Why do phasors work with s-domain transfer functions?

Question

In as few words as possible, here is my question:

Why can we get the output phasor by multiplying the input phasor by the s-domain transfer function evaluated at the (complex?) frequency we're dealing with?

An example will serve to better explain:

^{simulate this circuit – Schematic created using CircuitLab}

(Note: I am tacitly ignoring the initial conditions; for now I want to simplify the problem and only consider the steady-state response)

$$H(s)=\frac{V_C}{V_{in}}=\frac{1/(sC)}{sL+R+1/(sC)}=\frac{\frac 1{LC}}{s^2+s\frac R L +\frac 1{LC}}$$

If we wanted \$V_C\$­ as a phasor, we would simply perform

$$V_C = V_{in}\cdot H(j2\pi \cdot1000)=3\angle45˚\cdot \frac{\frac 1{LC}}{-4\pi^2 \cdot 1000^2+j2\pi \cdot 1000\frac R L +\frac 1{LC}}$$

This would result in some complex number, which we would interpret as representing a sinusoid in the time domain. (Note: Already I'm confused... the input voltage waveform is not a complex number in the s-domain; it's actually some function of s).

I can understand how phasors arise naturally when solving DEs in the time domain; you assume your output is of the form \$Ae^{j2\pi ft + \phi}\$ and the time-dependence cancels out in the equation. I can also understand that multiplying by the transfer function in the s-domain produces the correct output in the time domain (provided the system is LTI). I can even understand why Ohm's Law, KVL, and KCL work in the s-domain.

However, after all that I can't get my head around this "abuse of notation". Phasors and s-domain expressions shouldn't have any business hanging around each other! So what am I missing here?

Answer 1

Actually, phasors and s-domain expression do have some relationship. Recall that the 's' variable in the laplace transform is defined as:

$$ s = \sigma + \text{j}\omega$$

So, when you substitute \$s\$ by \$\text{j}\omega\$ in a transfer function, you are taking your function to the phasor domain (which will produce the steady-state solution only, not the transient response).

Now, remember that when using phasors, you're taking advantage of Euler's identity, that is, \$e^{\text{j}\omega t}=\cos(\omega t)+\text{j}\sin(\omega t)\$. Even though you have a real source, say it is \$v(t)=\text{V}_o\cos(\omega t)\$, you can use Euler's identity to express it as

$$ v(t)=\text{V}_oe^{\text{j}\omega t}$$

or more rigorously defined as

$$ v(t)=\Re{[\text{V}_oe^{\text{j}\omega t}]}$$

where \$\Re\$ means that you want the real part of the expression. That's the case since your source is a cosine, or the real part of Euler's identity.

Alternatively, if you source were a sine, then

$$ v(t)=\Im{[\text{V}_oe^{\text{j}\omega t}]}$$

where \$\Im\$ means that you want the imaginary part of Euler's identity.

That means, that once you solve your circuit, you need to take either the real (if your source was a cosine) or the imaginary (if your source was a sine) part of the complex valued solution.

Hope it helps!

Answer 2

The Laplace transform of \$f(t)\$ is defined as: $$F(s)=\int_{-\infty}^{\infty}e^{-st}f(t)dt$$ and the Fourier transform of \$f(t)\$ is defined as: $$F(j\omega)=\int_{-\infty}^{\infty}e^{-j\omega t}f(t)dt$$ Clearly, the substitution \$s\large\leftrightarrow\small j\omega \$ performs the transformation between the two domains.

For TF analysis we assume zero initial conditions, so, for example, \$cos(\omega t)\$ should be replaced with \$u(t)cos(\omega t)\$, and the corresponding Fourier transform is \$\large \frac{j\omega}{{\omega _0}^2-\omega ^2}\$, which is obtained from \$s\rightarrow j\omega\$ in the LT for \$cos(\omega t)\$; viz \$\large \frac{s}{s^2+{\omega _0}^2}\$

Answer 3

You can manually derive the general sinusoidal steady-state response for LTI systems and this relationship becomes clear. Let \$H(s)\$ be the system's transfer function, \$h(t)\$ be the impulse response (the inverse Laplace transform of the transfer function), and \$V_C(t)\$ be the steady-state response. With a generalized sinusoidal input \$g(t)=V_{in}\sin(\omega t+\theta)\$, the overall response, by the convolution theorem, is as follows: $$\int_0^t{h(\tau)g(t-\tau)d\tau}$$ $$=\int_0^\infty{h(\tau)g(t-\tau)d\tau}-\int_t^\infty{h(\tau)g(t-\tau)d\tau}$$ Since systems made purely of RLC components are stable in nature, it's intuitive that the second term decays as \$t\$ approaches \$\infty\$ and the first term doesn't, so you can tell the first term is the steady-state response \$V_C(t)\$. $$V_C(t)=\int_0^\infty{h(\tau)g(t-\tau)d\tau}=V_{in}\int_0^\infty{h(\tau)\sin(\omega (t-\tau)+\theta)d\tau}$$ Euler's formula: \$\sin(z)=\frac{e^{jz}-e^{-jz}}{2j}\$ $$=\frac{V_{in}}{2j}\int_0^\infty{h(\tau)\left(e^{j(\omega(t-\tau)+\theta)}-e^{-j(\omega(t-\tau)+\theta)}\right)d\tau}$$ $$=\frac{V_{in}}{2j}\left(e^{j(\omega t+\theta)}\int_0^\infty{h(\tau)e^{-j\omega\tau}d\tau}-e^{-j(\omega t+\theta)}\int_0^\infty{h(\tau)e^{j\omega\tau}d\tau}\right)$$ The transfer function evaluated at \$j\omega\$: \$\int_0^\infty{h(\tau)e^{-j\omega \tau}d\tau}=H(j\omega)\$ $$=\frac{V_{in}}{2j}\left(e^{j(\omega t+\theta)}H(j\omega)-e^{-j(\omega t+\theta)}H(-j\omega)\right)$$ $$\small{=\frac{V_{in}}{2j}((\cos(\omega t+\theta)+j\sin(\omega t+\theta))H(j\omega)-(\cos(\omega t+\theta)-j\sin(\omega t+\theta))H(-j\omega))}$$ $$=V_{in}\left(\frac{H(j\omega)-H(-j\omega)}{2j}\cos(\omega t+\theta)+\frac{H(j\omega)+H(-j\omega)}{2}\sin(\omega t+\theta)\right)$$ Since \$H(-j\omega)=\overline{H(j\omega)}\$, $$=V_{in}(\mathfrak{I}(H(j\omega))\cos(\omega t+\theta)+\mathfrak{R}(H(j\omega))\sin(\omega t+\theta))$$ $$V_C(t)=V_{in}|H(j\omega)|\sin(\omega t+\theta+\angle H(j\omega))$$ This implies that the output is simply the same sinusoid as the input, scaled by a factor and shifted by a phase angle completely determined by \$H(j\omega)\$. Specifically, if you were to transform this to a phasor, you get exactly the product \$(V_{in}\angle\theta)(H(j\omega))\$.

Answer 4

For the Laplace transform of a system with its input and output represented by phasors, you can refer to pages 44 to 47 of my paper "Timing, Synchronization and LLRF, The 6th OCPA Accelerator Topical School and Workshop, Beijing, China, October 14-18, 2019" available here

Research gate link here