Impedance of a Ladder Network

Description

Origin: HP-34C Student Engineering Applications, August 1979, p. 11

This program computes the input impedance of an arbitrary ladder network. Elements are added one at a time starting from the right. This first element must be in parallel
Suppose we have a network whose input admittance is Yin. Adding a shunt R, L or C, the input admittance becomes:
       Yin + (1/Rp + j0)
Ynew = Yin + (0 - j/(ωLp))
       Yin + (0 + jωCp)

Adding a series R, L or C, we have:
       (1/Yin + (Rs + j0))-1
Ynew = (1/Yin + (0 + jωLs))-1
       (1/Yin + (0 - j/(ωCs))-1

This program converts this admittance to an impedance for display.
Note: An erroneous entry may be corrected by entering the negative of the incorrect value.

Example:
f = 4 Mhz

Impedance of a Ladder Network

4 EEX 6 A 50 GSB 1 → 50.0000 (|Zin|, ohms)
2400 EEX CHS 12 GSB 3 → 15.7362 (|Zin|, ohms)
x↔y → -71.6559 (∠Zin, deg)

2.56 EEX CHS 6 B GSB 2 → 49.6509 (|Zin|, ohms)
x↔y → 84.2754 (∠Zin, deg)

796 EEX CHS 12 GSB 3 → 497.6942 (|Zin|, ohms)
x↔y → 0.9840 (∠Zin, deg)

Program Resources

Labels

Name Description Name Description
 A Store frequency  4 # - internal use -
 B Next element is added in series  5 # - internal use -
 1 Add resistance R [ohms]  6 # - internal use -
 2 Add impedance L [henrys]  9 # - internal use -
 3 Add capacity C [farads]

Storage Registers

Name Description
 0 ω
 1 Re
 2 Im

Flags

Number Description
0

Program

Line Display Key Sequence Line Display Key Sequence Line Display Key Sequence
000 022 42,21, 5 f LBL 5 044 40 +
001 42,21,11 f LBL A 023 45 0 RCL 0 045 33 R⬇
002 42 34 f REG 024 20 × 046 40 +
003 2 2 025 15 1/x 047 43 33 g R⬆
004 20 × 026 16 CHS 048 43, 6, 0 g F? 0
005 43 26 g π 027 0 0 049 32 4 GSB 4
006 20 × 028 34 x↔y 050 44 1 STO 1
007 44 0 STO 0 029 22 9 GTO 9 051 34 x↔y
008 43, 5, 0 g CF 0 030 42,21, 3 f LBL 3 052 44 2 STO 2
009 43 32 g RTN 031 43, 6, 0 g F? 0 053 34 x↔y
010 42,21,12 f LBL B 032 22 5 GTO 5 054 32 4 GSB 4
011 43, 4, 0 g SF 0 033 42,21, 6 f LBL 6 055 43, 5, 0 g CF 0
012 43 32 g RTN 034 45 0 RCL 0 056 43 1 g →P
013 42,21, 1 f LBL 1 035 20 × 057 43 32 g RTN
014 15 1/x 036 0 0 058 42,21, 4 f LBL 4
015 43, 6, 0 g F? 0 037 34 x↔y 059 43 1 g →P
016 15 1/x 038 42,21, 9 f LBL 9 060 15 1/x
017 0 0 039 45 2 RCL 2 061 34 x↔y
018 22 9 GTO 9 040 45 1 RCL 1 062 16 CHS
019 42,21, 2 f LBL 2 041 43, 6, 0 g F? 0 063 34 x↔y
020 43, 6, 0 g F? 0 042 32 4 GSB 4 064 42 1 f → R
021 22 6 GTO 6 043 43 33 g R⬆ 065 43 32 g RTN