Comments and trivia quiz about the new AFH for the 15c

[J-F Garnier]

The 15c CE is now available, and by a mere coincidence the corresponding Advanced Function Handbook (AFH) appeared in the hpcalc literature archive.

This revised manual is very nicely done and quite close to the original from 1982, globally much better than the Limited Edition version (2012). Thanks to the editors!

One section caught my attention. An example of "Solving a Large System of Complex Equations" in the AFH page 128 is still solved within the original 64-register HP-15C limitations:

Quote:Because this system is too large to solve [on the original HP-15C] using the standard method for a
system of complex equations, this alternate method (described in the owner’s handbook) is used [...].

Please refer to the AFH, page 128, for a description of the problem.

I thought it would be a good exercice to solve it directly, since the HP 15c CE now can do it.
Note that the AFH skips a big part of the process - entering the matrix coefficients - with an ellipsis:
100 STO A                          Stores Re(a11).
150 ENTER 800 ENTER 3 / - STO A    Stores Im(a11).
...                                [ the reader is supposed to fill the rest ! ]
150 ENTER 8 ENTER 3 / - STO A      Stores Im(a44).

Since this is a big part of the job, and a major source of errors, I'm detailing the full process below:

Code:

9 DIM (i)   ; allocate registers R0..9 only, to make room for the matrices
MATRIX 0 ; clear matrices
; pre-calculates imaginary parts in registers R2-R4
15 EEX 3 STO 0  ; w in R0 (temporary)
EEX CHS 2 * STO 2 ; w.L in R2
RCL 0 25 EEX CHS 8 * 1/x STO 3 ; 1/(w.C1) in R3
RCL 0 25 EEX CHS 6 * 1/x STO 4 ; 1/(w.C2) in R4
; create and fill complex matrix A
4 ENTER 8 DIM A 
MATRIX 1 ; init indexes
USER ; user mode on, up to the end
; row 1
100 STO A  ; Re(a11)
RCL 2 RCL 3 - STO A ; Im(a11)
0 STO A ; Re(a12)
RCL 3 STO A ; Im(a12)
0 STO A STO A ; a13=0
STO A STO A ; a14=0
; row 2
STO A ; Re(a21)=0
RDN STO A ; Im(a21)=Im(a12)
EEX 6 STO A ; Re(a22)
RCL 2 RCL 3 - STO A ; Im(a22)
EEX 6 ENTER CHS STO A ; Re(a23)
0 STO A ; Im(a23)
STO A STO A ; a24=0
; row 3
STO A STO A ; a31=0
RDN STO A ; Re(a32)=Re(a23)
0 STO A ; Im(a32)
EEX 6 STO A ; Re(a33)
RCL 2 RCL 4 - STO A ; Im(a33)
0 STO A ; Re(a34)
RCL 2 CHS STO A ; Im(a34)
; row 4
0 STO A STO A ; a41=0
STO A STO A ; a42=0
STO A ; Re(a43)
RDN STO A ; Im(a43)=Im(a34)
EEX 5 STO A ; Re(a44)
RCL 2 RCL 4 - STO A ; Im(a44)
; create and fill complex matrix B
8 ENTER 1 DIM B
MATRIX 1 ; init indexes
10 STO B  ; Re(b11)
;
; core of the complex system solving:
RCL MATRIX B
RCL MATRIX A
Py,x ; A in partitioned form
MATRIX 2 ; A~ form, 8x8 matrix
RESULT C ; result in matrix C
/ ; "divide" matrix B by A
Cy,x ; back to complex form
;
; display the results in C
MATRIX 1 ; init indexes
SCI 4
RCL C ; Re(I1)= 1.9552 -04
RCL C ; Im(I1)= 4.0964 -03
RCL C ; Re(I2)= -1.4489 -03
RCL C ; Im(I2)= -3.5633 -02
RCL C ; Re(I3)= -1.4541 -03
RCL C ; Im(I3)= -3.5633 -02
RCL C ; Re(I4)= 5.3446 -05
RCL C ; Im(I4)= -2.2599 -06
; compute the output voltage and phase
x<>y ->P
EEX 5 * ; amplitude= 5.3494 00
x<>y ; phase= -2.4213 00  (assuming DEG mode)

The results (amplitude= 5.3494, phase= -2.4213) are in line with the AFH results obtained with the alternate method, with a minor difference in the output phase.
This confirms (if it was still needed...) that the extended 15c works fine with 3 matrices spread over a total of 64+8+8=80 registers, with various matrix conversions and operations.


Now, a little quiz, for the fun:

- If you look at page 17 of the new AFH, you will see a nice picture in the old HP style.
This picture was not in the original AFH, can you find where it comes from? (AFH editors, please don't spoil the game too quickly...).
Hint: with a little thinking, no need to open all your old manuals. I found it in the 2nd manual I checked.

- The new AFH is very well revised for clarity, and very pleasant to read. I would love to have a paper copy.
However (sorry AFH editors), there is (at least) one minor but non-trivial typo, and this typo is already present in the original AFH !
The typo makes a math expression wrong.
Can you find it? (AFH editors, now you are welcome!)

J-F

Edit: the new AFH (ed.3) has now been corrected for the typo, but you can still try to find it in the original.