7 votes

How to "draw" the function/wave to send symbols using QAM?

The baseband QAM signal is complex, and the only way to draw it is by doing two drawings, one for the in-phase (real) component, and one for the quadrature (imaginary) component. The passband QAM ...
MBaz's user avatar
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7 votes
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MATLAB: Adding Noise with Regard to Signal to Noise Ratio (SNR) or EbNo?

These stuffs $E_s/N_0, E_b/N_0 \textrm{ and SNR}$ are convertible. \begin{align} E_s/N_0 &= E_b/N_0 + 10\log_{10}(k) \\ E_s/N_0 &= 10\log_{10}(T_{sym}/T_{samp}) + \mathrm{SNR} \end{align} ...
AlexTP's user avatar
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6 votes
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How do I convert a real baseband signal to a complex baseband signal?

To convert a real signal sampled at rate $2B$ to its complex baseband representation (sampled at rate $B$), you want to map the frequency content in the range $[0, B)$ in the real signal to the range $...
Jason R's user avatar
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5 votes
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Sample rates, Samples per Symbol, and Digital Pulse Shaping

Yes the OP is correct in that you can implement pulse shaping in less than 2 samples per symbol for exactly the reasons that was outlined. However importantly we must also keep in mind having excess ...
Dan Boschen's user avatar
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4 votes

IQ modulation principle

If you have a complex baseband signal and you modulate it with a cosine then you shift the signal in frequency but your passband signal is still complex-valued, so it cannot be transmitted over a ...
Matt L.'s user avatar
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4 votes

Why would we simulate baseband rather than passband

One very obvious advantage of baseband simulation of communication systems is the low sampling rate it requires. Considering a narrowband communication system having a few MHz of bandwidth to a few ...
Fat32's user avatar
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4 votes
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How is the maximum theoretical data rate of a channel equal to $2B\log_2(V)$ bits/sec.?

Each sample is at one of $V$ total levels. The number of bits needed to encode $V$ levels is $\log_2(V)$. If you send $2B$ of these samples per second, then the total number of bits sent per second (...
Peter K.'s user avatar
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3 votes

Autocorrelation and Power Spectral Density (Discrete)

First of all lets state more correctly that a discrete-time 1D auto-correlation sequence (ACS), $\phi_{XX}[\kappa]$, of a single parameter $\kappa$, of a discrete-time random process $X[n,s]$ will ...
Fat32's user avatar
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3 votes
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Relation between constellation SNR and baseband SNR

I think that the question by OP xvan has several misconceptions in it and these propagate into the answer that he has provided for his own question. Define \begin{align}\phi_1(t) &= \sqrt{\frac ...
Dilip Sarwate's user avatar
3 votes
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Why is Complex Baseband better than Real Representation?

The efficiency is in spectral utilization, at the expense of hardware complexity. Spectrum is such an expensive resource and the demand for data so great that this is a compelling trade to make. ...
Dan Boschen's user avatar
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3 votes
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Is BPSK relevant, or does it even exist, in wireless data transmission?

You are mostly correct. The main problem is that the introductory textbook definition of BPSK as a train of square pulses multiplied by a carrier is so simplistic that it hides what is really going on....
MBaz's user avatar
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2 votes
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QPSK on baseband question

Yes, you are correct, though your formula puts the constellation points at 0, 90, 180, and 270 degrees, where usually the constellation points are represented as being at 45, 135, 225, and 315 degrees....
Jim Clay's user avatar
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2 votes

How can I generate offset QPSK symbols from QPSK symbols

You already have a serial stream of bits, say one bit every $T$ seconds. Think of the bits as being numbered consecutively $b_0, b_1, b_2, \ldots$ and modulate the even-numbered bits onto the inphase ...
Dilip Sarwate's user avatar
2 votes

How can I generate offset QPSK symbols from QPSK symbols

The idea of OQPSK is very simple: Consider your QPSK as two BPSK, one orthogonal to the other. It helps remembering that you're free to rotate the constellation, so that the constellation points are ...
Marcus Müller's user avatar
2 votes

Converting complex baseband signal to passband signal

Your signal $z$ is centered at 20 MHz, but because your sampling rate is 30.72 MHz, 20 MHz is beyond the Nyquist frequency of $f_s/2$ = 15.36 MHz and therefore aliases to 20-30.72 = -11.72 MHz. The ...
Ill-Conditioned Matrix's user avatar
2 votes
Accepted

Why would we simulate baseband rather than passband

In OFDM or any other system, if you understood what is the difference between the baseband and passband, then you would decide by yourself if there is a difference or no. However, its already ...
Zeyad_Zeyad's user avatar
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2 votes
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IQ modulation: baseband to bandpass transformation

If you have a complex baseband signal $$x(t)=x_R(t)+jx_I(t)=|x(t)|e^{j\phi(t)}\tag{1}$$ and you multiply it with a (real-valued) sinusoid, the resulting signal is obviously still complex. What ...
Matt L.'s user avatar
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2 votes
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Baseband representation of a LFM chirp signal

The one element missing from your problem is the time duration of the pulse - I'll assume it is $T$. The general form of a chirp is $$ p(t)=\exp\left(j2\pi(f_0t+\tfrac{1}{2}\alpha t^2)\right), $$ ...
David's user avatar
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2 votes
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Nyquist Maximum data rate formula for PCM

The first document (from mason.gmu.edu) gives a confusing explanation of the formulas. The second document is what's the standard way of referring to the channel capacity Now the two formulas are: $...
Fat32's user avatar
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2 votes
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why don't we use Broadband technology to transmit digital signals ,and use Baseband technology to transmit analog signals?

"Broadband technology" is not inherently coupled with "analog signal transmission", as "baseband technology" is not coupled with "digital signal transmission". ...
V.V.T's user avatar
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2 votes
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How to recover data/information from real complex baseband signal?

In an actual received signal you will also have to address the frequency, phase and time offsets between the transmitter and receiver since they run off of independent clocks that aren’t otherwise ...
Dan Boschen's user avatar
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2 votes
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Digital Phase Modulation baseband or passband?

Assuming $g(t)$ is a baseband signal, $s_m(t)$ is passband. You can see that it is by multiplying the two complex exponentials, expanding the result into sine and cosine terms, and then taking the ...
MBaz's user avatar
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2 votes

Spectrum of a baseband digital signal

The spectral shape mainly hails from overlaying an f0- and f1-shifte sinc as formed by the upsampling filter – which is a boxcar filter in your case. So, basically, you're doing a bad version of ...
Marcus Müller's user avatar
2 votes
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The baseband sampling frequency when the negative spectrum is considered

As Hilmar points out in a comment, the answer depends on whether the baseband signal is real or complex. Consider a real baseband signal $r(t)$ with bandwidth $B$. It must be sampled at rate $2B$. ...
MBaz's user avatar
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1 vote

Restore real signal from its complex representation if sampled under Nyquist frequency

You simply can't. The analytic signal contains the exact same amount of information as the original real-valued signal, and if either is undersampled, you get the loss of information that ...
Marcus Müller's user avatar
1 vote
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Frequency Support of Window Functions

A window function $w(t)$ satisfies $w(t)\ge 0$. The value of a window's frequency response $W(f)$ at DC ($f=0$) equals its integral $$W(0)=\int_{-\infty}^{\infty}w(t)dt>0\tag{1}$$ which is clearly ...
Matt L.'s user avatar
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1 vote

Meaning of negative frequencies in Baseband non sinusoidal, non periodic signal

It means the same thing. The negative frequencies are there to cancel out the complex portion of the signal in the time domain, so that the time domain signal is constrained to the real axis. The ...
Andy Walls's user avatar
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1 vote
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Bandwidth computation for NRZ baseband transmission scheme

The first statement says that for the given binary sequence of $$a_k = \{ 1,0,1,0,1,0,...\}$$ the transmitted NRZ line-signal would oscillate at the maximum rate : $$V = \{+A,-A,+A,-A,...\}.$$ which ...
Fat32's user avatar
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1 vote
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Phase changes of BPSK modulated carrier after moving to baseband

There is no change - the complex envelope of the signal (magnitude and phase) at any given carrier frequency is no different from when that carrier frequency = 0 (DC, at baseband). To hold this ...
Dan Boschen's user avatar
  • 48.8k
1 vote

Why bit rate is same in a lower frequency spectrum and higher spectrum?

Because bit rate depends on the bandwidth and not on the carrier frequency. Of course at higher frequencies you have more bandwidth, and thus you can transmit more data. But 1 MHz in lower frequencies ...
BlackMath's user avatar
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