# What happens, when an unlimited bandpass-signal gets mixed?

In theoretical books there's always a bandpass-limited signal being mixed, that's been low-pass filtered beforehand.

But what happens, if the RF-signal from the antenna isn't band- or lowpass-filtered?

I know for example from decimation (lowpass-filtering and downsampling) that there would be aliasing, which isn't a form of mixing but nonetheless.

what happens, if the RF-signal from the antenna isn't band- or lowpass-filtered?

If the system is analog, lots of uncontrolled harmonics. If the system is digital, then yes, you will get aliasing.

Aliasing itself is not a form of mixing. But sampling is.

When an analog signal $x(t)$ is sampled at distinct times $\frac{1}{Fs}$ seconds apart, it is effectively mixed with a train of impulses at frequency $Fs$. Sample-and-hold turns these spikes into "steps". This is what is causing the "mirroring" of the spectrum.

In fact, by tuning your $Fs$ carefully, you can "shift" a signal into a baseband by undersampling it.

I know for example from decimation (lowpass-filtering and downsampling) that there would be aliasing, which isn't a form of mixing but nonetheless.

Probably it is clear by now that it is. Take an analog signal $x(t)$, multiply it with a train of impulses $u_s(t)$ at some frequency $Fs$, now multiply the result by another train of impulses $u_d(t)$ that is equivalent to "taking out" every other pulse. If you don't first low pass $x(t)$ for the spectrum you are trying to "squeeze it in" post mixing by $u_s(t) \cdot u_d(t)$, you will get aliasing. It's the same concept applied twice.

Hope this helps.

You are not mentioning sampling in your question, so I assume your question is purely in the analog world, with real signals (a simple mixer), although the same concepts described would apply digitally with the complication of Nyquist boundary issues.

When you multiply two real signals, the output is the sum and the difference of the frequencies (and phase!) of the two input signals. This is clear from the $cos(\alpha)cos(\beta)$ trigonometric relationship expanded as follows:

$$cos(\alpha)cos(\beta)= \frac{1}{2}cos(\alpha +\beta) +\frac{1}{2}cos(\alpha-\beta)$$

So for our case with an input frequency band centered on $\omega_1$ mixed with a signal at $\omega_2$:

$$cos(\omega_1 t+ \phi_1)cos(\omega_2 t+ \phi_2) = \frac{1}{2}cos((\omega_1+\omega_2t) t+ (\phi_1+\phi_2))+ \frac{1}{2}cos((\omega_1-\omega_2t) t+ (\phi_1-\phi_2))$$

Showing that the result is the sum and difference of the frequency and the phase of the two signals.

Thus we see the implications of filtering or not filtering as follows:

Often our interest with mixing is with frequency translation (but there are many other applications, such as phase detectors, frequency discriminators, attenuators, and building blocks for image reject mixers, vector modulators, etc). I am assuming the frequency translation application, and in that we have a single band of frequencies at the input that we want to translate to another band by multiplying it with a single tone (The Local Oscillator or LO).

With that we need to consider the implications of filtering at both the input and the output.

Filtering on the output:

The output filtering requirement is easy to see from the expanded formula above. Since the mixer output produces the sum and difference of the frequency of the two input signals (our band of interest and the LO), we need to filter out the one that we are not interested in. (The higher frequency if we are doing a down-converter, or the lower frequency if an up-converter). SIDE NOTE - You can get improved linearity performance by using a terminating (absorptive) filter for the rejected image instead of the traditional filter that will be normally be reflective, by absorbing the unwanted signal instead of reflecting it back into the mixer.

Filtering on the input:

On the input side, we can see that there are two solutions that will map to the same frequency band of interest at the output of the mixer: Referring to our output frequency band the IF (intermediate frequency), and our input frequency band the RF (radio frequency) we can see that the following two solutions will work:

$$IF = LO - RF$$ $$IF = LO + RF$$

Therefore there is an image frequency that will land on our same IF output if we do not bandpass filter the input to only pass one of them.

NOTE!! Even if you know that you do not have any signals present in the image frequency location, if you do not filter that location out you can get a 3 dB noise figure penalty for not doing so (meaning the SNR achieved could be 3 dB worst!). The only time this does not occur without filtering is when you have no active gain prior to the mixer input. This is because without filtering, the noise in the band of interest is equal to the (uncorrelated) noise in the image band, so you will have a 2x or 3dB noise penalty if you do not filter the noise first. If there is no active gain prior to the mixer, the filtering will not change the thermal noise floor that will be present at the input and output of the mixer; but if there is any gain, the filter can reduce the amplified noise back down to the noise floor, while the signal + noise in band is amplified by the gain.

Meaning, always filter!!

Image reject mixers do not have this problem so are a reasonable alternative to filtering in cases where the filter requirements would be too stringent.

Also an interesting aside to note: you can view an A/D converter as a mixer with an infinite amount of LO's, each at a harmonic of the sampling rate; which is a way to see aliasing in an A/D converter for someone that understands the (simpler) case of mixing described above.

If you use a quadrature mixer, then the IQ result is simply the original spectrum, both above and below the complex LO, shifted up or down by the frequency of the LO (including into the range of negative frequencies in the complex domain). In the case of sampled signal, the spectrum shift will circularly wrap around the sampling rate frequency. Of course, any harmonic distortions and phase errors in the mixer's quadrature local oscillator will also add noise to the resulting shifted spectrum.

There is no difference between the cases of filtered versus unfiltered signals in "theoretical" books. The modulation theorem of Fourier transform theory places no restriction on the signal that is being translated in frequency. That is, $\mathcal F\left[2x(t)\cos(2\pi f_0t)\right]$ equals $X(f+f_0) + X(f-f_0)$ for all choices of $x(t)$, whether filtered or unfiltered. What does happen when $X(f)$ is band-limited to $[-B,B]$ is that the support of $X(f+f_0)$ is $[-f_0 -B, -f_0+B]$ while the support of $X(f-f_0)$ is $[f_0-B, f_0+B]$ and in the special case when $-f_0+B < f_0 -B$, that is, $f_0 > 2B$, the spectra $X(f+f_0)$ and $X(f-f_0)$ have disjoint supports. This leads to the pretty pictures one sees in "theoretical" books which show $X(f)$ as a blob sitting at the origin while $X(f+f_0)$ and $X(f-f_0)$ are the same blob sitting at $-f_0$ and $+f_0$ respectively with no overlap whatsoever. But, the result $$\mathcal F\left[2x(t)\cos(2\pi f_0t)\right] = X(f+f_0) + X(f-f_0)$$ holds for all $x(t)$ and when $x(t)$ is not bandlimited, $X(f+f_0)$ and $X(f-f_0)$ might be nonzero for all $f$. For example, the Fourier transform of the Gaussian pulse $e^{-\pi t^2}$ is $e^{-\pi f^2}$ which is a Gaussian pulse in the frequency domain and both $e^{-\pi (f+f_0)^2}$ and $e^{-\pi (f-f_0)^2}$ have support $(-\infty,\infty)$ regardless of the choice of $f_0$.

"Practical" books, on the other hand, which describe actual results obtained when a real transmitter circuit (often involving a Class C amplifier) modulates a nonbandlimited $x(t)$ onto the carrier $2\cos(2\pi f_0t)$ will likely show that the spectrum of the transmitted signal is not $X(f+f_0)+X(f-f_0)$ as the theoreticians think, but something which is closer to a filtered version of $x(t)$ being modulated onto $2\cos(2\pi f_0t)$ where the filtering is determined by the tuned circuit that is the load of the Class C amplifier, as well as the frequency response of the antenna. This is merely an illustration of the adage that in theory, theory and practice are the same, but in practice they are different.