Oscilloscopes Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB)

Question 1Question →

Match each oscilloscope front-panel control or display feature on the left with its corresponding measurement function on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Volts/Div matches vertical scale (amplitude/voltage measurement), Time/Div matches horizontal sweep rate (period/frequency measurement), Trigger matches waveform stabilization, and Graticule grid matches the calibrated measurement overlay.

Each oscilloscope feature is mapped directly to its core operational role: the Volts/Div knob scales the vertical axis for voltage amplitude, Time/Div scales the horizontal axis for time intervals, the Trigger synchronizes the horizontal sweep for a stable trace, and the Graticule acts as the calibrated grid overlay.

Step-by-Step Solution

1

Identify the vertical axis control and its function

Volts/Div controls vertical scaling to measure voltage values such as peak or peak-to-peak amplitude.

Vertical displacement on an oscilloscope screen is directly proportional to signal voltage.

2

Identify the horizontal axis control and its function

Time/Div controls horizontal sweep time to measure period and calculate frequency (

f = \frac{1}{T}

).

Horizontal displacement represents elapsed time across the screen.

3

Identify the synchronization mechanism

The Trigger control stabilizes the signal display.

Without proper triggering, the trace starts at random points of the cycle, causing the waveform to scroll continuously.

4

Identify the physical reference markings

The Graticule grid provides the reference divisions for counting units.

The graticule lines allow the operator to multiply division counts by the scale settings.

Key Concept

Basic oscilloscope controls and display anatomy

Estimated Time:45s

Question 2Question →

An electronics technician connects an oscilloscope to test an alternating current circuit. The vertical scale (Volts/Div) is set to $5\text{ V/div}$ , and the displayed waveform has a total vertical span of $4\text{ divisions}$ from the highest peak to the lowest trough. What is the peak-to-peak voltage ( $V_{\text{p-p}}$ ) of the signal?

10\text{ V}

20\text{ V}

4\text{ V}

0.8\text{ V}

Show answer & explanation

Answer: $20\text{ V}$

Answer

20\text{ V}

Peak-to-peak voltage (

V_{\text{p-p}}

) is the total voltage difference between the highest positive peak and the lowest negative trough of a waveform. Multiplying the total vertical displacement of

4\text{ divisions}

by the vertical sensitivity setting of

5\text{ V/div}

gives

4 \times 5 = 20\text{ V}

.

Step-by-Step Solution

1

Identify the given vertical division count and the Volts/Div scale setting.

Vertical span =

4\text{ divisions}

; Vertical scale =

5\text{ V/div}

.

The vertical axis on an oscilloscope measures voltage according to the specified Volts/Div sensitivity.

2

Calculate the peak-to-peak voltage (

V_{\text{p-p}}

) by multiplying the total vertical divisions by the Volts/Div setting.

V_{\text{p-p}} = 4\text{ divisions} \times 5\text{ V/div} = 20\text{ V}

.

Each vertical grid square represents

5\text{ V}

, so

4

full squares equal

20\text{ V}

from minimum trough to maximum peak.

Key Concept

Calculating peak-to-peak voltage on an oscilloscope using graticule divisions and the Volts/Div scale setting

Estimated Time:45s

Question 3Question →

An oscilloscope displays a periodic alternating voltage waveform on its screen. One complete cycle of the waveform occupies $4\text{ horizontal divisions}$ across the graticule. If the oscilloscope's horizontal time base (Time/Div) control is set to $15\text{ ms/div}$ , what is the period of the waveform in milliseconds?

Show answer & explanation

Answer: 60

Answer

60\text{ ms}

The period (

T

) of a signal displayed on an oscilloscope is found by multiplying the number of horizontal divisions for one complete cycle by the Time/Div control setting:

4\text{ divisions} \times 15\text{ ms/div} = 60\text{ ms}

.

Step-by-Step Solution

1

Identify the number of horizontal graticule divisions occupied by one full waveform cycle.

4\text{ divisions}

On an oscilloscope display, the horizontal axis represents time, and one full cycle corresponds to the period (

T

).

2

Multiply the division count by the horizontal scale factor (

15\text{ ms/div}

).

4\text{ div} \times 15\text{ ms/div} = 60\text{ ms}

Multiplying the span in divisions by the time allocated per division gives the total duration of one cycle.

Key Concept

Calculating waveform period from oscilloscope horizontal division settings

Estimated Time:45s

Question 4Question →

An electronics technician is troubleshooting an intermittent circuit fault using an analog/digital oscilloscope. Match each oscilloscope control feature on the left with its correct functional operation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Trigger Level matches setting the voltage threshold to stabilize the sweep; AC Coupling matches inserting a series capacitor to block DC bias; Timebase (SEC/DIV) matches controlling horizontal sweep speed per division; Vertical Position matches shifting the trace vertically without changing signal amplitude.

Each control corresponds directly to its front-panel function: Trigger Level establishes the voltage threshold for horizontal sweep synchronization; AC Coupling blocks DC bias using an internal capacitor to view small AC variations; Timebase dictates horizontal sweep speed per division; and Vertical Position shifts the trace vertically for convenient graticule alignment.

Step-by-Step Solution

1

Identify the synchronization function of the oscilloscope.

Trigger Level sets the precise voltage threshold required to synchronize the horizontal sweep generator with the incoming signal.

Without proper trigger settings, periodic waveforms scroll or overlap asynchronously on the screen.

2

Identify the input filtering configuration.

AC Coupling engages an internal series capacitor to block steady DC levels while passing time-varying AC signals.

This allows small AC fluctuations (such as power supply ripple) to be viewed at high vertical sensitivity without driving the trace off-screen.

3

Identify the horizontal time-domain scaling control.

Timebase (SEC/DIV) calibrates the horizontal rate of trace movement across graticule divisions.

Adjusting SEC/DIV allows the user to expand or compress the horizontal waveform display to measure period and frequency.

4

Identify the trace alignment control.

Vertical Position relocates the waveform baseline up or down across the vertical graticule.

This provides visual alignment against graticule reference lines without modifying vertical gain (Volts/Div).

Key Concept

Oscilloscope Front-Panel Controls and Signal Conditioning

Estimated Time:1m 15s

Question 5Question →

An audio technician connects an oscilloscope to analyze a recurring tone from a signal generator. The horizontal Time/Div control is set to $250\text{ }\mu\text{s/div}$ . On the graticule display, exactly two complete cycles of the sine wave span $8\text{ horizontal divisions}$ . What is the frequency of the audio signal?

500\text{ Hz}

1\text{ kHz}

4\text{ kHz}

100\text{ Hz}

Show answer & explanation

Answer: $1\text{ kHz}$

Answer

The frequency of the audio signal is

1\text{ kHz}

(

1{}000\text{ Hz}

).

One complete sine wave cycle occupies

\frac{8\text{ divisions}}{2} = 4\text{ divisions}

. Multiplying by the horizontal timebase of

250\text{ }\mu\text{s/div}

gives a period of

T = 1{}000\text{ }\mu\text{s} = 1\text{ ms} = 0.001\text{ s}

. The frequency is

f = \frac{1}{T} = \frac{1}{0.001\text{ s}} = 1{}000\text{ Hz} = 1\text{ kHz}

.

Step-by-Step Solution

1

Determine the number of horizontal divisions occupied by one complete cycle.

\text{Divisions per cycle} = \frac{8\text{ divisions}}{2\text{ cycles}} = 4\text{ div/cycle}

The period of a periodic wave is defined by the horizontal duration of one full cycle.

2

Calculate the period (

T

) of the waveform by multiplying divisions per cycle by the horizontal timebase scale.

T = 4\text{ div} \times 250\text{ }\mu\text{s/div} = 1000\text{ }\mu\text{s} = 1 \times 10^{-3}\text{ s}

The Time/Div control establishes the time value represented by each graticule division.

3

Calculate the frequency (

f

) using the reciprocal formula

f = \frac{1}{T}

.

f = \frac{1}{1 \times 10^{-3}\text{ s}} = 1000\text{ Hz} = 1\text{ kHz}

Frequency is the inverse of the period in seconds.

Key Concept

Calculating waveform period and frequency from oscilloscope horizontal timebase settings and graticule divisions.

Estimated Time:1m 0s

Question 6Question →

A radio technician is testing the output of an RF signal generator using an oscilloscope. The horizontal timebase control is set to $250\text{ }\mu\text{s/div}$ . On the screen graticule, a continuous sinusoidal waveform is observed where exactly $2.5\text{ complete cycles}$ span across $8\text{ horizontal divisions}$ . What is the frequency of this signal, in hertz ( $\text{Hz}$ )?

Show answer & explanation

Answer: 1250

Answer

1250 Hz

The total duration of the

8\text{ horizontal divisions}

at

250\text{ }\mu\text{s/div}

is

2000\text{ }\mu\text{s}

(

2\text{ ms}

). Dividing by the

2.5\text{ complete cycles}

gives a period of

T = 800\text{ }\mu\text{s} = 0.0008\text{ s}

per cycle. Taking the reciprocal gives the frequency:

f = 1 / 0.0008\text{ s} = 1250\text{ Hz}

.

Step-by-Step Solution

1

Calculate the total elapsed time across the observed graticule divisions.

Total time

= 8\text{ div} \times 250\text{ }\mu\text{s/div} = 2000\text{ }\mu\text{s}

The horizontal axis represents time, and the timebase setting indicates the duration per division.

2

Calculate the period (

T

) of a single cycle.

T = \frac{2000\text{ }\mu\text{s}}{2.5\text{ cycles}} = 800\text{ }\mu\text{s} = 800 \times 10^{-6}\text{ s} = 0.0008\text{ s}

The period is the time required to complete one single cycle.

3

Calculate the frequency (

f

) from the period.

f = \frac{1}{T} = \frac{1}{0.0008\text{ s}} = 1250\text{ Hz}

Frequency in hertz is the reciprocal of the period in seconds (

f = 1/T

).

Key Concept

Calculating waveform period and frequency from oscilloscope horizontal timebase settings and graticule divisions.

Estimated Time:1m 30s

Question 7Question →

An oscilloscope is used to measure an alternating signal. The horizontal sweep rate (Time/Div) is set to $8\text{ }\mu\text{s/div}$ , and one complete cycle of the waveform spans exactly $5\text{ horizontal divisions}$ across the graticule. What is the frequency of this signal in kilohertz ( $\text{kHz}$ )?

Show answer & explanation

Answer: 25

Answer

The frequency of the signal is

25\text{ kHz}

.

One full waveform cycle occupies

5

horizontal divisions at

8\text{ }\mu\text{s/division}

, giving a period of

T = 40\text{ }\mu\text{s}

(

40 \times 10^{-6}\text{ s}

). The frequency is the reciprocal of the period (

f = \frac{1}{T} = \frac{1}{40 \times 10^{-6}\text{ s}} = 25{}000\text{ Hz}

), which equals

25\text{ kHz}

.

Step-by-Step Solution

1

Multiply the number of horizontal graticule divisions per cycle by the time base setting to find the period (

T

).

T = 5\text{ divisions} \times 8\text{ }\mu\text{s/div} = 40\text{ }\mu\text{s} = 40 \times 10^{-6}\text{ s}

Horizontal divisions represent the time dimension on an oscilloscope display.

2

Calculate the frequency (

f

) in hertz using the reciprocal relationship

f = \frac{1}{T}

.

f = \frac{1}{40 \times 10^{-6}\text{ s}} = 25{}000\text{ Hz}

Frequency is defined as the number of cycles per second.

3

Convert the resulting frequency from hertz (

\text{Hz}

) to kilohertz (

\text{kHz}

).

\frac{25{}000\text{ Hz}}{1{}000} = 25\text{ kHz}

The question specifies the final answer unit as kilohertz.

Key Concept

Calculating waveform period and frequency from oscilloscope horizontal Time/Div settings and graticule division counts

Estimated Time:1m 0s

Question 8Question →

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

12\text{ V}_{\text{p-p}}\text{ and }100\text{ kHz}

120\text{ V}_{\text{p-p}}\text{ and }25\text{ kHz}

120\text{ V}_{\text{p-p}}\text{ and }100\text{ kHz}

60\text{ V}_{\text{p-p}}\text{ and }100\text{ kHz}

Show answer & explanation

Answer: $120\text{ V}_{\text{p-p}}\text{ and }100\text{ kHz}$

Answer

120\text{ V}_{\text{p-p}}\text{ and }100\text{ kHz}

The true peak-to-peak voltage is calculated by multiplying the

6\text{ vertical divisions}

by

2\text{ V/div}

and accounting for the

10\times

probe factor, yielding

120\text{ V}_{\text{p-p}}

. The period for one cycle is

2\text{ divisions} \times 5\text{ }\mu\text{s/div} = 10\text{ }\mu\text{s}

, which gives a frequency of

f = 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

V_{\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

V_{\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

\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\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 = \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/T

.

Key Concept

Oscilloscope graticule measurement incorporating 10X probe scaling and period-to-frequency conversion

Estimated Time:2m 0s

Question 9Question →

A technician is evaluating an AC ripple voltage on a DC power supply using an oscilloscope. The vertical sensitivity control (Volts/Div) is set to $50\text{ mV/div}$ . If the displayed waveform occupies a total of $6\text{ vertical divisions}$ from its lowest trough to its highest crest, what is the peak voltage ( $V_{\text{pk}}$ ) of the ripple?

300\text{ mV}

150\text{ mV}

25\text{ mV}

1.5\text{ V}

Show answer & explanation

Answer: $150\text{ mV}$

Answer

The peak voltage of the ripple is

150\text{ mV}

.

Multiplying the total vertical deflection (

6\text{ divisions}

) by the vertical sensitivity (

50\text{ mV/div}

) yields a peak-to-peak voltage of

300\text{ mV}

. Peak voltage (

V_{\text{pk}}

) measures the amplitude from the reference centerline to the positive crest, which equals half of the peak-to-peak voltage:

\frac{300\text{ mV}}{2} = 150\text{ mV}

.

Step-by-Step Solution

1

Calculate the peak-to-peak voltage (

V_{\text{p-p}}

)

V_{\text{p-p}} = 6\text{ divisions} \times 50\text{ mV/div} = 300\text{ mV}

The total vertical span from crest to trough represents the full peak-to-peak voltage.

2

Calculate the peak voltage (

V_{\text{pk}}

)

V_{\text{pk}} = \frac{V_{\text{p-p}}}{2} = \frac{300\text{ mV}}{2} = 150\text{ mV}

Peak voltage is the amplitude measured from the center baseline to either peak, which is half of the peak-to-peak value.

Key Concept

Determining peak voltage from peak-to-peak vertical graticule measurements on an oscilloscope.

Estimated Time:1m 0s

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