DTo send/read memory contents

When sending or reading memory contents, additional codes must be added to append the memory channel as follows.

Additional code: 0000–0101 (0100=P1, 0101=P2)

DBand stacking register

To send or read the desired band stacking register’s contents, combined codes of the frequency band and register codes as follows are used.

For example, when sending/reading the oldest contents in the 21 MHz band, the code “0703” is used.

• Frequency band code

Code

Frequency band

Frequency range (unit: MHz)

01

1.8

1.800000–

1.999999

02

3.5

3.400000–

4.099999

03

7

6.900000– 7.499999

04

10

9.900000–10.499999

05

14

13.900000–14.499999

06

18

17.900000–18.499999

07

21

20.900000–21.499999

08

24

24.400000–25.099999

09

28

28.000000–29.999999

10

50

50.000000–54.000000

12

GENE

Other than above

 

 

 

 

• Register code

Code

Registered number

01

1 (latest)

02

2

03

3 (oldest)

 

 

DCodes for memory keyer contents

To send or read the desired memory keyer contents, the channel and character codes as follows are used.

• Channel code

Code

Channel number

01

M1

02

M2

03

M3

 

 

04

M4

 

 

CONTROL COMMAND 14

• Character’s code

Character

ASCII code

Description

0–9

30–39

Numerals

 

 

 

 

 

A–Z

41–5A

Alphabetical characters

 

 

 

 

 

space

20

Word space

 

 

 

 

 

/

2F

Symbol

 

 

 

 

 

?

3F

Symbol

 

 

 

 

 

,

2C

Symbol

 

 

 

 

 

.

2E

Symbol

 

 

 

 

 

^

5E

e.g., to send

 

, enter ^4254

BT

 

 

 

 

 

2A

Inserts contest number (can be

 

 

used for 1 channel only)

 

 

 

 

 

DCodes for memory name, opening message and CLOCK2 name contents

To send or read the desired memory name settings, the character codes, instructed codes for memory keyer contents as above, and follows are used.

Character’s code— Alphabetical characters

Character

ASCII code

Character

ASCII code

a–z

61–7A

 

 

 

 

Character’s code— Symbols

Character

ASCII code

Character

ASCII code

!

21

#

23

 

 

 

 

$

24

%

25

 

 

 

 

&

26

¥

5C

 

 

 

 

?

3F

22

 

 

 

 

27

`

60

 

 

 

 

+

2B

2D

 

 

 

 

:

3A

;

3B

 

 

 

 

=

3D

<

3C

 

 

 

 

>

3E

(

28

 

 

 

 

)

29

[

5B

 

 

 

 

]

5D

{

7B

 

 

 

 

}

7D

7C

_

5F

7E

 

 

 

 

 

@

40

 

 

 

 

 

 

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