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# Simultaneous Two Way Data Transmission Via Coaxial Line

IP.com Disclosure Number: IPCOM000048373D
Original Publication Date: 1982-Jan-01
Included in the Prior Art Database: 2005-Feb-08
Document File: 2 page(s) / 33K

IBM

## Related People

Krumrein, W: AUTHOR

## Abstract

Three examples are shown of a combined transmitter/ receiver resistance network connected to a coaxial transmission line. Transmission is based on injecting a current into the bridge-like resistance network. The receiving voltage is taken from one diagonal of the bridge and is not affected by the injected transmission current. The network loads the coaxial line with its characteristic resistance Z.

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Simultaneous Two Way Data Transmission Via Coaxial Line

Three examples are shown of a combined transmitter/ receiver resistance network connected to a coaxial transmission line.

Transmission is based on injecting a current into the bridge-like resistance network. The receiving voltage is taken from one diagonal of the bridge and is not affected by the injected transmission current. The network loads the coaxial line with its characteristic resistance Z.

In Fig. 1, two polarities are used for transmitting binary digits. At point B, either a negative current I(N) or a positive current I(P) is injected into the resistance bridge. The receiving voltage is taken from points A and C and fed to differential amplifier 1. A signal representing the received bit can be taken from terminal 2. The resistance values should be as follows: R1=2/3 Z; R2=2/3 Z; R3=2/3 Z; R4=2 Z.

The network of Fig. 1 is advantageous in that currents I(N) and I(P) need not be of exactly the same magnitude. Major attenuation by transmission line 3 can be tolerated, as the receiving voltage U(R) is either negative or positive and thus easily detectable.

Fig. 2 shows a different network. Current source I(S) may be of a single polarity or be chosen as in Fig. 1. As in Fig. 1, a differential amplifier is connected between points A and C for reception. The resistance values should be: R1=Z; R2=R3; R5=Z/2/-R2/2/ divided by 2 R2.

As an optimum value, R5 may be chosen to equal Z, so that R2=(square root of...