Noncoherent Detection

## Random time delay model in the passband

So far we have considered the problem of optimally detecting a transmitted signal waveform when the only disturbance resulting from the channel is additive white Gaussian noise. However, in practice there are other kinds of disturbances that are introduced by the channel in addition to just additive white Gaussian noise. Often the channel introduces a random delay to the transmitted signals. There are two main sources of such a delay - i) The time taken for an electromagnetic wave to travel the distance from the transmitter to the receiver in wireless communications. Often this may change since a person may be at different locations at different times ii) The receiver and the transmitted clocks may be not synchronized. When the receiver samples the output of the matched filter, it is expecting the time origin to begin somewhere which turns out to be different. As a result, the received passband signal will be given by

(1)
$$r(t) = s(t-T_d) + n(t)$$

If $T_d$ is known at the receiver, we can can compensate for $T_d$ and since $n(t)$ is wide-sense stationary, shifting $n(t)$ does not change the statistics of the noise. Hence, this case is identical to that of coherent detection and ideas developed in the previous section apply here directly. Our interest in this section is in the case when $T_d$ is not known at the receiver. In this case, we would like to know what the structure of the optimal detector is and what the ensuing probability of error will be.

## Equivalent Baseband model

To understand these, it is easier to work with the baseband model. Let us first understand the effect of the random time delay on the baseband model. To see this, consider the passband received signal when $s_m(t)$ is transmitted.

(2)
\begin{align} r(t) = \Re\{(s_{ml}(t-T_d) + n_l(t-T_d) )e^{j 2 \pi f_0 (t-T_d)} \} = \Re\{(s_{ml}(t-T_d) e^{j 2 \pi f_0 T_d} + \tilde{n}_l(t) )e^{j 2 \pi f_0 t) \} \end{align}

where $\tilde{n}_l(t)$ has the same statistics as $n_l(t)$ since $n_l(t)$ is circularly symmetric. The equivalent baseband signal is that $r_l(t) = s_{ml}(t-T_d) e^{j 2 \pi f_0 T_d} + n_l(t)$. Notice that the channel introduces a time shift to $s_l(t)$ and a phase shift $\phi = 2 \pi f_0 T_d$ to the complex baseband equivalent signal. Typically, we are interested in cases where $T_d << T_s$, the symbol duration and hence $s_{ml}(t) \approx s_{ml}(t-T_d)$; however, $f_0 T_d$ cannot be ignored since $f_0$ is very large. Hence, the net effect is that there is a phase rotation introduced to the baseband equivalent signal. If you are not convinced of this here is an example below to illustrate this

Thus, we can form an equivalent baseband representation where the baseband equivalent signal $r_l(t)$ and the projections of $r_l(t)$ on to the baseband equivalent basis functions $\phi_{li}(t)$ are given by

(3)
\begin{eqnarray} r_l(t) & = & e^{j \phi} s_{ml}(t) + n_l(t) \\ \nonumber \underline{r}_l & = & e^{j \phi} \underline{s}_{ml} + \underline{n}_l \end{eqnarray}

where $\phi$ is a random variable. We will assume that the (prior) distribution of $\phi$, namely $p_{\phi}(\phi)$ is known to the receiver but the realization is unknown.

## Non-coherent Detection

In the above expression we observe that if $\phi$ is known we can perform coherent detection.However in practise $\phi$ is never known at the reciever.If we make no attempt to estimate it and try to detect signals then it is called Non-coherent detection.We will assume that the (prior) distribution of $\phi$, namely $p_{\phi}(\phi)$ is known to the receiver but the realization is unknown.

### MAP detector

Consider,

(4)
\begin{align} P(\underline{s}_m | \underline{r}) = \frac {P(\underline{r}|\underline{s}_m) P(\underline{s}_m)} {P(\underline{r})} \end{align}

Assume $P(\underline{s}_m = \frac{1}{M}$. Therefore we have the decision,

(5)
\begin{align} \hat{m} = \mbox{ arg } \max_m P(\underline{r}|\underline{s}_m) \end{align}
(6)
\begin{eqnarray} P(\underline{r}|\underline{s}_m) &=& \int P(\underline{r}, \phi|\underline{s}_m) d\phi \\ &=& \int P(\underline{r}|\underline{s}_m, \phi) P(\phi| \underline{s}_m) d\phi \\ \end{eqnarray}
(7)
\begin{align} P(\underline{r}|\underline{s}_m) = \int_0^{2\pi} P(\underline{r}|\underline{s}_m, \phi) P(\phi) d\phi \end{align}

We will now study about the optimal detector for $P(\phi) = \frac{1}{2\pi}$ and over an AWGN channel.

In baseband let 2Em be the energy and the noise variance be $\sigma^2$ in each dimension.

(8)
\begin{eqnarray} P(\underline{r}|\underline{s}_m) &=& \frac{1}{\pi^N(2\sigma^2)^N} \int_0^{2\pi} e^{ \frac{-\| \underline{r}_l - e^{j\phi}\underline{s}_{ml} \|^2 }{2\sigma^2}} \frac{1}{2\pi} d\phi \\ &=& \mbox{ const }\int_0^{2\pi} e^{- \left( \frac{\| \underline{r}_l \|^2 +\| \underline{s}_{ml} \|^2 -2 \Re < \underline{r}_l,\underline{s}_{ml}> }{2\sigma^2}} \right) \\ &=& \mbox{ const. } e^ {-\frac{E_m^2}{2\sigma^2}} \int_0^{2\pi} e^{ \left( \frac {\Re < \underline{r}_l,\underline{s}_{ml}> }{\sigma^2}} \right) d\phi \end{eqnarray}

Let, $<\underline{r}_l , \underline{s}_{ml}> = |\underline{r}_l , \underline{s}_{ml}| e^{j\theta}$, then $\Re <\underline{r}_l , \underline{s}_{ml}> =|\underline{r}_l , \underline{s}_{ml}| \cos(\theta-\phi)$. Therefore

(9)
\begin{align} P(\underline{r}|\underline{s}_m) = \mbox{ const. } e^ {\frac{E_m^2}{\sigma^2}} \int_0^{2\pi} e^{ \left( \frac {|\underline{r}_l . \underline{s}_{ml}| \cos(\theta-\phi) }{\sigma^2}} \right) d\phi \end{align}

Note that $\theta$ has no effect on the integral since $\cos(\theta-\phi)$ is periodic with period $2\pi$ and we are integrating over one period.
Using the defination of the modified Bessel function of zeroth order and firt kind

(10)
\begin{align} I_0(x) = \frac{1}{2\pi} \int_0^{2\pi} e^{x\cos\phi} d\phi \end{align}

Bessel function figure

(11)
\begin{align} P(\underline{r}|\underline{s}_m) = \mbox{ const. } e^ {\frac{E_m^2}{\sigma^2}} I_0 \left( \frac {|\underline{r}_l . \underline{s}_{ml}| }{\sigma^2}} \right) \end{align}

For equal Em , we can compare $|\underline{r}_l . \underline{s}_{ml}|$

(12)
\begin{align} \hat{m} = \mbox{ arg } \max_m |\underline{r}_l . \underline{s}_{ml}| \end{align}

A more general form of MAP detector is given by,

(13)
\begin{align} \hat{m} = \mbox{ arg } \max_m \left[ P(\underline{s}_m) e^ {\frac{E_m^2}{\sigma^2}} I_0 \left( \frac {|\underline{r}_l . \underline{s}_{ml}| }{\sigma^2}} \right) \right] \end{align}

In passbad we compute the corect statistics using the following relations.

(14)
\begin{eqnarray} \Re \left[ \int r_l(t) s_{ml}^*(t) \right] &=& 2 \int r(t) s_m(t) dt \\ \Im \left[ \int r_l(t) s_{ml}^*(t) \right] &=& 2 \int r(t) \hat{s}_m(t) dt \end{eqnarray}

### Optimal Non-coherent Detection of FSK signals

(15)
\begin{align} s_m(t) = g(t) cos( 2\pi f_c t + 2\pi(m-1)\Delta f t) \end{align}
(16)
\begin{align} S_{ml}(t) = g(t) e^{j2\pi(m-1) \Delta f t} \end{align}

let g(t) be a rectangular pulse from 0 to T with amplitude $\sqrt{\frac{E_m}{T}}$.Suppose we wanted orthogonality between si(t) and sk(t) or si(t) and sk(t), the frequency separation between the signals should be $\frac{1}{T}$ but not$\frac{1}{2T}$.

## Probability of error for M-ary orthogonal signals

Consider a set of M orthogonal signals with vector representation given by , $s_{1l}= \left( \srqt{E_s},0,0,\ldots,0 \right) , \ldots , s_{Ml}= \left(0,0,\ldots,0, \srqt{E_s} \right)$. In praokis 5th edition they use es for the passband energy and hence Es = 2 es.

Let Sl1 be transmitted and $\underline{r}_l = e^{j\phi}\underline{s}_{l1} + \underline{n}_l$. Ecah com[ponent of $\underline{n}$ is complex and the real and imaginary parts ave variance N0.

(17)
\begin{align} |R_1| = {|\underline{r}_l , \underline{s}_{1l}| = | E_s e^{j\phi} + \underline{n}_l.\underline{s}_{1l}| \end{align}
(18)
\begin{align} |R_m| = {|\underline{r}_l , \underline{s}_{ml}| = |0 + \underline{n}_l.\underline{s}_{1l}| \hspace{5mm} 2<m<M \end{align}

$\Re(R_1) \sim N(E_s\cos(\phi), E_sN_0), \Im(R_1) \sim N(E_s\sin(\phi), E_sN_0)$ , $\Re(R_2) \sim N(0, E_sN_0),\Im(R_2) \sim N(0, E_sN_0)$

Distribution of |R1| is called Ricean with parameter S= Es and $\sigma^2= E_sN_0$

(19)
\begin{align} P_{|R_1|}(r_1) = \left\{ \begin{array} {lc} \frac{r_1}{\sigma^2}I_0\left( \frac{sr_1}{\sigma^2} \right) e^ {-\frac{r_1^2 + s^2}{2\sigma^2}} & r_1>0 \\ 0 & r_1<0 \end{array} \end{align}
(20)
\begin{align} P_{|R_1|^2}(r_1) = \left\{ \begin{array} {lc} \frac{1}{2\sigma^2}I_0\left( \frac{s}{\sigma^2} \sqrt{r_1} \right) e^ {-\frac{r_1 + s^2}{2\sigma^2}} & r_1>0 \\ 0 & r_1<0 \end{array} \end{align}

For $m\not= 1$ Rm has a Rayleigh distribution,

(21)
\begin{align} P_{R_m}(r_m) = \left\{ \begin{array} {lc} \frac{r_m}{\sigma^2} e^ {-\frac{r_m^2} {2\sigma^2}} & r_m>0 \\ 0 & r_m<0 \end{array}. \end{align}

$P_{R_m^2}(r_m)$ is exponentially distributed.

Probability of Error Caluculations

(22)
\begin{eqnarray} P_c &=& P \left[ R_2 < R_1,R_3 < R_1,R_4 < R_1,\ldots, R_m< R_1 \right] \\ &=& \int_0^\infty P \left[ R_2 < r_1,R_3 < r_1,R_4 < r_1,\ldots, R_m< r_1 | R_1=r_1 \right] P_{R_1}(r_1) dr_1 \\ &=& \int_0^\infty \left( P \left[ R_2<r_1 \right]^{M-1} \right) P_{R_1}(r_1) dr_1 \end{eqnarray}
(23)
\begin{align} P \left( |R_2|<|r_1|\right) = P\left( R_2^2 < r_1^2 \right) = 1- e^{-\frac{r_1^2}{2\sigma^2}} \end{align}

Using binomial expansion,

(24)
\begin{align} \left(1- e^{-\frac{r_1^2}{2\sigma^2}}\right)^ {M-1} = \sum_{n=0}^{M-1} (-1)^n {M-1 \choose n} e^{-\frac{nr_1^2}{2\sigma^2}} \end{align}
(25)
\begin{eqnarray} P_c &=& \sum_{n=0}^{M-1} (-1)^n {M-1 \choose n} \int_0^\infty e^{-\frac{nr_1^2}{2\sigma^2}} \frac{r_1}{\sigma^2}I_0\left( \frac{sr_1}{\sigma^2} \right) e^ {-\frac{r_1^2 + s^2}{2\sigma^2}} dr_1 \\ &=& \sum_{n=0}^{M-1} (-1)^n {M-1 \choose n} \int_0^\infty \frac{r_1}{\sigma^2}I_0\left( \frac{sr_1}{\sigma^2}\right) e^{-\frac{(n+1)r_1^2 + s^2}{2\sigma^2}} dr_1 \\ &=& \sum_{n=0}^{M-1} (-1)^n {M-1 \choose n} e^{-\frac{ns^2}{2(n+1)\sigma^2}} \int_0^\infty \frac{r_1}{\sigma^2}I_0\left( \frac{sr_1}{\sigma^2}\right) e^{-\frac{(n+1)r_1^2 + \frac{s^2}{n+1}}{2\sigma^2}} dr_1 \end{eqnarray}

Now we substitute $s\prime= \frac{s}{\sqrt{n+1}}$, $r\prime = r_1 \sqrt{n+1}$. Then the above integral becomes,

(26)
\begin{align} \int_0^\infty \frac{r_1}{\sigma^2}I_0\left( \frac{sr_1}{\sigma^2}\right) e^{-\frac{(n+1)r_1^2 + \frac{s^2}{n+1}}{2\sigma^2}} dr_1 = \frac{1}{n+1} \int_0^\infty \frac{r\prime}{\sigma^2}I_0\left( \frac{s\prime r\prime}{\sigma^2} \right) e^ {-\frac{r\prime^2 + s\prime^2}{2\sigma^2}} = \frac{1}{n+1} \end{align}
(27)
\begin{align} P_c = \sum_{n=0}{M-1} \frac{ (-1)^ {n+1}}{n+1} {M-1 \choose n} e^{ -\frac{n}{n+1}\frac{2E_s}{2N_0}} \end{align}
(28)
\begin{align} P_e = \sum_{n=1}{M-1} \frac{ (-1)^ {n+1}}{n+1} {M-1 \choose n} e^{ -\frac{n}{n+1}\frac{E_b \log_2^M}{2N_0}} \end{align}

we note that Es and Eb are pasband energies.Let us interpret this result for the binary FSK M=2.

(29)
\begin{align} P_e = \frac{1}{2} e^{\frac{-E_b}{2N_0}} \end{align}

Remember for coherent detection,

(30)
\begin{align} P_e = Q \left( \sqrt{\frac{E_b}{N_0}}\right) \approx \frac{1}{2} e^{-\frac{E_b}{N_0}} \end{align}

So Non-coherent detection asymptotically performs the same as coherent detection. For any finite SNR, $P_{e_{coh}}< P_{e_{non-coh}}$ since $Q \left( \sqrt{\frac{E_b}{N_0}}\right) \leq \frac{1}{2} e^{\frac{-E_b}{2N_0}}$

## Speacial case- Binary equal energy signals

Let $E_b = \|s_1\|^2 = \|s_2\|^2$ and $\rho= \frac{\underline{s}_1^H \underline{s}_2}{E_b}$. In general it can be shown that

(31)
\begin{align} P_e = Q(a,b) - \frac{1}{2}e^{-\frac{a^2+b^2}{2}} I_0(ab) \end{align}
(32)
\begin{align} a= \frac{E_b}{2N_0} \left( 1- \sqrt{1-|\rho|^2} \right) , b^2 = \frac{E_b}{2N_0} \left( 1+ \sqrt{1+|\rho|^2} \right) \end{align}

$Q(a,b) = \int_a^\infty x e^{-\frac{a^2+x^2}{2}} I_0(ax) dx$ is called the Marcum Q-function

Some conclusions,

1. As $|\rho| \rightarrow 1$ , $P_e \rightarrow \frac{1}{2}$. Antipodal signals cannot be non-coherently detected
2. Pe is minimized when $\rho = 0$. Therefore Othogonal signals are the best for Non-coherent detection.

page revision: 42, last edited: 28 May 2010 14:28