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Grace HMW2 1
Jessy Grace
ET372 HMW2
G00170423
Grantham University
08/24/2021
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Grace HMW2 2
Digital Signal Conditioning
1. A “1” signal is required to trigger an alarm, if the fluid level in a tank is more than 3m
deep. The level sensor gives an output of 9.3mV for every centimeter increase in depth.
What is the required alarm voltage?
(9.3mV/cm) (3m or 300cm) =V
(9.3x10
-3
/cm) (300cm) = 2.79V
2. In problem 1, if there are waves with amplitude of 35 cm due to pumping, what is the
value of R2 to give a dead band with a 5-cm safety margin to prevent the comparator
output from going low? Assume R1 = 5k ohms, and output high = 5V
R1= 5kohm
Vout= 5V
Vref= (9.3mV/cm) (35cm) = 325.5mV
Deadband (9.3mV/cm) (
+
-
5cm) = 46.5mV
Vb= 325.5mV + 46.5 mV= 372mV
R2=


󰇛󰇜󰇛󰇜


3. An 8 bit DAC has a reference voltage of 5V. What would be the voltage corresponding to
a binary word of 10010011?
Vout=
V
R
N
10
= a
7
2
7
+a
6
2
6
+a
5
2
5
+a
4
2
4
+a
3
2
3
+a
2
2
2
+a
1
2
1
+a
0
2
0
N
10
= (1)2
7
+ (0)2
6
+ (0)2
5
+ (1)2
4
+ (0)2
3
+ (0)2
2
+ (1)2
1
+ (1)2
0
= 147
2
n
= 2
8
= 256


5V= 2.87V
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Grace HMW2 3
4. What is the conversion time for the dual slope converter shown in the figure below if R =
1M ohm, C = 50nF, VR = 5V, integration time of 50 ms, and the input voltage is 8.2V?
What is the capacitor discharge time?




Given Values
R= 1M ohm
C=50nF
VR=5v
Integration time= 50ms (50x10
-3
)
Input Voltage= 8.2V








=


TD=


(50x10
-3
) = 30.49ms
5. A temperature between -10 and +250 is converted into a 0- to 5.0 Volt signal. This signal
is fed to an 8-bit ADC with a 5 Volt reference. What is the actual measurement range of
the system? What is the resolution?
Actual measurement range of the system
Percent Error:
LSB =


 
Actual measurement = 5-0.4% = 4.98 volts
Resolution Q = E_sfr / 2^n
Q =




Unformatted Attachment Preview

Grace HMW2 1 Jessy Grace ET372 HMW2 G00170423 Grantham University 08/24/2021 Grace HMW2 2 Digital Signal Conditioning 1. A “1” signal is required to trigger an alarm, if the fluid level in a tank is more than 3m deep. The level sensor gives an output of 9.3mV for every centimeter increase in depth. What is the required alarm voltage? (9.3mV/cm) (3m or 300cm) =V (9.3x10-3/cm) (300cm) = 2.79V 2. In problem 1, if there are waves with amplitude of 35 cm due to pumping, what is the value of R2 to give a dead band with a 5-cm safety margin to prevent the comparator output from going low? Assume R1 = 5k ohms, and output high = 5V R1= 5kohm Vout= 5V Vref= (9.3mV/cm) (35cm) = 325.5mV Deadband→ (9.3mV/cm) (+- 5cm) = 46.5mV Vb= 325.5mV + 46.5 mV= 372mV 𝑅1∗𝑉𝑜𝑢𝑡 (5𝑘𝑜ℎ𝑚)(5𝑣) R2= 𝑉𝑏 → 372𝑚𝑉 = 67204 𝑜𝑟 67.2𝑘𝑜ℎ𝑚𝑠 3. An 8 bit DAC has a reference voltage of 5V. What would be the voltage corresponding to a binary word of 10010011? 𝑁 Vout=2𝑁 VR N10= a727+a626+a525+a424+a323+a222+a121+a020 N10= (1)27+ (0)26+ (0)25+ (1)24+ (0)23+ (0)22+ (1)21+ (1)20= 147 2n= 28= 256 147 5V= 2.87V 256 Grace HMW2 3 4. What is the conversion time for the dual slope converter shown in the figure below if R = 1M ohm, C = 50nF, VR = 5V, integration time of 50 ms, and the input voltage is 8.2V? What is the capacitor discharge time? 𝑉𝑖𝑛 𝑉𝑟𝑒𝑓 = 𝑇𝑐 𝑇𝐷 Given Values R= 1M ohm C=50nF VR=5v Integration time= 50ms (50x10-3) Input Voltage= 8.2V 8.2𝑉 5𝑉 𝑇𝐷 5𝑉 5𝑉 = → = → TD= (50x10-3) = 30.49ms −3 −3 50x10 𝑇𝐷 50x10 8.2𝑉 8.2𝑉 5. A temperature between -10 and +250 is converted into a 0- to 5.0 Volt signal. This signal is fed to an 8-bit ADC with a 5 Volt reference. What is the actual measurement range of the system? What is the resolution? Actual measurement range of the system Percent Error: 1 1 LSB = 28 𝑥100 = 256 𝑥100 = 0.4% Actual measurement = 5-0.4% = 4.98 volts Resolution Q = E_sfr / 2^n Q= 5𝑉 28 5 = 256 = 19.5𝑚𝑉 Name: Description: ...
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