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According to theory, summing two signals in the logarithmic scale equals multiplying. Also according to theory, multiplying two sine waves results in two others (a+b; a-b).

So, if I take the signal from the antenna, convert it to the logarithmic scale and then take the logarithmic scale of the local oscillator, sum both and then anti-log, can i get the same result as a mixer?

Besides, can I correlate the resulting signal instead of converting it back to linear scale?

I say that because it looks like that operating the whole chain in logarithmic scale and never going back to linear scale, would result in a device with a way higher dynamic range, so, immunity to nearby transmitters...

Just need to find a way to demodulate FM right from the log scale signal.

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4 Answers 4

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According to theory, suming two signals in the logarithmic scale equals multiplying

To cite an old mathematician's insult: Your theorem seems to leave space for exceptions.

Namely, you're omitting that your signals need to all be strictly positive to be even representable on a logarithmic scale. For most physical signals, that is not the case!

Also according to theory, multiplying two sine waves results in two others (a+b; a-b).

Um, no. Multiplying two sine waves leads to two cosine waves. That makes a difference! Especially in mixing, where sines and cosines are orthogonal.

Also, sines are excellent examples for signals that are non-positive half of the time and hence can't be expressed by a logarithm.

So, if i take the signal from the antenna, convert it to the logarithmic scale and then take the logarithmic scale of the local oscillator, sum both and then anti-log, can i get the same result as a mixer ?

For the aforementioned reason of it being impossible to take the logarithm of a negative number: No.

You can, however, add a constant offset \$C_{RX}>A\$ to make the received signal \$r(t)=A\sin(2\pi f_{RF}t)\$ always positive:

$$s(t) = r(t) + C_{RX} = A\sin(2\pi f_{RF}t)+C_{RX}$$

You can then apply a logarithm to it, and then you can also do the same to your local oscillator \$o(t)= B\sin(2\pi f_{LO}t)\$ with some other constant \$C_{LO}>B\$:

$$m(t) = o(t) + C_{LO} = B\sin(2\pi f_{LO}t)+C_{LO}$$

When now adding \$\log(m(t)) + \log(s(t))\$, you of course get

$$ x(t) = \log(m(t)) + \log(s(t)) = \log(B\sin(2\pi f_{LO}t)) + \log( A\sin(2\pi f_{RF}t)+C_{RX}),$$

and by properties of the logarithm, that is of course

$$ \begin{align} x(t) &= \log\left[(B\sin(2\pi f_{LO}t)+C_{LO}) \cdot( A\sin(2\pi f_{RF}t)+C_{RX})\right]\\ &=\log\left[ \frac{AB}2\left(\cos\left(2\pi \left(f_{LO}-f_{RX}\right)t\right)-\cos\left(2\pi \left(f_{LO}+f_{RX}\right)t\right)\right) \\ + C_{LO}A\sin(2\pi f_{RF}t) \\ + C_{RX}B\sin(2\pi f_{LO}t)\\ + C_{LO}C_{RF} \right] \end{align} $$

As you can see, you've now got a sum of different frequency components in your logarithm:

$$ \begin{align} e^{x(t)} &= \frac{AB}2\left(\cos\left(2\pi \left(f_{LO}-f_{RX}\right)t\right)-\cos\left(2\pi \left(f_{LO}+f_{RX}\right)t\right)\right)\\ &+ C_{LO}A\sin(2\pi f_{RF}t) \\ &+ C_{RX}B\sin(2\pi f_{LO}t)\\ & + C_{LO}C_{RF} \end{align} $$

which clearly means you introduced signal components that your ideal product mixer would not have had. Because \$C_{RF}> A\$ and \$C_{LO}>B\$, it follows that the amplitude of the intended component, \$\frac{AB}2< {C_{RF}C_{LO}}\$, and that means that your unintended intermodulation products with the amplitudes \$ C_{LO}A\$ and \$ C_{LO}A\$ are more than four times as high in power than what you wanted to get.

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So, if i take the signal from the antenna, convert it to the logarithmic scale...

The signal will be bipolar in nature so, how are you going to deal with converting a negative portion of the signal to a log value. After all, a negative log value represents a signal that is between 0 and 1 ; it doesn't represent a negative signal value.

can i get the same result as a mixer ?

No, not one bit.

can i correlate the resulting signal instead of converting it back to linear scale ?

There is no resulting signal because it doesn't work.

I say that because it looks like that operating the whole chain in logarithmic scale and never going back to linear scale, would result in a device with a way higher dynamic range

That also is not true.

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  • \$\begingroup\$ What if i DC bias the signal ? \$\endgroup\$
    – Jorge Aldo
    Oct 14, 2023 at 17:02
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    \$\begingroup\$ Have you looked at this at all in detail? I mean; have you used (say) a spreadsheet to see the problems? @JorgeAldo if we are done here, please take note of this: What should I do when someone answers my question. If you are still confused about something then leave a comment to request further clarification. \$\endgroup\$
    – Andy aka
    Oct 14, 2023 at 17:26
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This works perfectly well when you have an analytical logarithm \$\ln z\$ handy.

This is actually an interesting insight; it's one of the form where, it is completely and utterly wrong to make the assertion, but by roundabout means, the math works out anyway.

For analytical signals, \begin{eqnarray} z_1(t) & = & e^{\alpha_1+j\omega_1 t } \\ z_2(t) & = & e^{\alpha_2+j\omega_2 t } \end{eqnarray} if we simply take the logarithm of each (making suitable nods to which branch cut we are using and all that), we can add the results together, and complex exponentiate again, \begin{eqnarray} \log z_1(t) & = & \alpha_1+j\omega_1 t \\ \log z_2(t) & = & \alpha_2+j\omega_2 t \end{eqnarray}

$$ \exp \left( \log z_1(t) + \log z_2(t) \right) = e^{\alpha_1+\alpha_2+j(\omega_1 + \omega_2) t} $$ Which, clearly, contains the sum frequencies, and the amplitude is the product of the two starting signals. Seems quite handy.

(Analytical log is usually written \$\textrm{Lg}\$ but Mathjax doesn't have that operator handy; please understand the above to mean this.)

Unfortunately, our circuits are constrained by having exclusively real-valued signals in our wires, so it's rather awkward to apply this in practice. We have to go to the trouble of constructing the real and complex components of such functions ourselves; which in turn means we need to spend extensive effort computing an analytical logarithm, and not only that, but somehow constructing unambiguously a phase (a complete, complex-valued, analytical signal \$z(t) = a + jb = e^{\alpha + j\omega t}\$) from arbitrary input signals (usually done with a lengthy kernel such as the Hilbert transform).

Whereas, simply multiplying real signals in the time domain affords the sum and difference frequencies, and usually other higher-order products due to the nonlinear devices we use to approximate this operation; all of these are easily separated with strategic design (avoiding overlapping sums/products i.e. image frequencies), and frequency-selective filters.

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Logarithmic multipliers are sometimes used in digital signal processing systems.

Or you can use a logarithmic ADC, then a digital adder.

One way to do it is to split the signal in to + and -, and handle the two signs in separate channels.

Any non-linear function can be used as part of a multiplier. One advantage of a pure multiplier is that it leaks only a small amount of other frequency components. If you push DC bias into your multiplier, you have a more difficult filtering task at the output. With signals that aren't hi-fi analog, sometimes that doesn't matter.

At it's roughest, least accurate approximation, a signed-channel multiplier is just a signed rectifier -- you multiply by a +1, -1 to restore a chopped signal to a signed rectified signal. If you just have a square-wave chopped signal to start with, and are using phase-locked direct demodulation (instead of IF), that's perfect.

If your modulated signal has been bandwidth limited (either during or after modulation), you have something like sine-wave modulation instead of square-wave modulation to start with, and even in the direct-detection case, your signed rectified signal will look like a rectified sine wave with envelope shaping. It will have sharp zeros at the original zero crossing points.

You can reduce the harmonic content at the digital-rectification stage by compressing the signal (logarithm), sharpening the peaks, flattening the zeros (add logarithm of sine wave), then expanding it out again (anti-logarithm). Now, instead of a rectified sine-wave with envelope shaping, you've shaped the rectified sine-wave to have sharper peaks and flatter zeros -- that is, you've made the signal more sine-wave shaped, with lower harmonic or cross-modulation content.

If you have a coherent fixed-frequency signal, you can do that digital shaping using fixed components. If you have an unknown/variable frequency, you can calculate the shaping values at runtime. In theory, that is approximating the logarithm of the heterodyne sine wave. In practice, it is 'fast efficient calculation of the shaping values'.

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