G(D)=1, 1+D+D3+D 1+D3+D The numbers on the different trajectories are the E/No values in dB. 3 0.8 0.6 1.5 0.4 0.5 0.2 0.5 0 09 0.4 0.6 0.8 IA Figure 4.24:EXIT function of a 16-state convolutional code 4.5.2 EXIT Chart In a complete turbo decoder this extrinsic information is now exchanged between the decoders,where output extrinsic LLRs become a pripori LLRs for the next decoder.In our analysis these iterations are captured by a sequence of applications of component decoder EXIT functions,that is where T is the EXIT function of decoder DEC1,and T2 of decoder DEC2.The behavior of this exchange can be visualized in the EXIT chart,where the EXIT functions of both nen decoder ever,the of information.As shown in Figure 4.25,for sufficiently large values of E/No (in this case for E No=0.8dB)these two curves leave open a channel,and they have no intersect point other than (1,1).Since the iterations start with zero LLRs,ie.,with zero mutual information at 4-504-50 3 4 3 4 1 ( ) 1, 1 DD D G D D D            The numbers on the different trajectories are the Eb/N0 values in dB. Figure 4.24: EXIT function of a 16-state convolutional code. 4.5.2 EXIT Chart In a complete turbo decoder this extrinsic information is now exchanged between the decoders, where output extrinsic LLRs become a pripori LLRs for the next decoder. In our analysis these iterations are captured by a sequence of applications of component decoder EXIT functions, that is 1212 (1) (2) (2) (1) (1) (2) 0 TT TT EA E A EA    II II II      where T1 is the EXIT function of decoder DEC1, and T2 of decoder DEC2. The behavior of this exchange can be visualized in the EXIT chart, where the EXIT functions of both component decoders are plotted, however, the transfer curve of decoder #2 is reflected about the /4 line to reflect that IE is the input mutual information, and IA is the output mutual information. As shown in Figure 4.25, for sufficiently large values of Eb/N0 (in this case for Eb/N0 = 0.8dB) these two curves leave open a channel, and they have no intersect point other than (1, 1). Since the iterations start with zero LLRs, i.e., with zero mutual information at