# Performance of Differential-Phase OFDM Systems over Selective Multipath Fading Channels
## 3 Problem Statement
### The Received OFDM Signal over Multipath Fading Channels
Let's rewrite \(r_{ki}\) from (20) as  
\[r_{ki}=\mathcal{J}_{1}+\mathcal{J}_{2}+\mathcal{J}_{3}+\mathcal{J}_{4}+ \mathcal{J}_{5}+\tilde{n}_{k,i} \tag{23}\]  
where \(\mathcal{J}_{1}\) and \(\mathcal{J}_{2}\) represent desired components, \(\mathcal{J}_{3}\), \(\mathcal{J}_{4}\), and \(\mathcal{J}_{5}\) represent undesired components due to ICI and ISI, and \(\tilde{n}_{k,i}\) is an AWGN term. (1) \(\mathcal{J}_{1}:\;\;n=k,\;\;N_{p}\leq N_{g},\;\;1\leq p\leq P_{1},\;\;0\leq m \leq N-1\)  
\[r_{k} = \frac{1}{N}\sum_{i=1}^{\infty}\sum_{p=1}^{P}\sum_{n=0}^{N-1}\sum_ {m=0}^{N-1}\alpha_{p}\,e^{i\phi_{p}}\,u_{n,i}\,f(m)\,e^{-j2\pi nN_{p}/N}\,e^{j2 \pi(n-k)m/N}+\bar{n}_{k} \tag{18}\]0 \[r_{k} = \frac{1}{N}\sum_{i=1}^{\infty}\sum_{p=1}^{P}\sum_{n=0}^{N-1}\sum_ {m=0}^{N-1}\alpha_{p}\,e^{i\phi_{p}}\,u_{n,i}\,f(m)\,e^{-j2\pi nN_{p}/N}\,e^{j2 \pi(n-k)m/N}+\bar{n}_{k} \tag{18}\]1  
\[r_{k} = \frac{1}{N}\sum_{i=1}^{\infty}\sum_{p=1}^{P}\sum_{n=0}^{N-1}\sum_ {m=0}^{N-1}\alpha_{p}\,e^{i\phi_{p}}\,u_{n,i}\,f(m)\,e^{-j2\pi nN_{p}/N}\,e^{j2 \pi(n-k)m/N}+\bar{n}_{k} \tag{18}\]2 (**a**) Received frame from the shortest multipath component and (**b**) Received frame from the \(p\)-th path with \(\tau_{p}>T_{g}\). \[r_{k} = \frac{1}{N}\sum_{i=1}^{\infty}\sum_{p=1}^{P}\sum_{n=0}^{N-1}\sum_ {m=0}^{N-1}\alpha_{p}\,e^{i\phi_{p}}\,u_{n,i}\,f(m)\,e^{-j2\pi nN_{p}/N}\,e^{j2 \pi(n-k)m/N}+\bar{n}_{k} \tag{18}\]3 (**a**) Received frame from the shortest multipath component and (**b**) Received frame from the \(p\)-th path with \(\tau_{p}<T_{g}\). This term represents a desired component since \(n=k\) and it is interference free because \(N_{p}<N_{g}\). (2) \(\mathcal{J}_{2}:\)\(n=k,\)\(N_{p}>N_{g},\)\(1+P_{1}\leq p\leq P,\)\(N_{p}-N_{g}\leq m\leq N-1\)  
\[r_{k} = \frac{1}{N}\sum_{i=1}^{\infty}\sum_{p=1}^{P}\sum_{n=0}^{N-1}\sum_ {m=0}^{N-1}\alpha_{p}\,e^{i\phi_{p}}\,u_{n,i}\,f(m)\,e^{-j2\pi nN_{p}/N}\,e^{j2 \pi(n-k)m/N}+\bar{n}_{k} \tag{18}\]4 \[r_{k} = \frac{1}{N}\sum_{i=1}^{\infty}\sum_{p=1}^{P}\sum_{n=0}^{N-1}\sum_ {m=0}^{N-1}\alpha_{p}\,e^{i\phi_{p}}\,u_{n,i}\,f(m)\,e^{-j2\pi nN_{p}/N}\,e^{j2 \pi(n-k)m/N}+\bar{n}_{k} \tag{18}\]5  
This term also represents a desired component since \(n=k\), yet, the block is missing some information as it will interfere with the following block. (3) \(\mathcal{J}_{3}:\)\(n\
eq k,\)\(N_{p}\leq N_{g},\)\(1\leq p\leq P_{1},\)\(0\leq m\leq N-1,\) [ICI]  
\[r_{k} = \frac{1}{N}\sum_{i=1}^{\infty}\sum_{p=1}^{P}\sum_{n=0}^{N-1}\sum_ {m=0}^{N-1}\alpha_{p}\,e^{i\phi_{p}}\,u_{n,i}\,f(m)\,e^{-j2\pi nN_{p}/N}\,e^{j2 \pi(n-k)m/N}+\bar{n}_{k} \tag{18}\]6  
This terms represents ICI between adjacent subchannels. \(\mathcal{J}_{3}=0\) due to the orthogonality between subcarriers. (4) \(\mathcal{J}_{4}:\)\(n\
eq k,\)\(N_{p}>N_{g},\)\(1+P_{1}\leq p\leq P,\)\(N_{p}-N_{g}\leq m\leq N-1,\) [ICI]  
\[r_{k} = \frac{1}{N}\sum_{i=1}^{\infty}\sum_{p=1}^{P}\sum_{n=0}^{N-1}\sum_ {m=0}^{N-1}\alpha_{p}\,e^{i\phi_{p}}\,u_{n,i}\,f(m)\,e^{-j2\pi nN_{p}/N}\,e^{j2 \pi(n-k)m/N}+\bar{n}_{k} \tag{18}\]7 \[r_{k} = \frac{1}{N}\sum_{i=1}^{\infty}\sum_{p=1}^{P}\sum_{n=0}^{N-1}\sum_ {m=0}^{N-1}\alpha_{p}\,e^{i\phi_{p}}\,u_{n,i}\,f(m)\,e^{-j2\pi nN_{p}/N}\,e^{j2 \pi(n-k)m/N}+\bar{n}_{k} \tag{18}\]8  
This term is due to the presence of ICI because of the excess delays that exceed the guard period at \(n\
eq k\). The above four terms are received from the current signaling interval (the \(i\)-th OFDM frame). The last term \(\mathcal{J}_{5}\) represents the received symbols from the previous frame (i.e., the \((i-1)\)-th frame); thus, \(\mathcal{J}_{5}\) is given by  
(5) \(\mathcal{J}_{5}:\)\(\forall n,\)\(N_{p}>N_{g},\)\(1+P_{1}\leq p\leq P,\)\(0\leq m\leq N_{p}-N_{g}-1\) and \(u_{n,(i-1)},\) [ISI]  
\[r_{k} = \frac{1}{N}\sum_{i=1}^{\infty}\sum_{p=1}^{P}\sum_{n=0}^{N-1}\sum_ {m=0}^{N-1}\alpha_{p}\,e^{i\phi_{p}}\,u_{n,i}\,f(m)\,e^{-j2\pi nN_{p}/N}\,e^{j2 \pi(n-k)m/N}+\bar{n}_{k} \tag{18}\]9 \[r_{k}=\sum_{i=1}^{\infty}r_{ki} \tag{19}\]0 \[r_{k}=\sum_{i=1}^{\infty}r_{ki} \tag{19}\]1