08 Sampling Problem Set
SAMPLING | PRACTICE PROBLEMS
Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: Given an analog signal measured with a sampling period of 0.03 ππ , what is the highest frequency that can be accurately reported most close to: A. 4 Γ 10* B. 12 Γ 10* C. 17 Γ 10* D. 23 Γ 10*
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SOLUTION 1: The formula for SAMPLING RATE can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The FREQUENCY or SAMPLING RATE is the rate at which something occurs or is repeated over a particular period of time or in a given sample. The frequency is inversely proportional to the SAMPLING PERIOD, or the number of samples taken in a regular time interval.
The frequency or sample rate is calculated as the inverse of the sampling period, shown in the following formula as: π. =
1 π₯π‘
Where: β’ π. is the sampling frequency given in units of hertz (Hz) β’ π₯π‘ is the sample period or number of samples taken in a regular time interval Plugging in the value for the given sampling period
π. =
1 = 33 Γ 10* π»π§ 0.03 Γ 102*
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The formula for NYQUIST FREQUENCY can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Nyquistβs (Shannonβs) sampling theorem states that in order to accurately reconstruct the analog signal from the discrete sample points, the sample rate must be larger than twice the highest frequency contained in the measured signal. Denoting this frequency, which is called the NYQUIST FREQUENCY, as π5 , the sample theorem requires that: π. > 2π5 Where: β’ π. is the frequency of sampling β’ π5 is the Nyquist frequency, representing the highest frequency contained in the measure signal Therefore, the sampling frequency must be greater than the Nyquist rate for accurate reproduction. Using the formula for the Nyquist frequency, we plug in the calculated value for the sampling frequency and calculate the highest frequency that can be accurately reported as: π. 33 Γ 10* π»π§ π5 < = = 16.5 Γ 10* π»π§ 2 2
Therefore, the correct answer choice is C. ππ Γ πππ
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PROBLEM 2: A 16 bit analog-to-digital converter has a resolution of 1.52588 Γ 102@ π with the lowest voltage being measured as half the magnitude of the highest voltage. If both the high and low voltages are positive, what is the value of the highest voltage closest to: A. 15 B. 20 C. 25 D. 30
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SOLUTION 2: The formula for VOLTAGE RESOLUTION can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When converting an analog signal to digital form, the resolution of the conversion is an important factor. For a measured analog signal over the nominal range πB , πC , where πB is the low end of the voltage range and πC is the nominal high end of the voltage range, the voltage resolution is given by:
πE =
πC β πB 2G
Where: β’ π is the number of conversion bits of the A/D converter with typical values of 4, 8, 10, 12, or 16. β’ πB is the low end of the voltage range β’ πC is the high end of the voltage range β’ π is the voltage resolution As we are given the voltage resolution and number of conversion bits, we can re-write the expression for voltage regulation in terms of the unknown voltage limits, given by the expression: πC β πB = 2G πE
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Plugging in the given values for the voltage regulation and number of conversion bits, we can derive an equation to express the difference between the upper and lower voltage limits as: πC β πB = 2
I*
1.52588 Γ 102@ π = 10 π
Now that we have one equation written for our two unknowns, we need to use additional information given in the problem to write an additional equation, so we are able to solve for the two unknowns. In the problem statement, we are told that the lowest voltage being measured is half the magnitude of the highest voltage. As both the voltage limits are positive, we can derive the following equation to represent the relationship between the upper and lower voltage limits: πC = |2πB | As πC and πB are both positive, this equation becomes: πC = 2πB We can then plug in this equation for the high voltage limit, into the first equation we wrote to put everything in terms of the lower voltage limit: 2πB β πB = 10 π
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We can now solve for the lower voltage limit: πB = 10 π As we have the value for the lower voltage limit, we can plus this into any of our two equation to solve for the higher voltage limit. Plugging the value for the lower voltage limit into the second equation, we can finally solve for the higher voltage limit: πC = 2πB = 2 10 π = 20 π
Therefore, the correct answer choice is B. ππ
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PROBLEM 3: An analog-to-digital conversion process has a resolution of approximately 1.52588 Γ 102@ π. Given that the voltage range is 0 π to 10 π, what is the number of conversion bits most close to: A. 4 B. 8 C. 16 D. 32
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SOLUTION 3: The formula for VOLTAGE RESOLUTION can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When converting an analog signal to digital form, the resolution of the conversion is an important factor. For a measured analog signal over the nominal range πB , πC , where πB is the low end of the voltage range and πC is the nominal high end of the voltage range, the voltage resolution is given by:
πE =
πC β πB 2G
Where: β’ π is the number of conversion bits of the A/D converter with typical values of 4, 8, 10, 12, or 16. β’ πB is the low end of the voltage range β’ πC is the high end of the voltage range β’ π is the voltage resolution Plugging in the given values, we can re-write the expression for voltage resolution as:
1.52588 Γ 102@ =
10 β 0 π 2G
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Solving for π, we calculate the number of conversion bits as: π = 16
Therefore, the correct answer choice is C. ππ
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PROBLEM 4: A uncertainty analysis is performed given three measurements, π₯I , π₯M , πππ π₯P represented by the function π = 3π₯I β 5π₯M + 7π₯P . Using the data point (3, 5, 7), the uncertainty
values
for
a
set
of
Β±0.03, Β±0.05, πππ Β± 0.07, respectively.
strain
gage
measurements
A. 0.34 B. 0.56 C. 0.67 D. 0.79
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given
as
What is the estimated uncertainty of the
calculation most close to:
are
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SOLUTION 4:
The formula for the KLINE-MCCLINTOCK EQUATION can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The first step in this problem is to calculate the partial derivative of each term given in the uncertainty function. If you are not familiar with partial derivatives, then please take the time to go back and review this topic in the subject of mathematics. Given the function π = 3π₯I β 5π₯M + 7π₯P , we will calculate the partial derivative with respect to the variable π₯I first: ππ =3 ππ₯I Next, we will calculate the partial derivative with respect to the variable π₯M : ππ = β5 ππ₯M Lastly, we will calculate the partial derivative with respect to the variable π₯P : ππ =7 ππ₯P
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As the partial derivatives for each uncertainty variable do contain any terms with a variable, we do not have to worry about plugging in any values for π₯I , π₯M , ππ π₯P . The uncertainty in π , π€[ , depends on more than one measurement and can be estimated using the Kline-McClintock equation. If the uncertainty in the independent variables are all given with the same odds, then the uncertainty in the results have these odds is:
π€[ =
πΏπ π€I πΏπ₯I
M
πΏπ + π€M πΏπ₯M
M
πΏπ + β― + π€G πΏπ₯G
M
Plugging in the given uncertainty variables, and calculated uncertainty measurements represented by the partial derivatives, we find the calculated measurement uncertainty is:
π€[ =
0.03 3
M
+
0.05 β5
M
+
0.07 7
M
= 0.56
Therefore, the correct answer choice is B. π. ππ
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PROBLEM 5: Given a frequency of 20 ππ»π§, the sampling frequency must be higher than which of the following frequencies in order to capture all audible frequencies with the most accuracy? A. 20 B. 26 C. 38 D. 40
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SOLUTION 5: The formula for NYQUIST FREQUENCY can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Nyquistβs (Shannonβs) sampling theorem states that in order to accurately reconstruct the analog signal from the discrete sample points, the sample rate must be larger than twice the highest frequency contained in the measured signal. Denoting this frequency, which is called the NYQUIST FREQUENCY, as π5 , the sample theorem requires that: π. > 2π5 Where: β’ π. is the frequency of sampling β’ π5 is the Nyquist frequency, representing the highest frequency contained in the measure signal
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Therefore, the sampling frequency must be greater than the Nyquist rate for accurate reproduction. Using the formula for the Nyquist frequency, we plug in the calculated value for the sampling frequency and calculate the highest frequency that can be accurately reported as: π. = 2 π5 = 2 20 = 40 ππ»π§ 2 20 = 40 ππ»π§
Therefore, the correct answer choice is D. ππ
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PROBLEM 6: Whenever the Florida Gators football team scores a touchdown, the crowd roars to a sampled frequency of 300 ππ»π§. If this is an accurate sample frequency, what is the highest frequency that can be accurately reported most close to: A. 150 Γ 10* B. 225 Γ 10* C. 250 Γ 10* D. 300 Γ 10*
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SOLUTION 6: The formula for NYQUIST FREQUENCY can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Nyquistβs (Shannonβs) sampling theorem states that in order to accurately reconstruct the analog signal from the discrete sample points, the sample rate must be larger than twice the highest frequency contained in the measured signal. Denoting this frequency, which is called the NYQUIST FREQUENCY, as π5 , the sample theorem requires that: π. > 2π5 Where: β’ π. is the frequency of sampling β’ π5 is the Nyquist frequency, representing the highest frequency contained in the measure signal Therefore, the sampling frequency must be greater than the Nyquist rate for accurate reproduction. Using the formula for the Nyquist frequency, we plug in the calculated value for the sampling frequency and calculate the highest frequency that can be accurately reported as:
π5
Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: Given an analog signal measured with a sampling period of 0.03 ππ , what is the highest frequency that can be accurately reported most close to: A. 4 Γ 10* B. 12 Γ 10* C. 17 Γ 10* D. 23 Γ 10*
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SOLUTION 1: The formula for SAMPLING RATE can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The FREQUENCY or SAMPLING RATE is the rate at which something occurs or is repeated over a particular period of time or in a given sample. The frequency is inversely proportional to the SAMPLING PERIOD, or the number of samples taken in a regular time interval.
The frequency or sample rate is calculated as the inverse of the sampling period, shown in the following formula as: π. =
1 π₯π‘
Where: β’ π. is the sampling frequency given in units of hertz (Hz) β’ π₯π‘ is the sample period or number of samples taken in a regular time interval Plugging in the value for the given sampling period
π. =
1 = 33 Γ 10* π»π§ 0.03 Γ 102*
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The formula for NYQUIST FREQUENCY can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Nyquistβs (Shannonβs) sampling theorem states that in order to accurately reconstruct the analog signal from the discrete sample points, the sample rate must be larger than twice the highest frequency contained in the measured signal. Denoting this frequency, which is called the NYQUIST FREQUENCY, as π5 , the sample theorem requires that: π. > 2π5 Where: β’ π. is the frequency of sampling β’ π5 is the Nyquist frequency, representing the highest frequency contained in the measure signal Therefore, the sampling frequency must be greater than the Nyquist rate for accurate reproduction. Using the formula for the Nyquist frequency, we plug in the calculated value for the sampling frequency and calculate the highest frequency that can be accurately reported as: π. 33 Γ 10* π»π§ π5 < = = 16.5 Γ 10* π»π§ 2 2
Therefore, the correct answer choice is C. ππ Γ πππ
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PROBLEM 2: A 16 bit analog-to-digital converter has a resolution of 1.52588 Γ 102@ π with the lowest voltage being measured as half the magnitude of the highest voltage. If both the high and low voltages are positive, what is the value of the highest voltage closest to: A. 15 B. 20 C. 25 D. 30
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SOLUTION 2: The formula for VOLTAGE RESOLUTION can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When converting an analog signal to digital form, the resolution of the conversion is an important factor. For a measured analog signal over the nominal range πB , πC , where πB is the low end of the voltage range and πC is the nominal high end of the voltage range, the voltage resolution is given by:
πE =
πC β πB 2G
Where: β’ π is the number of conversion bits of the A/D converter with typical values of 4, 8, 10, 12, or 16. β’ πB is the low end of the voltage range β’ πC is the high end of the voltage range β’ π is the voltage resolution As we are given the voltage resolution and number of conversion bits, we can re-write the expression for voltage regulation in terms of the unknown voltage limits, given by the expression: πC β πB = 2G πE
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Plugging in the given values for the voltage regulation and number of conversion bits, we can derive an equation to express the difference between the upper and lower voltage limits as: πC β πB = 2
I*
1.52588 Γ 102@ π = 10 π
Now that we have one equation written for our two unknowns, we need to use additional information given in the problem to write an additional equation, so we are able to solve for the two unknowns. In the problem statement, we are told that the lowest voltage being measured is half the magnitude of the highest voltage. As both the voltage limits are positive, we can derive the following equation to represent the relationship between the upper and lower voltage limits: πC = |2πB | As πC and πB are both positive, this equation becomes: πC = 2πB We can then plug in this equation for the high voltage limit, into the first equation we wrote to put everything in terms of the lower voltage limit: 2πB β πB = 10 π
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We can now solve for the lower voltage limit: πB = 10 π As we have the value for the lower voltage limit, we can plus this into any of our two equation to solve for the higher voltage limit. Plugging the value for the lower voltage limit into the second equation, we can finally solve for the higher voltage limit: πC = 2πB = 2 10 π = 20 π
Therefore, the correct answer choice is B. ππ
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PROBLEM 3: An analog-to-digital conversion process has a resolution of approximately 1.52588 Γ 102@ π. Given that the voltage range is 0 π to 10 π, what is the number of conversion bits most close to: A. 4 B. 8 C. 16 D. 32
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SOLUTION 3: The formula for VOLTAGE RESOLUTION can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When converting an analog signal to digital form, the resolution of the conversion is an important factor. For a measured analog signal over the nominal range πB , πC , where πB is the low end of the voltage range and πC is the nominal high end of the voltage range, the voltage resolution is given by:
πE =
πC β πB 2G
Where: β’ π is the number of conversion bits of the A/D converter with typical values of 4, 8, 10, 12, or 16. β’ πB is the low end of the voltage range β’ πC is the high end of the voltage range β’ π is the voltage resolution Plugging in the given values, we can re-write the expression for voltage resolution as:
1.52588 Γ 102@ =
10 β 0 π 2G
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Solving for π, we calculate the number of conversion bits as: π = 16
Therefore, the correct answer choice is C. ππ
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PROBLEM 4: A uncertainty analysis is performed given three measurements, π₯I , π₯M , πππ π₯P represented by the function π = 3π₯I β 5π₯M + 7π₯P . Using the data point (3, 5, 7), the uncertainty
values
for
a
set
of
Β±0.03, Β±0.05, πππ Β± 0.07, respectively.
strain
gage
measurements
A. 0.34 B. 0.56 C. 0.67 D. 0.79
Made with
given
as
What is the estimated uncertainty of the
calculation most close to:
are
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SOLUTION 4:
The formula for the KLINE-MCCLINTOCK EQUATION can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The first step in this problem is to calculate the partial derivative of each term given in the uncertainty function. If you are not familiar with partial derivatives, then please take the time to go back and review this topic in the subject of mathematics. Given the function π = 3π₯I β 5π₯M + 7π₯P , we will calculate the partial derivative with respect to the variable π₯I first: ππ =3 ππ₯I Next, we will calculate the partial derivative with respect to the variable π₯M : ππ = β5 ππ₯M Lastly, we will calculate the partial derivative with respect to the variable π₯P : ππ =7 ππ₯P
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As the partial derivatives for each uncertainty variable do contain any terms with a variable, we do not have to worry about plugging in any values for π₯I , π₯M , ππ π₯P . The uncertainty in π , π€[ , depends on more than one measurement and can be estimated using the Kline-McClintock equation. If the uncertainty in the independent variables are all given with the same odds, then the uncertainty in the results have these odds is:
π€[ =
πΏπ π€I πΏπ₯I
M
πΏπ + π€M πΏπ₯M
M
πΏπ + β― + π€G πΏπ₯G
M
Plugging in the given uncertainty variables, and calculated uncertainty measurements represented by the partial derivatives, we find the calculated measurement uncertainty is:
π€[ =
0.03 3
M
+
0.05 β5
M
+
0.07 7
M
= 0.56
Therefore, the correct answer choice is B. π. ππ
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PROBLEM 5: Given a frequency of 20 ππ»π§, the sampling frequency must be higher than which of the following frequencies in order to capture all audible frequencies with the most accuracy? A. 20 B. 26 C. 38 D. 40
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SOLUTION 5: The formula for NYQUIST FREQUENCY can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Nyquistβs (Shannonβs) sampling theorem states that in order to accurately reconstruct the analog signal from the discrete sample points, the sample rate must be larger than twice the highest frequency contained in the measured signal. Denoting this frequency, which is called the NYQUIST FREQUENCY, as π5 , the sample theorem requires that: π. > 2π5 Where: β’ π. is the frequency of sampling β’ π5 is the Nyquist frequency, representing the highest frequency contained in the measure signal
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Therefore, the sampling frequency must be greater than the Nyquist rate for accurate reproduction. Using the formula for the Nyquist frequency, we plug in the calculated value for the sampling frequency and calculate the highest frequency that can be accurately reported as: π. = 2 π5 = 2 20 = 40 ππ»π§ 2 20 = 40 ππ»π§
Therefore, the correct answer choice is D. ππ
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PROBLEM 6: Whenever the Florida Gators football team scores a touchdown, the crowd roars to a sampled frequency of 300 ππ»π§. If this is an accurate sample frequency, what is the highest frequency that can be accurately reported most close to: A. 150 Γ 10* B. 225 Γ 10* C. 250 Γ 10* D. 300 Γ 10*
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SOLUTION 6: The formula for NYQUIST FREQUENCY can be referenced under the topic of INSTRUMENTATION, MEASUREMENT, AND CONTROLS on page 127 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Nyquistβs (Shannonβs) sampling theorem states that in order to accurately reconstruct the analog signal from the discrete sample points, the sample rate must be larger than twice the highest frequency contained in the measured signal. Denoting this frequency, which is called the NYQUIST FREQUENCY, as π5 , the sample theorem requires that: π. > 2π5 Where: β’ π. is the frequency of sampling β’ π5 is the Nyquist frequency, representing the highest frequency contained in the measure signal Therefore, the sampling frequency must be greater than the Nyquist rate for accurate reproduction. Using the formula for the Nyquist frequency, we plug in the calculated value for the sampling frequency and calculate the highest frequency that can be accurately reported as:
π5
