## 2 Numerical Analysis of Generated Harmonics  
The regulation in the output frequency at the discrete level is ensured with the proper arrangement of a non-inverted and inverted form of the input half-cycles at the output. The half frequency output that is one-half of the input frequency with pulse voltage adjustment control, is shown in Figure 4. In this waveform, the second and third half cycles of the first and second cycles of the input voltage are inverted at the output without any voltage adjustment. Their second and fourth half-cycles remain non-inverted at the output, but their magnitude is one-half with respect to the input. The values of the output pulses are described in mathematical form, as can be seen in Equation (1). \[v_{o}(\omega t)=\left\{ \begin{array}{ll}kV_{m}\text{Sin}(\omega t)&0\leq \omega t\leq\pi\\ -V_{m}\text{Sin}(\omega t)&\pi\leq\omega t\leq 3\pi\\ kV_{m}\text{Sin}(\omega t)&3\pi\leq\omega t\leq 4\pi\end{array} \right. \tag{1}\]  
Figure 4: Output voltage for one-half frequency operation with respect to input for voltage (**a**) regulated form and (**b**) non-regulated form. The pulse selective technique, as described in [18], is useful for analyzing the output voltage and input current in terms of their rms values and harmonic indexes. In this regard, first the output voltage is taken for this purpose. Here '\(k\)' is the magnitude scaling factor of the output voltage-controlled pulses and its value may be adjusted from '0' to '1'. \[V_{o}=\frac{V_{m}}{2}\sqrt{1+k^{2}} \tag{2}\]  
The Fourier series form of the output voltage may be formulated as  
\[v_{o}(\omega t)=\frac{(k-1)V_{m}}{2}\sin(\omega t)+\sum_{n=1,3,5,..}^{\infty} \frac{-4(1+k)V_{m}}{\pi(n^{2}-4)}\sin(\frac{n\pi}{2})\sin(\frac{n\omega t}{2}) \tag{3}\]  
The detailed analysis of this equation shows that the most troublesome harmonic exists at the source voltage frequency level. Its values can only be made zero once the amplitude of the output pulse number '1' and '4' becomes to the input voltage level (\(k=1\)) as may see from the plot of Figure 4b. The mathematical verification of this observation is ensured by computing the harmonic factor of this output voltage waveform. The value of the harmonic factor is computed with the value of \(k=0.5\) and \(k=1\), respectively. \[V_{o(k=0.5)}=\frac{\sqrt{5}V_{m}}{4} \tag{4}\]  
\[v_{o1(k=0.5)}(\omega t)=\frac{2V_{m}}{\pi}\sin(\frac{\omega t}{2}) \tag{5}\]  
It has the following rms value. \[V_{o1(k=0.5)}=\frac{\sqrt{2}V_{m}}{\pi} \tag{6}\]  
The harmonic factor in this scenario is calculated as  
\[H.F=\sqrt{\frac{V_{o(k=0.5)}^{2}-V_{o1(k=0.5)}^{2}}{V_{o1(k=0.5)}^{2}}}\approx 73 \,\% \tag{7}\]  
With a similar method, for the scaling factor of \(k=1\), the computed harmonic factor is 62% and is lower than for a value of \(k=0.5\). There is an improvement in the harmonic factor of the output voltage in Figure 4b, where the amplitudes of all output half cycles are equal (\(k=1\)). On the other side, if one examines the instantaneous variation of the input currents, it is clear that it is non-sinusoidal and sinusoidal for a first and second case, as shown in Figure 4a,b, respectively. Similar numeric analysis used for the output voltage can also be employed for input currents to compute its harmonic and power factor. In this process, the displacement angle of the input current is assumed to be zero. The rms and instantaneous value of the input current shown in Figure 4a are computed as  
\[I_{\mathrm{s}}=\frac{I_{m}}{2}\sqrt{1+k^{2}} \tag{8}\]  
\[i_{\mathrm{s}}(\omega t)=\frac{(1+k)I_{m}}{2}\sin(\omega t)+\sum_{n=1,  
\[I_{s1(k=0.5)}=\frac{3I_{m}}{4\sqrt{2}} \tag{11}\]  
\[H.F=\sqrt{\frac{I_{s(k=0.5)}^{2}-I_{s1(k=0.5)}^{2}}{I_{s1(k=0.5)}^{2}}}\approx 33 \,\% \tag{12}\]  
\[V_{o}=\frac{V_{m}}{2}\sqrt{1+k^{2}} \tag{2}\]0  
By employing the same procedure, the value of harmonic and power factor for \(k=1\) is zero and unity, respectively. It is concluded from the previous investigation that power quality indexes of the outputs of Figure 4b are improved.