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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 ">!"

Peter K.
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Dan Boschen
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  • 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 $\pi$. – MBaz Jun 08 '16 at 14:15
  • 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. – Matt L. Jun 08 '16 at 14:23
  • 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. – Dan Boschen Jun 08 '16 at 14:27
  • Please define phase shift. Normally, an ideal phase shifter has a frequency response $$H(j\omega)=e^{j\theta\text{sign}(\omega)}$$ where $\theta$ is the phase shift. However, this doesn't make sense for a DC signal. – Matt L. Jun 08 '16 at 14:36
  • I am defining phase shift as a change in phase of the signal. So perhaps the question is clearer if I drop the word "shift" as no change is involved and just ask how do you implement a 9V battery that has a 45 degree phase? – Dan Boschen Jun 08 '16 at 14:42
  • To clarify, a DC signal can have a phase, and the question is how do you implement it? This gives insight to someone that is new to certain aspects of DSP - the answer is very simple but if I say anymore than that I will give it away. – Dan Boschen Jun 08 '16 at 14:52
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    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. – Hilmar Jun 08 '16 at 15:26
  • Yes makes perfect sense; see Matt's answer. The reason for this question is people often assume "phase" with "delay" which of course would not apply to a DC signal. Yes phase can be defined against a reference, and in this case that still holds as the 9V battery with 45 degree phase is compared to a 9V battery that has 0 degree phase. The idea of a constant value (such as DC) with a phase has practical application in DSP, for example in the FFT implementation. – Dan Boschen Jun 08 '16 at 15:33
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    The dsp-puzzle tag is a great idea. – A_A Jun 08 '16 at 16:55
  • Thank you all for the good discussion and humoring my mental exercise with complex signals! – Dan Boschen Jun 09 '16 at 01:03

5 Answers5

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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.

Matt L.
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  • Yes, that's all I could come up with. :-) – Peter K. Jun 08 '16 at 15:25
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    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. – Dan Boschen Jun 08 '16 at 15:38
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    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. – MBaz Jun 08 '16 at 16:38
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    @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. – Matt L. Jun 08 '16 at 16:44
  • @MattL. Let's not get too philosophical :) Personally, I think a passband QAM signal is a phyisical reality, which was built from an extremely convenient, mathematical description of a QAM complex envelope (baseband) signal. In any case, I think this is a distiction that is useful to make in a classroom (which is the context for the problem). – MBaz Jun 08 '16 at 16:50
  • It does exist in practice, exactly as Matt has answered- we often use a constant value with a phase shift in implementation (using two signals), used to rotate a signal for example- everything we are describing is an interpretation when you think about it. The mental exercise is in viewing signals as exponentials instead of sine waves and what the exponential frequency means in physical implementation. This is a very helpful mental exercise in bridging an analytical solution (using exponentials which is often easier and more compact instead of sines and cosines) to an implementation. – Dan Boschen Jun 08 '16 at 16:51
  • Here is a good practical example: using an IQ mixer to shift a 10MHz signal 45 degrees. With 10MHz on the LO port and assuming the IQ mixer's IF port goes to DC, apply DC Voltage X to the I port, where X is max voltage for the "linear" range of the mixer, and 0V to the Q port (less any DC offset). The output with this setting we will call the reference 10MHz out with 0 phase. Then apply .707X to both the I and Q ports, and the output will have shifted 45 degrees (with some minor mag change since the mixer will not be truly linear). We multiplied the 10MHz with a DC at 45 deg- the math is easy! – Dan Boschen Jun 08 '16 at 17:28
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    @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. – MBaz Jun 08 '16 at 18:33
  • I totally understand and agree with you MBaz! It is all just interpretations and a matter of tools that allow us to model physical systems. For that matter we could view cosine(omega t) as existing only on paper too! – Dan Boschen Jun 08 '16 at 18:36
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    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. – Marcus Müller Jun 08 '16 at 20:56
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    @MarcusMüller Not really all that different from realizing a single 9V battery with 4 separate 1.5V cells, is it? So for us to describe a single complex number (even on paper or purely theoretical space) we require the use of two real values (either magnitude and phase, or I + jQ). So why then does it trouble us to implement a complex number with two waveforms in the same fashion and not be able to call that pair of waveforms representative of a single complex waveform (well it actually does, and that is how we implement radios!). – Dan Boschen Apr 04 '20 at 13:53
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    So key to this, if I want to implement a constant source with a 45° phase shift (which is a phase rotation, I never liked the ubiquitous use of shift as it leads many to confuse a constant rotation which is the point of this with a constant time delay), I would implement such a rotation using two real values, and label one I and the other Q just as Matt answered. – Dan Boschen Apr 04 '20 at 13:55
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    @DanBoschen good argument; I don't even know my point of view anymore :) – Marcus Müller Apr 04 '20 at 16:12
  • @MarcusMüller That is why I love this question, and use it early when bringing newcomers over to the complex world! Our "reality" is all analogies anyway so just a matter of what we are used to and can relate to. That's why I also find the questions similar to "do imaginary numbers exist" so interesting. – Dan Boschen Apr 04 '20 at 16:20
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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.

hotpaw2
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    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. – hotpaw2 Jun 09 '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?

M529
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After going over this problem in your course, I now understand you have to think outside the box! You create a complex output by having two sources within your battery (one for the In-Phase component and the other for the Quadrature). Using Euler's formula $Ae^{j\theta}$ we see that $9e^{j(\pi/4)}$ which translates to $9/\sqrt2$ volts on each source! Now I'm curious on what real-world use case this would be viable for?

Aaron Hoover
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9 V DC batteries do not exist. I think you mean a 9 volt 5 nanohertz battery, which are sold in many stores and have output of cos(2 π t f) with f = 5 nHz until they are empty. Compared to this 0° battery, a 45° 9 V battery would begin at 6 V output and rise to 9 V after about half a year of use, after which it would fall until the battery is empty.

jpa
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  • Nice! I’m thinking $9e^{-2\pi f t}$ for your zero degree case. To follow your train of thought, what made your complex case go up before down? – Dan Boschen Apr 11 '23 at 10:25
  • @DanBoschen I'm interpreting 45° as phase lag, while 0° would be maximum at t=0. But I agree that 45° phase lead would make sense, which would just be a bit empty to start with. – jpa Apr 11 '23 at 12:00
  • Ah I got it, given you used a slow sinusoid to model the decay. (I like the thought --- if it was $9e^{-2\pi ft + j\pi/4}$ it would be 9V at 45 degrees starting at t=0 and decaying with a real frequency (exponential decay) :) – Dan Boschen Apr 11 '23 at 12:10