Hot answers tagged

45

I had heard that tape is still the best medium for storing large amounts of data. well, "best" is always a reduction to a single set of optimization parameters (e.g. cost per bit, durability, ...) and isn't ever "universally true". I can see, for example, that "large" is already a relative term, and for a small office, the ...


8

As you have realized, the hard part of doing digital communications is carrier, symbol and frame synchronization, and channel estimation/equalization. The bad news is that you can't get around these problems. The good news is that implementing these is not that hard, as long as you limit yourself to narrowband BPSK. I know, because I have done this myself, ...


8

I am currently implementing acoustic FSK modulation and demodulation. I am not a signal processing guy… Since you say you have matched filters, and you mention non-coherent detection, I think you're pretty much of a digital communication person already – the step to being a DSP person is pretty small :) The fully-fledged synchronizer SDR approach So, the ...


8

The $BT$ product is the bandwidth-symbol time product where $B$ is the $-3\textrm{ dB}$(half-power) bandwidth of the pulse/filter and $T$ is the symbol duration. For different applications you will find varying recommended values. In GSM telephony for instance, a $BT=0.3$ is recommended. In satellite communications with GMSK, for near-earth missions the ...


8

High-capacity tape formats use helical scanning. Here I look at why that gives a higher capacity than linear recording with four tracks (one stereo track on each side) like in a compact cassette (C-cassette). Figure 1. Stereo C-cassette player/recorder linear head (left) and a helical scanning head from a data drive (right), same scale. There is a too-short-...


6

This is actually a really tough problem because of the channel characteristics. Most computer speakers have fairly limited bandwidth, have significant non-linearities and the room acoustics are often time variant. Life becomes A LOT easier if you can just run a cable from the headphone output of one PC into the line input of the other.


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

For continuous-phase FSK, frequencies $f_1$ and $f_2$ and phases can be chosen so that a frequency spacing of $\frac{1}{2T}$ gives orthogonality. Note that it is not sufficient to have just $\Delta f = \frac{1}{2T}$; both $f_1$ and $f_2$ have to be integer multiples of $\frac{1}{2T}$ or $\frac{1}{4T}$ etc. Look for Sunde's FSK and minimum-shift keying (MSK) ...


5

With integer modulation indices, each symbol is actually an integer number of full oscillations. You DFT that, and get a sharp, discrete spectrum of tones, convolved with the pulse shape. With non-integer indices, you get "cutoff" oscillations. That is, we need to dsitinguish between two cases: Continuous-Phase FSK: your next oscillation starts with a ...


4

After playing around in GNURadio, this waveform looks like 2 level ASK bursts (OOK bursts plus some residual carrier). The carrier appears to be at 12.4 kHz. At the 44.1 ksps sample rate, there appears to be about 7.6 samples/symbol or a symbol period of about 173.5 microseconds (assuming the symbols are not Manchester encoded). See the top plot in the ...


4

In the end, I used DTMF (Dual Tone Multi Frequency signaling). The original DTMF has 16 signals each using a combination of 2 frequencies. But here I only used "1"(697 Hz and 1209 Hz) and "0" (941Hz and 1336 Hz) An outline of how the code works: The sender converts text to binary, then transmit "0" / "1" DTMF signals (here the timing is 0.3s for tone ...


4

Typical cheap cassette tape recorders and players in the 1970's used audio circuitry that did not have a completely flat frequency response and added a lot of phase distortion (mostly inaudible to most consumers). In addition, the computers did not use high sample rate low quantization DACS and ADCs (which cost a fortune back then), sometimes only a simple ...


4

You have already selected your answer, but I would like to put a few more lines. First of all, objecting to Marcus, I think that your first assumption is correct! We can store one hour of almost CD quality (14-16-bits, 44100Hz, stereo) analog audio into these commercial music casettes; i.e magnetic reels. So on a very rough basis you have almost a 600 MB (...


3

The confusion comes from the fact that what is tagged as "transmitted" isn't the real transmitted waveform but its baseband representation, which are $\left\{+1,-1\right\}$ symbols in 1 dimension. For 2-FSK, that representation would be $\left\{\left(+1,0\right), \left(0,+1\right)\right\}$ symbols in 2 dimensions. A common convenient way to handle such 2 ...


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

The answer to your question is yes, these two representations are equivalent. The first important thing to note is that in your formula, $d_k$ must be either $1$ or $-1$, because the frequency difference for MSK must be $\Delta f=1/2T$. I'll add some information showing the equivalence of the two representations. Expanding your first formula gives $$\begin{...


3

Since the comments on the question and the OP's responses don't seem to be converging, here goes. Pick a data rate $R$ bits/second (e.g. $10$ kilobits/second) that is acceptable to your client or boss. The FSK signal will then convey one bit every $R^{-1} = T$ seconds (every $T= 100$ microseconds if you chose the $R =10$ kbps data rate suggested as an ...


2

People have been decoding FSK signals for ages with very weak microprocessors. The usual method is called zero crossing detection. You basically just measure time (count the number of samples) between the times when the signal crosses the zero volt line. The number of samples multiplied by the sampling rate tells you the rate as a time period. One divided ...


2

You can make a very simple frequency discriminator by multiplying your received signal with a delayed version of itself, followed by a low pass filter. Adjust the delay to set the frequency range of your discriminator. This works because a multiplier followed by a low pass filter is a phase detector (multiply two sinusoids with same frequency and different ...


2

There's a time vs. frequency trade-off that can work against you. At a given signal-to-noise ratio, it takes longer (and often more compute power) to reliably determine which of two frequencies are being sent when they are close together than when they are far apart or orthogonal within the time interval. So if you use 10 times more frequencies within the ...


2

The three main reasons appear to be these: 1) In AFSK, it's not just jumping back and forth between frequencies. The tones must also be continuous phase. In other words, when the tone changes, there can't be any jump in phase. For example, if you're sending a 1200 hz tone and the waveform is at its peak when you switch to 2200 hz, the waveform is still at ...


2

So, ignoring noise, the input into the receiver multiplier coming from the channel is either $\cos(2\pi f_0t)$ or $\cos(2\pi f_1t)$. The other multiplier input is $\cos(2\pi f_0t)-\cos(2\pi f_1t)$. All signals are defined from 0 to $T$. Let's assume the transmitted signal has frequency $f_0$. Then you have an integrator: \begin{align} &\int_0^T \cos(2\...


2

One approach could be to perform an FM-detection step (e.g. an atan2() operation followed by a first-order difference) to transform the waveform to measurements of the approximate received frequency versus time. Your FSK signal should then look like a binary-modulated baseband waveform. Then you can apply a nonlinearity to the signal to induce a discrete ...


2

The frequency resolution is independent of the sampling rate. Consider that in the extreme of an infinite sampling rate (continuous time), the Fourier Transform of a single FSK symbol at either 1070 or 1270 Hz carrier will be a Sinc function with the first null at $1/T$ away from the carrier where $T$ is the symbol duration. For a 300 bps data rate, the ...


1

Before I address your question, I'll suggest that using a frequency domain technique (FFTs) to demodulate inherently time domain data (a time series of sequential data symbols) won't be very practical. Symbol timing recovery is inherently a time domain process. ISI encountered in common modulations like GFSK/GMSK means you really have more than 2 ...


1

$$A\cos(2\pi f+\theta_1)+B\cos(2\pi f+\theta_2)=C\cos(2\pi f+\theta_3)$$ where $$C=|u|\quad\textrm{and}\quad \theta_3=\arg\{u\}$$ with $$u=Ae^{j\theta_1}+Be^{j\theta_2}$$ The constant $C$ can be written as $$C=\sqrt{A^2+2AB\cos(\theta_1-\theta_2)+B^2}$$


1

After thinking for a while i came up with my own proof. I hope it is helpful to you and correct! Definition: Frame-Structure The frame structure used in the following is one of one-periodic signals, with a constant header and a uniform distributed payload. The ratio between payload and total frame is given as $$ p_{\%} = \frac{|payload|}{|frame|} $$ the ...


1

In communications engineering this problem is called "Unequal error protection" and you can find many references on google. It can be a very involved topic. However, a simple method is to group your bits of each sample into N groups (e.g. if you hve a 16-bit sample, you have N=4 groups, of 4 bit each). Then, you can encode each group with a different code ...


1

That looks very nice: I don't 100% agree with your phrasing on this one: however, my quadrature demod is capturing large amounts of noise during the idle periods. Well, the quadrature demod takes successive samples and gives you the phase advance between them. If you only capture silence, or more precisely, the noise between packets, the phase of each ...


1

(I assume maintaining the same symbol duration) nope! That is usually not the case, because it would be unfair – if you send the same number of symbols per second, you'd get twice as many bits per second using 4-FSK rather than 2-FSK. Therefore, the comparison is typically done with identical Bit Energy rather than identical Symbol duration (which would ...


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