In state of the art SDR Radios, RF is received and sent directly to an analog-to-digital (A/D) converter. In other words the RF is digitized and processed by digital circuits from that point.

There are no mixers. local oscillators or intermediate frequencies, avoiding distortion and other undesired effects of mixing. This is a major benefit of SDR receivers.

Direct conversion, RF to digital is the modern trend in amateur radio receivers and transmitters.

HINT: The correct answer is the only one with language about digital. i.e analog-to-digital and digitized

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An adaptive filter is used in digital signals processing (DSP) to remove unwanted "audio" noise in single-sideband (SSB).

Hint: "adaptive is for audio"

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Hint: Transforms.

Extract from here:

Single-Sideband (SSB) Modulation is an efficient form of Amplitude Modulation (AM) that uses half the bandwidth used by AM. This technique is most popular in applications such as telephony, HAM radio, and HF communications, i.e., voice-based communications. This example shows how to implement SSB Modulation using a Hilbert Transformer.

Hint: think Single-Side(B)and: Hil(B)ert

— serif

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SSB has the mathematical form of quadrature amplitude modulation (QAM) in the special case where one of the baseband waveforms is derived from the other.

Hint: The only answer with "Signal(s)" in it as "Signal" is in the question.

https://en.wikipedia.org/wiki/Single-sideband_modulation#Mathematical_formulation

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This is a fundamental mathematical limitation of digital signal processing, called the Nyquist theorem. In order to properly reproduce a sampled signal, it must be sampled at a rate (called the Nyquist rate) at least twice as high in frequency as the highest frequency component of the signal.

In addition to this, if you have an analog-to-digital (ADC) converter that samples at \(x\) Hz, the input analog signal must have all frequencies above \(x/2\) Hz filtered out or else those frequencies will "alias" down into the desired frequencies.

A more common example of this might be digital audio, which is sampled at 44.1 kHz, allowing 22.05 kHz (well above human hearing range) to be the highest pitch that can be reproduced. If you tried to sample audio that contained a 23 kHz tone, it would alias down as noise at 21.1 kHz.

A decent rundown of the Nyquist theorem can be found here.

Hint: A wavelength is composed of a positive half and a negative half. Digitally, this is minimally represented as one 1 and one 0 - two bits. Therefore the sampling frequency must be at least two times the frequency of the signal.

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To sample 1 mV out of a 1 V (1000 mV) signal requires a granularity of 1000 mV / 1 mV = 1000:1 resolution.

9 bits allows for 512:1 resolution (\(2^9 =512\)), which is less than adequate and 10 bits allows for 1024:1 resolution (\(2^{10} =1024\)), which is slightly more than adequate, so the minimum number of bits required is 10 bits.

Hint: 1V, 1mV, and 10bits are all multiples of 10.

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Converting signals from the time domain to the frequency domain is correct.

Fourier (FOUR-ee-yay) was a mathematician that developed a formula to convert a time-domain signal (amplitude with respect to time) into frequency-domain (relative amplitude or power with respect to frequency). The Fourier transform is rather complicated, but a computer can calculate a short-cut version based on digital samples called the "Fast" Fourier Transform, or FFT.

The display on a spectrum analyzer is the output of the FFT function.

Distractor Converting analog signals to digital form might trick you, because you usually start out with an analog signal before you perform an FFT, and the FFT's output is digital. This choice is wrong because the FFT itself needs digital input. There's an extra step, sampling the analog signal into digital, before the FFT sees it.

Test Tip: The question mentions "fast," which suggests the answer containing the word "time."

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Decimation refers to destruction (to decimate) of information. Specifically to remove digital samples from a stream of samples of an analog signal. If not done properly, it can lead to aliasing noise.

A decimator, which is what a system component that does this process is called, is typically employed to reduce the data capacity requirements when a signal is sampled at a relatively high sample rate (oversampled). In these situations, an analog signal is sampled at a very high rate, and then filtered in the digital domain to remove noise or high-frequency components, which can be significantly more efficient than analog filters, and then decimated to a lower effective sample rate to reduce the amount of data required to reliably regenerate the signal.

Memory Hint: In your head think "de-cimation" = "de-sample". Remove samples.

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Digital sampling of an analog signal must be done at twice the rate of the highest frequency component of the signal. This is called the Nyquist theorem or Nyquist rate. If samples are removed from a stream of samples making the effective sample rate lower than the Nyquist rate of the signal, all frequency components higher than the nyquist frequency of the new sample stream must be removed (filtered out) before the samples are dropped.

For example, if you had an analog signal with 10kHz as the highest frequency component of the signal, it must be sampled at least 20,000 times per second. If you then were to drop every 2nd (every other) sample (10,000 samples per second effective rate), you would first have to filter out every frequency component above 5kHz in the original signal or else they would alias down as noise into regenerated signal.

Hint: To Decimate Something is to Remove It from Existance

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The Nyquist Sampling Theorem states that to faithfully represent an analog signal in discrete time, the sample rate must be greater than twice the highest frequency component of interest. Applying this to a Direct Digital Conversion receiver, it means that the receive bandwidth must be less than half of the sample rate. Therefore, the sample rate limits the maximum receive bandwidth of a Direct Digital Conversion SDR (Software Defined Radio).

Recall that

\[\text{BW} = \frac{f}{Q}\]

\[\text{bandwidth} = \frac{\text{frequency}}{\text{Q}}\]

Hint: Receiver Analog-To-digital = "RATe"

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If there is no noise floor (ideally), there will be nothing to compare the intelligence of the signal to. Using a voltage level along with sampling, the intelligence (signal) can be detected.

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Finite Impulse Response (FIR) filters and Infinite Impulse Response (IIR) filters are both discrete-time filters, but one major difference is that IIR filters include a feedback path whereas FIR filters do not. Without feedback the effects of any input signal cannot persist longer than the fixed delay of the filter. In an FIR filter all input samples are treated equally. They all proceed through the same tapped delay line and fall off the end after a fixed amount of time. Because of this, all frequency components of the input signal are delayed by the same amount of time.

Stupid test hint: A finite value is a fixed amount. So, you should pick the answer with the word amount in it :)

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Hint: Only one answer has the word "delay" in it.

From: https://en.wikipedia.org/wiki/Filter_(signal_processing)#Quartz_filters_and_piezoelectrics "Engineers realized that a large number of crystals could be collapsed into a single component, by mounting comb-shaped evaporations of metal on a quartz crystal. In this scheme, a "tapped delay line" reinforces the desired frequencies as the sound waves flow across the surface of the quartz crystal. The tapped delay line has become a general scheme of making high-Q filters in many different ways."

More taps increase the steepness of the filter roll-off while increasing calculation time (delay) and for high-order filters, limiting bandwidth.

Study hints:

• Both the question and the answer have the word "filter" in them.
• Tap the digital (clock) to delay (snooze) the alarm.

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Tap - A FIR "tap" is simply a coefficient/delay pair. The number of FIR taps, (often designated as "N") is an indication of:
1) the amount of memory required to implement the filter,
2) the number of calculations required, and
3) the amount of "filtering" the filter can do; in effect, more taps means more stopband attenuation, less ripple, narrower filters, etc.

Silly memory trick: The capital M in the answer "More taps" resembles an opposed element knife sharpener

http://dspguru.com/dsp/faqs/fir/basics

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