Interview Questions.

Question 1. What Is Can?

Answer :

CAN is a multi-grasp broadcast serial bus trendy for connecting digital control unit (ECU).
Controller–area network (CAN or CAN-bus) is a automobile bus general designed to permit micro controllers a devices to speak with each other inside a automobile with out a bunch computer. 
CAN is a message-based totally protocol, designed specially for automotive programs but now extensively utilized in other regions together with industrial automation and scientific gadget.
The Controller Area Network (CAN) bus is a serial asynchronous bus used in instrumentation programs for industries such as cars.
Question 2. What Are The Can Frame Works?

Answer :

SOF – 1 Dominant
Arbitration Field – eleven bit Identifier, 1 bit RTR (or) eleven bit, 1SRR, 1IDE, 18 bit, 1RTR
Control Field – IDE, r0, four bits (DLC)
Data Field – (0-8) Bytes
CRC Field – 15 bits, Delimiter (1 bit recessive)
ACK Field – 1 bit, Delimiter (1 bit recessive)
EOF – 7 bits recessive
IFS – three bits recessive
Types of frames – Data, remote, Error body and Overload body
Types of mistakes – ACK blunders, Bit mistakes, Stuff error, Form mistakes, CRC mistakes
Error frame – zero-12 superposition flags, eight recessive (Delimiter)
Overload frame – 0-12 superposition flags, eight recessive (Delimiter)
CCNA Interview Questions
Question three. Why Can Is Having one hundred twenty Ohms At Each End?

Answer :

To limit the reflection reference, to reduce noise. To make certain that reflection does now not reason conversation failure, the transmission line must be terminated.

Question four. Why Can Is Message Oriented Protocol?

Answer :

CAN protocol is a message-primarily based protocol, now not an address based totally protocol. This method that messages are not transmitted from one node to every other node based totally on addresses. Embedded inside the CAN message itself is the concern and the contents of the facts being transmitted. All nodes inside the machine receive every message transmitted at the bus (and will renowned if the message become nicely acquired). It is as much as every node in the device to decide whether the message obtained need to be without delay discarded or saved to be processed. A single message may be destined for one specific node to receive, or many nodes based totally at the way the network and gadget are designed. For instance, an automobile airbag sensor can be connected thru CAN to a protection system router node best. This router node takes in other safety gadget facts and routes it to all different nodes at the safety system network. Then all the different nodes on the protection machine network can receive the cutting-edge airbag sensor statistics from the router on the same time, renowned if the message became obtained properly, and decide whether or not to make use of this information or discard it.

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Question five. Can Logic What It Follows?

Answer :

Wired AND common sense

Network Security Interview Questions
Question 6. What Is Can Arbitration?

Answer :

CAN Arbitration is nothing but the node trying to take manipulate on the CAN bus.

Question 7. How Can Will Follow The Arbitration?

Answer :

CSMA/CD + AMP (Arbitration on Message Priority)

Two bus nodes have were given a transmission request. The bus get admission to technique is CSMA/CD+AMP (Carrier Sense Multiple Access with Collision Detection and Arbitration on Message Priority). According to this algorithm both network nodes wait till the bus is unfastened (Carrier Sense). In that case the bus is loose both nodes transmit their dominant start bit (Multiple Access). Every bus node reads returned bit by bit from the bus at some point of the entire message and compares the transmitted price with the received fee. As long because the bits are equal from each transmitters nothing takes place. The first time there was a difference – in this situation the 7th little bit of the message – the arbitration manner takes area: Node A transmits a dominant stage, node B transmits a recessive degree. The recessive stage may be overwritten through the dominant degree. This is detected through node B because the transmitted value isn't identical to the received price (Collision Detection). At this factor of time node B has misplaced the arbitration, stops the transmission of any in addition bit straight away and switches to receive mode, due to the fact the message that has gained the arbitration should probable be processed by way of this node (Arbitration on Message Priority)

For instance, recall 3 CAN gadgets each trying to transmit messages:

Device 1 – deal with 433 (decimal or 00110110001 binary)
Device 2 – deal with 154 (00010011010)
Device 3 – cope with 187 (00010111011)
 Assuming all 3 see the bus is idle and begin transmitting at the same time, that is how the arbitration works out. All 3 gadgets will power the bus to a dominant state for the start-of-frame (SOF) and the two most sizeable bits of each message identifier. Each device will monitor the bus and decide achievement. When they write bit eight of the message ID, the tool writing message ID 433 will word that the bus is in the dominant kingdom while it changed into trying to let it be recessive, so it will expect a collision and give up for now. The closing devices will maintain writing bits till bit 5, then the device writing message ID 187 will word a collision and abort transmission. This leaves the tool writing message ID 154 closing. It will preserve writing bits at the bus until whole or an blunders is detected. Notice that this approach of arbitration will usually purpose the bottom numerical cost message ID to have precedence. This same method of bit-clever arbitration and prioritization applies to the 18-bit extension in the extended format as nicely.

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Question eight. What Is The Speed Of Can?

Answer :

40m @1Mbps and if the cable length increases will lower the speed, due to RLC at the cable.

Question 9. If Master Sends 764 And Slave Sends 744 Which Will Get The Arbitration?

Answer :

Starts from MSB, first nibble is equal, Master sends 7, slaves additionally sends 7 the message with extra dominant bits will benefit the arbitration, lowest the message identifier better the priority.

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Question 10. Standard Can And Extended Can Difference?

Answer :

Number of identifiers may be accommodated for preferred frame are 2power11.

Number of identifiers greater evaluate to base body, for prolonged body are 2power29.

IDE bit – 1 for extended body.

IDE bit –0 for Standard body.

Question eleven. What Is Bit Stuffing?

Answer :

CAN uses a Non-Return-to-Zero protocol, NRZ-5, with bit stuffing. The idea behind bit stuffing is to offer a assured part on the sign so the receiver can resynchronize with the transmitter earlier than minor clock discrepancies between the 2 nodes can motive a hassle. With NRZ-5 the transmitter transmits at most five consecutive bits with the equal price. After five bits with the same value (0 or one), the transmitter inserts a stuff bit with the other kingdom.

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Question 12. What Is The Use Of Bit Stuffing?

Answer :

Long NRZ messages cause troubles in receivers

Clock waft approach that if there aren't any edges, receivers lose tune of bits
Periodic edges allow receiver to resynchronize to sender clock
CCNA Interview Questions
Question thirteen. What Are The Functions Of Can Transceiver?

Answer :

The transceiver provides differential transmit capability to the bus and differential receive functionality to the CAN controller. Transceiver provides an advanced interface among the protocol controller and the bodily bus in a Controller Area Network (CAN) node.

Typically, every node in a CAN gadget have to have a device to transform the digital signals generated through a CAN controller to alerts appropriate for transmission over the bus cabling (differential output). It additionally offers a buffer among the CAN controller and the excessive-voltage spikes that may be generated at the CAN bus by way of outdoor sources (EMI, ESD, electric transients, and so on.).

The can transceiver is a device which detects the signal stages which are used at the CAN bus to the logical signal tiers identified via a microcontroller.

Question 14. Functionality Of Data Link Layer In Can?

Answer :

LLC (Logical Link Control) : Overload manipulate, notification, Message filtering and Recovery control capabilities.

MAC (Medium Access Control) : Encapsulation/ de-capsulation, error detection and manipulate, stuffing and de-stuffing and serialization/de-serialization.

Question 15. What Is Meant By Synchronization?

Answer :

Synchronization is timekeeping which calls for the coordination of events to operate a gadget in unison.

PRTG Interview Questions
Question sixteen. What Is Meant By Hard Synchronization And Soft Synchronization?

Answer :

Hard Synchronization to be achieved at each aspect from recessive-to-dominant side for the duration of Bus Idle. Additionally, Hard Synchronization is required for every received SOF bit. An SOF bit may be obtained each at some point of Bus Idle, and additionally all through Suspend Transmission and on the stop of Interframe Space. Any node disables Hard Synchronization if it samples an aspect from recessive to dominant or if it starts to send the dominant SOF bit.

Two styles of synchronization are supported:

Hard synchronizationis accomplished with a falling facet at the bus even as the bus is idle, that is interpreted as a Start of frame (SOF). It restarts the inner Bit Time Logic.
Soft synchronizationis used to prolong or shorten a chunk time while a CAN frame is obtained.
Question 17. What Is The Difference Between Function And Physical Addressing?

Answer :

Functional addressing is an addressing scheme that labels messages primarily based upon their operation code or content material. Physical addressing is an addressing scheme that labels messages based totally upon the bodily address location in their source and/or destination(s).

Border Gateway Protocol (BGP) Interview Questions
Question 18. What Happens If I Have To Send More Than eight-bytes Of Data?

Answer :

The J1939 fashionable has described a method of speaking extra than 8 bytes of statistics by way of sending the records in packets as specified in the Transport Protocol (TP). There are  styles of TP, one for broadcasting the statistics, and the opposite for sending it to a selected address.

DTC includes 4 components – SPN, FMI, OC and CM.

A DTC is a aggregate of 4 unbiased fields: the Suspect Parameter Number (SPN) of the channel or function that can have faults; a Failure Mode Identifier (FMI) of the precise fault; the occurrence count number (OC) of the SPN/FMI combination; and the SPN conversion approach (CM) which tells the receiving mode how to interpret the SPN. Together, the SPN, FMI, OC and CM shape a range of that a diagnostic device can use to recognize the failure that is being mentioned.

Network Security Interview Questions
Question 19. What Is Kwp2000?

Answer :

 KWP 2000(ISO14230) is a Diagnostic communications general. Specifies possible device configurations using the K & L traces. As 9141-2 however restricted to the physical traits. Specifies possible gadget configurations using the K & L lines.

5 Baud wake up as 9141- 2

New fast initialization method

Question 20. What Is Obdii?

Answer :

On-Board Diagnostics in an car context is a popular time period referring to a vehicle’s self-diagnostic and reporting capability.

Enhanced Interior Gateway Routing Protocol (EIGRP) Interview Questions
Question 21. Why Diagnostic Standards?

Answer :

As structures were given greater complex the link between reason and symptom have become much less apparent. This supposed that electronic structures needed to have a few level of self diagnosis and to speak to the outside world. Initially many structures used their very own protocols which meant that garages needed to have a large quantity of equipment – even to diagnose a single vehicle.

Question 22. What Is Meant By Verification And Validation?

Answer :

Verification and Validation (V&V) is the method of checking that a software machine meets specifications and that it fulfills its meant reason. It is generally a part of the software program trying out method of a venture.

According to the Capability Maturity Model (CMMI-SW v1.1),

Verification: The manner of comparing software to determine whether the products of a given improvement section fulfill the conditions imposed on the begin of that phase.

Validation: The manner of evaluating software throughout or at the cease of the development process to determine whether or not it satisfies distinct requirements.

Verification shows conformance with specification; validation shows that this system meets the customer’s wishes

Question 23. Can You Have Two Transmitters Using The Same Exact Header Field?

Answer :

No ,that would produce a bus struggle

Unless you have middle ware that guarantees handiest one node can transmit at a time
 For example use a low priority message as a token to emulate token-passing

Fix Protocol Interview Questions
Question 24. Can Bit Timing?

Answer :

According to the CAN specification, the bit time is split into 4 segments. The Synchronization Segment, the Propagation Time Segment, the Phase Buffer Segment 1, and the Phase Buffer Segment 2. Each section consists of a specific, programmable variety of time quanta (see Table 1). The duration of the time quantum (tq), that's the simple time unit of the bit time, is defined by the CAN controller’s gadget clock fsys and the Baud Rate Prescaler (BRP) :tq = BRP / fsys. Typical machine clocks are : fsys = fosc or fsys = fosc/2.

The Synchronization Segment Sync_Seg is that a part of the bit time where edges of the CAN bus stage are expected to arise; the gap between an aspect that occurs out of doors of Sync_Seg and the Sync_Seg is known as the segment mistakes of that edge. The Propagation Time Segment Prop_Seg is meant to compensate for the bodily delay instances in the CAN network. The Phase Buffer Segments Phase_Seg1 and Phase_Seg2 surround the Sample Point. The (Re-)Synchronization Jump Width (SJW) defines how some distance a resynchronization may also pass the Sample Point in the limits defined by using the Phase Buffer Segments to atone for aspect phase mistakes.

 Two styles of synchronization exist : Hard Synchronization and Re synchronization. A Hard Synchronization is accomplished once on the start of a body; inside a frame simplest Re synchronizations occur.

Hard Synchronization After a tough synchronization, the bit time is restarted with the stop of Sync_Seg, regardless of the threshold section blunders. Thus tough synchronization forces the edge which has brought on the difficult synchronization to lie inside the synchronization phase of the restarted bit time.
Bit Re synchronization Re synchronization results in a shortening or lengthening of the bit time such that the placement of the pattern point is shifted with regard to the edge.
CCNP Interview Questions
Question 25. Formula For Baudrate Calculation?

Answer :

The baud rate is calculated as:

baud charge (bits in line with 2d) = 18.432 x 10^6 / BRP / (1 + TSEG1 + TSEG2)

Question 26. What Happen When Two Can Nodes Are Sending Same Identifier At A Same Time?

Answer :

Two nodes at the network aren't allowed to send messages with the same identity. If  nodes try and send a message with the equal identification at the same time arbitration will no longer work. Instead, one of the transmitting nodes will hit upon that his message is distorted out of doors of the arbitration subject. The nodes will then use the mistake dealing with of CAN, which in this case ultimately will lead to one of the transmitting node being switched off (bus-off mode).

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Question 27. What Is The Difference Between Bit Rate And Baud Rate?

Answer :

The difference among Bit and Baud rate is complex and intertwining. Both are established and inter-associated. But the simplest explanation is that a Bit Rate is what number of statistics bits are transmitted per 2d. A baud Rate is the range of times in step with 2d a signal in a communications channel changes.Bit costs measure the wide variety of data bits (that is zero′s and 1′s) transmitted in a single second in a communique channel. A figure of 2400 bits in line with 2d approach 2400 zeros or ones may be transmitted in a single 2d, as a result the abbreviation “bps.” Individual characters (for instance letters or numbers) which are also called bytes are composed of several bits.A baud rate is the variety of instances a sign in a communications channel adjustments nation or varies. For example, a 2400 baud fee way that the channel can exchange states as much as 2400 times in step with second.

The term “alternate state” method that it is able to alternate from 0 to 1 or from 1 to zero up to X (in this situation, 2400) instances in keeping with 2nd. It also refers back to the real nation of the relationship, which includes voltage, frequency, or section degree).The major difference among the 2 is that one trade of kingdom can transmit one bit, or slightly extra or much less than one bit, that relies upon on the modulation approach used. So the bit rate (bps) and baud rate (baud in line with 2d) have this connection:bps = baud in step with 2d x the wide variety of bit according to baudThe modulation approach determines the quantity of bit according to baud. Here are  examples:When FSK (Frequency Shift Keying, a transmission method) is used, every baud transmits one bit. Only one alternate in country is needed to send a piece. Thus, the modem’s bps rate is same to the baud fee. When a baud rate of 2400 is used, a modulation technique referred to as section modulation that transmits 4 bits per baud is used. So:2400 baud x four bits in line with baud = 9600 bpsSuch modems are capable of 9600 bps operation.

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