# 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.
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$\mu$sec, 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.