Thesis - Department of Electronic & Computer Engineering

IFor example, let I 2= … = I Nand 1– I-------------- 2= δ 1, v 1( 0) = … = v N( 0) = 0 ,then we haveI 1τX ˙N() t = [ a – ( N – 1)b– 1]X N() t + I 1[ 1 + ( N – 1) ( 1 – δ 1)]τY ˙ N() t = [ a + b – 1 ]Y N() t + I 1δ 1.Further suppose a+ b = 1, it becomesτX ˙ N() t=– NbX N() t + I 1[ 1 + ( N – 1) ( 1 – δ 1)]τY ˙N() t = I 1δ 1tDefine y - and to be the solution toτ= y s δ 1 y=( N – 1)1 – -----------------δN 1-------------------------------- ( 1 – e .b– Nby )Hence the response timet s=y sτFor example, if we choosea = b = 0.5, then we get the WTA model:τv˙ i()t = – vi () t + I i + 0.5max( v i ()0 t , )–0.5 max( v j ()0 t , ).∑j≠iI 1Further fix the largest input I 1 = 1.0 and τ = 1ms. For δ 1 = 0.1 and N = 2 , theo-1.0retically, v 1 () t = ----------- = --------------- = 2.0 at steady state and t . These can1 – a 1 – 0.5s = 19ms14