HARMONIC VOLTAGES AND CURRENTS 419two parts are,
Part 1. For the 180◦rectangle waveform,bn 180 =4
πn, the fundamentalb 1180 =4
π
Part 2. For the 60◦rectangle waveform,bn 60 =2
πn(
cos2 πn
6−cos4 πn
6−cos8 πn
6+cos10 πn
6)
The value of the fundamental coefficientb 160 is,
b 160 =1
π( 4 )
1
2
=
2
πThe magnitude of the two parts is divided by√
3 to obtain the primary line current of the
delta-star transformer. The result is then added to the line current of the star-star transformer. The
total magnitude of the supply line harmonic coefficientbnsumis given by,
bnsum=1
πn[
4
√
3
+cosπn
6+
1
√
3
cos2 πn
6−1
√
3
cos4 πn
6−cos5 πn
6−cos7 πn
6−1
√
3
cos8 πn
6+
1
√
3
cos10 πn
6+cos11 πn
6]
and
isum(ωt)=imaxn∑=∞n= 1bnsumsinnωtThe value of the fundamental coefficientb1sumis,
b1sum=1
π(
4
√
3
+
4
√
3
2
+
2
√
3
)
=
4
√
3
πThe fundamental coefficients from the 180◦, 120◦and 60◦waveforms are found to be in the
ratio 2:
√
3:1 respectively. The fundamental coefficient of the supply current is double the magnitude
of the 120◦waveform coefficient, which is the desired result.
The 180◦waveform contains triplen harmonics forntaking odd values. The 60◦waveform
also contains the same triplen harmonics but with opposite signs, which therefore cancel those in the
180 ◦waveform. None of the waveforms contain even harmonics.
The following harmonics are contained in the waveform,
n= 12 k± 1Wherek= 1 , 2 , 3 ,...,∞. The lowest harmonic present is the eleventh.