Why is the phase of an imaginary signal 90 degrees?

[Manoj Sahu]

Like we write "R+jX" where jX is considered to be imaginary term and it said to have 90 degree phase shift.

 

[cnh1995]

Like we write "R+jX" where jX is considered to be imaginary term and it said to have 90 degree phase shift.

The imaginary part denotes the reactive component of the phasor.
In Z=R+jX, X is the reactive part of the impedance i.e. reactance. The phase difference between voltage across the reactance and current through the reactance is 90 degrees.

 

[Manoj Sahu]

The imaginary part denotes the reactive component of the phasor.
In Z=R+jX, X is the reactive part of the impedance i.e. reactance. The phase difference between voltage across the reactance and current through the reactance is 90 degrees.

Actually you didn't the question. What I was asking is "why 90 only not more or less?"

 

[anorlunda]

Actually you didn't the question. What I was asking is "why 90 only not more or less?"

See the PF Insights Article https://www.physicsforums.com/insights/case-learning-complex-math/

Any angle $\theta$ can be expressed as the sum of two components B+jC. where B and C are "orthogonal" (look up the meaning of orthogonal). But you can't do that if say C was 44 degrees from B. For example,there would be no sum of B and C able to add up to 50 degrees.

If $\sqrt{B^2+C^2}=1$ then multiplying by B+jC is "pure rotation" with no change in magnitude, only phase.

 

[NascentOxygen]

Actually you didn't the question. What I was asking is "why 90 only not more or less?"

Consider the behaviour of inductance we characterise as: v(t)=L.di/dt
If i(t) is sinusoidal, then v(t) being the derivative of a sinusoid is cosinusoidal, i.e., a sinusoid but advanced by 90°. There's the 90° you are asking about. When you calculate impedance as v(t) ÷ i(t) the 90° carries through.

Try a similar approach for capacitance and see what it leads to.

 

[jim hardy]

Easier way to think of it.
Just as in precision marching - two well executed "Column Left March" commands will exactly reverse your direction.

If you multiply a sinewave signal by [ -1] you have reversed its phase, that is shifted it by 180 degrees.

So - how would you phase shift it by only 90 degrees ? Easy >> multiply it by [√-1] which we know is an imaginary number .

So the reason it's 90 degrees not 89 or 181 is 90 is half of 180.
We assign to "operator j" the value [√-1] .

old jim

 

[Babadag]

In my opinion the use of imaginary sqrt(-1)=j it is a conventional mode to treat an electric sinusoidal variable as voltage and current. It presents a lot of advantages with respect the simply algebraic calculations.
If an a.c. variable voltage it is v1=sqrt(2).V1.cos(ꞷ.t+ φ) [that means it exists a vo of the same frequency but start at φ=0 vo=sqrt(2).Vo.cos(ꞷ.t)] then we could attach to it an imaginary value sqrt(2).V1.sin(ꞷ.t+ φ).j and we can represent it on Descartes coordinates plan [which rotates with ꞷ radians/sec] as V1ےφ.
Why Descartes preferred 90o between coordinates you’ll find here[I hope]:
https://en.wikipedia.org/wiki/Cartesian_coordinate_system

 

[K Murty]

Hi,

as others have stated, at least for an inductors and capacitors, with my understanding it is that these elements oppose instant changes in currents and voltages, hence they result in voltage and currents taking more time to change, as compared to resistances, whose I V response is instantaneous. This property due to the physics of such elements, results in the phase shifts.

$$V_{L} = L \cdot \frac{di}{dt}$$

If $$I(t) = I_{m} \sin ( \omega t)$$
Then $$\frac{di}{dt} = I_{m} \omega \cos(\omega t)$$
$$V_{L} = L \omega I_{m} \cos(\omega t)$$

It is a definition, or can be proven at least that:
$$V_{L} = L \omega I_{m} \cos(\omega t) = \Re( L \omega I_{m} \cdot e^{j (\omega t)} )$$
Comparing with current, we have to express the sine as a cosine:

$$i(t) = \Re( I_{m} \cdot e^{j (\omega t - \frac{\pi}{2} )} )$$

We can see a difference in the angle between the curves.
$$j = e^{j (\frac{\pi}{2}) } \\ -j = e^{j -(\frac{\pi}{2}) } \\ -1 = e^{j(\pi)}$$
In polar form, when multiplied into another exponential, means a rotation (anticlockwise) of 90 degrees. -J means a clockwise rotation of 90, and -1 you can see. Thus when you take the derivative, say of:
$$\frac{d}{dt} e^{j (\omega t )} = {j \omega} e^{j (\omega t )} = {\omega} e^{j (\omega t + \frac{\pi}{2}) }$$

Hence when in phasor domain, we have:
$$\underline{V} = j \omega L \underline{I}$$
This captures the inductive reactance, and the phase shifts in the phasor domain, as a phasor is an exponential that keeps the peak value and phase of a signal, with the time domain exponential factored out. Of course, this is only possible for sinusoidal currents and voltages, as you probably know. I have read the term imaginary signal once or twice I think, but the most important aspect is that complex numbers are used to represents the amplitude and phase of a signal, which can be clearly seen in the mathematics with little to moderate pains...below, is an addition of sin(x) + cos(x)

$$\Re \left( e^{jx} \right) = \cos(x) \\ \Re \left( e^{j(x - \frac{\pi}{2} )} \right) = \sin(x) \\ Hence: \Re \left( e^{jx} + e^{j(x - \frac{\pi}{2} )} \right)$$

$$\Re \left( e^{jx} \cdot ( e^{j0} + e^{j -\frac{\pi}{2} } \right)$$

Above we factored out the angular frequency, its always kind of hidden that way, the term inside the brackets are your phasors, and phasor addition has to be done by converting to polar form, which is easy and obvious given they are exponentials.$$\Re \left( e^{jx} \cdot ( e^{j0} + e^{j -\frac{\pi}{2} }) \right) \\e^{j0} +e^{j -\frac{\pi}{2}} =\underline{ 1 - 1j }_{\text{ rectangular or cartesian form}} = \sqrt{2} \cdot e^{j - \frac{\pi}{4}}$$

$$\Re \left(e^{jx} \cdot \sqrt{2} \cdot e^{j - \frac{\pi}{4}} \right) = \Re \left( \sqrt{2} \cdot e^{j(x - \frac{\pi}{4} )}\right) \\\Re \left( \sqrt{2} \cdot e^{j(x - \frac{\pi}{4} )}\right) = \sqrt{2} \cos(x - \frac{\pi}{4})$$

$$\sqrt{2} \cos ( x - \frac{\pi}{4}) = \sqrt{2} \sin( x + \frac{\pi}{4} )$$

Now, of course we have used the cosine definition for phasors, but other texts also use the sine one. In one book I have by an Indian author, he avoids the use of imaginary numbers in the introduction to phasors, by referring to the imaginary part as the y component and summing up the components of the vector in the x and y directions, which was also quite nice to see, but later he adopts the normal approach.