6

What is the advantage of performing the FSK using IQ modulation? You only need one RF oscillator operating at a single frequency, instead of having 2 (or more in the case of M-ary FSK) oscillators operating at separate frequencies for each bit/symbol. Since you only have one oscillator, you don't have to worry about discontinuities in the phase of the ...


5

Wow, I'm honored by Matt L. doing what I'm often doing: Referring people to GNU Radio. The project actually has a list of recommended literature, but I don't know how well that'd fit you. It's probably still worth looking into. Then, regarding QPSK: Well, it's one of the basic constellations, and you'd probably be best off reading a textbook intro to ...


5

First, the answer to why you do not see a negative voltage is that the output being digital will range from 0 to the maximum digital voltage at the output (+Vs). This will have a DC offset of +Vs/2 which is simply filtered out with a high pass filter (series cap) resulting in a bipolar waveform with negative voltage after the series cap. UPDATE: Based on ...


5

Both amplitude and frequency modulated radio signals nowadays use quadrature modulation and demodulation as a mean to transfer and receive radio signals. This question is ill-defined – a signal doesn't use quadrature modulation/demodulation, transmitters/receivers do. So, your question is, if we try to "rescue" it, is Do modern FM receivers/...


5

In general, it means several waveforms will coexist in 5G ecosystem. In the context of LTE-based 5G (up to now), several subcarrier spacings (hence CP) coexist. The first and foremost motivation for this is to support not only mobile broadband but also other use-cases such as mMTC and URLLC. Some of benefits of mixed numerology are flexible scheduling and ...


4

I waited a bit to see if someone else takes the challenge, but since there are no answers yet, I'm providing mine now.


4

How about just using $\dfrac{d\phi}{dt}$ to compute the instantaneous frequency? GNU Octave code (I just made up a sample rate of 2 Msps): pkg load signal; x = csvread('nonlinear_freq_fit.csv'); y = hilbert(x); z = y - mean(real(y)); dphi_dt = diff(unwrap(arg(z))); Fs = 2e6; f_inst = (Fs/2)/pi * dphi_dt; k = [0:length(f_inst)-1]; t = k/Fs; plot(k, f_inst); ...


3

Assuming an FM modulation of a sine wave of frequency $f_m$. Theoretically, the modulated signal is an infinite sum of cosines at the frequencies $f_c + nf_m$ ($n\in \mathbb{Z}$) which the amplitudes depend on the coefficients $J_n(\beta)$ (of course in addition to the amplitude of the carrier wave), where $\beta$ is the modulation index and $f_c$ is the ...


3

With GMSK the frequency modulation index is 0.5. With GFSK, the frequency modulation index is larger than 0.5 (the signaling tones are farther apart). The receiving techniques used for GMSK as an FSK work for GFSK.


3

I just skimmed https://github.com/SaucySoliton/PiFmRds/blob/master/src/pi_fm_rds.c#L454 and from what that code looks like, it initializes a clock generator to run at an adjustable clock. Then, it uses the audio amplitudes to modify that clock's frequency in real time. PWM doesn't seem to be involved, aside from the program using the PWM unit to generate ...


3

If $t$ is time, $s(t)$ is the (appropriately scaled) signal, $\omega_0$ is the angular frequency, and $\phi_0$ is a phase offset, then phase modulation is $$ t \mapsto \sin(\omega_0 t + \phi_0 + s(t)) $$ Different signs for $s(t)$ and/or $\phi_0$ may be used, depending on conventions and context. Frequency modulation is more complex to write down, because ...


3

There are many ways of demodulating the FSK signal Indeed! What is the advantage of performing the FSK using IQ modulation? Depends. Generally, IQ is the only shape you have your signal in, so using that is not much of a question – a direct downconversion system has IQ signals, and that's what you'll use. Using such a system has a lot of advantages, ...


3

What if they did all use the same carrier? What you're describing is a single-frequency network (SFN). These are in common use for things that are not stupid FM broadcast. The whole truth is that reception from different senders in an SFN just look like heavy multipath propagation, where the sent signal just takes multiple paths of different length to the ...


3

There's simply no inherent relation between carrier frequency and data rate. I assume a classical comms engineering education here, but since that most often introduces the concept of equivalent baseband early, a good explanation would be: This becomes obvious when you think about the complex baseband representation of a passband being the same, ...


2

Both amplitude and frequency modulated radio signals nowadays use quadrature modulation and demodulation as a mean to transfer and receive radio signals. The use of a quadrature transmitter / receiver is not needed unless you are using a quadrature signal. A simple car radio would likely use something much simpler. Radios that are capable of quadrature ...


2

Checking the book you linked to, this equation refers to FM modulation over a (weakly) non-linear channel, modeled as a memoryless third-order non-linearity (Eq. 82). The third-order non-linearity produces signals at twice and at three times the carrier frequency $f_c$ (Eq. 84). Note that also the frequency deviations of these additional signal components at ...


2

In U.S. broadcast FM, the various subcarriers (pilot tone, stereo L-R, RDS, etc.) are modulated into the upper sidebands (above the 15 kHz L+R/mono audio, but below 100 kHz, for instance 38 kHz DSB for the stereo subcarrier) of the baseband signal before that entire multiplex is used to wideband frequency modulate a single, much higher frequency, FM carrier....


2

Integration was incorrect. I was integrating it with $dx= 1$. Doing m(j)*1/Fsoperation in the for loop, I have made $dx= 1/Fs$ now. As can be seen, frequency changes for different amplitudes now. Moreover, resultis a constant, it should be changed toresult2 so that it gives result with respect to time $t$.


2

Yes. Pulse compression is really just running the returned signal through a pulse matched filter, which is equivalent to cross correlation. If you view it as a pulse matched filter, matched filters are optimal for detection of a signal in AWGN. If you view it as a cross correlation, the output of the correlation will peak when the signal best matches with ...


2

Why not try your hand at working out the differences for yourself? Since you are interested in cosines, take, for example, the signal $x(t) = \exp(j(2\pi t + \theta))$ which is a complex sinusoid of period $1$ and sample it $16$ times per second to get $16$ samples $x[n]$, $n = 0, 1, \dots, 15$, where $x[n] = x\left(\frac{n}{16}\right) = \exp\left(j\left(\...


2

If the signal frequency is constant and exactly integer periodic in the FFT width, then the transform basis vector for each FFT result bin is the same as a sinusoidal (complex exponential) IQ down-modulator mixer input that starts with a phase of zero at sample 0. Thus the atan2() results for a signal within that FFT result bin should be identical. If you ...


2

Your question is based on a misunderstanding. The RF signal is not directly sampled by the sound card. Instead, the earphone cable just doubles as an antenna for a dedicated FM receiver; the soundcard hasn't got anything to do with FM reception. In fact, it doesn't even "get" radio frequencies – they're filtered out before and fed to the FM receive chain.


2

If the two nodes run the same clock i would know perfectly where the signal starts from the end of the first chirp, after the fixed period of pause. In this case the signal i try to demodulate would have no phase shifting, correct?! Wrong. The signal travels a distance. The distance it travels inherently means you get a phase shift – a full period ($2\pi$) ...


2

The short answer to your question is yes and the method is called beamforming which you alluded to in your question by mentioning multiple aerials, and knowing the propagation time from two distinct transmission locations broadcasting on the same frequency to each receiving aerial is key to beamforming. The number of required aerials depends on the ...


2

The problem comes from sampling the phase. For the given phase function, the changes in the phase from one value $t_i$ to the next one $t_{i+1}$ can become too large, so phase unwrapping will not work properly. If you simply removed the factor $2\pi$ in the exponent, the problem would be solved already (note that the factor shouldn't be there anyway ...


2

An important thing to consider is that the pitch of musical notes does not directly correspond to frequency peaks in a spectrogram. This is because the energy of typical musical sounds are partially to mostly in the overtones and harmonics, at frequencies well above the note pitch (depending on the source of the music). So it would help to investigate ...


2

What your plot shows is noise from varying size discontinuities in the waveform. When you change a modulating frequency, you have to make sure the resulting FM waveform does not have discontinuities. You have to do this by incrementally changing the phase of the argument to the sin() function, not by naively multiplying the modulating frequency by time, ...


2

It sounds like you are asking how to demodulate FM. This involves the use of a Frequency Discriminator which converts Frequency to Magnitude. A simple discriminator can be done either in the digital or analog form with a delay and multiply as demonstrated in the graphic below. The output is also sensitive to the amplitude of the input, so it is important to ...


2

This is what an NCO (Numerically Controlled Oscillator) does. See this post for more details on the implementation and design considerations Numerically Controlled Oscillator (NCO) for phasor implementation? By combining this with a D/A converter for an analog output we get what is commonly referred to as a Direct Digital Synthesizer (DDS). With an NCO ...


2

I'll assume AM DSB modulation for simplicity, but the same idea applies to quadrature modulation. I'll assume that the received signal has been properly bandpass filtered. I'll also ignore noise and carrier synchronization. So, the received signal is $$r(t) = s(t)\cos(2\pi f_c t).$$ We want to recover the baseband signal $s(t)$ from $r(t)$. The bandwidth of ...


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