Hot answers tagged

7

Let's first start by developing some intuition of what a complex signal looks like. We can use GNU Radio to generate a signal that's just an unmodulated carrier, and then put that into a UI that will display the real and imaginary components over time: The result for 80 Hz is this: Notice how the real part is 90 degrees behind the imaginary part. If you ...


7

There is a range of possibilities, depending on what kind of signal you want to find. I'll start from easy and move up to hard. I'm assuming that you are using an FFT to get your spectra. RFI. An earlier poster referenced some papers on finding RFI. I don't know precisely how that is defined, but lets assume that it is unintentional RF from things like ...


6

Complexity of your question It's hard to make general statements here, because computational complexity is a property of the implementation of a receiver, which usually is a choice given the properties of the transmission. For example, assume you have 2-FSK. You can either demodulate that by having say, 2 bandpass filters applied to a passband signal, and a ...


6

The 7300's "IF output" is a USB digital signal at 48kSPS, the same as the USB AF output (and you can only have one or the other at a given time). It places the dial frequency at an IF frequency of 12kHz, and you can get a maximum bandwidth of around 16kHz (+/- 8kHz from the dial frequency) which is dependent on the selected mode and filter settings,...


5

From https://link.springer.com/chapter/10.1007/978-3-322-92773-6_9 : Spectral subtraction is a method for restoration of the power or the magnitude spectrum of a signal observed in additive noise, through subtraction of an estimate of the average noise spectrum from the noisy signal spectrum. The noise spectrum is estimated, and updated, from the periods ...


5

I like your PLL approach, because it doesn't try to "recover the original signal from noise", but actually goes ahead and detects what you're actually interested in, the presence of a specific frequency, and uses that to generate a "perfect" tone. Much cleverer than spending hundreds on the best thinkable crystal filter on the market! (I'm always baffled ...


4

Some quick definitions: a sinusoid with angular frequency $\omega$ and phase $\varphi$ at time $t$ is: $$ \cos(\omega t + \varphi ) $$ Now let's consider a scenario: we have an ideal mixer with an LO with $\omega = 1$ and variable phase, and we want to produce an output at $\omega = 0.3$. We know we can do that with an input to the mixer with either: $\...


4

A negative frequency is just a positive frequency "in the opposite direction". Imagine I have a transparent wheel, which has a black disc inside of it at one point near the edge. Now imagine I shine a light through the wheel's diameter from the side, so that the shadow of the disc appears on the wall. If I spin the wheel, you can watch the shadow ...


4

Ringing in narrow filters may already have an answer here: https://dsp.stackexchange.com/questions/2170/why-do-i-see-ringing-in-the-output-of-a-digital-filter-with-a-narrow-transition "looking into the future", or more realistically, a filter delay, seems to be a requirement to reliably estimate the shape and time location of the falling (and rising) edges ...


4

Completely independent. It is dependent on frequency only. The amplitude of the signal is only relevant when looking at the number of bits used (and the dynamic range). See here for more details.


4

The answer can be fond in The ARRL Handbook 2019, Vol 3, although it's spreaded across different chapters. In short, SNR is typically calculated for the noise floor of 2500 Hz SSB signal. Particularly this is how WSJT-X calculates negative SNRs for FT8 and other modes. Now the trick is that by deviding the bandwidth in half you decrease the noise floor by 3 ...


4

What does this noise reduction do? That should depend on the device. Generally, it's to be assumed that it applies (analog and/or digital) signal processing to improve the perceived Signal-to-Noise ratio. In simple cases, this might simply mean reducing the bandwidth of an analog receiver. Sure, voice will not sound as crisp, but if that buys one a ...


4

512 samples should take 256 microseconds, so your times look about right. The exact time it takes to execute those lines of code will depend on the rest of what your computer is busy with. And it's unlikely that the elapsed time counter will be accurate to better than 10 us anyway (but I may be wrong, try it yourself by timing a delay loop). Whats slightly ...


4

The RTL-SDR interfaces to your computer via USB. The USB interface and driver will buffer the data and send it to your software at the requested sample rate on average (baring underflow due to software glitches), but with varying timing due to interface jitter. The RTL2832U USB is likely sending larger chunks of data over USB in blocks of far more than 512 ...


3

I've heard that 2 days in the lab can save 2 hours of reading but in this case it turned out to be easy to test. It is the sample rate of the SDR which changes the bandwidth of the signal. Since I am using the PLUTO SDR I can't go below around 500ksps. So I tried with a symbol rate of 300 kbaud at a sample rate of 600k. Able to demod successfully. I asked ...


3

As hotpaw2 said, the thing is that while it looks like random noise, the actual UWB signal is correlated, in some specific way known to the receiver. I'll explain this using an example (keeping it short, I'm at GRCon): Imagine your UWB transmitter using a scheme where it generates a wideband noise signal. The way it encodes a "0" bit is that it turns on ...


3

Even if all the transmitted UWB pulses are identical, and initially start at an unknown time, the pattern variations in the delta time between pulses could be used to code information. As with any spread (spectrum or otherwise) system, a longer sequence with more coding might allow more (orthogonality coded?) signals to occupy a channel at some given ...


3

Well, this is a well-studied field (Radio Frequency Interference detection and mitigation). There are tons of literature about it. The noise you receive is theoretically Additive White Gaussian Noise (AWGN). That means that its Probability Density Function (PDF) is Gaussian, and the pdf of its samples' power is exponential. By setting a false alarm ...


3

Sounds like your SDR transmitter has a frequency error of 40-50 ppm (or, less likely, your transmitter is fine but your spectrum analyzer has the error). Try transmitting pure carrier, without modulation, and see if you get the same amount of offset. If so, you might be able to adjust some calibration in hardware or firmware to fix it, or connect an external ...


3

I don't understand the parameters of your equations. But ignoring the equations and answering the text of the question: beam forming depends on frequency and phase. If the phase of the output of two beam forming elements is equal in magnitude but opposite in phase at the angles and frequency of measurement, the received output power output from that 2 ...


2

(This answer is from the perspective of GNU Radio programming; I'm not familiar with what SatNOGS is doing.) There are two problems to solve to use a given data file. First, you need to decode the file format into samples of a signal. The OGG File Source presumably solves this problem for you. Converting the file to a raw format is also an option, but not ...


2

Theoretically, the rate is independant. In reality, exceeding certain power levels will cause real world digitizers to increase in non-linearities (clipping, saturation, thermal damage, and etc.), which will introduce harmonics and other spectra above any Nyquist rate based on linear system assumptions, if not complete system failure.


2

Quantization noise is highly dependent on the signal source distribution and its amplitude, the number of ADC bits, and the use of dithering. For a high resolution ADC that is digitizing a full amplitude sine wave, the maximum noise contribution is: $$\text{SNQR}_\text{dB}=1.76+Q \cdot 6.02 \tag 1$$ where $Q$ is the number of bits. Here we can observe ...


2

An ideal frequency mixer simply multiplies its inputs. One input is the RF signal we wish to shift in frequency, and the other is the local oscillator (LO). For the moment, let's consider just an ideal frequency mixer where the LO is a pure sinusoid. Since mixers are used to shift frequencies, we can use the Fourier transform to better understand their ...


2

...it looks like a "quadrature mixer" is really just two separate mixers, with one fed by an LO to which a [constant] phase delay has been applied relative to the other's LO. Yes ... and. Quadrature versions of the modulating signal may comprise the "baseband" inputs to the quadrature mixers: Here's where "the phase shift is 90 degrees for all frequencies" ...


2

Negative frequencies in the real (single DOF measurement) world are just what we call positive frequencies that happen to below some other frequency. Above baseband, that LSB signal isn't really negative, just lower than some reference frequency (the carrier). The in phase signal alone isn't in phase to anything. It just has a related phase to a second (...


2

The data is AM modulated on the 2.4 kHz subcarrier, with 256 different levels representing a single value from 0 to 255. It's a scanline every 1/2 second from the cameras with sync and telemetry data added to the beginning and end. Each line is 2080 data points (words) long, so it broadcasts at 4160 baud. The sync lines at the beginning let you know when a ...


2

This is a basic question and needs an answer. Regenerative receiver or audion. Single LC-filter with undamping: Q-multiplier. The frequency transfer of such a filter can be obtained from a simulation. See figure. As you can see: the AM demodulated audio response is affected when selectivity is too high. The example shows 1 MHz mediumwave. Audio bandwidth of ...


2

A low-pass filter is one way to do it. Due to the interpolation done in the repeat block, you are representing a symbol with 1000 samples, and your sample rate is 2 million samples per second (the samp_rate variable). So your square wave has a frequency of: $$ \require{cancel} {2000000\ \cancel{\text{samples}} \over \text{second}} {1\ \text{symbol} \over ...


2

The Head block does that: End the operation of the flow graph after an amount of samples have passed. Since your sample rate is fixed, that's the way you want to go.


Only top voted, non community-wiki answers of a minimum length are eligible