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 $45^\circ$)
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Sign up to join this communityHow 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 $45^\circ$)
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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.
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.
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?