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```4.1
Remainder
147 2
73 2
1
36 2
1
18 2
0
92
0
42
1
22
0
12
0
0
1
4.9 To find the decimal value for 10010001, we must first subtract 1 from the LSB thus
giving:
10010000
Next, we invert the 1's and 0's as follows:
01101111
Finally, we evaluate the decimal:
 N10  027   126   125   024   123   122   121  120 
 111
 N10  111
4.23 Quantization error is computed using equation 4.1.
Vru  Vrl
2N
For 8 bits this becomes
8  (8)
input res error  0.5
 0.0313V and this is 0.42% of the 7.5V input.
28
For 12 bits this becomes
8  (8)
input res error  0.5
 0.00195V and this is 0.026% of the 7.5V input.
212
For 16 bits this becomes
8  (8)
input res error  0.5
 0.000122 V and this is 0.0016% if the 7.5V input.
216
input res error  0.5
4.27 Since the signal from the transducer varies between 15mV (0.015V) and the
A/D converter input range is 10V, we can select a gain of 100 which will yield an input
of 1.5V. A gain of 100 is chosen such that the amplified signal is not saturated (i.e.
greater than the input range).
The quantization error from Eq. 4.1 is as follows:
V  V 
Quantization Error  0.5 ru N rl Volts
 2

 10   10 
 0.5

212


 2.44  10 3 Volts
The transducer voltage is 3.75mV but after a gain of 100 it becomes 0.375V.
Thus, the quantization error as a percent reading is as follows:
 2.44  10 3Volts 

 100  0.651%
0.375Volts 

If the transducer output were attenuated by a factor of 2/3 (to 10 mV) the gain
could be set to 1000 without saturating the A/D converter. The 3.75 mv output
would then become 3.7510-3(2/3)1000 = 2.5 V at the input to the A/D
converter and the resolution error would be reduced to 0.098%.
CHAPTER 5:
5.3
V(t) = 25t
0 t 1
T = 1sec
o = 2f
= 2(1/T)
= 2(1/1)
By Eq. 5.5.
2 T
bn   f (t ) sin n o tdt
T 0
Thus,
2 1
b1   25t sin( 2t )dt
1 0
= 2(-3.9789)
= -7.9578
2 1
25t sin( 2  2t )dt
1 0
= 2(-1.9894)
= -3.9788
b2 
ao is the average, Eq. 5.3
1 T
a o   f t  cos n o tdt
T 0
Thus,
25
0
1
2
1 1
25t cos(0)dt
1 0
= 12.5
a0 
and an 
2 T
f t cos no tdt
T 0
2 1
25t cos(2t )dt
1 0
=0
a1 
2 1
25t cos(2  2t )dt
1 0
=0
a2 
5.16
The function has frequencies of 5 and 20 Hz. The minimum sampling rate would then
be 40 Hz to avoid aliasing. Sampling at 30 Hz would not produce aliases for the 5 Hz
signal but would for the 20 Hz signal. The alias frequency can be evaluated using the
folding diagram in Section 5.1.
fN = 30/2 = 15 Hz. f/fN = 20/15 = 1.3333. From the folding diagram, fa/fN = 0.666. Thus fa
= 0.66615 = 10 Hz. This is the difference between the sampling rate and the signal
frequency.
5.19
The maximum frequency in the signal of problem 5.11 is 20 Hz. The sampling rate
should exceed twice this value, or 40 Hz.
5.26
Use Appendix A-3. fN = 1500/2 = 750 Hz. f/fN = 1000/750 = 1.333. From Figure A.1, fa/fN
= 0.666. Thus fa = 0.666*750 = 500 Hz.
147 2 73 2 36 2 18 2 9 2 4 2 2 2 1 2 0 11001001
From the remainders, we get the 8-bit number as follows:
N2 = 1 0 0 1 0 0 1 1
4.9 To find the decimal value for 10010001, we must first subtract 1 from the LSB thus giving:
10010000
Next, we invert the 1's and 0's as follows:
01101111
Finally, we evaluate the decimal:
4.23 Quantization error is computed using equation 4.1.
For 8 bits this becomes
and this is 0.42% o
For 12 bits this becomes
For 16 bits this becomes
and this is 0.0016% if the 7.5V input. V err
4.27
is chosen such that the amplified signal is not saturated (i.e. greater than the input range).
The quantization error from Eq. 4.1 is as follows:
ts
The transducer voltage is 3.75mV but after a gain of 100 it becomes 0.375V. Thus, the quantization
error as a percent reading is as follows:
to 1000 without saturating the A/D converter. The 3.75 mv output would then become
-3
on error
```
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