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Difficulty: HardOscilloscopes

A communications technician connects an oscilloscope to monitor an unmodulated carrier signal from a transmitter stage. A standard 10×10\times passive attenuation probe is connected to Channel 1, while the oscilloscope input channel remains set to standard 1×1\times scale. On the display, the sinusoidal trace spans 6 vertical divisions6\text{ vertical divisions} from peak to peak with the vertical sensitivity set to 2 V/div2\text{ V/div}. Horizontally, exactly 4 complete cycles4\text{ complete cycles} span across 8 divisions8\text{ divisions} with the horizontal timebase set to 5 μs/div5\text{ }\mu\text{s/div}. What are the true peak-to-peak voltage (Vp-pV_{\text{p-p}}) and frequency (ff) of the input signal?

  1. A
    12 Vp-p and 100 kHz12\text{ V}_{\text{p-p}}\text{ and }100\text{ kHz}
  2. B
    120 Vp-p and 25 kHz120\text{ V}_{\text{p-p}}\text{ and }25\text{ kHz}
  3. 120 Vp-p and 100 kHz120\text{ V}_{\text{p-p}}\text{ and }100\text{ kHz}Answer
  4. D
    60 Vp-p and 100 kHz60\text{ V}_{\text{p-p}}\text{ and }100\text{ kHz}

Answer

120 Vp-p and 100 kHz120\text{ V}_{\text{p-p}}\text{ and }100\text{ kHz}
The true peak-to-peak voltage is calculated by multiplying the 6 vertical divisions6\text{ vertical divisions} by 2 V/div2\text{ V/div} and accounting for the 10×10\times probe factor, yielding 120 Vp-p120\text{ V}_{\text{p-p}}. The period for one cycle is 2 divisions×5 μs/div=10 μs2\text{ divisions} \times 5\text{ }\mu\text{s/div} = 10\text{ }\mu\text{s}, which gives a frequency of f=1/(10 μs)=100 kHzf = 1 / (10\text{ }\mu\text{s}) = 100\text{ kHz}.

Step-by-Step Solution

1
Calculate the displayed peak-to-peak voltage from the vertical graticule divisions and Volts/Div setting
Vp-p(displayed)=6 divisions×2 V/div=12 VV_{\text{p-p(displayed)}} = 6\text{ divisions} \times 2\text{ V/div} = 12\text{ V}
Each vertical division represents 2 volts on the scope screen.
2
Apply the 10X attenuation probe multiplier to obtain the true input voltage
Vp-p(true)=12 V×10=120 Vp-pV_{\text{p-p(true)}} = 12\text{ V} \times 10 = 120\text{ V}_{\text{p-p}}
A 10X attenuation probe divides the circuit voltage by 10 before reaching the scope input, so the observed display value must be multiplied by 10 when the scope is not set to auto-scale.
3
Determine the horizontal division span for one complete cycle
Divisions per cycle=8 divisions4 cycles=2 div/cycle\text{Divisions per cycle} = \frac{8\text{ divisions}}{4\text{ cycles}} = 2\text{ div/cycle}
One full period corresponds to exactly one complete waveform cycle.
4
Calculate the period (T) of the waveform using the horizontal timebase setting
T=2 div×5 μs/div=10 μs=10×10−6 sT = 2\text{ div} \times 5\text{ }\mu\text{s/div} = 10\text{ }\mu\text{s} = 10 \times 10^{-6}\text{ s}
Multiplying the division width of one cycle by the timebase gives the duration of one period.
5
Calculate the frequency (f) using the reciprocal of the period
f=1T=110×10−6 s=100000 Hz=100 kHzf = \frac{1}{T} = \frac{1}{10 \times 10^{-6}\text{ s}} = 100{}000\text{ Hz} = 100\text{ kHz}
Frequency is the inverse of the period: f=1/Tf = 1/T.

Key Concept

Oscilloscope graticule measurement incorporating 10X probe scaling and period-to-frequency conversion
Estimated Time:2m 0s
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Topics

Oscilloscopes

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