The phase difference between two signals of equal frequency may be found from an elliptical Lissajous figure as the inverse sine of
Answer: A
Where the ellipse crosses the vertical axis the horizontal signal is zero, and the height at that point relative to the peak height is the sine of the phase angle. The method is quick and needs no calibration of either axis.
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Measuring phase difference from the time base display rather than from a Lissajous figure has the advantage that
Answer: B
With both waveforms swept together it is plain which crosses zero first, whereas an ellipse of a given shape corresponds to a lead or a lag equally. The time between corresponding points, as a fraction of the period, gives the angle.
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A waveform occupies 4.6 divisions peak to peak with the vertical control at 0.5 V per division. Its peak to peak value is
Answer: C
The product is \( 4.6 \times 0.5 = 2.3 \) V peak to peak. The peak value is half of that and the root mean square value of a sinusoid is the peak divided by root two, that is 0.81 V, so it is worth being clear which quantity is wanted.
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A sinusoid measures 2.3 V peak to peak on the screen. Its root mean square value is about
Answer: D
The peak is 1.15 V and the root mean square value is \( 1.15/1.414 = 0.81 \) V. An oscilloscope always shows the instantaneous waveform, so every root mean square figure taken from it involves a factor that depends on the shape of the wave.
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One cycle of a waveform occupies 5 divisions with the time base at 2 ms per division. The frequency is
Answer: A
The period is \( 5 \times 2 = 10 \) ms and its reciprocal is 100 Hz. Spreading the waveform over as much of the screen as possible reduces the fractional error in reading the number of divisions, so one cycle across ten divisions is the ideal.
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A pulse rises across 1.4 divisions with the time base at 50 ns per division. Its rise time as displayed is
Answer: B
The displayed rise time is \( 1.4 \times 50 = 70 \) ns. If the instrument's own rise time is appreciable the true value is the square root of the difference of the squares, which matters when the two are within a factor of three of each other.
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An oscilloscope of 3.5 ns rise time displays a pulse whose rise appears to be 5 ns. The true rise time of the pulse is about
Answer: C
The rise times combine in quadrature, so the true value is \( \sqrt{5^2 - 3.5^2} = \sqrt{25 - 12.25} = 3.57 \) ns. With the two so close the instrument is contributing as much as the pulse, and a faster oscilloscope is really needed.
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An oscilloscope has a bandwidth of 100 MHz. Its rise time is about
Answer: D
Correct answer: 3.5 ns
\[ t_r = \frac{0.35}{BW} = \frac{0.35}{100\times10^{6}} = 3.5\ \mathrm{ns}. \] The constant 0.35 comes from the response of a single-pole system between the ten and ninety per cent points, and the relation is why bandwidth and rise time are two ways of quoting the same limitation. Note also that the instrument and the signal add in quadrature, so a scope must be several times faster than the edge it is to measure honestly.
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A tube has a deflection sensitivity of 0.5 mm/V. A signal of 40 V peak to peak produces a deflection of
Answer: A
Correct answer: 20 mm
\[ 0.5\ \mathrm{mm/V} \times 40\ \mathrm{V} = 20\ \mathrm{mm}. \] The signal is quoted peak to peak and the sensitivity applies to the total voltage swing, so the answer is the full peak-to-peak deflection on the screen. Halving it to work in peak volts would give 10 mm, which is offered here.
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A Lissajous figure shows 3 tangencies to a horizontal line and 2 to a vertical line. If the horizontal input is 100 Hz, the vertical input is
Answer: B
The vertical frequency is to the horizontal as the horizontal tangencies are to the vertical ones, so \( f_y = 100 \times 3/2 = 150 \) Hz. Counting tangencies rather than crossings avoids the ambiguity that arises when the figure touches itself.
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