Differential Detection

Differential (Non-Coherent) Detection of PSK signals

Intuition : since the phase shift doesnot change over two symbols durations,let us transmit information by changing the phase of the carrier from the phase at the previous symbol duration depending on what info we need to transmit.

To analyse this scheme let us consider the following format . At each time we transmit either

\begin{align} s_k(t) = \[ \left\{ \begin{array}{ccc} A\cos(\omega_c t) & K \leq t < (k+1)T_s \\ \mbox{or} \\ -A\cos(\omega_c t) & K \leq t < (k+1)T_s \end{array} \] \end{align}

We make ,

\begin{eqnarray} s_k(t) &=& s_{k-1}(t) \mbox{ if info bit $a_k = 0$} \\ &=& -s_{k-1}(t) \mbox{ if info bit $a_k = 1$} \end{eqnarray}

Now let us look at symbols transmitted using two consecutive time instants {ak-1,ak}.Suppose $A\cos(\omega_c t)$ were transmitted at time k-1.(receiver does not know this)

There are two possibilities


Notice that these two signals are orthogonal.So we can use our previous result. However notice that the energy is twice Eb !.
To keep the notation simpler let us focus on k=1. the Receiver structure will be :

\begin{eqnarray} Z_0 &=& ( \begin{array}{cc} r_0^* & r_1^*) \end{array} \left( \begin{array}{c} 1 \\ 1 \end{array} \right) = r_0^* + r_1^* \hspace{6mm} |z_0|^2 = |r_0|^2 +|r_1|^2 + 2 \Re \left[ r_0^* r_1^*\right] \\ Z_1 &=& ( \begin{array}{cc} r_0^* & r_1^*) \end{array} \left( \begin{array}{c} 1 \\ -1 \end{array} \right) = r_0^* - r_1^* \hspace{6mm} |z_0|^2 = |r_0|^2 +|r_1|^2 - 2 \Re \left[ r_0^* r_1^*\right] \end{eqnarray}

We make a decision,

\begin{eqnarray} \mbox{if }|z_0|^2 > |z_1|^2 \equiv \left{ \Re \left[ r_0^* r_1 \right] > 0 \mbox{ bit 0 was sent } \\ \mbox{if }|z_0|^2 < |z_1|^2 \equiv \left{ \Re \left[ r_0^* r_1 \right] < 0 \mbox{ bit 1 was sent } \end{eqnarray}

Suppose the signal transmitted at time K-1(here K-1=) were $-A\cos(\omega_c t)$ then the possibilities are


Our standard receiver structure is still the same .However

\begin{eqnarray} Z_0 &=& ( \begin{array}{cc} r_0^* & r_1^*) \end{array} \left( \begin{array}{c} 1 \\ 1 \end{array} \right) = - r_0^* - r_1^* \\ Z_1 &=& ( \begin{array}{cc} r_0^* & r_1^*) \end{array} \left( \begin{array}{c} 1 \\ -1 \end{array} \right) = -r_0^* + r_1^* \end{eqnarray}

We make decision by comparing $|z_0^2$ and $|z_1^2$ is a similar way i.e.

\begin{eqnarray} \mbox{if }|z_0|^2 > |z_1|^2 \equiv \left{ \Re \left[ r_0^* r_1 \right] > 0 \mbox{ bit 0 was sent } \\ \mbox{if }|z_0|^2 < |z_1|^2 \equiv \left{ \Re \left[ r_0^* r_1 \right] < 0 \mbox{ bit 1 was sent } \end{eqnarray}

Therefore the receiver structure and Probability of error do not depend on what was transmitted during the (K-1)th instant. This is known as differential detection.Transmitter and receiver can be implemented as follows.

Differential encoding

Form a new symbol $b_k= b_{k-1}\oplus a_k$ and then use BPSK on $b_k$.


Probability of error$P_e = \frac{1}{2} e^{-\frac{E}{2N_0}$. Remember that E=2Eb, so

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

DPSK is asymptotically equivalent to BPSK!!!

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