Electrical Engineering Students’ Achievement in Measurement Accuracy of Digitized Signals – Work in Progress

Abstract

The current paper presents of a continuous longitudinal research dealing with different aspects of accuracy and the way the students understand it. Previous studies have shown that understanding the accuracy concept is not obvious. A previous study exposed students’ difficulties in calculating measurement accuracy of digitized signals. After writing a special study unit on measurement accuracy and measurement error and adding it to the curriculum of Digital Electronics course, two questions questionnaire entered to the final test on digital electronics course. The students’ grades showed an improvement in understanding the accuracy concept, but not sufficient. The study unit included explanation, tutorial and homework. A replication of the previous experiment in Fall semester 2015 showed a substantial improvement in the students’ achievements regarding the accuracy concept. This time the students got explanation and example but no homework concerning the accuracy concept. Nevertheless, they had the chance to see and solve the previous questionnaire so the questions regarding accuracy did not surprise them. Solving old test questionnaires is a common mode of learning among our students therefore this may be the explanation of the improvement in the students’ grades.

Appendix – Questionnaire Solution

Answer 1.

The noise amplitude can be either 0.06 mV or −0.06 mV. Therefore, signal + noise range is between N + S(max) = 200.06 mV to N + S(min) = 199.94 mV.

The digital output for 200.06 mV is: $N\hbox{max} = \left[ {\frac{200.06}{500} \times 2^{14} } \right] = 6555$ and for 199.94 mv is: $N\hbox{min} = \left[ {\frac{199.94}{500} \times 2^{14} } \right] = 6551$.

The digital output for 200 mV is: $N = \left[ {\frac{200}{500} \times 2^{14} } \right] = 6553$. The Error for both extremes is: $Err = \frac{6555 - 6553}{6553} \times 100\% = \frac{6553 - 6551}{6553} \times 100\% = 0.03\%$.