Fourier Analysis of Three Phase Inverter

Fourier Analysis of Three Phase Inverter

The wave form of V_R is shown below:

Since, the waveform has half wave symmetry(it is odd function), we get only odd harmonic components and the fourier series is given by

$$VR = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;1,...\infty ,2} {{1 \over m}\;\sin (m\theta )} $$

Now, we can write V_Y and V_B in terms of fourier components,

$$\eqalign{
& VY = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;1,...\infty ,2} {{1 \over m}\;\sin (m(\theta - \pi /3))} \cr
& VB = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;1,...\infty ,2} {{1 \over m}\;\sin (m(\theta + \pi /3)} ) \cr} $$

For the analysis,we can break these series in three sub-series as follows:

1. 1^st Sub-Series: m = 1, ∞,6(i.e 1^St, 7^th,13^th and so-on)
2. 2^nd Sub-Series: m = 3, ∞,6(i.e 3^th, 9^th,15^th and so-on)
3. 3^rd Sub-Series: m = 5, ∞,6(i.e 5^th, 11^th,17^th and so-on)

For the first sub-series, m = 1, ∞,6 or m = 1 + 6k where k = 0, ∞

$$\eqalign{ & VR = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;1,...\infty ,6} {{1 \over m}\;\sin (m\theta )} \cr & VY = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;1,...\infty ,6} {{1 \over m}\;\sin (m\theta - \pi /3)} \cr & VB = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;1,...\infty ,6} {{1 \over m}\;\sin (m\theta + \pi /3)} \cr} $$

Subsitiuting the value of m = 1 + 6k in expression of V_Y in sine term, we get

Sin[m (wt - 2π/3)] = Sin[wt -(1 + 6k) 2π/3]

= Sin[wt -2π/3 + 4πk]

= Sin[wt -2π/3]

Again, substitute the value of m = 1 + 6k in expression of V_B on sine term, we get

Sin[m ( wt + 2π/3)] = Sin[wt +(1 + 6k) 2π/3]

= Sin[wt +2π/3 + 4πk]

= Sin[wt +2π/3]

Thus, first sub-series represents positive sequence system voltage and can be represented by the phasor diagram as shown below:

B) For the second sub-series, m = 3, ∞,6 or m = 3k where k = 0, ∞

$$\eqalign{ & VR = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;3,...\infty ,6} {{1 \over m}\;\sin (m\theta )} \cr & VY = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;3,...\infty ,6} {{1 \over m}\;\sin (m\theta - \pi /3)} \cr & VB = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;3,...\infty ,6} {{1 \over m}\;\sin (m\theta + \pi /3)} \cr} $$

Subsitiuting the value of m = 3k in expression of V_Y in sine term, we get

Sin[m ( wt - 2π/3)] = Sin[wt - 3k 2π/3]

= Sin[wt -2πk]

= Sin(wt)

Similarly, substituting this value of m = 3k in expression of V_B in sine term, we get

Sin[m ( wt + 2π/3)] = Sin[wt + 3k 2π/3]

= Sin[wt +2πk]

= Sin(wt)

Thus, second sub-series represents zero sequence system voltage and can be represented by phasor diagram as below;

C) For the third sub-series, m = 5, ∞,6 or m = (5 + 6k) where k = 0, ∞

$$\eqalign{ & VR = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;5,...\infty ,6} {{1 \over m}\;\sin (m\theta )} \cr & VY = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;5,...\infty ,6} {{1 \over m}\;\sin (m\theta - \pi /3)} \cr & VB = {{4\;VS} \over {\pi 2}}\sum\limits_{m\; = \;5,...\infty ,6} {{1 \over m}\;\sin (m\theta + \pi /3)} \cr} $$

Subsitiuting the value of m = 5 + 6k in expression of V_Y in sine term, we get

Sin[m ( wt - 2π/3)] = Sin[wt -(5 + 6k) 2π/3]

= Sin[wt -10π/3 + 4πk]

= Sin[wt - 2π + 2π/3]

= Sin(wt + 2π/3) ( Equivalent to sine term of V_B)

Again, substitute the value of m = 1 + 6k in expression of V_B on sine term, we get

Sin[m ( wt + 2π/3)] = Sin[wt +(5 + 6k) 2π/3]

= Sin[wt +10π/3 + 4πk]

= Sin[wt +2π - 2π/3]

= Sin(wt - 2π/3) ( Equivalent to sine term of V_y)

Therefore, the 3^RD sub-series represents negative sequence system voltage and can be represented by the equation below;

The above results for three sub-series can be summarized as

$$\eqalign{
& VR = {{4\;VS} \over {\pi 2}}[\sum\limits_{m\; = \;1,...\infty ,6} {{1 \over m}\;\sin (m\theta )} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;|\;\; + \;\;\;\;\sum\limits_{m\; = 3\;,...\infty ,6} {{1 \over m}\;\sin (m\theta )\;\;\;\;\;\; + \;\;\;\;\;} \sum\limits_{m\; = \;5,...\infty ,6} {{1 \over m}\;\sin (m\theta )]} \cr
& VY = {{4\;VS} \over {\pi 2}}[\sum\limits_{m\; = \;1,...\infty ,6} {{1 \over m}\;\sin (m(\theta - 2\pi /3))} \;\;\;\; + \;\;\;\;\;\sum\limits_{m\; = 3\;,...\infty ,6} {{1 \over m}\;\sin (m\theta )\;\;\;\;\;\; + } \;\;\;\;\;\sum\limits_{m\; = \;5,...\infty ,6} {{1 \over m}\;\sin (m(\theta + 2\pi /3)]} \cr
& VB = {{4\;VS} \over {\pi 2}}[\sum\limits_{m\; = \;1,...\infty ,6} {{1 \over m}\;\sin (m(\theta + 2\pi /3))} \;\;\;\; + \;\;\;\;\;\sum\limits_{m\; = 3\;,...\infty ,6} {{1 \over m}\;\sin (m\theta )\;\;\;\;\;\; + } \;\;\;\;\;\sum\limits_{m\; = \;5,...\infty ,6} {{1 \over m}\;\sin (m(\theta - 2\pi /3)]} \cr
& \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \Downarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \Downarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \Downarrow \cr
& \;\;\;\;\;\;\;\;\;Positive\;Sequence\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;Zero\;Sequence\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;Negative\;Sequence\;\;\;\; \cr} $$

Line to Line Voltage

Fourier Components of line to line voltage can be calculated as follows:

V_RY = V_R - V_Y

The zero sequence system cancels, therefore triplines are absent in line to line voltage.

$$\eqalign{
& VRY = {{4\;VS} \over {\pi 2}}[\sum\limits_{m\; = \;1,...\infty ,6} {{1 \over m}\;\sin (m\theta )} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;|\;\; + \;\;\;\;\sum\limits_{m\; = 3\;,...\infty ,6} {{1 \over m}\;\sin (m\theta )\;\;\;\;\;\; + \;\;\;\;\;} \sum\limits_{m\; = \;5,...\infty ,6} {{1 \over m}\;\sin (m\theta )]} - \cr
& \;\;\;\;\;\;\;\;\;{{4\;VS} \over {\pi 2}}[\sum\limits_{m\; = \;1,...\infty ,6} {{1 \over m}\;\sin (m(\theta - 2\pi /3))} \;\;\;\; + \;\;\;\;\;\sum\limits_{m\; = 3\;,...\infty ,6} {{1 \over m}\;\sin (m\theta )\;\;\;\;\;\; + } \;\;\;\;\;\sum\limits_{m\; = \;5,...\infty ,6} {{1 \over m}\;\sin (m(\theta + 2\pi /3)]} \cr
& \;\;\;\;\; = \;\sqrt 3 {{4\;VS} \over {\pi 2}}[\sum\limits_{m\; = \;1,...\infty ,6} {{1 \over m}\;\sin (m(\theta + \pi /6))} \;\;\; + \;\sum\limits_{m\; = \;5,...\infty ,6} {{1 \over m}\;\sin (m(\theta - \pi /6)]} \cr} $$

The following analysis can be done with other line to line voltage and same results will be obtained in other line to line voltages.

Hence, the following conclusion can be derived from the above expression in comparison to that of V_R :

1)There are no triplines(i.e 3kf_o where k = 1,to infinite).

2) The amplitude of remaining components is √3 times greater than that of V_R.

3) The negative sequence components lag by π/3 w.r.t positive sequence components.

Load Neutral Voltage

The potential of load neutral with respect to point ‘O’ is a rectangular square wave(similar to that of V_R) having frequency 3 times that of V_R and amplitude 1/3 of V_R. Hence, the Fourier components of V_N are given by;

$$VN = {{4\;VS} \over {2\pi }}\sum\limits_{m\; = \;3,...\infty ,6} {{1 \over m}\;\sin (m\theta )} $$

This is identical to zero sequence component of V_R.