Wireless Ad Hoc and Sensor Networks

438 Wireless Ad Hoc and Sensor NetworksCASE 1 The outgoing traffic f i() ⋅ is known. Now, define the traffic rateinput, ui( k), asu( k) = Sat ( T − 1[ f ( u ( k)) + ( −k ) e ( k)])i p i i+1 1 bv bi(9.3)where k bv is a gain parameter. In this case, the buffer occupancy error atthe time k + 1 becomese ( k+ 1) = Sat [ k e ( k) + d( k)]bi p bv bi(9.4)The buffer occupancy error will become zero as k →∞, provided0< < 1.k bvCASE 2 The outgoing traffic f i() ⋅ is unknown and has to be estimated.In such a case, we define the traffic rate input, ui( k), asu( k) = Sat [ T − 1(ˆ f ( u ( k)) + ( −k ) e ( k))]i p i i+1 1 bv bi(9.5)where f ˆi ( u i+1 ( k))is an estimate of the unknown outgoing traffic fi( ui+1( k)).In this case, the buffer occupancy error at the time instant k + 1 becomesebi( k + 1) = Satp ( kbvebi( k) + f ̃ i( ui( k)) + d( k)),where ̃f + 1i( u i+ 1( k )) = f i( u i+1( k )) −f ˆ ( u ( k))represents the estimation error of the outgoing traffic.i i+1THEOREM 9.3.1 (IDEAL CASE)Consider the desired buffer length, q id , to be finite, and the disturbancebound, d M , to be equal to zero. Let the virtual-source rate for Equation 9.2 begiven by Equation 9.3. Then, the buffer occupancy feedback system is globally2asymptotically stable provided 0< k bvmax < 1.PROOF Let us consider the Lyapunov function candidate J = [ ebi( k) ]2.Then, the first difference is∆J = [ e ( k+ )] −[ e ( k)]bi1 2 bi2(9.6)Substituting error at time k + 1 from Equation 9.4 in Equation 9.8 yields( ) ≤−( − )∆J = k −1 [ e ( k)] 1 k e ( k)2 bv bi 2 2 2bvmaxbi(9.7)The first difference of the Lyapunov function candidate is negative forany time instance k . Hence, the closed-loop system is globally asymptoticallystable.