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I am reviewing about using complex number in solving electrical linear circuits.

I see that the input sinusoidal signal \$ V_m \cdot \cos(\omega t) \$ is replaced \$ V_m \cdot e^{jwt}\$.

Then solve for the complex output voltage with the complex input voltage above.

Finally, output voltage is the real part of "complex output voltage" that is determined above.

What I am confused is that why is it possible to do that?

The real input voltage is replaced by a complex input voltage and the output voltage is also the real part?

Why the imaginary part added to the input doesn't affect the real part of the output voltage?

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up vote 3 down vote accepted

Why the imaginary part added to the input doesn't affect the real part of the output voltage?

Short answer: because of the linearity.

A system may be described as an operator \$\operatorname{L}\$, which takes a function and outputs the transformed function: $$y(t) = \operatorname{L}(x(t))$$

A system is linear if for any input \$x(t)\$ and any complex number \$\alpha\$ $$\operatorname{L}(\alpha x(t)) = \alpha \operatorname{L}(x(t))$$ and for any two inputs \$x_1(t)\$, \$x_2(t)\$ $$\operatorname{L}(x_1(t) + x_2(t)) = \operatorname{L}(x_1(t)) + \operatorname{L}(x_2(t))$$

Suppose that you feed a linear system with the input $$z(t) = \cos(\omega t + \varphi) + j\sin(\omega t + \varphi)$$

Then, using the properties of a linear operator $$\operatorname{L}(\cos(\omega t + \varphi) + j\sin(\omega t + \varphi)) = \operatorname{L}(\cos(\omega t + \varphi)) + j\operatorname{L}(\sin(\omega t + \varphi))$$

Thus, we can "complexify" a real signal, compute the complex solution, and take the real part. The linearity guarantees that we can do so.


It may be helpful to look at a simple counterexample. Consider a nonlinear system that outputs the square of its input. If we feed such system with \$\cos(t)\$, then the output will be \$\cos^2(t)\$.

Let's try to use the "comlexification" technique and represent \$\cos(t)\$ as \$e^{jt}\$. Then the system output will be \$(e^{jt})^2 = e^{j\cdot 2t}\$. But \$\operatorname{Re}(e^{j\cdot 2t}) = \cos(2t)\$, which is clearly incorrect result. We have got the incorrect result because the system under consideration is not linear.

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Generally, a sinusoidal waveform may be represented as a complex number by using Euler's identity and freezing time at \$\small t=0\$.

Consider a sinusoid with amplitude, \$\small V_m\$ a phase angle, \$\small \phi\$, which may be written: $$V= V_m\: e^{j(\omega t+\phi)}=V_m[cos(\omega t+\phi) +jsin(\omega t+\phi)]$$

This represents a vector of magnitude , \$\small V_m\$, that rotates as time increases from \$\small t=0\$. The real part of this complex number is the horizontal component of the vector, and the imaginary part is the vertical component of the vector

Rotating vectors are not very convenient things draw or do maths with, so instead we freeze them at \$\small t=0\$ and call them \$phasors\$. Thus, in phasor form, the above equation becomes:

$$V= V_m[cos(\phi) +jsin(\phi)]=V_me^{j\phi}$$

For example, a sinusoidal voltage signal of amplitude 3V with phase angle, \$\small \phi=0\$, would be written \$\small V= 3+j0\frac{}{}=3\$; and a sinusoidal current signal of amplitude 5A and phase angle, \$\small \phi=-30^o\$, would be represented as \$\small I=4.33-j2.5\$,

The advantage of complex notation is that we can do circuit analysis if we also express resistances and reactances in their complex forms: $$R\rightarrow R+j0 = R$$ $$L\rightarrow 0+j\omega L = j\omega L$$ $$C\rightarrow 0+\frac{1}{j\omega C}= \frac{1}{j\omega C}= \frac{-j}{\omega C}$$

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The complex number representation is just a mathematically "easy" way to keep track of the phase of a quantity (ie voltage or current) and how that interacts with impedances that change its phase.

In your example though, if the input does not see complex impedances (only real impedances, aka resistors) the phase will not change from input to output.

But I think you may have confused the purpose of the real part as compared to the magnitude of the quantity.

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