Homework 1 in Microcontrollers and practical robotics (0)

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Homework 01

Homework 1. a)123/2= 61,5; 61/2=30,5; 30/2=15; 15/2=7,5; 7/2= 3,5; 3/2=1,5 (1111011) 0.456*2=0.912; 0,912*2=1,824; 0,824*2=1,648; 0,648*2=1,296; 0,296*2=0,592; 0,592*2=1,184; 0,184*2=0,368; 0,368*2=0,736; (0,736*2=1,472; 0,472*2=0,944; 0,944*2=1,888) So b)123/16= 7,6875 ; 0,6875*16=11 ; 7/16 is less than one, then hex is 7B 0,456*16= 7,296 ; 0,296*16=4,736; 0,736*16=11,776; 0,776*16=12,416; 0,416*16=6,656; 0,656*16=10,496; 0,496*16=7,936; 0,936*16=14,976; 0,976*16=15,616; 0,616*16=9,856; 0,856*16=13,696; 0,696*16=11,136; 0,136*16=2,176; 2,176*16=2,816; 0,816*16=13,056 Hex is 7B.74BC6A7EF9DB22D c) 123,456/5=24,6912/5=4,93824/5=0,987648 0,987648*5=4,93824; 0,93824*5=4,6912; 0,6912*5=3,456; 0,456*5=2,28; 0,28*5=1,4; 0,4*5=2; 0*5=0 base-5 is d)BCD is 2. a) Unsigned 16-bit binary is 0000000001101011. Hexadecimal is 6B b) Unsigned 16-bit binary is 0000000010110101. Hexadecimal is B5 3. a) Signed two's c

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Mikrokontrollerid ja robootika

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oData transparency: In bit and byte oriented protocols, there is a problem if a control character (for ETX (End of Text) ·Same as ETB, only no more blocks will follow. ITB (End of > Differences with HDLC ­ length of protocol field (1B or 2B) byte-oriented protocols) or the start-of-frame flag (for bit-oriented protocols) appears in the actual data. Intermediate Transmission Block) ·Same as ETB, except that the receiving statio Differs from HDLC because of multiaccess MAC that provides · Maximum payload length (default: 1500) This was not likely to happen in ASCII text, but is very likely with binary data. This is known as a data will not acknowledge after the error checking. EOT (End of Transmission) framing/error detection: · Type of CRC (2B or 4B) transparency problem an can be rectified with byte stuffing (for byte-orien

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