ME 365
Homework Set 2
Out September 1 , 2011
Due September 8, 2011
Problem No. 1
In this question two analog to digital converters (ADC) will be considered.
The
information about these ADC’s are given in the following table:
ADC
Number of bits
Nominal Range
ADC1:
8 bits
0 to 10 volts
ADC2:
12 bits
10 to +10 volts
(a)
Determine the output code (decimal integer) for both of the ADCs for the given
input voltages.
i. 3.5 volts
ii.
2.35 volts iii. 11.35 volts
Solution:
i.
for ADC1, 3.5V is clipped to 0V and therefore it is coded as 0
for ADC2, Q
2
=20/2
12
=0.0049V
code=round[(3.5V(10V))/Q
2
]=1327
ii.
for ADC1, Q
1
=10/2
8
=0.0391V
code=round[2.35V/Q
1
]=60
for ADC2, code=round[(2.35V(10V))/Q
2
]=2520
iii.
for ADC1, 11.35 is clipped to V
ADCmax
=10VQ
1
=9.9609V and it corresponds
to the maximum positive code 2
8
1=255
Similarly for ADC2, 11.35 is clipped to V
ADCmax
=10VQ
2
=9.9951V and it
corresponds to the maximum positive code 2
12
1=4095
(b)
ADC1 produced the follow output code (1110 0010)
2
.
What is the range of input
voltages that could produce this code, assuming positive codes only?
Solution:
(1110 0010)
2
=1*2
7
+1*2
6
+1*2
5
+1*2
1
=(226)
10
The estimated voltage from this output code is 226*Q
1
=8.8366V
Therefore, the range of input voltage that could produce this code is
(8.8366Q
1
/2, 8.8366+Q
1
/2)=(8.8171V, 8.8562V)
Note:
If you did not use the rounded number of Q
1
, but use this
(226*(10/2
8
)(10/2
8
)/2, 226*(10/2
8
)+(10/2
8
)/2 )
You will get this range (8.8086V, 8.8477V)
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Homework Set 2
Out September 1 , 2011
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 Fall '07
 MERKLE
 Digital Signal Processing, Signal Processing, 0.0015mm, 3.5 volts

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