10 votes
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What is the connection between analog signal to noise ratio and signal to noise ratio in the IQ plane in a quadrature demodulation system?

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) \...
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7 votes

FSK and IQ modulation

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 ...
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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 ...
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6 votes

Why do SDR receivers emit I/Q samples?

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 ...
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5 votes
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Quadrature modulation for FM and AM

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 ...
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5 votes

Sending complex signal over a single wire

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 ...
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4 votes

FSK and IQ modulation

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 ...
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4 votes
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Quadrature modulation

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 ...
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4 votes

Quadrature modulation

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 ...
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4 votes
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How to reconstruct RF signal using IQ data

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 ...
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4 votes
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Difference between these two digital down conversion methods

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, ...
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4 votes

In-Phase and Quadrature components of low-pass equivalent of a band-pass signal?

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) \\ \...
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4 votes

Quadrature demodulation, homodyne detection, lock-in detection - what's the difference?

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 ...
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4 votes

How to test I/Q modulation with only one branch (I or Q)?

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 ...
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Quadrature component - condition on equivalent lowpass signal

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 ...
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4 votes
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In case of using a Doppler radar, do we have to use IQ demodulation to find the direction of a target?

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 ...
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I Q sampling and baseband version of analytic signal

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 $\...
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4 votes
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Fast phase calculation

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 (...
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4 votes
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I/Q sampling with just one ADC

Is this an entirely stupid idea? No, but you've just came to the conclusion that instead of sampling complex, with Nyquist rate being the bandwidth, you should do twice as many samples. That simply ...
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3 votes

What type of information can be contained in I/Q (quadrature) data?

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 ...
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3 votes
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Splitting a Complex FIR Filter into Real Filters

I'd say there are two ways to look at this problem. If you're receiving a conventional $M$-QAM signal with AWGN noise, then the I and Q streams are actually two independent, real, $\sqrt{M}$-PAM ...
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3 votes

Play IQ data in the soundcard

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 ...
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3 votes

Play IQ data in the soundcard

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'...
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3 votes

IQ imbalance amplitude

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 ...
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3 votes
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IQ signal FM descrimination using phase derivative

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() ...
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3 votes
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Obtain I/Q Components from a Real Signal On the Fly - Hilbert Transform or Digital Downconversion?

[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 ...
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3 votes
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Why is In-phase Quadrature sampling not used to record and store digital audio?

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 ...
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Quadrature Detector with a single ADC

The sampling offset is a time delay between the I and Q paths. The Fourier Transform of a time delay $T$ is $e^{-j\omega T}$ and thus we see that this will impart a linear phase that is proportional ...
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2 votes
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What is the advantage of measuring IQ data non-digitally?

IQ demodulation is normally done with the "I" component being sampled at a rate that is at or near (including below) the carrier frequency of the signal, or the center of the band of interest. Since ...
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2 votes

Difference between these two digital down conversion methods

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 ...
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