Fourier analysis allows for a time domain signal (Anything we can measure as a function of time) to be transformed into the frequency domain. Letting anyone determine what frequency components exist in the signal.

Hint: Squares have Four(ier) sides

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An easy way to remember this one is that analog-to-digital conversion is always an approximation, because the digital version of something is only an approximation of the analog version. The other answers are unrelated to analog-to-digital conversion.

Read more about Successive Approximation ADCs on Wikipedia.

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For the mathematically educated, don't overthink this one.. The trick here, is to notice that the question is deliberately abstract, rather than specific. It's not talking about voltages or electromagnetic waves. The word "signal" is generic, and it could be anything: voltage, pressure in sound waves, etc. Similarly, the word "amplitude" is a generic word that means "amount of something". So, you want to plot "some kind of signal" (whatever it is) over time? Then you will want to show its amplitude (whatever it is) over time.

• Power at intervals of phase. This is just wrong in every way. It's the only answer that doesn't even mention time.
• Frequency at different times. This one isn't totally crazy. This is how frequency modulation works. You vary frequency over time. But that's a very specific and unique case, and certainly isn't a general example of how to describe any signal in the time domain.
• Discrete impulses in time order. You wouldn't describe a signal as a series of impulses unless it consists of a series of impulses, and again, that's not very general.

Note for the pedants: yes, you can decompose any function into a series of delta functions (impulses). Stop overthinking it.

Hint: If you're in the time domain! You have Ample time.

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When an analog signal is sampled by an analog to digital converter the digital value is not infinitely precise, but is truncated to a value which can be represented by the digital output. This quantization error prevents us from hearing signals less than 1 least significant bit in peak-to-peak amplitude.

If we add a small amount of uncorrelated noise to the analog signal before it is sampled, this combined signal can cause bit transitions in the output which are statistically proportional to the weak signal. This noise is called "dither" because it causes the least significant bit to fluctuate randomly. By averaging over time the dither can be eliminated and the weak signal recovered.

Unrelated Hint: "dither" when installing satellite dishs is a 'small' adjustment

Specific knowledge hint: Dithering in visual media refers to introducing noise into an image with a limited color pallette to "mix" colors. https://en.wikipedia.org/wiki/Dither

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The name "true-RMS" is a big hint if you understand RMS. RMS means "root mean square", and it's a way of determining an equivalent DC voltage for an AC voltage. That is, a typical wall (mains) voltage in the United States is AC with an RMS voltage of 120V, and it will deliver the same power into a resistive load as 120V DC, even though the AC voltage swings between -170V and +170V.

For a sine wave with zero bias (centered at 0V), the RMS voltage is simply 0.707 times the maximum voltage; 120 = 0.707 * 170. Because signals like that are very common, some meters will simply measure the peak voltage and multiply by 0.707. That's cheap and easy, but it doesn't work if the voltage has a bias or isn't sinusoidal. Measuring the peak and multiplying by 0.707 won't give you the true-RMS. For those signals, you really need to measure the voltage over time, square it, average it, and then take the square root. That's more complex and more expensive, but it does give you the "true RMS" for all signals, whether they are sinusoidal or not.

Example: a square wave with -5V and +5V has an RMS of 5V; the peak and RMS voltages are the same.

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Recall that PEP means the peak envelope power.

The peak envelope of a sinusoidal waveform is its peak-to-peak value, which is twice its peak value. After performing the integration, the average value of a sinusoidal waveform is \(2 \times \frac{V_p}{\pi}\) \(≈\) \(0.637 \times V_p\). The ratio of the peak envelope of a sinusoidal waveform to its average value is therefore \(\frac{2 \times V_p}{\left(\frac{2 \times V_p}{\pi}\right)} = \pi\), or \(3.14\), which is closer to "2.5 to 1" than the other answers.

If the root mean square (RMS) value of the same waveform is considered instead of its average, the ratio of the peak envelope of a sinusoidal waveform to its RMS value is \(\frac{2 \times V_p}{V_p \times \frac{\sqrt{2}}{2}}\) \(=\) \(2 \times \sqrt{2}\) \(≈\) \(2.8\), which is even closer to "2.5 to 1" than the other answers.

A true SSB signal is not a simple sinusoid, but made of many superimposed sinusoids, making this an approximation.

Hint: this is probably best solved on the test by a bit of logic. We know it’s surely not 1:1 (since peak is almost always higher than average), but it’s unlikely to be 13 or 25 to 1 because that would be a very strange voice (see E8A07, it’s “speech characteristics” that drive the difference). So 2.5 is the most likely,

Silly hint: PEP-to-average contains 2 hyphens, single-sideband contains 1. 2 to 1 is close to the correct answer of 2.5 to 1

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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 direct conversion or flash ADC is optimized for speed at expense of almost every other parameter. The ADC requires a comparator for every possible output code. This limits the number of output bits, since each additional bit will double the complexity of the device.

Since the output can be determined essentially as fast as a compare and a priority encoder can run; flash ADC's can be produced capable of gigahertz sampling rates. High sample rates are required for a software defined radio, since the sample rate of the ADC is often a limiting factor in the bandwidth of the radio.

** Test Tip = Remember that 'The Flash' runs at a 'Very High Speed'

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In binary encoding, the number of levels that can be encoded by a certain number of bits is 2 to the power of the number of bits.

2 bits, \(2^2\) = 4 levels

6 bits, \(2^6\) = \(2 \times 2 \times 2 \times 2 \times 2 \times 2\) = 64 levels

For 8 bits, \(2^8=256\). So an 8-bit encoded value can be one of 256 numerical values (typically 0-255). The answer is 256.

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There are often times when signals are passed through a low-pass filter before being converted to a digital signal. The purpose of a low-pass filter being used in conjunction with a Digital-To-Analog (D2A) converter is to remove unwanted harmonics from the output caused by the discrete analog levels generated.

Hint: Filters generally limit or remove.

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In an analog-to-digital converter, the goal is to find the most accurate binary number representation of the input signal. The most accurate output will, by definition, be the one with the least distortion.

Word association hint: Digital and Distortion both start with 'D'

Hint: think harmony between analog and digital

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