SIGNALS AND EMISSIONS
Waveforms: measurement, peak-to-peak, RMS, average; Electromagnetic Waves: definition, characteristics, polarization
Which of the following is the easiest voltage amplitude parameter to measure when viewing a pure sine wave signal on an analog oscilloscope?
-
Correct AnswerPeak-to-peak voltage
-
RMS voltage
-
Average voltage
-
DC voltage
Tags: none
What is the relationship between the peak-to-peak voltage and the peak voltage amplitude of a symmetrical waveform?
-
0.707:1
-
Correct Answer2:1
-
1.414:1
-
4:1
peak to peak is 2x peak voltage in a symmetrical waveform.
Last edited by jmsian. Register to edit
Tags: none
What input-amplitude parameter is valuable in evaluating the signal-handling capability of a Class A amplifier?
-
Correct AnswerPeak voltage
-
RMS voltage
-
Average power
-
Resting voltage
A class A amplifier features an amplifying element (Transistor, MOSFET, Tube, etc) that must remain in conduction (not saturated or cut off) during the entire wave form.
- RMS voltage does not represent the entire input amplitude range.
- Neither does average power.
- Resting voltage (or quiescent operation) also does not tell us anything except the DC state of the amplifier.
ONLY when looking at peak voltage will one be able to evaluate the capability of a class A amplifier.
Last edited by kevinruggles. Register to edit
Tags: none
What is the PEP output of a transmitter that develops a peak voltage of 30 volts into a 50-ohm load?
-
4.5 watts
-
Correct Answer9 watts
-
16 watts
-
18 watts
-
PV = 30V
RMS V = (0.5)((2)^(1/2))*PV
RMS V = 15*sqrt (2)
-
I = V/R = (.3)*sqrt (2)
-
P = I^2*R = ((.3)*sqrt (2))^2 * 50 = 9
OR
PV = 30V RMS E = PV/sqrt(2), = 21V
Power = E*I
You know E and R. E=IR, R=E/I. So, Power also = E^2/R
((21)^2)/50 = 9 watts
or, the easiest way to remember this is to take the PV and multiply it by 0.707. This gives you the approximate RMS voltage. Square the product, then divide it by the resistance to determine the power.
PV = 30V Vrms = 0.707PV given as 300.707 = 21.21 Vrms 21.21*21.21 = 449.86 449.86/50 = 8.997W (round up to 9W for the closest answer)
Last edited by boothe. Register to edit
Tags: none
If an RMS-reading AC voltmeter reads 65 volts on a sinusoidal waveform, what is the peak-to-peak voltage?
-
46 volts
-
92 volts
-
130 volts
-
Correct Answer184 volts
We are given the RMS value of the sinusoidal wave form. We must first compute the peak value:
Peak = RMS / (1/sqrt(2)), which is to say: ** Peak Voltage = RMS Voltage/(0.707)**
Which for our problem is: Peak = 65V/0.707 = 91.94V
But this is not the answer! The question asks for Peak-to-Peak, which is 2*Peak.
For our problem: Peak-To-Peak = 2* 91.94V = 183.88 Volts
Last edited by floottooter. Register to edit
Tags: none
What is the advantage of using a peak-reading wattmeter to monitor the output of a SSB phone transmitter?
-
It is easier to determine the correct tuning of the output circuit
-
Correct AnswerIt gives a more accurate display of the PEP output when modulation is present
-
It makes it easier to detect high SWR on the feed line
-
It can determine if any flat-topping is present during modulation peaks
Tags: none
What is an electromagnetic wave?
-
Alternating currents in the core of an electromagnet
-
A wave consisting of two electric fields at right angles to each other
-
Correct AnswerA wave consisting of an electric field and a magnetic field oscillating at right angles to each other
-
A wave consisting of two magnetic fields at right angles to each other
Tags: none
Which of the following best describes electromagnetic waves traveling in free space?
-
Electric and magnetic fields become aligned as they travel
-
The energy propagates through a medium with a high refractive index
-
The waves are reflected by the ionosphere and return to their source
-
Correct AnswerChanging electric and magnetic fields propagate the energy
Tags: none
What is meant by circularly polarized electromagnetic waves?
-
Waves with an electric field bent into a circular shape
-
Correct AnswerWaves with a rotating electric field
-
Waves that circle the Earth
-
Waves produced by a loop antenna
In electrodynamics, circular polarization of an electromagnetic wave is a polarization in which the electric field of the passing wave does not change strength but only changes direction in a rotary manner.
https://en.wikipedia.org/wiki/Circular_polarization
Memory aid: circularly: like a wheel rotating
Last edited by robertov416. Register to edit
Tags: none
What type of meter should be used to monitor the output signal of a voice-modulated single-sideband transmitter to ensure you do not exceed the maximum allowable power?
-
An SWR meter reading in the forward direction
-
A modulation meter
-
An average reading wattmeter
-
Correct AnswerA peak-reading wattmeter
the key here is you need to find allowable power (watts). Knowing this you can eliminate the two answers which do not refer to a wattmeter. Next thing to look at is what kind of allowable power and that is maximum, and of the two remaining answers the peak-reading wattmeter will give you your maximum watts reading, the average will only give you an average of the maximum and minimum watts out.
Last edited by beringhill. Register to edit
Tags: none
What is the average power dissipated by a 50-ohm resistive load during one complete RF cycle having a peak voltage of 35 volts?
-
Correct Answer12.2 watts
-
9.9 watts
-
24.5 watts
-
16 watts
Average power dissipation is computed based on the RMS value, not peak or peak-to-peak values.
To answer this question, first determine the RMS voltage of the RF cycle:
RMS = Peak Voltage / sqrt(2) = 35V/1.414 = 24.745V
Then, use that value in Ohm's law to compute the average power: P = E^2 / R
P = 24.745^2 / 50 = 612.5V / 50 Ohms = 12.25 Watts
Last edited by orochimaru. Register to edit
Tags: none
What is the peak voltage of a sinusoidal waveform if an RMS-reading voltmeter reads 34 volts?
-
123 volts
-
96 volts
-
55 volts
-
Correct Answer48 volts
RMS voltage is simply Peak Voltage (Vp) divided by the square root of 2. The square root of two is 1.414. If the RMS voltage is 34 then Vp = Sqrt of 2 * RMS or 48 = 1.414 * 34
Last edited by k9lpk. Register to edit
Tags: none
Which of the following is a typical value for the peak voltage at a standard U.S. household electrical outlet?
-
240 volts
-
Correct Answer170 volts
-
120 volts
-
340 volts
Which of the following is a typical value for the peak-to-peak voltage at a standard U.S. household electrical outlet?
-
240 volts
-
120 volts
-
Correct Answer340 volts
-
170 volts
Vpeak = Vrms * sqrt(2)
sqrt(2) ~= 1.414
US Household Vrms = 120V
so,
Vpeak = 120 x 1.414 = 169.7V
Peak to peak voltage = 2 x Vpeak = 339V
So, correct answer, rounding up, is 340V
Last edited by bennettj1087. Register to edit
Tags: none
Which of the following is a typical value for the RMS voltage at a standard U.S. household electrical power outlet?
-
Correct Answer120V AC
-
340V AC
-
85V AC
-
170V AC
Don't get tripped-up by thinking this question is looking for something "fancy" like "PEAK voltage of a standard U.S. household electrical power outlet" (170V AC) or "PEAK TO PEAK voltage of a standard U.S. household electrical power outlet" (340V AC)
The value of "120V AC" commonly associated with standard U.S. household electrical power outlets is, in fact, the RMS value, which typically goes unsaid.
Last edited by kc5qnj. Register to edit
Tags: none
What is the RMS value of a 340-volt peak-to-peak pure sine wave?
-
Correct Answer120V AC
-
170V AC
-
240V AC
-
300V AC
(A). The RMS value of a 340 volt peak-to-peak pure sine wave is 120V AC.
In a sine wave, the RMS value is given as:
a / sqrt(2) given: a as amplitude
For this question: (340 / 2) * 0.707 ~= 120
Divide 340 in half to account for half wave and 0.707 is an approximation of sqrt(2)
Last edited by hucker. Register to edit
Tags: none
View Privacy Policy | Get help with HamStudy.org™