A-D and D-A Converters

The input to digital circuuit is X_2,X_1,X_0 and output is B_1,B_0

B_0=\bar{X_2}\bar{X_1}X_0+X_2X_1X_0
\;\;=X_0(\bar{X_2}\bar{X_1}+X_2X_1)
\;\;=X_0(\overline{X_2\oplus X_1})

Temperature range of -40^{\circ}C \; to \; 55^{\circ}C
So. Total range in 95^{\circ}C
Since, resolution is 0.1^{\circ}C
So, number of steps will be 950
To have 950 steps, we need at least 10 bits.

The reference voltage is 5 V.
The number of bits in ADC are 8.
So, the resolution will be =\frac{5}{2^8-1}=\frac{5}{255}
The applied input is 3.5 V.
The succesive approximation ADC start working from the MSB so,

After one clock:
SAR will toggle it's MSB from 0\rightarrow 1. So, output of SAR will be 1000 0000.

After second clock:
SAR will toggle its 7th bit from 0\rightarrow 1 but 1100 0000 will result in value greater than 3.5. So, output of SAR after 2nd clock will be 1000 0000.

After third clock:
SAR will toggle it's 6th bit from 0\rightarrow 1 and output will be 1010 0000.

Hex Number (AB.CD)
1010\;\;1011\;.\; 1100\;\;1101
For finding its octal number, we can add one zero in both extreme and grouping.
010\;101\;011\;.\;110\;011\;010
Its equavalent octal number is (253.632)_8

Initial stage of the counter =(111)_2=(7)_10
So output will be equal to 7 V.
Next state of counter =(110)_2=(6)_10
So output should be = 6 V
But output is 3 V that meansLSB of counter is connected to MSB of DAC and MSB of counter is connected to LSB of DAC.
Similarly next state oc counter =(101)_2=(5)_10
Input to DAC =(101)_2=(5)_10
So, output = 5V
When counter goes to (100)_2 then input to DAC =(001)_2=(1)_10
So output = 1 V
So connections are not proper.

Even when V_1 \gt V_2, the (o/p) 'V_0' is high and for the next case (V_1 \lt V_2) (o/p) is low.It is 1 bit quantizer. Since, it has two states which can be represented by 1 bit.