Could any student of electrical or control engineering give me

>>9065006
>For me it should be A = (N + 1) / 2 * (u (n) + u (n-1)).
i agree that there should be an effect of the adder. but i disagree with your formula, though i have to admit that i never bothered getting used to the block diagrams of filters.

the path leading to the first adder always returns the last value that leaves the adder back to it.

lets assume we start the system. then the output of the adder a(n) is zero for n < 0.
let u(n), the entry sequence, be equal to 1 e.g. u(0) = u(1) = ... = 1.

then we have:
a(0) = u(0) + a(-1) = u(0) = 1
a(1) = u(1) + a(0) = 2;
a(2) = u(2) + a(1) = 3;
and so on.

as you can see a(n) is monotonically incrasing.
but going by your formula this cannot occur as u(n) + u(n-1) is always either 1 or 2 for my entry sequence.

>>9065741
addendum:
now i recall that the block diagrams are rather unintuitive as a recursion such the one at the first adder usually just means that the entry sequence is lead back with some delay (here a delay of 1). as this is closer to the solution presented in your image i guess thats how it is (even though its a super retarded way of depicting it).

tl;dr:
you are correct IMO

>>9065006
oh wait.
h(n) are the impulse responses and u(n) is the unit step i.e. u(n) = 1 for n >= 0.

i think my first interpretation is correct.

to calculate all h's(n) you need to set the input signal to x(n) = delta(n) i.e. 1 for n = 0 and zero else.

then the first adder creates the unit signal u(n).

so yes, the paper is correct.