By Gary Goodman from HP document collection. Dated 24/3/85
The required count is calculated from Φ2
and the audio frequency required.
Count (decimal)
{ Φ2 / ( 2 x F ) } - 2
for 1000Hz and a 1MHz Φ2 :
Count = { 1,000,000 / ( 2 X 1000 ) } - 2 = 498
498d → $01F2
Run as XEQALPHATI
ALPHA
Prompt F
for required audio tone frequency.
Enter freq
R/S
Result is the 2-byte value for T1CL & T1CH
e.g.
XEQALPHATI
ALPHA
1000
R/S
Result:
$001F2
The Counter should be set to
T1CH = $01 T1CL = $F2
Entered as follows (make sure you're in DECS mode in Ladybug while entering the program - otherwise the 1000000
won't be interpreted correctly). When viewing/editing the program later it will look different!
LBL ¬TI DECS WSIZE 20 ¬ F PROMPT STI 01 1000000 ENTERI LDI 01 2 x / 2 - HEXS RTN
XEQALPHABFILALPHA
MHZ ?
5 R/S
N
Order/poles = N
Positive for HPF
Negative for LPF
5 pole LPF = -5
5CHSR/S
COEFFICIENTS 0.618
(at this point you can RCL 01 etc. up to RCL 05 to see the calculated Butterworth Coefficients)
R/S
LPF 393.5 -12
The calculated component values are now in Reg 01 - Reg 05
XEQALPHABFILALPHA
MHZ ?
1 R/S
N
Order/poles = N
Positive for HPF
Negative for LPF
7 pole HPF = 7
7R/S
COEFFICIENTS 2.247
(at this point you can RCL 01 etc. up to RCL 07 to see the calculated Butterworth Coefficients)
R/S
HPF 17.88 -06
The calculated component values are now in Reg 01 - Reg 07
Page created Thu May 26 18:35:00 2022 by John Pumford-Green
Page last updated: 06/03/25 06:49 GMT
— John Pumford-Green 24/04/23 16:07