# Tag Info

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Because each step in the processing chain is linear we consider a case with only noise and no coherent signal. Denote the noise $\xi(t)$. The $I$ and $Q$ signals are \begin{align}\ I(t) &= \xi(t) \cos(\Omega t) \\ Q(t) &= - \xi(t) \sin(\Omega t) \, . \end{align} We express the effect of the filter as a convolution with the time response function $h$, ...

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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 ...

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The SDR (or any general digital signal processing system) takes the received RF signal, and downconverts it from the carrier frequency to the baseband. Now, the real bandpass signal from the antenna does not necessarily have a symmetric spectrum around the carrier frequency, but it can be arbitrary. If the downconverter now shifts the spectrum to the center ...

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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 ...

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Both amplitude and frequency modulated radio signals nowadays use quadrature modulation and demodulation as a mean to transfer and receive radio signals. This question is ill-defined – a signal doesn't use quadrature modulation/demodulation, transmitters/receivers do. So, your question is, if we try to "rescue" it, is Do modern FM receivers/...

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As you already know, the two methods are theoretically identical if there's no noise. If there is out-of-band noise then the low pass filters in method A will further suppress the out-of-band noise, whereas there's no such noise suppression with method B. This would be one advantage of method A over method B. Note, however, that the phase splitter in method ...

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Note that signals sent over wires (and over the air, and over any medium) are always real. What quadrature means is that, on a passband channel (wired or otherwise), you can transmit two signals at the same time. The first signal, which we'll call $s_I(t)$, is mixed with a carrier $c_I(t)=\cos(2 \pi f_c t + \phi)$; the second signal, $s_Q(t)$, is mixed with ... 4 i will call the bandpass signal: \begin{align} y(t) &= \Re \Big\{ (x(t) + j\hat{x}(t)) \, e^{j \omega_0 t} \Big\} \\ \\ &= x(t) \cos(\omega_0 t) - \hat{x}(t) \sin(\omega_0 t) \\ \end{align} the complex-valuedx(t) + j \hat{x}(t)$is the low-pass equivalent to the bandpass signal. the real part$x(t)$is the in-phase component and the ... 4 They are essentially the same but may have key differences varying across implementation. The process is identical in terms of multiplying a signal by a sine wave at the center frequency of the signal (the carrier) which moves it to DC, and then low pass filtering that signal to maximize the SNR for the detection (presence of the signal at the carrier). ... 4 The FSW is superheterodyne. This means that it will not have a DC offset or I/Q imbalance, but the IF path will appear as an asymmetrical lowpass (because it is really a shifted bandpass), which is then compensated for in the calibration. The first question: have you run the calibration properly? leave both machines running for half an hour in a stable ... 4 Your idea with the Hilbert transform doesn't work. The only signal (apart from$x(t)=0$) for which the Hilbert transform is zero, is a constant signal. A band pass signal$x(t)$can be written in terms of its complex envelope$x_{LP}(t): \begin{align}x(t)&=\textrm{Re}\left\{x_{LP}(t)e^{j\omega_0t}\right\}\\&=\textrm{Re}\{x_{LP}(t)\}\cos(\... 3 IQ data samples can contain double the bandwidth of information within the same frequency bandwidth and/or sample rate of single channel (non-IQ or strictly-"real") data samples. When heterodyning a real RF signal to a strictly real signal at another frequency (usually a lower IF frequency or to baseband, for easier further filtering, processing, or ... 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 I'm replying to Oleg's recent comments here in an "Another Answer" block because the "add a comment" capability prevents me from providing a complete reply to Oleg. Here's my reply: You can indeed take the sqrt(I^2 + Q^2). (That is one step in one method of AM demodulation.) But you realize that if you do so all of your computed sqrt samples will be ... 3 If the spectrum of your I/Q samples is centered at zero then you'll have to perform either AM or FM demodulation before routing any real-valued audio samples to a sound card. For AM demodulation you'll need to implement a complex-input "envelope detector" which produces a real-valued audio signal riding on a DC bias. (In a few days check the web page: www.... 3 What it looks like you are experiencing is carrier offset and not IQ imbalance, at least from this view that you show. Looking at the scale of the two axis it looks balanced and it is just that each axis is drawn at a different scale? See my example plot below of QPSK before and after proper carrier recovery. Note that it also helps to show the correct ... 3 The RF signal r(t) is obtained from the complex baseband (IQ) signal s(t)=x(t)+jy(t) in the following way:r(t)=\text{Re}\{s(t)e^{j\omega_0t}\}\tag{1}$$where \omega_0 is the carrier frequency (in rad/s). There are two other equivalent representations of (1), where I use s(t)=x(t)+jy(t)=a(t)e^{j\phi(t)}:$$\begin{align}r(t)&=x(t)\cos(\... 3 I'd say there are two ways to look at this problem. If you're receiving a conventionalM$-QAM signal with AWGN noise, then the I and Q streams are actually two independent, real,$\sqrt{M}$-PAM signals. You can filter and process them independently. The same is true over the wireless channel, as long as the fading is flat and you do the appropriate ... 3 As you already know, the quadrature error, also called quadrature skew, describes how far off the actual angle between I and Q is from the ideal 90 degrees. It is one component of modulation error ratio. There seem to be lots of people talking about how to measure quadrature skew, but few people talking about how to compensate. The two major approaches to ... 3 Depends on context. For strictly real-valued signals it sometimes refers to creating an analytic signal (see: http://en.m.wikipedia.org/wiki/Analytic_signal ) from a signal with no imaginary component, and taking the magnitude (or complex with both magnitude and phase) envelope of that hypothetical now complex creation. 3 If we take the complex baseband signal as$S(t) = I(t) + jQ(t)$(which of course means that we have two separate wires on which the$I(t)$and$Q(t)$voltages appear), then we have a nice signal representation for the transmitted signal $$s(t) = I(t) \cos(2\pi f_c t) - Q(t) \sin(2\pi f_ct) = \operatorname{Re}\left(S(t)e^{j2\pi f_c t}\right).$$ If we ... 3 The idea is that you can transmit two signals,$I(t)$and$Q(t)$, over the same bandwidth and at the same time, and still recover each independently of the other. The math is pretty simple. If the transmitted signal is$s(t)=I(t)\cos(2\pi f_ct)-Q(t)\sin(2\pi f_ct)$, then (ignorning noise, ignoring a factor of 1/2 and assuming coherent reception) the receiver ... 3 Radar designer here: It sounds like you’re talking about pulse-Doppler (PD) radar systems. For PD radars, the process is essentially as you described: Generate a waveform (typically at IF) and then convert to RF Transmit the waveform at RF Receive the waveform at RF, and eventually mix it down to IF. Apply IQ demodulation (for digital receivers a Hilbert ... 3 I simply use unwrapped atan2(IQ(i)) - atan2(IQ(i-1)) to estimate a discrete derivative, then low pass filter to below 15 kHz. Although with a shallow slope, the 1st order approximation to atan() given by Boschen will work just as well. Your noise might be due to not unwrapping the phase delta, or to not low pass filtering after doing the phase ... 3 The problem in your derivation is the way you discard the high frequency components. Note that$\hat{s}(t)$, the Hilbert transform of the bandpass signal$s(t)$, has frequency components around$\omega_c$as well as around$-\omega_c$. So you can't just discard one of the two terms$\hat{s}(t)e^{j\omega_ct}$and$\hat{s}(t)e^{-j\omega_ct}$. The low pass ... 3 [Update: I mentioned a possible +3dB processing gain by including the Hilbert Transform prior to DDC for the case of a real IF signal in the first version of this post, which @MattL questioned so I dug into this further and confirmed that there is no such processing gain, so the only advantage to doing that would be to simplify filtering since it would ... 3 It is because the audio signals are real and already at baseband. In contrast radio frequency signals are often represented as complex numbers once they are brought back to baseband. Real signals can be represented as a single stream of real numbers, while for complex numbers two streams of real numbers are required to represent them (as in$I+jQ\$). When ...

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Below summarizes efficient phase estimators for this application updated to include both a phase range of +/- 30 degrees and +/- 60 degrees. This is given in two parts, estimators for a real IF (intermediate frequency) signal, and estimators for a baseband complex signal. At the end are additional considerations related to acquisition. Efficient Phase ...

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Here's another difference between the two methods. If the two lowpass digital filters in Method A are identical then their frequency-domain passband magnitude responses will be identical. And that's a good thing if your final goal is phase- or FM-demodulation. In Method B any practical Hilbert transformer will have ripples (fluctuations) in its frequency-...

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