Q#19 (Past Exam Paper – June 2014 Paper 13 Q5)
The display on a cathode-ray oscilloscope shows the signal produced by an electronic circuit.
The time-base is set at 5.0 ns per division and the Y-gain at 10 V per division.
What is the frequency of the signal?
A 2.0 × 10-8 Hz
B 2.5 × 10-2 Hz
C 5.0 × 107 Hz
D 3.1 × 108 Hz
Solution:
Answer: C.
Considering the 2 peaks, it can be observed that a complete wave occupies 4 divisions (horizontally).
Time-base = 5.0 ns / div
1 division - - > 5.0 ns
4 divisions - - > 4 × 5.0 = 20 ns
So, the period T = 20 ns
Frequency = 1 / T = 1 / (20×10-9) = 5.0×107 Hz
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