Sample Homework Question no.3

Course: Electrical and Computer Engineering
Level: Undergraduate

Question:

Note:Could you please answer these questions to the best of you knowledge and in your own words. There is no attachment.

    a) A signal of strength 1V is to sent along 4Km of copper cable. If the signal strength has fallen to 0.5 V at a distance of 1Km along the cable, what will it be at the destination? Show your work.

    b) A CAN network uses an arbitration process known as 'wired-AND'. Briefly explain how this process avoids unwanted conflicts in the network.

    c) In a CAN network Device X and Device Y start to transmit at the same time. Device X has identifier 011 1011 1100 and Device Y has identifier 011 0111 1000. Find the state of the network as the first few bits are sent, and hence determine which device obtains priority.

Answer:

1.

2. A CAN network uses an arbitration process known as 'wired-AND ', which means that all the modules connected to the bus are connected in a wired-and structure. The wired-and connection means that when one of the module send a zero then the status of the bus is equal to zero ( Bus=M1 AND M2AND ….AND Mn). With this kind of connection when a module Mi see the status 0 of the bus Mi understand that he must stop the transmission and do it, then only the Module which transmitted the zero will continue to transmit and take the priority which avoids unwanted conflicts.

3. The priority of a CAN message is determined by the identifier. The identifier contains 11 bits D10 D9 …. D1 D0. The test of which module take the priority is done bit by bit to avoid any error. The Module who sends the lowest identifier has the highest priority. This is due to the fact that the transmission of identifier begins by the bit D10 to the bit D0 and is due to the fact that the modules are connected with wired-AND connection.

Device X has identifier 011 1011 1100 and Device Y has identifier 011 0111 1000.

Lets say the identifier X_id=X10 X9 ….X0=011 1011 1100 and Y_id=Y10 Y9 ….Y1.

X10=Y10=0 no decision of priority yet, X9=Y9=1 (no priority decision), X8=Y8=1 (no priority decision), X7=1, Y7=0 ( then Y7 will put the status of bus the dominant state =0) then the device Y will continue to send the message and the device X will stop the transmission (We see that. X_id >Y_id ).

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