Understanding \$ j \omega \$ axis in transfer function

Question

I understand the transfer function up to the point of \$ H(\omega) = \lvert H(\omega) \rvert \angle H(\omega) \$ where the magnitude and phase are decoupled into two separate functions, equivalent of saying \$ H(\omega) = { Y_0 \over X_0} \cdot e^{j(\angle y- \angle x \; \text {in radian})} \$. I can understand this in polar equation on the complex plan.

Then all of a sudden the transfer function jumps into \$ H(j\omega) = \lvert H(j\omega) \rvert \angle H(j\omega) \$ swapping every \$ \omega \$ with \$ j\omega \$. I'm lost. I searched here and there but I found no explanation on how and why \$ j \$ axis is transformed into \$ j\omega \$ axis, or why \$ j\omega \$ axis is added to the complex plane as a third axis if it's a different axis.

• Does \$ j\omega \$ axis mean \$ j \$ axis multiplied by a frequency magnitude of \$ \omega \$? If so what does it mean and why an axis has a magnitude? That make no sense to me whatsoever to give magnitude to an axis as axis should only be directional (imagine if someone ask me to graph a line on the x-2y axis instead of a x-y axis)
• Or, is \$ j\omega \$ an entirely different axis from \$ j \$ axis - meaning a number on \$ j\omega \$ axis has no relation to a number on \$ j \$ axis? If that's the case could someone help explain what are their difference and when should I use which?
• Since \$ \omega \$ is the angular velocity, could \$ j \omega \$ represent the rotational axis, a third axis that is orthogonal to \$ j \$ axis and the real axis like a cross product sort of thing?

[EDIT]

Just to clarify my thought process here. If I have 2D plane with two axis, x and y, I can define a value \$ z = ax + by \$, where a is a position on the x-axis and b is a position on the y-axis. We can call the value z on the x,y plane. We are not going to call it x,by or x,\$ \omega y\$ plane.

Same logic goes here where the complex frequency s is a value on \$ \sigma \$ or real axis and on \$ j \$ imaginary axis. So a value of s is \$ s = a\sigma + bj\$, where a and b are positions on the real and imaginary axis. So s is on \$ \sigma,j \$ plane, not \$ \sigma,\omega j \$ plane. What is \$ j \omega \$ axis?

Answer 1

We start with $$H(s)$$ in the frequency domain, and since $$s =\sigma+j\omega$$ We can replace s with $$H(\sigma+j\omega)$$ But, in steady state the real part (sigma) is zero, so we get,

$$H(j\omega)$$

To clarify the "transient" (sigma) vs. "steady-state" (omega) , here is damped cosine, $$h(t)=e^{-0.8t}cos(5t)$$

Re-writing with only exponentials (using euler), $$h(t)=e^{-0.8t}(\frac{e^{j5t}+e^{-j5t}}{2})$$

Here you can see the exponential with the real value (-0.8) is the transient influence, and those with the imaginary (j5 and -j5) are steady-state. In this case the steady-state settles out to zero, so i'll give another example that does not.

Let, $$f(t)=e^{-0.8t}cos(5t)-cos(5t)$$

Re-writing with all exponentials,

$$f(t)=e^{-0.8t}(\frac{e^{j5t}+e^{-j5t}}{2})-(\frac{e^{j5t}+e^{-j5t}}{2})$$

For completeness, here are the Laplace transforms of the above two time functions h(t) and f(t),

$$\mathcal{L}[h(t)]=H(s)=\frac{s+0.8}{s^2+1.6s+25.64}$$

and,

$$\mathcal{L}[f(t)]=F(s)=\frac{0.8s^2+0.64s-20}{(s^2+25)(s^2+1.6s+25.64)}$$

and now they are in the s-domain and you can analyze them with frequency domain tools/methods.