# I Q sampling and baseband version of analytic signal

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?

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 $$-2s(t)\sin(\omega_ct)$$.

• Thanks! How did you derive that $\frac12\left[s_a(t)e^{-j\omega_ct}+s_a^*(t)e^{j\omega_ct}\right]=\textrm{Re}\left\{s_a(t)e^{-j\omega_ct}\right\}$ ?
– axk
Jan 21, 2020 at 22:14
• Got it. It is because $c_1^*c_2^*=(c_1 c_2)^*.$
– axk
Jan 21, 2020 at 22:32