# Energy and Spectrally Efficient Modulation Scheme for IoT Applications
## Appendix C  
The peak to average power ratio (PAPR) of the OSP-MSK transmitted signal \(S(t)\) can be expressed as follows:  
\[PAPR_{S(t)}=10 log_{10}\left[\frac{P_{peak}}{P_{avg}}\right]=10 log_{10}\left[ \frac{max\{|S(t)|^{2}\}}{\frac{1}{2T}\int_{0}^{2T}S^{2}(t)dt}\right], \tag{17}\]  
where \(P_{peak}\) and \(P_{avg}\) are the peak and the average power of the OSP-MSK transmitted signal, respectively. However, \(S^{2}(t)\) can be given as follows:  
\[S^{2}(t)=a^{2}+a^{2}sin\left(\frac{\pi t}{T}\right)sin(4\pi f_{c}t)+\frac{a^{ 2}}{2}sin\left(\frac{\pi t}{T}\right)+\frac{a^{2}}{2}sin(4\pi f_{c}t), \tag{18}\]where \(a=\sqrt{\frac{2E_{b}}{T}}\). Hence, to find the maximum of \(\{\left|S(t)\right|^{2}\}\), we let \(\frac{d}{dt}\left\{S^{2}(t)\right\}=0\). That is:  
\[\begin{array}{c}\frac{d}{dt}\left\{S^{2}(t)\right\}=\Big{(}\frac{\pi a^{2}}{T} \Big{)}cos\big{(}\frac{\pi t}{T}\big{)}sin\big{(}4\pi f_{c}t\big{)}+\big{(}4 \pi a^{2}f_{c}\big{)}sin\big{(}\frac{\pi t}{T}\big{)}cos\big{(}4\pi f_{c}t \big{)}\\ \hskip 14.226378 pt+\Big{(}\frac{\pi a^{2}}{2T}\Big{)}cos\big{(}\frac{\pi t}{T} \big{)}+\big{(}2\pi a^{2}f_{c}\big{)}sin\big{(}\frac{\pi t}{T}\big{)}cos\big{(} 4\pi f_{c}t\big{)}=0.\end{array} \tag{16}\]  
For simplicity, let \(t=\frac{T}{2}\) and \(f_{c}=\frac{1}{4T}\); therefore, the right and the left-hand sides of Equation (16) are equalized. As such, the peak power of \(S(t)\) can be obtained as follows:  
\[P_{peak}=S^{2}(t)_{|t=\frac{T}{2},\ f_{c}=\frac{1}{4T}}=3a^{2}=\frac{6E_{b}}{ T}. \tag{17}\]  
However, the average power of \(S(t)\) is given by:  
\[P_{avg}=\frac{1}{2T}\int_{0}^{2T}S^{2}(t)\text{dt}, \tag{18}\]  
which can be rewritten as follows:  
\[\begin{array}{c}P_{avg}=\frac{1}{2\pi}\ \Big{[}\int_{0}^{2T}a^{2}dt\\ +\int_{0}^{2T}a^{2}sin\big{(}\frac{\pi t}{T}\big{)}sin\big{(}4\pi f_{c}t\big{)} dt\\ +\int_{0}^{2T}\frac{a^{2}}{2}sin\big{(}\frac{\pi t}{T}\big{)}dt+\int_{0}^{2T} \frac{a^{2}}{2}sin\big{(}4\pi f_{c}t\big{)}dt\Big{]},\end{array} \tag{19}\]  
\[\begin{array}{c}P_{avg}=\frac{1}{2\pi}\ \Big{[}\int_{0}^{2T}a^{2}dt+\frac{a^{2}}{ 2}\int_{0}^{2T}cos\big{(}\frac{\pi}{T}-4\pi f_{c}\big{)}tdt\\ -\frac{a^{2}}{2}\int_{0}^{2T}cos\big{(}\frac{\pi}{T}+4\pi f_{c}\big{)}tdt\\ +\int_{0}^{2T}\frac{a^{2}}{2}sin\big{(}\frac{\pi t}{T}\big{)}dt+\int_{0}^{2T} \frac{a^{2}}{2}sin\big{(}4\pi f_{c}t\big{)}dt\Big{]}.\end{array} \tag{20}\]  
Therefore, the average power \(P_{avg}\) in Equation (20) is given as follows:  
\[P_{avg}=a^{2}=\frac{2E_{b}}{T}. \tag{21}\]  
Substituting equations (17) and (21) into Equation (14), the PAPR of the OSP-MSK transmitted signal is given by:  
\[PAPR_{S(t)}=10 log_{10}\bigg{[}\frac{P_{peak}}{P_{avg}}\bigg{]}=10 log_{10}(3)=4. 77\text{dB}. \tag{22}\]  
The value of the PAPR of the transmitted signal \(S(t)\) provided by Equation (22) is low compared to the high PAPR values of the OFDM system, which can be greater than 10 dB [27; 35]. However, it should be mentioned here that the low PAPR property of the OSP-MSK signal doesn't contradict with the use of nonlinear PAs at the TX side of the OSP-MSK modulation, since both \(S_{1}(t)\) and \(S_{2}(t)\) are constant envelope signals (i.e., unitary PAPR signals).