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Is it correct to say that if we have a radio signal $s(t)$ centered around the angular frequency of $\omega$ as $\omega\pm\omega_B/2$ (where $\omega_B$ is the bandwidth of the signal) and the corresponding analytic signal is $s_a(t)=s(t)+j \hat{s}(t)$ (where $\hat{s}(t)$ is the Hilbert transform of $s(t)$) then $I$ and $Q$ as received from an SDR receiver (RTL-SDR) can be represented as

$I=Re(s_a(t) e^{-j \omega t})$
$Q=Im(s_a(t) e^{-j \omega t})$

In other words is it correct that $I$ and $Q$ are the real and imaginary parts of the analytic signal shifted to baseband?

I'm trying to derive the above from the assumption that the $I$ component is a low-pass filtered version of $s(t) \cos\omega t$ and the $Q$ component is a low-pass filtered version of $s(t) \sin\omega t$.

Considering that

$s(t) \cos\omega t = (s_a(t)-j\hat{s}(t))\times {1\over{2}}(e^{-j\omega t}+e^{j\omega t})$
$s(t) \sin\omega t = (s_a(t)-j\hat{s}(t))\times {1\over{2}}j(e^{-j\omega t}-e^{j\omega t})$

and then expanding these and discarding high frequency components (both negative and positive) I arrived at the following

$I={1\over{2}}(s_a(t) e^{-j\omega t} - j\hat{s}(t)e^{j\omega t})$
$Q={1\over{2}}j(s_a(t)e^{-j\omega t} + j\hat{s}(t)e^{j\omega t})$

And I expected that $I+jQ$ should give me the shifted analytic signal $s_a(t)e^{-j\omega t}$ but when I substituted $I$ and $Q$ from above into $I+jQ$ I ended up with $-j\hat{s}(t)e^{j\omega t}$ which is the negative frequency replica of the signal shifted to baseband, which would be a mirrored version of $s_a(t) e^{-j \omega t}$ (i.e. of what I expected).

Is my initial statement about the relation between $I$ and $Q$ and the analytic signal correct and there's just a mistake in my math or am I making a wrong assumption somewhere along the way?

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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 filter filters out parts of both terms, but leaves other parts unchanged.

You could better represent $s(t)$ as

$$s(t)=\frac12\big[s_a(t)+s_a^*(t)\big]\tag{1}$$

where $s_a^*(t)$ is the complex conjugate of the analytic signal $s_a(t)$. This has the advantage that both components on the right-hand side of $(1)$ have only frequency components either at $\omega_c$ ($s_a(t)$) or at $-\omega_c$ ($s_a^*(t)$), but not at both frequencies.

So, for example, the in-phase component is easily derived as a low pass filtered version of $2s(t)\cos(\omega_ct)$:

$$\begin{align}I(t)&=\textrm{LP}\big\{2s(t)\cos(\omega_ct)\big\}\\&=\frac12\textrm{LP}\left\{\big[s_a(t)+s_a^*(t)\big]\left[e^{j\omega_ct}+e^{-j\omega_ct}\right]\right\}\\&=\frac12\left[s_a(t)e^{-j\omega_ct}+s_a^*(t)e^{j\omega_ct}\right]\\&=\textrm{Re}\left\{s_a(t)e^{-j\omega_ct}\right\}\tag{1}\end{align}$$

where $\textrm{LP}\{\cdot\}$ is the low pass filter operator.

In a similar way, the quadrature component $Q(t)$ can be shown to be a low pass filtered version of $-2s(t)\sin(\omega_ct)$.

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  • $\begingroup$ Thanks! How did you derive that $\frac12\left[s_a(t)e^{-j\omega_ct}+s_a^*(t)e^{j\omega_ct}\right]=\textrm{Re}\left\{s_a(t)e^{-j\omega_ct}\right\}$ ? $\endgroup$ – axk Jan 21 '20 at 22:14
  • $\begingroup$ Got it. It is because $c_1^*c_2^*=(c_1 c_2)^*.$ $\endgroup$ – axk Jan 21 '20 at 22:32

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