Re: simple filter question

[Doug Stewart]

On Fri, Jul 12, 2013 at 1:02 PM, ionone <address@hidden> wrote:

Hi

i can't understand what computations does Octave with this simple filter :
b = [1 2 3 4]
aa = [0.1 0.2]
A = filter(1, aa, b)

the result in Octave is : A = 10  0  30 -20

i thought Octave did this :

A(1) = 1/0.1 (this number is correct)
A(2) = 2/0.1 + 1/0.2 = 25 (differs from 0)
A(3) = 3/0.1 + 2/0.2 = 40 (differs from 30)
A(4) = 4/0.1 + 3/0.2 = 55 (differs from -20)
...
please help me understand what computations does Octave in this case
I've turned the equations in every possible way without getting it :/

many thanks.

Jeff



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y(now)=10*x(now)-2*y(1timeago)

y(1)=10*(1)-2*(0) =10

y(2)=10*(2)-2*(10)=0

y(3)=10*(3)-2*(0)=30

y(4)=10*(4)-2*(30)=-20

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DAS

https://linuxcounter.net/user/206392.html