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Altera_Forum's avatar
Altera_Forum
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13 years ago

continuous averaging using VHDL

I have a question related to VHDL programming. I want to calculate the continuous average. My example code is: process (clk, reset) begin if (reset = '1') then state<=idle; out-v...
Altera_Forum's avatar
Altera_Forum
Icon for Honored Contributor rankHonored Contributor
13 years ago

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If n is a power of 2 you don't need a multiply - just some shiftsand two adds

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Yes, in that case we convert:

y(n) = a*y(n-1) + (1-a)*x(n) ... two multipliers, one adder

...to y(n) = a*y(n-1) + x(n) -a*x(n)

i.e. y(n) = a*(y(n-1) -x(n)) + x(n); ... one multiplier, one subtractor, one adder

then (a) can be power of 2 but we get limitations on cutoff point