6

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 signal, though, is real, and it is a pulse-shaped carrier whose amplitude and phase depend on each symbol. Myself, I would "draw" it using Matlab or some other ...


6

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 $[-\frac{B}{2}, \frac{B}{2})$ in the resulting complex signal. This can be done in a couple different ways: Design a linear filter to approximate a Hilbert ...


5

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} where $k$ is the number of information bits per symbol, $T_{sym}$ is the signal's symbol period and $T_{samp}$ is the signal's sampling period. More details can ...


4

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 GHz of carrier frequency, it would be much easier (in terms of computation) to simulate it in the baseband sampling rate. The reliability and the quality of a ...


3

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 single channel (you would need two channels, one for the real part and one for the imaginary part). You can, however, transmit the same information over a single ...


3

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 exist as long as the process is stationary or at least WSS (wide sense stationary). And therefore the power spectral density (PSD), $\Phi_{XX}(e^{j2\pi fT})$, of ...


2

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. Functionally it is equivalent though.


2

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 carrier and the odd-numbered bits onto the quadrature carrier with each modulation lasting for $2T$ seconds. That is, bit $b_{2n}$ creates a BPSK signal $(-1)^{...


2

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 on the I and Q axis. Instead of choosing one point of a four-points constellation based on two bits, you now choose two separate constellations points from two ...


2

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 signal straddles $\pm f_s/2$, which is not what you want. To center a 20 MHz wide signal at 20 MHz, you must sample at least $2*(20+20/2)=60$ MHz, or roughly ...


2

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 mentioned in above comments, you shouldn't have any major difference between simulating base-band and pass-band. Now let me explain what are the base-band and pass-...


2

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 $$ p(t)=\exp\left(j2\pi(f_0t+\tfrac{1}{2}\alpha t^2)\right), $$ where $f_0$ is the frequency at the start of the chirp and $\alpha$ is the chirp rate. For you case $f_0 =-f_{limit}$. To find $\alpha$ we set the end frequency to ...


1

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 equivalence the baseband signal must be complex (often represented with I and Q data paths) unless the signal is completely on the real axis... for example If the ...


1

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 and 1 MHz at higher frequencies have no difference on the data rate. Other effects may need to be taken into consideration though at higher frequencies. For ...


1

If you are a bass and sing A 220 Hz and B 247 Hz at 120 beats per minute, you can sing data at a certain rate. If you are a soprano and sing two quarter notes A 880 Hz and B 988 Hz at the same BPM, your data rate isn't any higher. If you sing more notes (higher bandwidth) you could communicate a more complex score (e.g. a higher data rate). So bandwidth (...


1

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: $C = 2B ~\log_2(M) ~~~~~~~~~~~~,~~~\text{Nyquist}$ $C = B ~\log_2(1 + \text{SNR}) ~~~,~~~\text{Shannon-Hartley}$ Eventhough the first formula, (referred to as ...


1

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 2T}\cos(2\pi f_ct),~ 0 \leq t \leq T,\\ \phi_2(t) &= -\sqrt{\frac 2T}\sin(2\pi f_ct),~ 0 \leq t \leq T, \end{align} where $f_c$ is the carrier frequency are ...


1

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 happens in IQ-modulation is that you generate a real-valued band-pass signal containing the same information as $x(t)$: $$s(t)=\text{Re}\{x(t)e^{j\omega_ct}\}=x_R(t)...


1

One easy way would be to use GNU Radio, specifically GNU Radio Companion, to design your signal processing chain. GNU Radio has a large library of built-in signal processing blocks. All signal processing in GNU Radio is performed on the complex baseband version of the received signal. You could tune the USRP to 900 MHz, then have two bandpass filters-- one ...


1

There are many possible methods to isolate those 2 signals. You can heterodyne your IQ signal up or down by 25MHz by multiplying by a complex exponential of plus or minus that offset frequency, then baseband filter each IQ component separately with either a real FIR or real IIR low-pass filter. Or you can multiply a FIR low-pass filter kernel by those same ...


1

The allowed bandwidth of the data created and used to modulate the IQ signal needs to be below the cutoff frequency of the receiver post-heterodyne low pass filter, which needs to be below the carrier frequency, which needs to be below half the HF sampling rate to avoid aliasing. So try a lower data rate and/or a much higher frequency carrier, with a ...


1

Starting with a Bandpass signal - this is a signal that has a bandwidth $BW$ about a center frequency $f_c$, so the signal content goes from $f_c-1/2BW$ to $f_c+1/2BW$. For a real signal (i.e. not complex) the their is also the corresponding negative frequency component i.e. the signal also exists at $-f_c$. Basebanding often refers to shifting this signal ...


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