modulation - Contradiction between complex baseband and real-valued baseband

Suppose a direct-conversion transmitter+receiver (ideal transmitter) along with its filters. In complex baseband, the filters in the signal chain can be modeled as a complex-valued FIR filter ($y$, $c$, $x$ are complex-valued!):

$$y = \sum_{k=0}^K c_k x[n-k]$$

Rewriting this in cartesian coordinates gives:

$$y_i + j y_q = \underbrace{\sum_{k=0}^K \left(a_k x_i[n-k] - b_k x_q[n-k]\right)}_{y_i} + j \sum_{k=0}^K \left( b_k x_i[n-k] + a_k x_q[n-k] \right)$$

Now only considering the real part, it can be seen that it consists of two different filters with which the real and imaginary input signals are filtered.

Now consider the same system from a practical perspective: After the DAC and upconversion, the RF signal looks as follows:

$$x_{rf} = x_i \cos \omega_c t + x_q \sin \omega_c t$$

The output sequence $y_i$ is obtained by multiplying $x_{rf}$ with $\cos(\omega_c t +\phi)$ and filtering it with a filter called $H$:

$$y_i = H( x_{rf} \cos(\omega_c t +\phi) ) \\ \approx H(x_i/2 \cos\phi + x_q/2 \sin\phi) \\ = \frac{\cos\phi}{2} H(x_i + x_q \tan\phi) = G(x_i + x_q \tan\phi)$$

The filter $H$ (or $G$) is modeled as a FIR filter:

$$y_i = \sum_{k=0}^K \left( g_k x_i[n-k] + g_k \tan\phi x_q [n-k] \right)$$

From the equation above it can be seen that the real and imaginary part are filtered through only one filter (they differ just by the constant $\tan\phi$ !).

Where does this contradiction come from?

PS: I know that the first approach is the correct one because it gives the correct results. I do not understand why it is not consistent with my second approach

Answer

Ok after studying for hours I will attempt an answer to my own question. It does make sense now but I hope that it is correct.

Let us start with the second form:

$$y_i = \sum_{k=0}^K \left( g_k x_i[n-k] + g_k \tan\phi x_q[n-k] \right) = \sum_{k=0}^K \left( a_{i,k} x_i[n-k] + b_{i,k} x_q[n-k] \right)$$

Yes, from this is evident that $a_{i,k}=\beta b_{i,k}$, i.e. the i-output is a linear combination of the two I/Q inputs. This however, falls apart as soon as there is any filter in the RF path. Let us consider the a filter $h$ at RF:

$$y_{rf} = \int_{-\infty}^{\infty} x_{rf}(\tau) h(t-\tau)d\tau$$

Since the input signal has bandpass characteristics (from which the definition of complex baseband arises) it also has the following canonical representation:

$$h_c(t) = 2 h_i(t) \cos\omega_c t +2 h_q(t)\sin\omega_c t \\ = 2\mathrm{Re}\left\{ h_z(t) exp(j\omega t) \right\}$$

and

$$h_z(t) = h_i(t) + j h_q(t)$$

From this is is clear that I and Q parts are filtered with two different filters and hence the i-output is not a linear combination of the inputs any more.

Back to the first form: Indeed, solving only for the I channel gives $2K$ real valued coefficients, in the general case!

Solving for the Q channel gives again $2K$ coefficients ($4K$ in total). However, if I and Q are perfectly balanced, the coefficients obtained from the I channel and the Q channel will be identical. So there are really only $2K$ real-valued or $K$ complex valued coefficients, as expected.

However, if there is an I/Q imbalance, the input/output relationship can still be faithfully described but now in general all $4K$ real-valued coefficients are required!

Interestingly, the difference between the coefficients obtained from the I channel and the Q channel corresponds to the (frequency dependent) I/Q imbalance.