TI-65 Programs Part III: Impedance and Phase Angle of a Series RLC Circuit, 2 x 2 Linear System Solution, Prime Factorization (from TI-65 Manual)
TI-65 Programs Part III: Impedance and Phase Angle of a Series RLC Circuit, 2 x 2 Linear System Solution, Prime Factorization (from TI-65 Manual)
This is the third and final part of programs I will post today, this Fourth of July.
Click here for Part I: Digital Root, Complex Number Multiplication, Dew Point
Click here for Part II: Reynolds Number/Hydraulic Diameter, Escape Velocity, Speed of Sound/Resonant Frequencies in an Open Pipe
TI-65 Impedance and Phase Angle of a Series RLC Circuit
Formulas:
Impedance: Z = √(R^2 + (XL –XC)^2)
Phase Angle: Φ = atan ((XL – XC)/R)
Where:
R = resistance of the resistor in ohms (Ω)
L = inductance of the inductor in Henrys (H)
C = capacitance of the capacitor in Farads (F)
f = resonance frequency in Hertz (Hz)
XL = 2*π*f*L
XC = 1/(2*π*f*C)
Program:
CODE | STEP | KEY | COMMENT |
38 | 00 | * | Start with f |
2 | 01 | 2 | |
38 | 02 | * | |
2nd 17 | 03 | π | |
39 | 04 | = | Calculate 2*π*f |
12.0 | 05 | STO 0 | |
12.1 | 06 | STO 1 | |
38 | 07 | * | |
51 | 08 | R/S | Prompt for L |
39 | 09 | = | |
12.0 | 10 | STO 0 | Calculate XL |
13.1 | 11 | RCL 1 | |
38 | 12 | * | |
51 | 13 | R/S | Prompt for C |
39 | 14 | = | |
34 | 15 | 1/x | |
12.1 | 16 | STO 1 | Calculate XC |
51 | 17 | R/S | Prompt for R |
12.2 | 18 | STO 2 | |
2nd 33 | 19 | x^2 | |
59 | 20 | + | |
16 | 21 | ( | |
13.0 | 22 | RCL 0 | |
49 | 23 | - | |
13.1 | 24 | RCL 1 | |
17 | 25 | ) | |
2nd 33 | 26 | x^2 | |
39 | 27 | = | |
33 | 28 | √ | |
-2nd 16 | 29 | INV 2nd ENG | Remove ENG Notation |
51 | 30 | R/S | Display Z |
16 | 31 | ( | |
13.0 | 32 | RCL 0 | |
49 | 33 | - | |
13.1 | 34 | RCL 1 | |
17 | 35 | ) | |
28 | 36 | ÷ | |
13.2 | 37 | RCL 2 | |
39 | 38 | = | |
-24 | 39 | INV TAN | arctangent |
51 | 40 | R/S | Display Φ |
Input: f [RST] [R/S], L [R/S], C [R/S], R [R/S]
Result: Z [R/S] Φ
Test: f = 60 Hz, L = 0.25 H, C = 16 * 10^-6 F, R = 150 Ω
Result (in degrees mode): Z ≈ 166.18600 Ω, Φ ≈ -25.49760°
Source: Browne Ph. D, Michael. “Schaum’s Outlines: Physics for Engineering and Science” 2nd Ed. McGraw Hill: New York, 2010
TI-65 2 x 2 Linear System Solution
Let M = [ [a, b], [c, d] ], S = [ [ f ], [ g ] ]
Determinant: E = a*d – b*c
If E ≠ 0, the solutions to the system Mx = S:
x1 = d/E * f – b/E * g
x2 = -c/E * f + a/E * g
Memory Registers:
R0 = a
R1 = b
R2 = c
R3 = d
R4 = f
R5 = g
Hence [ [R0, R1], [R2, R3] ] * [ [x1], [x2] ] = [ [R4], [R5] ]. The determinant is stored in R6. Since so many storage registers are used, and storage registers eat up programming memory, the program will need to be short.
Program:
CODE | STEP | KEY | COMMENT |
13.0 | 00 | RCL 0 | Calculate det(M) |
38 | 01 | * | |
13.3 | 02 | RCL 3 | |
49 | 03 | - | |
13.1 | 04 | RCL 1 | |
38 | 05 | * | |
13.2 | 06 | RCL 2 | |
39 | 07 | = | |
12.6 | 08 | STO 6 | |
13.3 | 09 | RCL 3 | Calculate x1 |
38 | 10 | * | |
13.4 | 11 | RCL 4 | |
49 | 12 | - | |
13.1 | 13 | RCL 1 | |
38 | 14 | * | |
13.5 | 15 | RCL 5 | |
39 | 16 | = | |
28 | 17 | ÷ | |
13.6 | 18 | RCL 6 | |
39 | 19 | = | |
51 | 20 | R/S | Display x1 |
13.0 | 21 | RCL 0 | Calculate x2 |
38 | 22 | * | |
13.5 | 23 | RCL 5 | |
49 | 24 | - | |
13.2 | 25 | RCL 2 | |
38 | 26 | * | |
13.4 | 27 | RCL 4 | |
39 | 28 | = | |
28 | 29 | ÷ | |
13.6 | 30 | RCL 6 | |
39 | 31 | = | |
51 | 32 | R/S | Display x2 |
Input:
Store values:
a [STO] 0, b [STO] 1, c [STO] 2, d [STO] 3; f [STO] 4, g [STO] 5
Press [RST] [R/S]
If det(M) ≠ 0, x1 will be calculated. Press [R/S] to get x2.
Press [RCL] 6 to get the determinant of M.
Test: Solve
2*x1 + 3*x2 = 3.45
-6*x1 + x2 = 4.26
R0 = 2, R1 = 3, R2 = -6, R3 = 1, R4 = 3.45, R5 = 4.26
Results: x1 = -0.4665, x2 = 1.461. Determinant = 20 (stored in R6)
TI-65 Prime Factorization
This prime factorization comes straight from the Texas Instruments TI-65 Manual.
Program:
CODE | STEP | KEY | COMMENT |
12.1 | 00 | STO 1 | Store n in R1 |
0 | 01 | 0 | |
12.0 | 02 | STO 0 | Store 0 for comparisons |
3 | 03 | 3 | |
12.2 | 04 | STO 2 | Trail factor of 3 |
2nd 53.0 | 05 | LBL 0 | Test 2 as a factor |
13.1 | 06 | RCL 1 | |
28 | 07 | ÷ | |
2 | 08 | 2 | |
39 | 09 | = | |
2nd 28 | 10 | FRAC | Is frac(R1/2)≠0? |
-3rd 43 | 11 | INV x=m | x≠m |
0 | 12 | 0 | R1≠R0? |
2nd 54.1 | 13 | GTO 1 | Go to odd factors |
2 | 14 | 2 | |
12.28 | 15 | STO÷ | |
1 | 16 | 1 | STO÷ 1 |
51 | 17 | R/S | Display 2 if it is a factor |
2nd 54.0 | 18 | GTO 0 | GTO 0, test 2 again |
2nd 53.1 | 19 | LBL 1 | Odd factors loop begins here |
13.1 | 20 | RCL 1 | |
-3rd 42 | 21 | INV x<m | x≥m |
2 | 22 | 2 | Is R1≥R2? |
2nd 54.2 | 23 | GTO 2 | All factors found? No: GTO LBL 2 |
13.1 | 24 | RCL 1 | If complete, display 1 |
51 | 25 | R/S | (program execution ends here) |
2nd 54.1 | 26 | GTO 1 | |
2nd 53.2 | 27 | LBL 2 | Label 2 starts here |
13.1 | 28 | RCL 1 | |
28 | 29 | ÷ | |
13.2 | 30 | RCL 2 | |
39 | 31 | = | |
2nd 28 | 32 | FRAC | Is frac(R1/R2)≠0? |
-3rd 43 | 33 | INV 3rd x=m | x≠m |
0 | 34 | 0 | |
2nd 54.3 | 35 | GTO 3 | |
13.2 | 36 | RCL 2 | Display odd factor |
51 | 37 | R/S | |
12.28 | 38 | STO÷ | |
1 | 39 | 1 | STO÷ 1 |
2nd 54.1 | 40 | GTO 1 | |
2nd 53.3 | 41 | LBL 3 | Test next odd factor |
2 | 42 | 2 | |
12.59 | 43 | STO+ | |
2 | 44 | 2 | STO+ 2 |
2nd 54.1 | 45 | GTO 1 | |
Input: Enter n, press [RST] [R/S]. Each prime factor is displayed, keep on pressing [R/S] until you get 1 displayed.
Test 1: Factorize 102
Input: 102 [RST] [R/S]
Result: 2, press [R/S]
Result: 3, press [R/S]
Result: 17, press [R/S]
Result: 1
Final result: 102 = 2 * 3 * 17
Test 2: Factorize 168
Input: 168 [RST] [R/S]
Repeated presses of [R/S] gives: 2, 2, 2, 3, 7, 1
Final result: 168 = 2 * 2 * 2 * 3 * 7 = 2^3 * 3 * 7
Resource: Texas Instruments. “Texas Instruments Professional TI-65 Guidebook” 1986
This blog is property of Edward Shore, 2016.
TI-65 Programs Part III: Impedance and Phase Angle of a Series RLC Circuit, 2 x 2 Linear System Solution, Prime Factorization (from TI-65 Manual)
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