions from all the inputs to the output $\tilde{v_{DC}}$. Here the interest is focused on the transfer from $\tilde{\upbeta_d}$ to $\tilde{v_{DC}}$, which is the second element of the transfer matrix H(s): ${H}_{\tilde{\upbeta_d}\to \tilde{v_{DC}}}(s)={H}_{12}(s)$. Taking into account the expressions of matrices A, B and C given in Eqs. (5.85) and (5.86), quite laborious but straightforward algebra leads to the final form of the required transfer function:

$$ {H}_{\tilde{\upbeta_d}\to \tilde{v_{DC}}}(s)=-\frac{s\frac{\upbeta_{de}}{2C}}{s^3+{s}^2\left[\frac{1}{ RC}+\frac{r}{L_2}\right]+s\left[\frac{r}{ RC{L}_2}+\frac{{\left(1-{\upalpha}_e\right)}^2}{L_1C}-\frac{\upbeta_{de}^2}{2{L}_2C}\right]+\frac{{\left(1-{\upalpha}_e\right)}^2r}{L_1{L}_2C}}. $$

The reader is invited to solve the following problems.

Problem 5.2

Modeling a basic structure in the GAM sense

Let us consider the diagram from Fig. 5.31 and assume that the switching function acting on the switches takes value 1 if H ₁ is turned on and −1 if H ₂ is turned on. One should note that the two switches cannot be simultaneously turned on.

Fig. 5.31

Basic converter

• (a) By taking current i _a sinusoidal and voltage v ₀ continuous, draw on one plot v _a , i _a and i ₀ such that u is delayed by angle φ in relation to i _a .

• (b) Establish the generalized averaged diagram by employing the necessary topological transform but without computing the terms 〈v _a 〉₁ and 〈i ₀〉₀.

• (c) In the case where the rectifier current (voltage inverter) is not controlled (i.e., it is a diode rectifier), compute the terms 〈v _a 〉₁ and 〈i ₀〉₀, then give the expressions of these terms in the general case when the control input is the AC current-voltage phase lag φ.

Problem 5.3

Capacitive half-bridge series-resonant power supply

Given the circuit in Fig. 5.32,

Fig. 5.32

Series-resonant switching power supply

• (a) by using the results provided in Chap. 3, establish the equivalent exact topological diagram of the converter by defining the switching functions;

• (b) deduce its generalized averaged model equivalent diagram without developing the coupling terms;

• (c) develop the terms 〈e(t)〉₁ and 〈s(t)〉₁ and compute the coupling terms if the phase reference is the angle δ between the inverter commutations and those of the rectifier;

• (d) write the GAM equations of the converter and compute its equilibrium points. Using these results, build the small-signal model (its state-space representation) by taking as control input the variations of δ and as output the variations of v _Co .

Problem 5.4

A dynamical model of the resonant power supply depicted in Fig. 5.33 is sought. In this topology, both the inverter and the rectifier are based only on switches, without employing capacitive half-bridges.

Fig. 5.33

Resonant power supply

The following assumptions are adopted:

• operating frequency is quite close to resonance; this allows i _L and v _C fundamental harmonics prevalence in relation to their higher-order harmonics;

• operating frequency and output capacitor values are sufficiently high in order to ignore the output voltage v _S ripple (it is superposed on its average value);

• switches H _1h and H _1l are complementary and driven by the switching function u ₁ such that u ₁ = 1 if H _1h is turned on and u ₁ = 0 if H _1h is turned off; the switching function u ₂ plays the same role for the complementary switches H _2h and H _2l : u ₂ = 1 yields H _2h being turned on and u ₂ = 0 yields H _2h being turned off. Both switching functions have pulsation ω and duty ratio 0.5 such that u ₁ is the phase origin and u ₂ is delayed by an angle φ with respect to u ₁.

• (a) Write the differential equations that govern the evolutions of the capacitor C voltage v _C and of the inductance L current i _L in the resonant circuit. These equations must be written as depending on the inverter output voltage e(t) and on rectifier input voltage s(t). Express the dynamic of the output voltage v _S as a function of the output current i _s (t).

• (b) Write the voltages e(t) and s(t) as functions of E, u ₁, u ₂ and v _S . Express the currents i _e (t) and i _s (t) as functions of i _L , u ₁ and u ₂.

• (c) Deduce the switching model of the converter in Fig. 5.19 having the switching functions u ₁ and u ₂. Deduce also the equivalent topological diagram that makes apparent the coupling terms.

• (d) With the variable changes

  $$ \left\{\begin{array}{l}{u}_1^{\prime }=2{u}_1-1\\ {}{u}_2^{\prime }=2{u}_2-1\end{array}\right. $$

  • (i) compute the terms 〈u ^′₁ 〉₀, 〈u ^′₁ 〉₁, 〈u ^′₂ 〉₀ and 〈u ^′₂ 〉₁;

  • (ii) express the coupling terms and the converter switched model having these new switching functions;

  • (iii) give the dynamical expressions of 〈i _L 〉₀, 〈i _L 〉₁, 〈v _C 〉₀ and 〈v _C 〉₁ as functions of the previous coupling terms and make the associated computations. Knowing that the current i _L cannot have a DC component, show that 〈v _C 〉₀ has no dynamic and deduce its steady-state value;

  • (iv) give the expression of the output voltage dynamic v _S (identical with its average value 〈v _S 〉₀). Show that the term whose average is 〈v _C 〉₀ has no influence on v _S ;

  • (v) give the large-signal generalized averaged model of the entire converter by making all the associated computations.

• (e) Build the small-signal model resulted from the large-signal model obtained at point (v).

• (f) Compare by simulation the time behavior of the various models obtained.

Problem 5.5

The DC-DC parallel-resonant converter in Fig. 5.34 is controlled via a squared switching function u ₁. Develop the large-signal GAM of this structure.

Fig. 5.34

Parallel-resonant switching power supply

Problem 5.6

Figure 5.35 shows an electronic circuit that supplies a high-intensity discharge (HID) lamp. The switching function u is a high-frequency squared signal of 0.5 duty ratio. When one transistor is turned on the other one is turned off, and inversely,

Fig. 5.35

Electronic ballast feeding a HID lamp

• (a) develop the large-signal generalized averaged model;

• (b) deduce the small-signal averaged model;

• (c) compute the transfer function between the control input frequency and the lamp current.

Appendix

Justification of equation (5.5)

$$ {\left\langle x\cdot y\right\rangle}_k(t)={\displaystyle \sum_i{\left\langle x\right\rangle}_{k-i}(t)\cdot {\left\langle y\right\rangle}_i(t)} $$

This relation is the Fourier-series correspondent of the convolution theorem holding for the Fourier transforms of nonperiodical signals. Note that it is valid for signals having the same fundamental frequency.

Let us first consider that x and y are both DC signals, i.e., they are equal to their zero-order sliding harmonics. Then, the maximal-degree harmonic of their product will be zero as well. In this case it is obvious that

$$ {\left\langle x\cdot y\right\rangle}_0(t)={\left\langle x\right\rangle}_0(t)\cdot {\left\langle y\right\rangle}_0(t) $$

If now one supposes that x is a biased sinusoidal signal of pulsation ω (therefore containing harmonics of maximal degree equal to one) and y contains harmonics until second order (time variable is skipped), then

$$ x={\left\langle x\right\rangle}_0+{\left\langle x\right\rangle}_{-1}{e}^{-j\omega t}+{\left\langle x\right\rangle}_1{e}^{j\omega t}, $$

$$ y={\left\langle y\right\rangle}_0+{\left\langle y\right\rangle}_{-1}{e}^{-j\upomega t}+{\left\langle y\right\rangle}_1{e}^{j\upomega t}+{\left\langle y\right\rangle}_{-2}{e}^{-2j\upomega t}+{\left\langle y\right\rangle}_2{e}^{2j\upomega t}. $$

Their product will contain harmonics of maximal degree equal to three. One obtains successively

$$ \begin{array}{c}\kern-11.50em x\cdot y={\left\langle x\right\rangle}_0{\left\langle y\right\rangle}_0+{\left\langle x\right\rangle}_{-1}{\left\langle y\right\rangle}_1+{\left\langle x\right\rangle}_1{\left\langle y\right\rangle}_{-1}\\ {}\kern-4.30em +\left({\left\langle x\right\rangle}_{-1}{\left\langle y\right\rangle}_0+{\left\langle x\right\rangle}_0{\left\langle y\right\rangle}_{-1}+{\left\langle x\right\rangle}_1{\left\langle y\right\rangle}_{-2}\right){e}^{-j\upomega t}\\ {}\kern-5.9em +\left({\left\langle x\right\rangle}_{-1}{\left\langle y\right\rangle}_2+{\left\langle x\right\rangle}_0{\left\langle y\right\rangle}_1+{\left\langle x\right\rangle}_1{\left\langle y\right\rangle}_0\right){e}^{j\upomega t}\\ {}\kern3.1em +\left({\left\langle x\right\rangle}_{-1}{\left\langle y\right\rangle}_{-1}+{\left\langle x\right\rangle}_0{\left\langle y\right\rangle}_{-2}\right){e}^{-2j\upomega t}+\left({\left\langle x\right\rangle}_1{\left\langle y\right\rangle}_1+{\left\langle x\right\rangle}_0{\left\langle y\right\rangle}_2\right){e}^{2j\upomega t}\\ {}\kern-7.7em +{\left\langle x\right\rangle}_{-1}{\left\langle y\right\rangle}_{-2}{e}^{-3j\upomega t}+{\left\langle x\right\rangle}_1{\left\langle y\right\rangle}_2{e}^{3j\upomega t},\end{array} $$

which can be posed under the form

$$ \begin{array}{c}x\cdot y={\left\langle xy\right\rangle}_{-3}{e}^{-3j\upomega t}+{\left\langle xy\right\rangle}_{-2}{e}^{-2j\upomega t}+{\left\langle xy\right\rangle}_{-1}{e}^{-j\upomega t}+{\left\langle xy\right\rangle}_0\\ {}\kern-2em +{\left\langle xy\right\rangle}_1{e}^{j\upomega t}+{\left\langle xy\right\rangle}_2{e}^{2j\upomega t}+{\left\langle xy\right\rangle}_3{e}^{3j\upomega t},\end{array} $$

where

$$ \left\{\begin{array}{c}{\left\langle xy\right\rangle}_0={\left\langle x\right\rangle}_{-1}{\left\langle y\right\rangle}_1+{\left\langle x\right\rangle}_0{\left\langle y\right\rangle}_0+{\left\langle x\right\rangle}_1{\left\langle y\right\rangle}_{-1}\\ {}\kern-0.55em {\left\langle xy\right\rangle}_1={\left\langle x\right\rangle}_{-1}{\left\langle y\right\rangle}_2+{\left\langle x\right\rangle}_0{\left\langle y\right\rangle}_1+{\left\langle x\right\rangle}_1{\left\langle y\right\rangle}_0\\ {}\kern-5.58em {\left\langle xy\right\rangle}_2={\left\langle x\right\rangle}_1{\left\langle y\right\rangle}_1+{\left\langle x\right\rangle}_0{\left\langle y\right\rangle}_2\\ {}\kern-9.78em {\left\langle xy\right\rangle}_3={\left\langle x\right\rangle}_1{\left\langle y\right\rangle}_2.\end{array}\right. $$

(5.90)

The result given by (5.90) can be generalized to the product of two periodical signals having the same fundamental frequency x and y containing harmonics up to degree n, respectively m. Thus, one can formulate the following proposition about the expression of the kth-order harmonic of the product xy:

and must prove by mathematical induction that it is true for any n and m. To this end, one supposes that (n, m) is true and must prove that either (n+1, m) or (n, m+1) is true. One can note that adopting the variable change z = e ^jωt, signal x can be expressed as a sum of two n-degree polynomials in z and 1/z, whereas signal y is a sum of two m-degree polynomials in z and 1/z. Results concerning the general form of coefficients of products of polynomials can further be applied in order to get the expression of the harmonics of the two signals expressed as polynomials (Osborne 2000).