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In a standard mixer, we multiply two signals $x_1(t)$ and $x_2(t)$, each with units of volts $(V)$, and get at the output:

$$y(t)=x_1(t)\,\cdot\,x_2(t)$$

I have always thought that the output of mixing operation was another voltage signal. But how can this be? Looking at the units of the factors, we multiply a quantity of volts, $x_1(t)$, by another quantity of volts, $x_2(t)$. So shouldn't the units of $y(t)$ be $V\cdot V=V^2=W$?

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    $\begingroup$ $V^2$ isn't equal to $W$, $W = V \cdot A = V^2/ \Omega$ $\endgroup$ – Hilmar Nov 22 at 14:44
  • $\begingroup$ Ah right- I'm assuming that the output $y(t)$ is measured across an ideal $1\Omega$ resistor: $(x_1(t)x_2(t))/1\Omega$, so $V^2/\Omega=W$. But even so, if the units were $V^2$, the question remains the same: how can a mixer output be in units $V$ $\endgroup$ – Ben Granger Nov 22 at 16:18
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If the multiplier takes two voltages as input and returns a voltage as output, then there is necessarily a constant involved, with units of [1/V].

Take for example, AD633 (which was the first search result). The output is the product of the 2 inputs times a constant:

$V_{out} = \frac{V_1 \times V_2}{10V}$

So the output units are Volts.

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    $\begingroup$ Good answer. Also, it's worth noting that there must be a constant then you'd end up with different results depending on the units you used in your calculation. 0.5V * 0.5V = 0.25V but 500mV * 500mV = 250000mV = 250V. So something must be wrong there. But if we have our constant, 0.5V * 0.5V / 1V = 0.25V and 500mV * 500mV / 1000mV = 250mV $\endgroup$ – Jack B Nov 23 at 15:05
  • $\begingroup$ @JackB ... don't know what was in your mind, but from basic physics/math the product $500 mV \times 500 mV = 250000 \mu V^2$ and not $250000 mV$. $\endgroup$ – Fat32 Nov 24 at 13:33
  • $\begingroup$ @Fat32 Yes, the units in the first example are wrong. The point that I was trying to make was that if you don't have the right units, you get gibberish. The output of the mixer is in volts because that's what the mixer outputs - a signal in volts. But because the units don't match on each side of the equation, the result is wrong. And the fact that it depends on the units (mV or V) used is a big indicator that it is wrong. It was supposed to be a demo of why the units have to match. $\endgroup$ – Jack B Nov 24 at 14:49
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In addition to Juancho's answer for the general mixer, I would like to give an example for a more simpler frequency mixer most commonly used in communication systems to shift the frequency spectrum of a message signal up or down for transmission or reception etc.

The simplest understanding of a physical realisation of a mixer assumes an on-off switching applied to the input signal $x(t)$ ;

$$ z(t) = x(t) w(t).$$

This on-off switching (which typically realized by a transistor control base input) essentially creates an output voltage mathematically equivalent to the multiplication of the input signal $x(t)$ in units of Volts and a unitless switching waveform $w(t)$. Hence the output is in units of Volts again.

The resulting up-shifted (modulated) spectrums of $x(t)$ are available at different frequencies and bandpass filtered subsequently to extract the one of interest.

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    $\begingroup$ This is also the case for a diode ring mixer, where the output is the input multiplied by a unit square wave. $\endgroup$ – TimWescott Nov 22 at 23:30
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Mathematically speaking $V^2$ is perfectly alright to use. Physically I am not sure you can multiply electrical signals like that.

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In Power they use the term kilo volt amps (kva) to make a distinction with kilo watts.

Following that, perhaps volt volt would make sense.

In a lot of cases, a voltage ( or binary voltages/currents) is just physical way to encode a mathematical abstaction. When we multiply two numbers, we get a number. We don’t have to attach physical units to every product we compute.

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