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


2 Answers 2


As a sidenote:

When you demodulate your signal $x_{\mathrm{rf}}(t)$ you say that there is some phase difference $\phi$. This translates to crosstalk (i.e., the quadrature and in-phase components influence each other at the receiver). It's easy to see that if $\phi = 0$ (namely, we know exactly when to sample the received signal), there is no crosstalk at all:

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

In the baseband domain, this means that the filter $c_k$ is purely real (if it was purely imaginary, then there wouldn't be crosstalk either, you just have to invert the channels to retrieve the information correctly).

The answer:

Whether there is crosstalk or not, both of your equations get to the same result, but with different names. Using your notation, $a_k$ in the first equation would be equivalent to $g_k$ in the second one, and $b_k$ to $-g_k\tan(\phi)$. I believe that your confusion arose from thinking of $a_k$ and $b_k$ as two different filters (which is correct), instead of thinking of them as the real and imaginary part of a single filter ($c_k$).

  • $\begingroup$ Thanks, this makes sense. However, my point is: In the second case ($a_k=g_k$ and $b_k=-g_k \tan\phi$) the coefficients are NOT arbitrary, there is a relationship $b_k = \beta a_k$. In the first case $a_k$ and $b_k$ are arbitrary. In subsequent SIMULINK simulations I have seen that in absence of any RF filters both are indeed equivalent. But as soon as I add a bandpass filter at RF, $y_i$ cannot be described as linear combination of $x_i$ and $x_q$ any more! It requires that $a_k \neq \beta b_k$ (i.e., both branches have different filters). Why is this the case? $\endgroup$
    – divB
    Sep 14, 2018 at 19:23
  • $\begingroup$ @divB I'm not sure I understand your problem, but it's probably my fault and not yours. Maybe a simulation diagram showing what you meant in the comment would help. $\endgroup$
    – Tendero
    Sep 15, 2018 at 0:01

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:

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


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

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.


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