Sunday, September 22, 2019

Finite State Machine and Data Converters (ADC & DAC)



 1. The advantage of using a dual slope ADC in a digital voltmeter is that high Accuracy

Answer : 

The dual-slope ADC architecture was truly a breakthrough in ADCs for high resolution and high accuracy applications such as digital voltmeters (DVMs), etc. 

2.  The fastest ADC is  Flash type.



3. Which ADC has fixed Conversion time ?

Answer :

Successive  Approximation as SAR the conversion time is independent of the magnitude of the input sampled value.
     
      4.  Find the full scale output of a 4-bit DAC produce an output of  0.1V for a digital input 0001.  

      Answer : 1.5V 

      5. What is the output of the following circuit for the input b4=1, b3=0, b2=1, b0=0 and Vref =16V ?  
         

  
      Answer : -5V 

6. Output of the Mealy machine depends on  both present input and present state .

7. Compared to Moore FSM , a mealy FSM might have fewer states and have output generally one clock cycle earlier.

8. The number of comparators in a 4-bit flash ADC is 15

Answer :

 As the number of comparators is given by 2N-1, 16-1 =15

      9. Consider the circuit shown below . What would be the output sequence Z whne the input sequence x=01100. Assume the initial state {AB} = {00}
      

         Answer :
         JA = ((x’B)’(xB’)’)’ = x’B+xB’

   KA = x’B
   JB = (x’+A’)’ = xA
   KB = (x’+A’)’ = xA
   Z = (A’+B’)’ = AB

Output Sequence Z = 00101

10.  Let X is the input sequence whereas Z is the output sequence for the state machine given below . Which of the following option correctly describes the output Z sequence for the input sequence X given below.
X = 001101101011010 
Answer :
 Initial state à S0
·         When the input X  is 0, the state moves from S0àS2 and the output is 0
·         X = 0 , S2 à S2 , Z=0
·         X = 1, S2 à S4, Z = 0
·         X = 1, S4 à S3, Z= 1 ……….

The output Sequence Z = 00010110100101


     11. An 8-bit DAC has Vref  = -5V. What is the output voltage when Bin = 10110100? (Assume Rf = R/2)

      Answer :
         
        The output voltage is 3.516 V

     12. ___________ in  DACis defined as variation in analog step sizes between successive bits.

      Answer :

      Differential Non-Linearity Error : Analog step size changes with increasing digital input, measure of largest deviation between successive bits.

    13. The figure below shows a 3-bit Flash ADC circuit. What would be the encoded binary output D { D2D1D0}when V=8V and Vin = 3.45V

        Answer :
         Procedure :  Set the MSB of the Digit as 1 and evaluate the VDAC and compare with Vin.


·         If  Vin > VDAC     , SAR bit  is unchanged
·         If  Vin < VDAC     , SAR bit is Reset.

Ø  3-Bit ADC ,Vin = 3.45, Vref = 8 

Ø  SAR = 100   
VDAC = Vref/2^1 = 8/2 = 4  , Vin < VDAC     , SAR bit 3  is reset , 000

Ø  SAR = 010   
VDAC = 0+Vref/2^2 = 0+8/4 = 2  , Vin > VDAC     , SAR bit 2  is unchanged , 010

Ø  SAR = 011   
VDAC = 0+2+Vref/2^3 = 0+2+ 8/8 = 3  , Vin > VDAC  , SAR bit 1  is unchanged , 011

The answer is 011.

13. In a 5-bit successive approximation ADC with the reference voltage 1V, If an input voltage of 0.67V is applied , after 3 clock cycles the content of SAR is 

Answer :

Ø  5-Bit ADC ,Vin = 0.67, Vref = 1

Ø  SAR = 10000 (First cycle)   
VDAC = Vref/2^1 = 1/2 = 0.5  , Vin > VDAC     , SAR bit 5  is unchaged , 10000

Ø  SAR = 11000   (Second Cycle)
VDAC = 0.5+Vref/2^2 = 0.5+1/4 = 0.75  , Vin < VDAC  , SAR bit 4  is reset , 10000

Ø  SAR = 10100    (Third Cycle)
VDAC = 0.5+0+Vref/2^3 = 0.5+0+1/8 =0.625  , Vin > VDAC  , SAR bit 3  is unchanged , 10100

The answer , after 3 clock cycle the content of SAR is 10100

15. In a 5-bit successive approximation ADC with the reference voltage 1V, If an input voltage of 0.3V is applied , after 4 clock cycles the content of SAR is 

Answer :

Ø  5-Bit ADC ,Vin = 0.3, Vref = 1
Ø  SAR = 10000 (First cycle)   
VDAC = Vref/2^1 = 1/2 = 0.5  , Vin < VDAC     , SAR bit 5  is reset , 00000
Ø  SAR = 01000   (Second Cycle)
VDAC = 0+Vref/2^2 = 0+1/4 = 0.25  , Vin > VDAC  , SAR bit 4  is unchanged , 01000
Ø  SAR = 01100    (Third Cycle)
VDAC = 0+0.25+Vref/2^3 = 0+0.25+1/8 =0.325  , Vin < VDAC  , SAR bit 3  is reset , 01000
Ø  SAR = 01010    (Third Cycle)
VDAC = 0+0.25+0+Vref/2^4 = 0+0.25+0+1/16 =0.3125  , Vin < VDAC  , SAR bit 2  is reset , 01000

The answer , after 3 clock cycle the content of SAR is 01000







No comments:

Post a Comment

Microprocessor & Microcontroller - NPTEL (noc23_ee47)

  Microprocessor & Microcontroller - NPTEL (noc23_ee47) Week - 02