# Tag Info

2

tl;dr: you can't. That's basics! longer story: You ask about symbols. When the symbol rate is much larger than the carrier frequency, you've basically lost, and cannot transport that many symbols per second; you don't have the bandwidth; you get at most twice the carrier frequency in bandwidth, namely from 0 Hz to twice the carrier frequency. However, you'...

5

If you did a continuous on off keying of a 10101010... pattern, then you would see sidebands as described since this is simply an up-conversion of the Fourier Transform of a 50% duty cycle square wave (moved to any carrier frequency). However if the data pattern for this case of a rectangular on-off keying was random, the resulting spectrum would be ...

2

The total power of an amplitude modulated signal is \begin{align}\overline{s^2_{AM}(t)}&=\overline{\big(A+m(t)\big)^2\cos^2(2\pi f_ct)}\\&=\frac12\overline{\big(A+m(t)\big)^2}+\frac12\overline{\big(A+m(t)\big)^2\cos(4\pi f_ct)}\tag{1}\end{align} The second term on the right-hand side of $(1)$ is zero if $m(t)$ is a lowpass signal, and if $f_c$ is ...

1

First of all, you're understating your problem. The error probability depends not only on your noise power (density), but also on the actual distribution of noise. Also, you need to realize that bit error probability is not the same as symbol error probability, not even proportional, so at best, the formula you pasted (from an uncited source) is only an ...

4

This is unfortunately a difficult problem. A microphone doesn't have a single frequency response. It has a different one for each direction of incidence. Especially for second order microphones (dipoles, cardioids, etc.), the frequency response also depends heavily on the distance and radiation impedance of the source. So first you need to decide: what ...

1

It might help to watch my YouTube video explaining negative frequency: https://youtu.be/gz6AKW-R69s I'm a professor and I've been building what I think of as an on-line video textbook for Signals, Systems, and Digital Communications.

1

My goal is to make an approach to recognize modulation type of unknown signals. Cool! Automatic modulation classification is still an exciting topic, after it was (one of) the foundations from which in the 1980s Software Defined Radio arose, e.g. [1]. Do you think that problem is unsolvable and this is waste of time? Absolutely not! There's quite a few ...

4

So this seems to be a general block diagram for these types of modulations. No, this is a block diagram of a IQ upconverter with some unspecified digital data modulator on the input. You need to mentally divide the modulation from the tools you use to implement it. There's other architectures that can produce the same signal (and they're not uncommon). But ...

0

Depending on the length of each symbol, a different starting phase for a symbol might be required to match the ending phase of the previous symbol in order not to create discontinuities.

1

My first question is, why should we allow the signals have different phases? Counterquestion: Why should they not be allowed? Enforcing continuous phase can make systems more complex, without any benefit. I cant see a reason for this and as far as I know, different phases result in phase discontinuity of the transmit signal. Depending on how you do that ...

0

I think this is treating the $10^8\pi\cdot t$ as the carrier frequency and the $5 sin(2\pi\cdot 10^3 t)$ as the modulation. So the maximum phase difference of this signal to the carrier without modulation would be $\pm5$ radians. To compute the change in frequency you need to find the compute $\frac{d}{dt}$ of the argument of the cos (divided by $2\pi$) and ...

2

With no filtering or pulse shaping what you have is your signal with the default rectangular pulse shape $$\Pi(t) = \begin{cases}1 & \text{for} & -\frac T2\leq t\leq \frac T2\\0 & \text{otherwise}\end{cases}$$ Where $T$ is the symbol duration. The Fourier transform of the such a pulse result in a sinc function of the form  H(f) = T\...

0

Think of the signal in the IQ domain, ie the complex plane. The unfiltered signal can be thought of as jumping instantaneously between the individual constellation points (on the unit circle for QPSK), so that its always at unit amplitude. Once filtered the signal can be thought of as moving around the complex plane smoothly in loops, crossing the the ...

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