How to realize linear function (filter?) with saturation?

Initial signal equals ones, then minus ones (it can switch many times):

u = [ones(15,1);-ones(5,1)];

If u==1 then output should be .1; .2... up to 1 and stay 1 (saturation). After u==-1 the output should be -.1; -.2... up to -1.

I tried filter:

Nb = 10;
b = ones(1,Nb)/Nb;
y = filter(b,1,u);
stem([u, y])

It's wrong:

How to make transform function (maybe filter)?

Answer 1

[Jan]

u = [ones(15,1);-ones(5,1)];
Ramp = 0.1:0.1:1;     % Set of values
v = SaturatedRamp(u, Ramp);

function v = SaturatedRamp(u, R)
% Running index, starting from 1 at each change:
k    = [false; diff(u(:)) ~= 0];
s    = [1; find(k)];
I    = ones(numel(u), 1);
I(k) = 1 - diff(s);
I    = cumsum(I);

R = R(:);
I = min(numel(R), I);  % Limit indices to length of ramp
v = R(I) .* u(:);      % Copy sign of original data
end

[Alexander]

Thank you very much, Jan!

(I looked through function

ischange(A,'linear','Threshold',THR);

how it plots the results.)