Number Examples

I. Kaya

When ADC has 12 bits, how I can be processed by a 16 bits processor?

When ADC has 12 bits, how I can be processed by a 16 bits processor?

When ADC has 12 bits, how I can be processed by a 16 bits processor?

-0.42 Volt = 1101 1111 0111 in Q.11 format, it would take place in Q.15 arithmetic processor as ;

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1101 1111 0111 0000 which is on left side alignment or

1111 1101 1111 0111 on the right side alignment

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If 1.23 volt is 010111110111 in Q12 or Q.11 formats. In this case it is processed by a processor operating in Q.15 arithmetic as:

010111110111 0000 on the left side alignment or

0000 010111110111 on the right side alignment.

When ADC has 12 bits, how I can be processed by a 16 bits processor?

If we have an 8 bit number A= 0100 0011 b in Q.7 arithmetic, and B=10 1111 0011 b in Q.9 arithmetic, how we can add these numbers using Q.15 arithmetic processor.

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A = 0100 0011 0000 0000 red indicates sign bits

B= 10 1111 0011 00 0000 green indicates added bits to make it Q.15 arithmetic.

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0100 0011 0000 0000

+ 1011 1100 1100 0000

= 1111 1111 1100 0000

When data is -324.645 how it could be appearing in memory as in IEEE754 format?

Number is -324.645

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324. = 0001 0100 0100 in binary

0.645 *2= 1.29, 0.29*2=0.58, 0.58x2= 1.16, 0.16*2=0.32, 0.32*2=0.64, 0.64*2=1.28, 0.28*2=

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0.645 =1010010

So number is 0001 0100 0100.1010010 or 1.010 0010 0101 0010 x 2^8

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8 = 0000 1000, 8+127=0000 1000 + 0111 1111 = 1000 0111 = 135

In memory

1 1000 0111 0100 0100 1010010 = C3A25400 appears in memory