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How can you implement a $9\rm V$ battery with a phase of $45^\circ$? (As a black box with a DC Voltage of $9\rm V$ and a phase of $45^\circ$)

Please preface your answer with spoiler notation by typing the following two characters first ">!"

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  • $\begingroup$ Can you please clarify what do you mean by the phase of a signal with frequency 0 Hz? If $V_{DC}=|A|e^{j\phi}$, then the voltage is complex for any $\phi$ except 0 and $\pi$. $\endgroup$
    – MBaz
    Jun 8 '16 at 14:15
  • $\begingroup$ There's nothing wrong with puzzles, but I wonder if this is about DSP, and - like MBaz - I wonder how you define a phase shift in that case. $\endgroup$
    – Matt L.
    Jun 8 '16 at 14:23
  • $\begingroup$ It will be clearer once you see the answer, I just wanted to give people a chance to provide the answer, with DSP in mind. The point that a DC signal can have a phase shift is salient to understanding certain DSP processing so is a favorite question I like to ask my students in my class. $\endgroup$
    – Dan Boschen
    Jun 8 '16 at 14:27
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    $\begingroup$ This makes no sense. Phase can only be defined against so reference (i.e. phase between two things). DC cannot have phase. Can you describe an experiment or a setup at which the phase that you are ask for would be observable. Who would you know that there is a 45 degree phase or any other phase for that matter. $\endgroup$
    – Hilmar
    Jun 8 '16 at 15:26
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    $\begingroup$ The dsp-puzzle tag is a great idea. $\endgroup$
    – A_A
    Jun 8 '16 at 16:55
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OK, this is a slightly constructed situation, but as far as I can see, the following is the only thing that makes sense, kind of ...

If $A$ is your DC voltage, then a phase of $45$ degrees means multiplying it with $e^{j\pi/4}$, i.e., you get $Ae^{j\pi/4}=\frac{A}{\sqrt{2}}(1+j)$. So scale the voltage with $1/\sqrt{2}$, and - apart from the ground wire - use two wires connected to '+' coming out of that box. Stick a tag with $j$ on it on one of the two.

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    $\begingroup$ I like this question for the purpose of explaining why we need "two scope probes" to observe a complex signal in the lab- why there are both I and Q signal paths in the implementations of any complex signal processing at baseband. Matt was spot on. $\endgroup$
    – Dan Boschen
    Jun 8 '16 at 15:38
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    $\begingroup$ I agree this "kind of" makes sense, but I believe @DanBoschen's interpretation is incorrect, or at least incomplete. As he says, DC voltage with a phase is a perfectly well defined and useful theoretical concept; however, such a thing doesn't exist in practice. $Ae^{j\pi/4}$ is complex; the two wires represent something related, but different: the real and the imaginary parts of the complex quantity. You haven't created a DC voltage with a phase; you have just two DC voltages.You can think of them as defining a complex quantity, but that's just an interpretation, not physical reality. $\endgroup$
    – MBaz
    Jun 8 '16 at 16:38
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    $\begingroup$ @MBaz: In that sense complex quantities are never physical realities. Complex numbers are just a convenient tool to make one thing out of two (namely, real and imaginary part), and any processing involving complex signals can be realized with two wires, tagging one of them with $j$. As you know, digital communications without complex(-valued) processing would be much harder to explain and to write down. Is a QAM constellation a physical reality? I'd say yes. $\endgroup$
    – Matt L.
    Jun 8 '16 at 16:44
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    $\begingroup$ @DanBoschen I don't disagree with anything you've said, but that's not the way I interpret complex signals. I teach my students that a complex envelope exists only on paper and, in a sense, in software; quadrature signals only exist physically if they're passband. But that's just the way I see it; I'm not saying your way is wrong. $\endgroup$
    – MBaz
    Jun 8 '16 at 18:33
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    $\begingroup$ To add to this discussion myself: @MBaz is absolutely right, there can be a complex voltage with a phase of 45° – but it's not coming out of a something that we could call a "battery", ever, because it's not a "DC voltage", but two DC voltages, making up a complex signal. $\endgroup$
    – Marcus Müller
    Jun 8 '16 at 20:56
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Take a sinusoidal oscillator circuit with two outputs that are always 45 degrees out of phase from each other feeding two DC coupled 9V amplifier circuits. Stop the oscillator (set f = 0) when one output (whichever one you designate is "real") of the two is at 90 degrees (of a sinewave full swing). The two outputs together will have a phase relationship when considered as one system output.

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  • $\begingroup$ A complex signal can be considered as just 2 real signals that together and pairwise (labeling one as "imaginary") comply with (or approximate closely enough) the rules for complex multiplication and addition operations within some system. And if you can consider the pair to represent a complex number, you can consider the pair to have a phase angle. $\endgroup$
    – hotpaw2
    Jun 9 '16 at 16:55
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That's so easy, it's not even complex:

Take a rasp and rasp down the edges to the angle you desire - 45° it'll be in your case. Just make sure you don't spill any acid on the table, would you?

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  • $\begingroup$ Sehr gut- Ausgezeichnet! $\endgroup$
    – Dan Boschen
    Jun 9 '16 at 21:57

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