Feedback-linearization Control Applied to Power Electronic Converters

Abstract

Feedback linearization is a powerful instrument that transforms a generic nonlinear plant model into a linear one by using a nonlinear feedback that cancels the original plant nonlinearity. Usually the target plant is a pure integrator; hence it can be controlled (with zero steady-state error) by a simple proportional controller (Isidori 1989). However, this feature comes with a drawback: the linearized system is likely to be sensitive to parameters’ variations and/or to the operating point, as the nonlinear feedback derives from the system model.

Problems

Problem 11.1

Feedback-linearization control of series-resonant converter output voltage

Let us consider the series-resonant power converter whose circuit diagram is shown in Fig. 11.15, where primary power source is voltage source E. Resonant filter bank has inductor L, capacitor C and equivalent resistance r _L .

Fig. 11.15

Circuit diagram of series-resonant DC-DC converter

Output voltage is filtered using capacitor C ₀. Output resistance R ₀ is considered constant and known. Signals u ₁ and u ₂ are rectangular waveforms of fixed frequency and 0.5 duty ratio that are switching the DC variables E and v ₀. The input variable is the phase lag φ between these two switching signals, where u ₁ is taken as phase origin.

Control of output voltage v ₀ is the stated goal. For this purpose, the expression of the control input φ that linearizes the plant having v ₀ as output is to be obtained.

Solution

Figure 11.16 presents waveforms of signals u ₁ and u ₂ and emphasizes the angle lag between them, φ, which serves as control input in this application. Current i _L is lagged by angle α in relation to signal u ₁.

Fig. 11.16

Rectangular waveforms of switching signals in series-resonant converter case

By considering as state variables the ones corresponding to energy accumulations in the circuit, the converter switched state-space model is given by (see Sect. 5.7.2)

$$ \left\{\begin{array}{c}L\frac{d{i}_L}{ dt}=E\cdot {u}_1-{v}_C-{r}_L{i}_L-{v}_0\cdot {u}_2\\ {}\kern-10em C\frac{d{v}_C}{ dt}={i}_L\\ {}\kern-6.2em {C}_0\frac{d{v}_0}{ dt}={i}_L{u}_2-\frac{v_0}{R_0}.\end{array}\right. $$

(11.35)

One makes the following notations concerning the state variables’ averages in the sense of the first-order harmonic:

$$ \begin{array}{ccc}\hfill {\left\langle {i}_L\right\rangle}_1={x}_1+j{x}_2,\hfill & \hfill {\left\langle {v}_C\right\rangle}_1={x}_3+j{x}_4,\hfill & \hfill {\left\langle {v}_0\right\rangle}_0={x}_5\hfill \end{array}. $$

(11.36)

By averaging the state-space circuit representation in the sense of the first-order harmonic, one obtains (according to the developments presented in Chap. 5)

$$ \left\{\begin{array}{c}\frac{d{\left\langle {i}_L\right\rangle}_1}{ dt}=-j\upomega {\left\langle {i}_L\right\rangle}_1-\frac{r_L}{L}{\left\langle {i}_L\right\rangle}_1-\frac{{\left\langle {v}_C\right\rangle}_1}{L}+\frac{E}{L}{\left\langle {u}_1\right\rangle}_1-\frac{{\left\langle {v}_0\cdot {u}_2\right\rangle}_1}{L}\\ {}\kern-13.5em \frac{d{\left\langle {v}_C\right\rangle}_1}{ dt}=-j\upomega {\left\langle {v}_C\right\rangle}_1+\frac{{\left\langle {i}_L\right\rangle}_1}{C}\\ {}\kern-12em \frac{d{\left\langle {v}_0\right\rangle}_0}{ dt}=\frac{1}{C_0}{\left\langle i\cdot {u}_2\right\rangle}_0-\frac{{\left\langle {v}_0\right\rangle}_0}{R_0{C}_0}.\end{array}\right. $$

(11.37)

Term $\langle$ v ₀ ⋅ u ₂ $\rangle$ ₁ is developed by using results presented in Chap. 5. Thus, one can note that switching function u ₂ has zero average value; therefore relation (5.19) can be used to write further:

$$ {\left\langle {v}_0\cdot {u}_2\right\rangle}_1={\left\langle {v}_0\right\rangle}_0\cdot {\left\langle {u}_2\right\rangle}_1, $$

(11.38)

where $\langle$ u ₂ $\rangle$ ₁ is expressed by using expression (5.23) of a rectangular zero-averaged switching function delayed by angle φ + α:

$$ {\left\langle {u}_2\right\rangle}_1=\frac{2}{\uppi j}{e}^{-\left(\varphi +\upalpha \right)}. $$

(11.39)

Further, by replacing notations (11.36) and expressions (11.38) and (11.39) in Eq. (11.37) and by separating the real and the imaginary parts, one obtains

$$ \left\{\begin{array}{l}\overset{\cdotp }{x_1}=-\frac{r_L}{L}{x}_1+\upomega {x}_2-\frac{x_3}{L}-\mathrm{Re}\left(\frac{2}{j\uppi L}{\left\langle {v}_0\right\rangle}_0{e}^{-j\left(\varphi +\upalpha \right)}\right)\hfill \\ {}\overset{\cdotp }{x_2}=-\upomega {x}_1-\frac{r_L}{L}{x}_2-\frac{x_4}{L}-\mathrm{Im}\left(\frac{2}{j\uppi L}{\left\langle {v}_0\right\rangle}_0{e}^{-j\left(\varphi +\upalpha \right)}\right)\hfill \\ {}\overset{\cdotp }{x_3}=\frac{x_1}{C}+\upomega {x}_4\hfill \\ {}\overset{\cdotp }{x_4}=\frac{x_2}{C}-\upomega {x}_3\hfill \\ {}\overset{\cdotp }{x_5}=\frac{4}{\uppi {C}_0}\sqrt{x_1^2+{x}_2^2} \cos \varphi -\frac{{\left\langle {v}_0\right\rangle}_0}{R_0{C}_0}.\hfill \end{array}\right. $$

(11.40)

The new control variable $w=d{\left\langle {v}_0\right\rangle}_0/ dt=\overset{\cdotp }{x_5}$ is introduced. One can note that the last equation from (11.40) contains an explicit expression of the control input; therefore, the relative degree is r = 1. This last equation may be rewritten as

$$ w=\frac{4}{\uppi {C}_0}\sqrt{x_1^2+{x}_2^2} \cos \varphi -\frac{{\left\langle {v}_0\right\rangle}_0}{R_0{C}_0}, $$

which allows the angle φ to be computed as the linearizing control input

$$ \varphi = \arccos \left(\frac{\pi}{4}\cdot \frac{w{C}_0+{x}_5/{R}_0}{\sqrt{x_1^2+{x}_2^2}}\right). $$

(11.41)

Knowing that i _L (t) ≈ 2(x ₁ cos ωt − x ₂ sin ωt) (see relation (5.29) from Chap. 5), the quantity $\sqrt{x_1^2+{x}_2^2}$ can otherwise be expressed as I _L /2, where I _L represents the first-order harmonic’s magnitude of current i _L . Therefore, the final expression of the linearizing feedback is

$$ \varphi = \arccos \left(\frac{\uppi}{2}\cdot \frac{v{C}_0+{\left\langle {v}_0\right\rangle}_0/{R}_0}{I_L}\right). $$

(11.42)

Before declaring expression (11.42) as providing the linearizing feedback sought for, one must analyze the stability of the zero dynamics. One notes that the fourth-order dynamics composed of the first four equations from (11.40) must be analyzed. To this end, one puts $w=\overset{\cdotp }{x_5}=0$ and x ₅ ≡ $\langle$ v ₀ $\rangle$ ^*₀ in (11.41) to obtain the expression of the control input to be further replaced into these equations:

$$ {\varphi}_0= \arccos \left(\frac{\uppi}{4{R}_0}\cdot \frac{{\left\langle {v}_0\right\rangle}_0^{*}}{\sqrt{x_1^2+{x}_2^2}}\right). $$

(11.43)

The fourth-order nonlinear zero dynamics that result are:

$$ \left\{\begin{array}{c}\overset{\cdotp }{x_1}=-\frac{r_L}{L}{x}_1+\upomega {x}_2-\frac{x_3}{L}-\mathrm{Re}\left(\frac{2}{j\uppi L}{\left\langle {v}_0\right\rangle}_0^{*}{e}^{-j\left({\varphi}_0+\upalpha \right)}\right)\\ {}\overset{\cdotp }{x_2}=-\upomega {x}_1-\frac{r_L}{L}{x}_2-\frac{x_4}{L}-\mathrm{Im}\left(\frac{2}{j\uppi L}{\left\langle {v}_0\right\rangle}_0^{*}{e}^{-j\left({\varphi}_0+\upalpha \right)}\right)\\ {}\kern-15em \overset{\cdotp }{x_3}=\frac{x_1}{C}+\upomega {x}_4\\ {}\kern-14.7em \overset{\cdotp }{x_4}=\frac{x_2}{C}-\upomega {x}_3.\end{array}\right. $$

(11.44)

Stability of nonlinear dynamical system (11.44) will be analyzed by means of the Lyapunov approach. Notation $\mathbf{y}={\left[\begin{array}{cccc}\hfill {x}_1\hfill & \hfill {x}_2\hfill & \hfill {x}_3\hfill & \hfill {x}_4\hfill \end{array}\right]}^T$ is used to denote the state vector corresponding to the zero dynamics. The following Lyapunov candidate function is defined as a quadratic form of the state y:

$$ V\left(\mathbf{y}\right)=\frac{1}{2}{\mathbf{y}}^T\mathbf{Qy}, $$

(11.45)

with Q being a symmetric positive defined matrix containing characteristic values of the energy accumulation elements in the circuit (inductor’s inductance and capacitor’s capacity, respectively):

$$ \mathbf{Q}=\left[\begin{array}{cccc}\hfill L\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill L\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill C\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill C\hfill \end{array}\right]. $$

Function (11.45) can be declared a Lyapunov function – or else, an energy function – if its time derivative is negative whatever the value of state y is. This is proved next. By time-deriving expression (11.45) one obtains

$$ \frac{ dV\left(\mathbf{y}\right)}{ dt}=L{x}_1\overset{\cdotp }{x_1}+L{x}_2\overset{\cdotp }{x_2}+C{x}_3\overset{\cdotp }{x_3}+C{x}_4\overset{\cdotp }{x_4}. $$

(11.46)

To simplify writing, notation

$$ z=\frac{2}{j\uppi L}{\left\langle {v}_0\right\rangle}_0^{*}{e}^{-j\left({\varphi}_0+\upalpha \right)} $$

(11.47)

is adopted. Equation (11.46) is further developed by using expressions of the state variables’ time derivatives as given in (11.44); it becomes, after some simple algebra,

$$ \frac{ dV\left(\mathbf{y}\right)}{ dt}=-{r}_L{x}_1^2-{r}_L{x}_2^2-L{x}_1\mathrm{Re}(z)-L{x}_2\mathrm{Im}(z). $$

(11.48)

One must now get the expressions of Re(z) and Im(z). According to (11.47), variable z can be written

$$ z=-\frac{2}{\uppi L}{\left\langle {v}_0\right\rangle}_0^{*} \sin \left({\varphi}_0+\upalpha \right)-j\frac{2}{\uppi L}{\left\langle {v}_0\right\rangle}_0^{*} \cos \left({\varphi}_0+\upalpha \right), $$

therefore

$$ \begin{array}{cc}\hfill \mathrm{Re}(z)=-\frac{2}{\uppi L}{\left\langle {v}_0\right\rangle}_0^{*} \sin \left({\varphi}_0+\upalpha \right),\hfill & \hfill \mathrm{Im}(z)=\hfill \end{array}-\frac{2}{\uppi L}{\left\langle {v}_0\right\rangle}_0^{*} \cos \left({\varphi}_0+\upalpha \right). $$

(11.49)

At this point, we will establish a connection between complex components of the inductor current’s first-order harmonic, x ₁ and x ₂, and lag angle α. Developments presented in Sect. 5.5.1 of Chap. 5 – related to the extraction of real-time-varying signal from the components of the generalized averaged model – are used for this purpose. Thus, according to Eq. (5.33) and taking into account that α ∈ (0, π/2) is a lag angle (therefore α = − ψ), the following relations hold:

$$ \begin{array}{cc}\hfill \sin \left(-\upalpha \right)=- \sin \upalpha =\frac{x_1}{\sqrt{x_1^2+{x}_2^2}},\hfill & \hfill \cos \left(-\upalpha \right)= \cos \upalpha =-\frac{x_2}{\sqrt{x_1^2+{x}_2^2}};\hfill \end{array} $$

therefore the real and imaginary part of the first-order harmonic as a complex variable can respectively be expressed as

$$ \begin{array}{cc}\hfill {x}_1=- \sin \upalpha \cdot \sqrt{x_1^2+{x}_2^2},\hfill & \hfill {x}_2=- \cos \upalpha \cdot \sqrt{x_1^2+{x}_2^2}\hfill \end{array}. $$

(11.50)

Finally, expressions (11.49) and (11.50) are substituted into Eq. (11.48) to provide the function V’s time derivative in the form

$$ \begin{array}{c}\kern-5em \overset{\cdotp }{V}\left(\mathbf{y}\right)=-{r}_L{x}_1^2-{r}_L{x}_2^2\\ {}-\frac{2}{\uppi}{\left\langle {v}_0\right\rangle}_0^{*}\sqrt{x_1^2+{x}_2^2}\cdot \sin \upalpha \cdot \sin \left({\varphi}_0+\upalpha \right)\\ {}-\frac{2}{\uppi}{\left\langle {v}_0\right\rangle}_0^{*}\sqrt{x_1^2+{x}_2^2}\cdot \cos \upalpha \cdot \cos \left({\varphi}_0+\upalpha \right),\end{array} $$

or, equivalently,

$$ \overset{\cdotp }{V}\left(\mathbf{y}\right)=-{r}_L{x}_1^2-{r}_L{x}_2^2-\frac{2}{\uppi}{\left\langle {v}_0\right\rangle}_0^{*}\sqrt{x_1^2+{x}_2^2}\cdot \cos {\varphi}_0. $$

(11.51)

Expression (11.43) of the control input φ₀ is substituted into (11.51) to yield the final expression of $\overset{\cdotp }{V}\left(\mathbf{y}\right)$:

$$ \overset{\cdotp }{V}\left(\mathbf{y}\right)=-{r}_L{x}_1^2-{r}_L{x}_2^2-\frac{{\left\langle {v}_0\right\rangle}_0^{*2}}{2{R}_0}<0. $$

(11.52)

Relation (11.52) shows that the energy function V defined as a quadratic form of the zero dynamics state vector (see Eq. (11.45)) is strictly decreasing with time, and this allows a conclusion that the zero dynamics are in this case stable.

Once Eq. (11.42) of the necessary phase lag has been found as linearizing feedback, it will be used for obtaining u ₂. Indeed, u ₂ is obtained by delaying u ₁ by the phase lag φ given in (11.42).

The control block diagram, profiting from the feedback linearization, is given in Fig. 11.17, which shows that gain K _C is now sufficient for controlling output voltage v ₀.

Fig. 11.17

Series-resonant converter output voltage control block diagram based upon feedback linearization and gain control

The following problems are left to the reader to solve.

Problem 11.2

Voltage regulation of buck-boost converter using feedback-linearization control

Given the electrical circuit of a buck-boost DC-DC converter supplying a resistance load R, presented in Fig. 4.12 from Chap. 4, and its model given by Eq. (4.34) from the same chapter:

$$ \left\{\begin{array}{c}L\overset{\cdotp }{i_L}= Eu-{v}_C\left(1-u\right)-{r}_L{i}_L\\ {}\kern0.6em C\overset{\cdotp }{v_C}=-{i}_L\left(1-u\right)-\frac{v_C}{R},\end{array}\right. $$

where r _L is the inductor’s resistance and the other notations preserve their usual meaning, address the following points.

• (a) Design a feedback-linearization control in order to regulate the output voltage, v _C . Based upon the zero dynamics analysis, decide whether a direct or an indirect control approach is suitable.

• (b) For the numerical values L = 3 mH, C = 1200 μF, r _L  = 0.1 Ω, E = 12 V and rated R = 100 Ω compute the linear controller that ensures a voltage closed-loop bandwidth of 50 rad/s.

• (c) Implement the numerical simulation block diagram in Simulink^® in order to validate the closed-loop behavior imposed in part b. Analyze the dynamical behavior in response to load step variations.

Problem 11.3

Reactive power control of STATCOM using linearizing feedback

The electrical circuit of a static synchronous compensator (STATCOM) operating in full wave is given in Fig. 11.18. Resistance R _S represents the inductor resistance on a phase and load resistance R includes power switch losses.

Fig. 11.18

Electrical circuit of static synchronous compensator (STATCOM)

The reactive power control will be achieved by using the dq model in the sense of the first-order harmonic, namely by regulating the averaged value of the three-phase current q component at reference value i ^* _q . Output voltage v _C is uncontrolled. The control input is α, the phase lag between the switching function u ₁ and the first-phase grid voltage e ₁, which is taken as phase origin.

• (a) Using the approach presented in Chap. 5 – related to computation of the generalized averaged model (GAM) – prove that the first-order harmonic dq model of the STATCOM is the following (Petitclair et al. 1996):

  $$ \left\{\begin{array}{l}\overset{\cdotp }{{\left\langle {i}_q\right\rangle}_0}=-\frac{R_S}{L_S}{\left\langle {i}_q\right\rangle}_0-\upomega {\left\langle {i}_d\right\rangle}_0+\frac{2}{\uppi {L}_S}{\left\langle {v}_C\right\rangle}_0 \sin \upalpha \hfill \\ {}\overset{\cdotp }{{\left\langle {i}_d\right\rangle}_0}=\upomega {\left\langle {i}_q\right\rangle}_0-\frac{R_S}{L_S}{\left\langle {i}_d\right\rangle}_0-\frac{2}{\uppi {L}_S}{\left\langle {v}_C\right\rangle}_0 \cos \upalpha +\frac{E}{L_S}\hfill \\ {}\overset{\cdotp }{{\left\langle {v}_C\right\rangle}_0}=-\frac{3}{\uppi C}{\left\langle {i}_q\right\rangle}_0 \sin \upalpha +\frac{3}{\uppi C}{\left\langle {i}_d\right\rangle}_0 \sin \upalpha -\frac{1}{ RC}{\left\langle {v}_C\right\rangle}_0,\hfill \end{array}\right. $$

  where E is the grid voltage amplitude and ω is the grid voltage pulsation (ω = 2π ⋅ 50 rad/s).

• (b) Obtain the expression of the linearizing feedback ensuring regulation of $\langle$ i _q $\rangle$ ₀ at the value i ^* _q .

• (c) Analyze the zero dynamics stability by using the small-signal model.

• (d) Provide the global reactive power control structure and emphasize the relation between the imposed value of the reactive power Q ^* and i ^* _q .

Problem 11.4

Watkins–Johnson DC-DC converter controlled by feedback linearization

For a Watkins–Johnson converter – whose electrical circuit was presented in Fig. 4.37 in Chap. 4 and repeated in Fig. 8.34 in Problem 8.1 in Chap. 8 – it is required to regulate the output voltage at the setpoint v ^* _C . Its averaged model, given by relation (8.38) of Chap. 8, is given hereafter:

$$ \left\{\begin{array}{c}L\overset{\cdotp }{i_L}=\left(2\upalpha -1\right)E-\upalpha {v}_C\\ {}\kern4pt C\overset{\cdotp }{v_C}=\upalpha {i}_L-\frac{v_C}{R},\end{array}\right. $$

where notations used have their usual meaning, with duty ratio α being the control input and R being the load resistance. The following issues must be addressed.

• (a) Deduce the current value i ^* _L that corresponds to the voltage setpoint v ^* _C .

• (b) Compute the expression of a feedback-linearization control law issued from an indirect control approach, that is, from regulating current i _L to the setpoint i ^* _L computed in part a.

• (c) Knowing that the output voltage operating range is [−3E, E], design a two-loop control structure whose outer loop is dedicated to regulating the output voltage v _C by means of a PI controller and the inner loop is in charge with current i _L control. Design the inner-loop controller so that the closed-loop bandwidth is at least ten times larger than the voltage plant bandwidth for the entire operating range. An outer loop PI controller will result such that the closed-loop bandwidth will be two times larger than the voltage plant bandwidth for the operating point corresponding to v _C  = 0 V.