Closed form of optimal current waveform for class-F PA up to fourth harmonic

Abstract

In this paper, rigorous analytical derivation of the coefficients of optimal current waveform for class-F power amplifier (PA) up to fourth harmonic (dc, 1st, 2nd and 4th harmonic) is presented. The coefficients of the optimal current waveform along with related maximum attainable efficiency are provided in closed form. The results obtained are also of interest for the inverse class-F, in the case when optimal voltage waveform consists of dc, 1st, 2nd and 4th harmonic.

Appendix A

Full account of details of derivation of (17) and (18) (Section 2) is provided in this appendix.

From (11), (13) and (15) we obtain

$$ x_{0} =-\frac{x_{1} +\lambda_{2} x_{3} }{\lambda_{1} }, ~~~x_{2} =-\frac{x_{1} +x_{3}}{\lambda_{1}}, ~~~x_{4} =-\frac{\lambda_{2} x_{1} +x_{3} }{\lambda_{1} }. $$

(A.1)

Combining (A.1) and (12) yields

$$ x_{1} \left(2-{\lambda_{1}^{2}} +{\lambda_{2}^{2}}\right)+x_{3} (1+2\lambda_{2})=0, $$

(A.2)

whereas combining (A.1) and (14) yields

$$ x_{3} \left(2-{\lambda_{1}^{2}} +{\lambda_{2}^{2}}\right)+x_{1} (1+2\lambda_{2})=0. $$

(A.3)

According to (A.1), x ₁ and x ₃ are not simultaneously equal to zero, since otherwise, it would lead to x ₀ = x ₂ = x ₄ = 0, which together with x ₁ = x ₃ = 0 contradicts (8).

Equations (A.2) and (A.3) can be considered as a system of linear equations in terms of $(2-{\lambda _{1}^{2}} +{\lambda _{2}^{2}})$ and (1+2λ₂). Thus we have the following two cases: either (i) ${x_{1}^{2}} -{x_{3}^{2}} =0$ or (ii) ${x_{1}^{2}} -{x_{3}^{2}} \ne 0$. In what follows we show that case (i) leads to contradiction, while case (ii) leads to the optimal waveform given by (19).

(i) From x ₁ = ± x ₃, taking into account that at least one of x ₁ and x ₃ is not equal to zero, it follows that x ₁ ≠ 0 and x ₃ ≠ 0. Substitution of x ₁ = ± x ₃ into (A.1) leads to x ₀ = ± x ₄. Furthermore, inserting x ₁ = ± x ₃ and x ₀ = ± x ₄ into (9) implies x ₀ = x ₄ = 0, since x ₁ ≠ 0. This contradict our assumption that A ₄ = 2x ₀ x ₄ ≠ 0.

(ii) The determinant of the system (A.2)–(A.3) is non-zero and consequently it has trivial solution only: $2-{\lambda _{1}^{2}} +{\lambda _{2}^{2}} =0$ and 1+2λ₂ = 0, that is,

$$ \lambda_{2} =-1 / 2, ~~~{\lambda_{1}^{2}} =9 / 4. $$

(A.4)

Since A ₁>0, from (16) and (A.4) it follows that (17) holds. Substituting λ₁ = −A ₁ = −3/2 and λ₂ = −1/2 into (A.1) we obtain

$$ x_{0} =\frac{2x_{1} -x_{3} }{3},~~~x_{2} =\frac{2x_{1} +2x_{3} }{3},~~~x_{4} =\frac{-x_{1} +2x_{3} }{3}. $$

(A.5)

From (2), using (A.5), it follows that

$$ 2{x_{1}^{2}} +2{x_{3}^{2}} =1. $$

(A.6)

From (A.5), (A.6) and (5) it follows that

$$ 8x_{1} x_{3} =1. $$

(A.7)

Finally, substituting (A.5)–(A.7) into (4) and (6) leads to (18).