# Performance of Differential-Phase OFDM Systems over Selective Multipath Fading Channels
## 3 Problem Statement
### Signal-To-Interference Ratio (SIR)  
Since each received symbol contains desired and possibly undesired terms, the system performance is evaluated by taking these facts into considerations. Due to the nature of the WSS-UCS channels, we can show that \(\mathbb{E}[\mathcal{J}_{i}\mathcal{J}_{k;i}^{*}]=0\). Therefore, the average symbol energy in the \(k\)-th received symbol is evaluated as \(0.5\mathbb{E}[r_{k;i}r_{k;i}^{*}]\), where  
\[\mathbb{E}[r_{k;i}r_{k;i}^{*}] = \frac{\text{desired components}}{\mathbb{E}[\mathcal{J}_{1} \mathcal{J}_{1}^{*}]+\mathbb{E}[\mathcal{J}_{2}\mathcal{J}_{2}^{*}]}+\frac{ \text{undesired components}}{\mathbb{E}[\mathcal{J}_{3}\mathcal{J}_{3}^{*}]+ \mathbb{E}[\mathcal{J}_{4}\mathcal{J}_{4}^{*}]+\mathbb{E}[\mathcal{J}_{5} \mathcal{J}_{5}^{*}]+\sigma_{k}^{2}} \tag{29}\]The energy in the desired components \(\mathcal{J}_{1}\) and \(\mathcal{J}_{2}\) is given by  
\[\mathcal{S}=0.5\mathbb{E}[\mathcal{J}_{1}\mathcal{J}_{1}^{*}]+0.5\mathbb{E}[ \mathcal{J}_{2}\mathcal{J}_{2}^{*}] \tag{30}\]  
By recalling \(\mathcal{J}_{1}\) from (24) and \(\mathcal{J}_{2}\) from (25),  
\[0.5\mathbb{E}[\mathcal{J}_{1}\mathcal{J}_{1}^{*}] = \sum_{p=1}^{P_{1}}E_{s}\,\mathbb{E}[\alpha_{p}^{2}] \tag{31}\] \[= \sum_{p=1}^{P_{1}}E_{s}\,\rho(\tau)\,\delta(\tau-\tau_{p})\]  
and  
\[0.5\mathbb{E}[\mathcal{J}_{2}\mathcal{J}_{2}^{*}] = \sum_{p=1+P_{1}}^{P}E_{s}\,\mathbb{E}[\alpha_{p}^{2}]\left[\frac {N_{f}-N_{p}}{N}\right]^{2} \tag{32}\] \[= \sum_{p=1+P_{1}}^{P}E_{s}\,\rho(\tau)\,\delta(\tau-\tau_{p})\left[ \frac{N_{f}-N_{p}}{N}\right]^{2}\]  
The desired signal energy is  
\[\mathcal{S}=\sum_{p=1}^{P_{1}}E_{s}\,\rho(\tau)\,\delta(\tau-\tau_{p})+\sum_{p =1+P_{1}}^{P}E_{s}\,\rho(\tau)\,\delta(\tau-\tau_{p})\left[\frac{N_{f}-N_{p}}{ N}\right]^{2} \tag{33}\]  
Given that \(E[\mathcal{J}_{3}\mathcal{J}_{3}^{*}]=0\), the undesired terms contain the interference with energy given by  
\[\mathcal{I}=0.5\mathbb{E}[\mathcal{J}_{4}\mathcal{J}_{4}^{*}]+0.5\mathbb{E}[ \mathcal{J}_{5}\mathcal{J}_{5}^{*}] \tag{34}\]  
By recalling \(\mathcal{J}_{4}\) from (27) and \(\mathcal{J}_{5}\) from (28), the energy of the undesired terms is  
\[0.5\mathbb{E}[\mathcal{J}_{4}\mathcal{J}_{4}^{*}] = \sum_{p=1+P_{1}}^{p}\sum_{n=0,n\
eq k}^{N-1}E_{s}\,\mathbb{E}[ \alpha_{p}^{2}]\left[\frac{N_{g}-N_{p}}{N}\right]^{2}\left[\frac{\text{sinc}(( n-k)(N_{g}-N_{p})/N)}{\text{sinc}((n-k)/N)}\right]^{2} \tag{35}\] \[= \sum_{p=1+P_{1}}^{p}E_{s}\,\rho(\tau)\,\delta(\tau-\tau_{p})\left[ \frac{N_{g}-N_{p}}{N}\right]^{2}\left\{\,\sum_{n=0}^{N-1}\left[\frac{\text{sinc }((n-k)(N_{g}-N_{p})/N)}{\text{sinc}((n-k)/N)}\right]^{2}-1\right\}\]  
and  
\[\mathcal{S}=0.5\mathbb{E}[\mathcal{J}_{1}\mathcal{J}_{1}^{*}]+0.5\mathbb{E}[ \mathcal{J}_{2}\mathcal{J}_{2}^{*}] \tag{30}\]0  
since \(|u_{k,i}|=|u_{k,(i-1)}|=\sqrt{2E_{s}/T_{s}}\). By manipulating (35) and (36), the interference energy can be expressed as  
\[\mathcal{S}=0.5\mathbb{E}[\mathcal{J}_{1}\mathcal{J}_{1}^{*}]+0.5\mathbb{E}[ \mathcal{J}_{2}\mathcal{J}_{2}^{*}] \tag{30}\]1 \[\mathcal{S}=0.5\mathbb{E}[\mathcal{J}_{1}\mathcal{J}_{1}^{*}]+0.5\mathbb{E}[ \mathcal{J}_{2}\mathcal{J}_{2}^{*}] \tag{30}\]2Finally, the Signal to Interference ratio from the \(k\)-th subcarrier is  
\[\mathcal{S}=0.5\mathbb{E}[\mathcal{J}_{1}\mathcal{J}_{1}^{*}]+0.5\mathbb{E}[ \mathcal{J}_{2}\mathcal{J}_{2}^{*}] \tag{30}\]3  
In order to improve SIR, an increase in the number of components with no interference is needed, i.e., \(P_{1}\) paths. This can be achieved by increasing the guard period. Another way to improve the SIR is to increase the number of subchannels. Increasing \(N\) will increase the frame duration as \(T_{f}=NT_{s}+T_{g}\). However, the system may exhibit a lower data rate and may not succeed as orthogonality ensures the least frequency space between subcarriers.