Automotive Interview Questions and CAN interview questions and important C programming interview Questions are asked in Automotive interviews.This are the very important questions for automotive interviews.
21.Why Diagnostic Standards?
Answer:As systems got more complex the
link between cause and symptom became less obvious. This meant that
electronic systems had to have some level of self diagnosis and to
communicate to the outside world. Initially many systems used their
own protocols which meant that garages had to have a large number of
tools – even to diagnose a single vehicle.
22.What is meant by verification and
validation??
Answer:Verification and Validation
(V&V) is the process of checking that a software system meets
specifications and that it fulfills its intended purpose. It is
normally part of the software testing process of a project.
According to the Capability Maturity Model (CMMI-SW
v1.1),
-
Verification: The process of evaluating software to determine whether the products of a given development phase satisfy the conditions imposed at the start of that phase.
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Validation: The process of evaluating software during or at the end of the development process to determine whether it satisfies specified requirements.
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Verification shows conformance with specification; validation shows that the program meets the customer’s needs
23.Can you have two transmitters using the same
exact header field?
Answer:No – that would
produce a bus conflict
• Unless you have middle ware that
ensures only one node can transmit at a time
– For example use a low priority message as a token
to emulate token-passing
24. CAN physical layer voltage levels
25. CAN bit timing:
According to the CAN specification, the
bit time is divided into four segments. The Synchronization Segment,
the Propagation Time Segment, the Phase Buffer Segment 1, and the
Phase Buffer Segment 2. Each segment consists of a specific,
programmable number of time quanta (see Table 1). The length of the
time quantum (tq), which is the basic time unit of the bit time, is
defined by the CAN controller’s system clock fsys and the Baud Rate
Prescaler (BRP) :tq = BRP / fsys. Typical system clocks are : fsys =
fosc or fsys = fosc/2.
The Synchronization Segment Sync_Seg is
that part of the bit time where edges of the CAN bus level are
expected to occur; the distance between an edge that occurs outside
of Sync_Seg and the Sync_Seg is called the phase error of that edge.
The Propagation Time Segment Prop_Seg is intended to compensate for
the physical delay times within the CAN network. The Phase Buffer
Segments Phase_Seg1 and Phase_Seg2 surround the Sample Point. The
(Re-)Synchronization Jump Width (SJW) defines how far a
resynchronization may move the Sample Point inside the limits defined
by the Phase Buffer Segments to compensate for edge phase errors.
Two types of synchronization exist : Hard
Synchronization and Re synchronization. A Hard Synchronization is done
once at the start of a frame; inside a frame only Re synchronizations
occur.
• Hard Synchronization After
a hard synchronization, the bit time is restarted with the end of
Sync_Seg, regardless of the edge phase error. Thus hard
synchronization forces the edge which has caused the hard
synchronization to lie within the synchronization segment of the
restarted bit time.
• Bit Re synchronization Re synchronization leads to a
shortening or lengthening of the bit time such that the position of
the sample point is shifted with regard to the edge.
26. Formula for Baudrate
calculation?
The baud rate is calculated as:
baud rate (bits per second) = 18.432 x 10^6 / BRP / (1 + TSEG1 + TSEG2)
27. What happen when two CAN nodes are sending same
identifier at a same time?
Two nodes on the network are not allowed to send
messages with the same id. If two nodes try to send a message with
the same id at the same time arbitration will not work. Instead, one
of the transmitting nodes will detect that his message is distorted
outside of the arbitration field. The nodes will then use the error
handling of CAN, which in this case ultimately will lead to one of
the transmitting node being switched off (bus-off mode).
28. what is the difference between Bit Rate and
Baud Rate?
The difference between
Bit and Baud rate is complicated and intertwining. Both are dependent
and inter-related. But the simplest explanation is that a Bit Rate is
how many data bits are transmitted per second. A baud Rate is the
number of times per second a signal in a communications channel
changes.Bit rates measure the number of data bits (that is 0′s and
1′s) transmitted in one second in a communication channel. A figure
of 2400 bits per second means 2400 zeros or ones can be transmitted
in one second, hence the abbreviation “bps.” Individual
characters (for example letters or numbers) that are also referred to
as bytes are composed of several bits.A baud rate is the number of
times a signal in a communications channel changes state or varies.
For example, a 2400 baud rate means that the channel can change
states up to 2400 times per second. The term “change state” means
that it can change from 0 to 1 or from 1 to 0 up to X (in this case,
2400) times per second. It also refers to the actual state of the
connection, such as voltage, frequency, or phase level).The main
difference between the two is that one change of state can transmit
one bit, or slightly more or less than one bit, that depends on the
modulation technique used. So the bit rate (bps) and baud rate (baud
per second) have this connection:bps = baud per second x the number
of bit per baudThe modulation technique determines the number of bit
per baud. Here are two examples:When FSK (Frequency Shift Keying, a
transmission technique) is used, each baud transmits one bit. Only
one change in state is required to send a bit. Thus, the modem’s
bps rate is equal to the baud rate. When a baud rate of 2400 is used,
a modulation technique called phase modulation that transmits four
bits per baud is used. So:2400 baud x 4 bits per baud = 9600 bpsSuch
modems are capable of 9600 bps operation.
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