comparing diagram to formula of INL DNL

[yefj]

Hello , by definition DNK(K)=(V_out(K+1)-V_out(K)-Vlsb)/Vlsb
So for K=2
DNL(K)=(V_out(K+1)-V_out(K)-Vlsb)/Vlsb=((3/8)-(3/8)-(1/8))/(1/8)=-1 [LSB]
INK(K)=(V_out(K)-K*Vlsb)/Vlsb=(3/8)-2*(1/8))/(1/8)=1

In the place Where DNL is +1 i get -1 for some reason.I dont know how to interpet this large jump form DNL INL.
Howdo i methamichly derive the INL DNL of these K=2 case?
Thanks.

 

[yefj]

Hello , What is the calculation of the INL that makes -1 results?
Thanks.

 

[sutapanaki]

For the 1st case you calculate the DNL for code 2 (010), that is the step from code 1 do code 2. This step is 3/8-1/8=2/8 and then you look at the relative error of the step with respect to 1LSB=1/8. So, DNL(2)=(3/8-1/8-1/8)/(1/8)=1. INL is the difference between the ideal analog voltage at code 2 vs. the real one, all this with respect to 1LSB. So, real voltage is 3/8, ideal voltage is 2/8, difference is 1/8. INL(2)=(2/8-1/8)/(1/8)=1LSB

For the second case, code 5, result is not INL(5)=-1. INL there is 0. Both the actual voltage and the ideal voltage are at 5/8, so INL(5)=0.
Another way to see this is to calculate as INL(5)=DNL(1)+DNL(2)+DNL(3)+DNL(4)+DNL(5)=0+1+0+0-1=0