A-D and D-A Converters

Question 1
A 2-bit flash Analog to Digital Converter (ADC) is given below. The input is 0 \leq V_{IN} \leq 3 Volts. The expression for the LSB of the output B_0 as a Boolean function of X_{2},X_{1} \; and \; X_{0} is
A
X_{0}[\overline{X_{2}\oplus X_{1}}]
B
\bar{X_{0}}[\overline{X_{2}\oplus X_{1}}]
C
X_{0}[{X_{2}\oplus X_{1}}]
D
\bar{X_{0}}[X_{2}\oplus X_{1}]
GATE EE 2016-SET-1   Digital Electronics
Question 1 Explanation: 
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})
Question 2
A temperature in the range of -40^{\circ}C to 55^{\circ}C is to be measured with a resolution of 0.1^{\circ}C. The minimum number of ADC bits required to get a matching dynamic range of the temperature sensor is
A
8
B
10
C
12
D
14
GATE EE 2016-SET-1   Digital Electronics
Question 2 Explanation: 
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.
Question 3
An 8-bit, unipolar Successive Approximation Register type ADC is used to convert 3.5 V to digital equivalent output. The reference voltage is +5 V. The output of the ADC, at the end of 3rd clock pulse after the start of conversion, is
A
1010 0000
B
1000 0000
C
0000 0001
D
1000 0000
GATE EE 2015-SET-1   Digital Electronics
Question 3 Explanation: 
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.
Question 4
The Octal equivalent of HEX and number AB.CD is
A
253.314
B
253.632
C
526.314
D
526.632
GATE EE 2007   Digital Electronics
Question 4 Explanation: 
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
Question 5
It is required to design an anti-aliasing filter for an, 8 bit ADC. The filter is a first order RC filter with R = 1\Omega and C = 1F. The ADC is designed to span a sinusoidal signal with peak to peak amplitude equal to the full scale range of the ADC.

What is the SNR (in dB) of the ADC ? Also find the frequency (in decades) at the filter output at which the filter attenuation just exceeds the SNR of the ADC.
A
50 dB, 2 decade
B
50 dB, 2.5 decade
C
60 dB, 2 decade
D
60 dB, 2.5 decade
GATE EE 2006   Digital Electronics
Question 6
It is required to design an anti-aliasing filter for an, 8 bit ADC. The filter is a first order RC filter with R = 1\Omega and C = 1F. The ADC is designed to span a sinusoidal signal with peak to peak amplitude equal to the full scale range of the ADC.

The transfer Function of the filter and its roll off respectively are
A
1/(1 + RCs), - 20 dB/decade
B
(1 + RCs), - 40 dB/decade
C
1/(1+ RCs), -40 dB/decade
D
{RCs/(1+RCs)}, -20 db/decade
GATE EE 2006   Digital Electronics
Question 7
A student has made a 3-bit binary down counter and connected to the R-2R ladder type DAC [Gain=(-1K\Omega/2R)] as shown in figure to generate a staircase waveform. The output achieved is different as shown in figure. What could be the possible cause of this error ?
A
The resistance values are incorrect option.
B
The counter is not working properly
C
The connection from the counter of DAC is not proper
D
The R and 2R resistance are interchanged
GATE EE 2006   Digital Electronics
Question 7 Explanation: 
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.
Question 8
The voltage comparator shown in figure can be used in the analog-to-digital conversion as
A
a 1-bit quantizer
B
a 2-bit quantizer
C
a 4-bit quantizer
D
a 8-bit quantizer
GATE EE 2004   Digital Electronics
Question 8 Explanation: 
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.
Question 9
A sample-and-hold (S/H) circuit, having a holding capacitor of 0.1 nF, is used at the input of an ADC (analog-to-digital converter). The conversion time of the ADC is 1\musec, and during this time, the capacitor should not lose more than 0.5% of the charge put across it during the sampling time. The maximum value of the input signal to the S/H circuit is 5V. The leakage current of the S/H circuit should be less than
A
2.5 mA
B
0.25 mA
C
25 mA
D
2.5 mA
GATE EE 2001   Digital Electronics
Question 10
Among the following four, the slowest ADC (analog-to-digital converter) is
A
parallel comparator (i.e., flash) type
B
successive approximation type
C
integrating type
D
counting type
GATE EE 2001   Digital Electronics
There are 10 questions to complete.