The general method to analyze the harmonic characteristics of a non-sinusoidal periodic voltage (or current) is to directly perform Fourier decomposition on it. But to integrate the voltage or current in a cycle to calculate the Fourier coefficient, the operation process is complicated and the amount of operation is large. Because the converter equipment has discrete sampling and modulation switching characteristics, the piecewise integration can be replaced by a simple triangular transformation, which can simplify the analysis process of the converter equipment related waveforms. This method of harmonic analysis for steady-state operation of converter equipment is called switching function method.

Fourier analysis of the non-linear function model of converter f _i (i = 1,2,3,4,5,6) is performed according to the definition: f i = a 0 + ∑ n = 1 ∞ a n cos ⁡ n ω t + b n sin ⁡ n ω t (6)

Among them, a ₀ , a _n , b _n are Fourier coefficients, and using the orthogonality of trigonometric functions, the calculation formulas of a ₀ , a _n , b _n can be obtained as: a 0 = 1 T ∫ − T 2 T 2 f i ω t d ω t a n = 2 T ∫ − T 2 T 2 f i ω t cos ⁡ n ω t d ω t b n = 2 T ∫ − T 2 T 2 f i ω t sin ⁡ n ω t d ω t (7)

The DC side voltage waveform of the converter is the result of modulating the AC voltage by the non-linear function model of the converter, During normal operation, each converter valve is triggered and turned on in turn at equal intervals, and the output DC voltage u _d is formed by modulating the input three-phase AC voltages u _a , u _b , u _c by the non-linear function models of converter f ₁ ∼f ₆ : u d = f 1 − f 4 u a + f 3 − f 6 u b + f 5 − f 2 u c (8)

Fourier decomposition is performed on the proposed model f _i according to the principle of switching function method under the symmetrical triggering. a ₀ , a _n , b _n are calculated and substituted into Equation 6, and the sine and cosine terms of the same frequency are combined to obtain: f i = 1 3 + ∑ n = 1 ∞ ∑ j = 1 7 A j cos n ω t − φ j (9)

In the equation, A _j and φ _j (j = 1,2,3,4,5,6,7) are as follows, where A ₄ = -A ₃ , A ₆ = -A ₅ . when n is 1 in Equation 9, the denominator of A ₅ and A ₆ will be zero, the equation will become meaningless. At this time, only limit operation needs to be done, that is, use lim n → 1 A 5 = γ 2 π P 1 and lim n → 1 A 6 = − γ 2 π P 1 to replace A ₅ and A ₆ for operation. A 1 = 2 n π P 1 sin n γ 2 cos α + π 6 φ 1 = n α + γ 2 + 2 i − 1 6 π A 2 = − 2 n π P 1 sin n γ 2 cos α + γ + π 6 φ 2 = n α + γ 2 + 2 i + 3 6 π A 3 = − 1 n + 1 π P 1 sin n + 1 γ 2 φ 3 = n + 1 α + γ 2 + n 2 i − 1 + 1 6 π A 5 = − 1 n − 1 π P 1 sin n − 1 γ 2 φ 4 = n + 1 α + γ 2 + n 2 i + 3 + 1 6 π A 7 = 2 n π sin n 2 π − 3 γ 6 φ 5 = n − 1 α + γ 2 + n 2 i − 1 − 1 6 π φ 6 = n − 1 α + γ 2 + n 2 i + 3 − 1 6 π φ 7 = n α + γ 2 + 2 i + 1 6 π

Fourier decomposition is performed on f _i ^- and f _i ⁺ in the same way under asymmetrical triggering. f i − = − 1 2 π σ + γ + Q 1 P 1 − Q 2 P 4 − 2 π 3 + ∑ n = 1 ∞ ∑ j = 1 7 B j cos n ω t − θ j (10) f i + = 1 2 π σ − γ + Q 3 P 1 − Q 4 P 4 + 2 π 3 + ∑ n = 1 ∞ ∑ j = 1 7 C j cos n ω t − η j (11) Where Q ₁ , Q ₂ , Q ₃ , Q ₄ , B _j , C _j and θ _j , η _j (j = 1,2,3,4,5,6,7) are as follows, B ₂ = A ₂ , θ ₂ = φ ₂ , B ₄ = A ₄ , θ ₄ = φ ₄ , B ₆ = A ₆ , θ ₆ = φ ₆ , C ₁ = A ₁ , η ₁ = θ ₁ , C ₃ = A ₃ , η ₃ = θ ₃ , C ₅ = A ₅ , η ₅ = θ ₅ , η ₇ = θ ₇ : B 1 = 2 n π P 4 sin n γ 2 cos α + σ + π 6 θ 1 = n α + σ + γ 2 + 2 i − 1 6 π B 3 = − 1 n + 1 π P 4 sin n + 1 γ 2 θ 3 = n + 1 α + σ + γ 2 + n 2 i − 1 + 1 6 π B 5 = − 1 n − 1 π P 4 sin n − 1 γ 2 θ 5 = n − 1 α + σ + γ 2 + n 2 i − 1 − 1 6 π B 7 = 2 n π sin n 2 π − 3 γ − 3 σ 6 θ 7 = n α + γ + σ 2 + 2 i + 1 6 π Q 1 = γ ⁡ cos α + γ + π 6 − 2 ⁡ cos α + γ 2 + π 6 sin γ 2 Q 2 = γ ⁡ cos α + σ + π 6 − 2 ⁡ cos α + σ + γ 2 + π 6 sin γ 2 When n is 1 in Equation 10, the denominator of B ₅ and B ₆ will be zero, the equation will become meaningless. At this time, only limit operation needs to be done, that is, use lim n → 1 B 5 = − γ 2 π P 4 and lim n → 1 B 6 = − γ 2 π P 1 to replace B ₅ and B ₆ for operation. C 2 = − 2 n π P 4 sin n γ 2 cos α + σ + γ + π 6   η 2 = n α + σ + γ 2 + 2 i + 3 6 π C 4 = 1 ( n + 1 ) π P 4 sin ( n + 1 ) γ 2   η 4 = ( n + 1 ) α + σ + γ 2 + n ( 2 i + 3 ) + 1 6 π C 6 = 1 ( n − 1 ) π P 4 sin ( n − 1 ) γ 2   η 6 = ( n − 1 ) α + σ + γ 2 + n ( 2 i + 3 ) − 1 6 π C 7 = 2 π P 4 sin n ( 2 π − 3 γ + 3 α ) 2   Q 3 = γ ⁡ cos α + π 6 − 2 ⁡ cos α + γ 2 + π 6 sin γ 2 Q 4 = γ ⁡ cos α + σ + γ + π 6 − 2 ⁡ cos α + σ + γ 2 + π 6 sin γ 2

When n is 1 in Equation 11, the denominator of C ₅ and C ₆ will be zero, the equation will become meaningless. At this time, only limit operation needs to be done, that is, use lim n → 1 C 5 = γ 2 π P 1 and lim n → 1 C 6 = γ 2 π P 4 to replace C ₅ and C ₆ for operation.

First, according to Equation 7, calculate the values of a ₀ , a ₁ , b ₁ when n is 1, and substitute them into Equation 6. Combined with Equation 8, it is found that the result of modulating the AC voltage by the fundamental component of the non-linear function model of the converter does not change with time, that is, the DC component in the output voltage, and its expression is Equation 12. u z = 6 U m π [ sin 2 π − 3 γ 6 cos α + γ 2 + 1 P 1 ( sin γ 2 ( cos α + γ + π 6 cos α + γ + 4 π 3 − cos α + π 6 cos α + γ 2 + 2 π 3 ) + 3 4 sin ⁡ γ sin 2 α + γ + 7 π 6 + γ 2 ) ] (12)

When n is a positive integer that is not equal to 1, combine Equation 2 to obtain f ₁ -f ₄ , f ₃ -f ₆ , f ₅ -f ₂ and substitute them into Equation 8 to obtain:

When the harmonic order n of the non-linear function model of the converter is 5, 11, 17... 6k-1 (k = 1, 2, 3...) times, the DC side will generate n+1st harmonics: u n + 1 = − 3 2 U m ∑ j = 1 7 D j cos n + 1 ω t + ϕ j (13)

When the harmonic order n of the non-linear function model of the converter is 7, 13, 19... 6k+1 (k = 1, 2, 3...) times, the DC side will generate n-1st harmonics: u n − 1 = 3 2 U m ∑ j = 1 7 D j cos n − 1 ω t + ϕ j (14)

In Equations (13) and (14) D _j = A _j R _j , where R _j , ϕ _j are as follows: R j = sin ⁡ φ i = 1 , j − sin ⁡ φ i = 4 , j 2 + cos ⁡ φ i = 1 , j − cos ⁡ φ i = 4 , j 2 ϕ j = tan − 1 sin ⁡ φ i = 1 , j − sin ⁡ φ i = 4 , j cos ⁡ φ i = 1 , j − cos ⁡ φ i = 4 , j

From Eqs. 13, 14, it can be known that the n = 6k±1st harmonic of the non-linear function model of the converter modulates the AC voltage so that the DC side only contains the 6kth characteristic harmonic component in addition to the DC component. Therefore any 6kth harmonics on the DC side can be regarded as the superposition of u _n+1 at n = 6k-1 and u _n-1 at n = 6k+1.

If σ = 0 in Eqs. 4, 5, it can be seen the expression is the same as that when the trigger signal is symmetrical, so the symmetrical triggering operation is regarded as a special case of the asymmetrical triggering operation. In the case of asymmetrical triggering, the non-linear function model of converter has changed greatly, and various harmonic components have appeared. At this time, the voltage on the AC side may be affected by this, and various non-characteristic subharmonics will be generated on the DC side.

Randomly select one of the thyristors for discussion. For example, the firing angle of thyristor 3 lags behind the normal firing angle σ degrees. Affected by this, the conduction time of thyristor 1 will be extended by σ angle. The time domain waveform diagram of the non-linear function model of converter f _i (i = 1,2,3,4,5,6) in this case is shown in Figure 6.

According to the same principle of Equation 8, the expression of the output DC voltage u _d ^’ during asymmetrical triggering is Equation 15: u d ′ = f 1 + − f 4 u a + f 3 − − f 6 u b + f 5 − f 2 u c (15)

u _d ’ can also be expressed as: u d ′ = u d + ∆ u d (16) Where: ∆ u d = f 1 + − f 1 u a + f 3 − − f 3 u b

It can be seen that in the case of the asymmetrical triggering, the DC side voltage u _d ^’ will still contain the harmonic components in the symmetrical case u _d . Since the output voltage in the case of the symmetrical triggering has been analyzed before, we will only analyze the non-characteristic harmonics of the DC side Δu _d caused by the asymmetrical moment. The harmonics of the final output voltage can be summed up by the u _d and Δu _d .

According to equation 9 and 11, f ₁ ⁺ -f ₁ is obtained by calculation: f 1 + − f 1 = E + ∑ n = 1 ∞ ∑ j = 1 7 S j sin n ω t + ζ j (17) Where E, S _j and ζ _j (j = 1,2,3,4,5,6,7)are as follows: S j = C j ⁡ sin ⁡ η i = 1 , j − A j ⁡ sin ⁡ φ i = 1 , j 2 + C j ⁡ cos ⁡ η i = 1 , j − A j ⁡ cos ⁡ φ i = 1 , j 2 ζ j = tan − 1 C j ⁡ sin ⁡ η i = 1 , j − A j ⁡ sin ⁡ φ i = 1 , j C j ⁡ cos ⁡ η i = 1 , j − A j ⁡ cos ⁡ φ i = 1 , j E = 1 2 π σ − γ + Q 3 P 1 − Q 4 P 4

According to equation 9 and 10, f ₃ ^- -f ₃ is obtained by calculation: f 3 − − f 3 = F + ∑ n = 1 ∞ ∑ j = 1 7 T j sin n ω t + ξ j (18) Where F, T _j and ξ _j (j = 1,2,3,4,5,6,7) are as follows: T j = B j ⁡ sin ⁡ θ i = 3 , j − A j ⁡ sin ⁡ φ i = 3 , j 2 + B j ⁡ cos ⁡ θ i = 3 , j − A j ⁡ cos ⁡ φ i = 3 , j 2 ξ j = tan − 1 B j ⁡ sin ⁡ θ i = 3 , j − A j ⁡ sin ⁡ φ i = 3 , j B j ⁡ cos ⁡ θ i = 3 , j − A j ⁡ cos ⁡ φ i = 3 , j F = − 1 2 π σ + γ + Q 1 P 1 − Q 2 P 4

Substitute f ₁ ⁺ -f ₁ , f ₃ ^- -f ₃ into Equation 16 to get: ∆ u d = U m G 1 cos ω t − ψ 1 + U m G 2 ∑ n = 1 ∞ cos n + 1 ω t − ψ 2 + U m G 3 ∑ n = 1 ∞ cos n − 1 ω t − ψ 3 (19) Where G 1 = E 2 − E F + F 2 ψ 1 = − tan − 1 2 E − F 3 F G 2 = − 1 2 S j 2 + T j 2 + 2 S j T j cos ⁡ ζ j ⁡ cos ξ j + 2 π 3 + sin ⁡ ζ j ⁡ sin ξ j + 2 π 3 ψ 2 = tan − 1 S j ⁡ sin ⁡ ζ j + T j ⁡ sin ξ j + 2 π 3 S j ⁡ cos ⁡ ζ j + T j ⁡ cos ξ j + 2 π 3 G 3 = 1 2 S j 2 + T j 2 + 2 S j T j cos ⁡ ζ j ⁡ cos ξ j − 2 π 3 + sin ⁡ ζ j ⁡ sin ξ j − 2 π 3 ψ 3 = tan − 1 S j ⁡ sin ⁡ ζ j + T j ⁡ sin ξ j − 2 π 3 S j ⁡ cos ⁡ ζ j + T j ⁡ cos ξ j − 2 π 3

It can be seen from Equation 19 that when the harmonic order n of the non-linear function model of the converter is an even number of 2k (k = 1, 2, 3 … ), the DC side will generate 2k±1st odd order voltage harmonics; when n is an odd number of 2k+1 (k = 1, 2, 3 … ), the DC side will generate 2kth and 2k+2nd even order voltage harmonics. The calculation formula of DC side voltage harmonics can be derived as follows:` ∆ u z = U m G 3 cos ψ 3 ∆ u 1 = U m G 1 cos ω t − ψ 1 + G 3 cos ω t − ψ 3 ∆ u 2 k = U m G 2 cos 2 k ω t − ψ 2 + G 3 cos 2 k ω t − ψ 3 ∆ u 2 k + 1 = U m G 2 cos 2 k + 1 ω t − ψ 2 + G 3 cos 2 k + 1 ω t − ψ 3 (20)

In Equation 20, Δu _z represents the DC component of the DC side voltage, which is formed by Δu _d3 when n = 1; Δu ₁ represents the fundamental component of the DC side voltage, which is formed by the superposition of Δu _d1 and Δu _d3 when n = 2; Δu _2k represents the even-order harmonic component of the DC side voltage, which is formed by the superposition of Δu _d2 at n = 2k-1 and Δu _d3 at n = 2k+1; Δu _2k+1 represents the odd-order harmonic component of the DC side voltage, which is formed by the superposition of Δu _d2 at n = 2k and Δu _d3 at n = 2k+2.

Taking the PS9 power supply of the poloidal field of the EAST device as an example, the effective value of the three-phase voltage input by the power grid is 259.3 V, the amplitude U _m is 366.7 V, and the frequency is industrial frequency 50 Hz. Taking the power supply system under rapidly rising and changing current demands still operating in steady state as a reference, when the converter works in the inverter state, if the firing angle delay is too large, it will easily lead to commutation failure. Therefore, the range of firing angle α is generally between π/9 rad to 13π/18 rad. Correspondingly, the range of commutation overlap angle γ is generally between 8π/180 rad to π/10 rad, and the range of deviation angle σ is generally between 0 rad to 2π/9 rad (Wang Z M et al., 2020). As shown in Tab.1, select operating parameters at equal intervals according to the above range, and 8 cases are set to verify the harmonic analysis in this paper.

Based on equations (13) and (14), the harmonic voltages on the DC side of the converter under symmetrical triggering conditions can be obtained. Table 2 and Figure 7 show the comparison of calculation results of the DC side harmonic formula (Method A) and the direct Fourier decomposition (Method B) under the conditions of Cases 1 to 4.

It can be seen from Table 2 and Figure 7 that the output voltage only contains 6kth characteristic harmonics under the symmetrical triggering condition, and the values of the voltage harmonics are equal.

Based on Equation 20, the harmonic voltages on the DC side of the converter under asymmetrical triggering conditions can be obtained. Table 3 and Figure 8 show the comparison of calculation results of the DC side harmonic formula (Method A) and the direct Fourier decomposition (Method B) under the conditions of Cases 5 to 8.

It can be seen from Table 3 and Figure 8 that the output voltage not only contains 6kth characteristic harmonics but also includes various non-characteristic harmonics under the asymmetrical triggering condition, and the values of the voltage harmonics are equal.