2022-10-22 01:54:47 -04:00
|
|
|
|
The continuous form of a PI loop is
|
|
|
|
|
|
|
|
|
|
|
|
A(t) = P e(t) + I ∫e(t')dt'
|
|
|
|
|
|
|
|
|
|
|
|
where e(t) is the error (setpoint - measured), and the integral goes
|
|
|
|
|
|
from 0 to the current time 't'.
|
|
|
|
|
|
|
|
|
|
|
|
In digital systems the integral must be approximated. The normal way
|
|
|
|
|
|
of doing this is a first-order approximation of the derivative of
|
|
|
|
|
|
A(t).
|
|
|
|
|
|
|
|
|
|
|
|
dA(t)/dt = P de(t)/dt + Ie(t)
|
|
|
|
|
|
A(t_n) - A(t_{n-1}) ≅ P (e(t_n) - e(t_{n-1})) + Ie(t_n)Δt
|
|
|
|
|
|
A(t_n) ≅ A(t_{n-1}) + e(t_n)(P + IΔt) - Pe(t_{n-1})
|
|
|
|
|
|
|
|
|
|
|
|
Using α = P + IΔt, and denoting A(t_{n-1}) as A_p,
|
|
|
|
|
|
|
|
|
|
|
|
A ≅ A_p + αe - Pe_p.
|
|
|
|
|
|
|
|
|
|
|
|
The formula above is what this module implements. This way, the
|
|
|
|
|
|
controller only has to store two values between each run of the loop:
|
|
|
|
|
|
the previous error and the previous output. This also reduces the
|
|
|
|
|
|
amount of (redundant) computations the loop must execute each
|
|
|
|
|
|
iteration.
|
|
|
|
|
|
|
|
|
|
|
|
Calculating α requires knowing the precise timing of each control loop
|
|
|
|
|
|
cycle, which in turn requires knowing the ADC and DAC timings. This
|
|
|
|
|
|
is done outside the Verilog code. and can be calculated from
|
|
|
|
|
|
simulating one iteration of the control loop.
|
|
|
|
|
|
|
2022-10-22 01:55:56 -04:00
|
|
|
|
# Fixed Point Integers
|
|
|
|
|
|
|
2022-10-22 01:54:47 -04:00
|
|
|
|
A regular number is stored in decimal: 123056.
|
|
|
|
|
|
This is equal to
|
|
|
|
|
|
|
|
|
|
|
|
6*10^0 + 5*10^1 + 0*10^2 + 3*10^3 + 2*10^4 + 1*10^5.
|
|
|
|
|
|
|
|
|
|
|
|
A whole binary number is only ones and zeros: 1101, and is equal to
|
|
|
|
|
|
|
|
|
|
|
|
1*2^0 + 0*2^1 + 1*2^2 + 1*2^3.
|
|
|
|
|
|
|
|
|
|
|
|
Fixed-point integers shift the exponent of each number by a fixed
|
|
|
|
|
|
amount. For instance, 123.056 is
|
|
|
|
|
|
|
|
|
|
|
|
6*10^-3 + 5*10^-2 + 0*10^-1 + 3*10^0 + 2*10^1 + 1*10^2.
|
|
|
|
|
|
|
|
|
|
|
|
Similarly, the fixed point binary integer 11.01 is
|
|
|
|
|
|
|
|
|
|
|
|
1*2^-2 + 0*2^-1 + 1*2^0 + 1*2^1.
|
|
|
|
|
|
|
|
|
|
|
|
To a computer, a whole binary number and a fixed point binary number
|
|
|
|
|
|
are stored in exactly the same way: no decimal point is stored. It is
|
|
|
|
|
|
only the interpretation of the data that changes.
|
|
|
|
|
|
|
|
|
|
|
|
Fixed point numbers are denoted WHOLE.FRAC or [WHOLE].[FRAC], where
|
|
|
|
|
|
WHOLE is the amount of whole number bits (including sign) and FRAC is
|
|
|
|
|
|
the amount of fractional bits (2^-1, 2^-2, etc.).
|
|
|
|
|
|
|
|
|
|
|
|
The rules for how many digits the output has given an input is the
|
|
|
|
|
|
same for fixed point binary and regular decimals.
|
|
|
|
|
|
|
|
|
|
|
|
Addition: W1.F1 + W2.F2 = [max(W1,W2)+1].[max(F1,F2)]
|
|
|
|
|
|
|
|
|
|
|
|
Multiplication: W1.F1W2.F2 = [W1+W2].[F1+F2]
|
2022-11-11 21:57:58 -05:00
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
When multiplying two fixed point integers, where the decimal points
|
|
|
|
|
|
do not correspond to the same points, then:
|
|
|
|
|
|
|
|
|
|
|
|
* the output has the same number of bits as a normal addition/multiplication
|
|
|
|
|
|
* for multiplication, the LSB is interpreted as position `m+n`, where
|
|
|
|
|
|
`m` is the interpretation of the LSB of the first integer and `n` as
|
|
|
|
|
|
the LSB of the second.
|
