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I have a simple question that I cannot seem to wrap my head around. let's say I have a function Asin(wt), if I multiply a '-j' by this function to get:

-jAsin(w*t), I have been told that this is essentially a phase shift by negative 90 degrees.

So essentially the result is cos(w*t)

Thinking in terms of microwave 90degree hybrids entire S parameter matrix's claim that multiplying by 'j' will cause a phase shift.

I can't seem to prove mathematically why this is though as when I apply Euler's formula:

e^(jθ) = cos(θ) + j sin(θ)

I get -jAsin(wt) = Acos(wt)-Ae^(jw*t)

Can someone help me mathematically show that a multiplication of j will cause a phase shift? Or am I completely wrong altogether?

EDIT: this is my math so far:

Convert sin to cos with an argument: -jsin(wt) = e^(-jpi/2)cos(wt-pi/2)

cos(w*t-pi/2) is the Real[e^j(wt-pi/2)]

so e^(-j*pi/2)*e^j(wt-pi/2)=e^(j(wt-pi)). looking at only the real part of this

is -cos(wt)

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  • \$\begingroup\$ Can I assume that you already do understand that if you take (1+2j) as a vector on the complex plane and multiply that by j to get (-2+1j), that the new vector will be at 90 degrees counterclockwise with respect to the original vector? There's more to follow that and some reasoning with respect to their application in electronics. But if that first bit isn't clear, then the rest will be a bit of a slog. (I've not considered your starting statement. So I may find something wrong with your logic starting there when and if I take a moment for that.) First things first, though. \$\endgroup\$ Nov 10, 2023 at 1:22
  • \$\begingroup\$ @periblepsis Yes I understand this result, but surely the shift in the real complex plane does not translate to a phase shift of sine : sine(wt-theta). If I have -jsin(wt) this is a sine wave that is purely imaginary. How is that a real cosine? -cos(wt). Surely jsin(wt) != cos(wt) as this disagrees with Euler formula \$\endgroup\$ Nov 10, 2023 at 1:31
  • \$\begingroup\$ Let me jump forward a little as another probe. Try out this statement: Vectors are not just lists of numbers. All lists of numbers are vectors. But not all vectors are lists of numbers. Vectors can be functions. Does this also make sense to you? \$\endgroup\$ Nov 10, 2023 at 1:40
  • \$\begingroup\$ @periblepsis Sure, but I am not seeing what you are getting at. \$\endgroup\$ Nov 10, 2023 at 1:43
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    \$\begingroup\$ @Jirhska, you don't multiply the time-domain representation by j, you need to convert to a phasor representation first. See my latest edit. \$\endgroup\$
    – The Photon
    Nov 10, 2023 at 1:59

4 Answers 4

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I think that confusion could be related with the two possible methods for representing sinusoidal time varying signals associated with the complex plane:

1. AS COMPLEX NUMBERS

This form is close related to a two terms Fourier series decomposition for sinusiodal signals. The "vector" resulting from summing two complex exponentials, rotating in opposite directions, has a time varying magnitude and describes a trajectory along a straight line in complex plane.

First, consider the following plot for \$\sin(\omega t)\$:

enter image description here

The next animation shows an alternative representation for that "vector" summ (non centered at origin). This version is prefered in diagrams showing several Fourier terms, with multiple harmonics, which I'm gonna use on the next animations here:

enter image description here

The representation for \$cos(\omega t)\$ below, looks like that for \$\sin(\omega t)\$, but note that, for instance when \$\omega t = 0\$ rad, the positions are \$(1,0)\$ and \$(0,0)\$) respectively:

enter image description here

So far, I've shown only real valued signals. Next, it's worth to noting that \$-j\sin(\omega t)\$, does not represent a physical signal. For this reason, in most cases, the coeficients in Fourier series expansion appear as pairs of complex conjugates values that, when summed up, generate the real valued terms. A similar argument is applicable to Fourier Transform, too.

So:

$$-j\sin(\omega t)=\frac{1}{2}e^{-j\omega t} - \frac{1}{2}e^{j\omega t}$$

The animation below shows that, multiplying the real signal \$\sin(\omega t)\$ by \$-j\$, causes a shift by 90\${^\circ}\$ clockwise (-90\${^\circ}\$), transforming it in a signal with imaginary component only.

enter image description here

Therefore:

$$-j\sin(\omega t) \neq \cos(\omega t)$$

Now, let me show you a more elaborated signal that results from shifting \$\sin(\omega t\$) by 30\${^\circ}\$ counterclockwise. In other words, multiplying it by the complex \$e^{j\frac{\pi}{6}}\$ or \$\frac{\sqrt{3}}{2}+j0.5\$:

$$\left(\frac{\sqrt{3}}{2}+j0.5\right)\sin(\omega t)=$$ $$e^{j\frac{\pi}{6}}\sin(\omega t)=$$ $$e^{j\frac{\pi}{6}}\frac{1}{j2}\left(e^{j\omega t}-e^{-j\omega t} \right)=$$ $$\frac{1}{2}\left[e^{j(\omega t-\frac{\pi}{3})}-e^{-j(\omega t+\frac{\pi}{3})} \right]$$ $$\frac{1}{2}\left[e^{j(\omega t-\frac{\pi}{3})}+e^{-j(\omega t-\frac{2\pi}{3})} \right]$$

enter image description here

2. AS DECOMPOSITION OF ROTATING PHASORS

This representation is used in literature when discussing phasors related to AC circuits. The related time varying signals are always real, resulting from taking the real (\$Re\left\{.\right\}\$) or the imaginary component (\$Im\left\{.\right\}\$) from rotating phasors (here I take the real components). These phasors rotate in the same direction due the fact it are different entities - with no summation. The animation below shows the 90\${^\circ}\$ shifting between \$\cos(\omega t)\$ and \$\sin(\omega t)\$ (or \$\cos(\omega t-\frac{\pi}{2})\$):

enter image description here

Finally, a 30\${^\circ}\$ shifting between two waveforms:

enter image description here

I think this second form relating sinusiodal signals with the complex plane is the most useful and the one you are interested in.

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  • \$\begingroup\$ very nice, what did you use to do the cool animations? \$\endgroup\$
    – danmcb
    Nov 30, 2023 at 9:39
  • \$\begingroup\$ It can be done, for example, using desmos, Python (matplotlib, manim,...), Matlab, Scilab, Mathematica, Maple and others. \$\endgroup\$ Nov 30, 2023 at 17:51
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e^(jθ) = cos(θ) + j sin(θ)

I get -jAsin(wt) = Acos(wt)-Ae^(jw*t)

You might be better off going to

$$ \sin\theta = \frac{e^{j\theta}-e^{-j\theta}}{2j}$$

However this is not going to get you to \$j\sin\theta = \cos\theta\$ if that's what you mean by a phase shift. It can't get there because \$\cos\theta\$ is always real and \$j\sin\theta\$ is always complex, so they absolutely can't be equal.


That said, if your function \$A\sin(\omega t)\$ is meant to be a phasor representing a voltage or current in AC steady state, then all you need to know is that when multiplying complex numbers,

$$\angle AB = \angle A + \angle B$$

and therefore any time you multiply a phasor by \$j\$, which has angle \$90^\circ\$, you always advance the phase by \$90^\circ\$, regardless of what you started with.


Edit to add: In comments you asked,

can I be annoying and ask for a derivation OR a pointer to a derivation that allows us to add the angle?

The easy way is to write your phasors in exponential functions like

$$ a + jb = r e^{j\theta} $$

But be careful, the \$\theta\$ in this formula is not the argument of the sine function in your \$A\sin(\omega t)\$. You should be writing

$$ A \sin(\omega t) = A \sin(\omega t) e^{j0}$$,

then you multiply by \$j\$:

$$ (j)(A \sin(\omega t)) = (1 e^{j\frac{\pi}{2}})(A \sin(\omega t) e^{j0})$$ $$ = A \sin(\omega t) e^{j\frac{\pi}{2}}$$

Now you can see the phase term (the argument of the exponential) has increased from 0 to \$\frac{\pi}{2}\$, meaning the phase of the phasor has advanced in phase.

Of course you could also just recognize that \$jA\sin(\omega t)\$ is a purely imaginary number, and therefore is always 90 degrees advanced from \$A\sin(\omega t)\$ which is a purely real number.


Maybe the real answer:

It seems likely that what has really happened here is you are mixing up the time domain and phasor representations of your signal, and when you treat your signal \$A\sin(\omega t)\$ as a phasor, the phasor value should be \$1 + j0\$. The angle of this phasor is 0.

Now when you "multiply by j" you do this in the phasor domain and get \$j(1 + j0)\$, which is \$0 + j1\$. And this has an angle of \$\pi/2\$ radians or \$90^\circ\$, which is 90 degrees advanced from the phasor you started with.

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  • \$\begingroup\$ This is good, but can I be annoying and ask for a derivation OR a pointer to a derivation that allows us to add the angle? I listed some math in my Original question regarding only taking the real part of the [e^j(wt-pi/2)] which would show that jsinθ=cosθ. But I need to know why this is a reality. Maybe I am forgetting something elementary here. \$\endgroup\$ Nov 10, 2023 at 1:38
  • \$\begingroup\$ You can write a complex number \$a+jb\$ as \$re^{j\theta}\$. But when you start with \$A\sin(\omega t)\$ you don't have \$\theta = \omega t\$. You have \$r=A\sin(\omega t)\$ and \$\theta = 0\$ because you are startingn with a real number (\$b=0\$). \$\endgroup\$
    – The Photon
    Nov 10, 2023 at 1:46
  • \$\begingroup\$ maybe the question is "Why can a phasor be used to represent a time domain phase shift of a signal, AND why is it only for AC voltage/current functions?" \$\endgroup\$ Nov 10, 2023 at 2:02
  • \$\begingroup\$ That would be a separate question, but the answer is basically "because the math works out". \$\endgroup\$
    – The Photon
    Nov 10, 2023 at 2:07
  • \$\begingroup\$ I should add: 1. Please review the Wikipedia phasor article first, if you are considering submitting this "big" question. 2. The concept can be extended to non-AC-steady-state solutions in Laplace analysis. \$\endgroup\$
    – The Photon
    Nov 10, 2023 at 2:12
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In the complex plane, the unit vector j is expressed in polar coordinates as 1∠90°.

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  • \$\begingroup\$ yet this would be a phase shift to the function of sin(wt) not a phase shift to the argument of the sin(wt) input \$\endgroup\$ Nov 10, 2023 at 1:17
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    \$\begingroup\$ @Jirhska You're going to have to explain what you mean by that. How do you define "a phase shift to the function of sin(ωt)" and "a phase shift to the argument of the sin(ωt) input"? How are they different? \$\endgroup\$
    – Hearth
    Nov 10, 2023 at 1:55
  • \$\begingroup\$ @Hearth j*sin(wt) is a purely imaginary function so the real sin(wt) has been shifted 90degrees in the real complex plane. A phase shift to the argument would be sin(wt-pi/2) which would =cos(wt), \$\endgroup\$ Nov 10, 2023 at 1:57
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This is where complex numbers in polar form shine. Let \$r\angle\theta\$ represent the number \$ r e^{j \theta} = r (\sin \theta + j \cos \theta)\$. Then \$\left(a \angle \alpha\right) \cdot \left(b \angle \beta\right) = \left(a\cdot b\right) \angle \left(\alpha + \beta\right)\$. This is a useful thing to remember about complex multiplication in general: you multiply the magnitudes and add the arguments.

If you want to prove that statement, you can either use the definition of \$r\angle\theta\$ and the angle-sum identities for \$\sin\$ and \$\cos\$ (just expand both sides out until they're identical, remembering that \$j^2 = {-}1\$) or you can use the exponential identity \$e^x e^y = e^{x+y}\$ to show that \$\left(a\, e^{j \alpha}\right) \cdot \left(b\, e^{j \beta}\right) = ab\, e^{j\left(\alpha + \beta\right)}\$

Since \${-}j = 1\angle {-}90^{\circ}\$, it follows that \${-}j \cdot \left(r\angle\theta\right) = r\angle\left(\theta - 90^{\circ}\right)\$

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