A sine wave plus all its odd harmonics can be mathematically represented by

sin(θ) + sin(3θ)/3 + sin(5θ)/5 + ...

but this is equivalent (differing only by amplitude, and k = a constant multiple of pi) to

(4/pi) X [sin(kθ) + sin(3kθ)/3 + sin(5kθ)/5 + ...]

which, according to An Introduction to Harmonic Analysis (by Yitzhak Katznelson, Dover Publications, 1976), is the Fourier Expansion of a square wave, making answer (A) the correct one.

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A cosine wave, square wave, and sine wave all have symmetric rise-and-fall patterns because they are all sums of their sinusoid harmonics, according to Hugh L. Montgomery and Robert C. Vaughan (Multiplicative Number Theory, Cambridge, 2007, p. 536-537).

But a sawtooth wave is an asymmetric triangular wave, because its rise time differs from its fall time, making it the correct one.

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A square wave consists of the fundamental and odd harmonics because it is symmetric.

A sine and cosine wave consist of only the fundamental frequency, so there are no harmonics.

A sawtooth wave consists of the fundamental, odd, and even harmonics. Only asymmetric waves contain even harmonics and this is the only answer which is an asymmetric wave.

*Hint - Sawtooth wave = all harmonics (Sawz-All)

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According to Fundamentals of Electric Circuits (McGraw-Hill, 2012, p. 467), the RMS (root-mean-square) of a sinusoidal AC voltage is essentially the square root of the arithmetic mean (averages) of the squares of the original values, which, after performing the actual calculation of

Vrms = lim (T->∞) SQRT(integral(0 to T)(V(t)^2)dt / T)

yields Vrms = Vp/SQRT(2),

which is a useful quantity because it is identical to the value of DC voltage that is applied across a resistor to result in the same amount of output power that would heat the resistor, resulting in answer (C).

Because Vrms = Vp/SQRT(2), and the average peak AC value is Vp, the square of the average peak AC value is Vp^2, which does not equal Vp/SQRT(2), eliminating answer (A).

Also, the amount of heat in a given resistor as the corresponding peak AC voltage is

Pp = (Vp^2)/R, but the Prms = (Vrms^2)/R =

((Vp/SQRT(2))^2)/R = (Vp^2)/2R =

((Vp^2)/R)/2 = Pp/2,

or only half the heat given off by using the peak value, which eliminates answer (B).

Finally, the average AC value, because of its symmetric (positive and negative) nature, is zero, and the square root of zero is zero, so answer (D) is also incorrect.

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The term “RMS” stands for “Root-Mean-Squared”. Most books define this as the “amount of AC power that produces the same heating effect as an equivalent DC power”.

By Ohm's Law (\(V=IR\)), Voltage, Intensity (or Current) and Resistance are related. To know voltage, it is sufficient to know current and resistance. By having a complex waveform, there will not be any easy formula to convert AC voltage to RMS Voltage. By measuring the heating effect in a known resistor, we would find the current passing through it, with which we could find the RMS voltage applied to the circuit.

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PEP is Peak Envelope Power. it's the highest power passed to the antenna from the transmitter.

PEP-to-average power ratio is determined by the waveform shape made by the voice, thus the characteristics of the modulating signal

The first thing to remember is that PEP to average ratio is not a static value and is determined by the specific signal you are looking at - because of this, it can vary. Looking at the other answers:

• The frequency of the modulating signal - this is a signal used to move speech up to the RF frequency - It's not going to vary over time, so doesn't contribute to finding the ratio.
• The degree of carrier suppression - similar to the above answer, this is a part of the signal that doesn't vary, so won't vary.
• Amplifier gain - again, a mechanism that doesn't vary, so it won't contribute to calculating the answer.

Hint: average speech [KQ4AEY]

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A pulse width modulated signal can be used to encode data for transmission whereby the width of the pulses can be encoded to a specific value.

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Analog-->Digital, which is to "sample" the signal.

Sampling a signal actually involves recording an amplitude value at a specific point in time and storing the value off as a time:value in a lookup table. This long standing technology extends over many technologies include Compact Disks, MP3 files, etc... The higher the sampling rate, the higher the resolution. With today's resolution, I'm amazed as how some people can still tell the difference between music recorded on vinyl (analog) and CDs (digital).

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The waveform of a stream of digital data bits would look like a series of pulses with varying patterns on a conventional oscilloscope. (E8A15)

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