Why don't phasors give the transient state?

Question

I don't understand why phasor analysis doesn't tell us anything about the transient state. At exactly what part of the analysis is the transient part "lost"?

Answer 1

Phasor analysis allows us to analyze a circuit's response to a sinusoidal steady-state response at a given single frequency. We represent a time-domain voltage \$V(t) = V_{0}cos(\omega t + \phi)\$ in phasor form by transforming it into a complex exponential via Euler's formula...

$$A\cos(\omega t + \phi) = Re\{Ae^{j( \omega t~+~\phi)}\} = Re\{Ae^{j\phi}e^{j\omega t}\}$$

...and then ignoring the frequency/time dependence (since we assume everything in the circuit is excited by a steady sinusoid of the same frequency). Thus,

$$V= Ae^{j\phi}$$

This \$V\$ is what we call a phasor. We can represent any current or voltage as a phasor. In order to recover a time-domain representation from a phasor, you can multiply it by \$e^{j\omega t}\$ and then take the real part. Note that sometimes for the sake of brevity/familiarity in calculating power, we also convert the amplitude of the phasor into RMS values (divide magnitude by \$\sqrt{2}\$ for a sinusoid). Phasors allow us to use analogous DC analysis techniques to recover transfer functions of linear circuits (by using impedances). Using superposition, we can use Fourier analysis to analyze a circuit's complete steady-state response as a sum of its steady-state response due to different frequency components.

It's useful to note the relation of phasors to the Laplace representation of a circuit. The Laplace representation of a circuit uses the variable \$s = \sigma + j\omega\$. Note that for \$\sigma = 0\$, the transfer function of the Laplace representation of a circuit reduces down to the phasor representation. This is a good indication that the real part of \$s\$ represents a transient response (and this can be easily observed by noting that \$e^{at}\$ for any real \$a\$ will result in a real value that either exponentially grows or decays). Note that the Laplace representation is a more general representation of a circuit that includes both transient and steady-state responses. Likewise, it's good to note that the Fourier transform is just a special case of the more general Laplace transform (the case where \$\sigma = 0\$ in \$s = j\omega\$).

Answer 2

Mathematically, nothing gets lost. Phasor analysis gives you the value of every voltage and current in the circuit for all time, in the form of cosine functions:

$$V_a = A_1 \cos (\omega t + \phi_1)$$ $$V_b = A_2 \cos (\omega t + \phi_2)$$ $$I_c = A_3 \cos (\omega t + \phi_3)$$ $$etc.$$

You specify \$\omega\$ and the amplitude and phase for at least one voltage or current, and phasor analysis gives you \$A\$ and \$\phi\$ for all the rest.

The problem isn't that the transient behavior gets lost, it's that you never put it in! By definition, phasor analysis works on unchanging eternal sinusoids at a single frequency applied to a linear time-invariant system -- the so-called sinusoidal steady state. "Steady state" is the opposite of "transient". You can extend this to cover exponential growth and decay, but again, this is eternal growth and decay. The math only works because your voltages and currents are complex exponentials, which aren't distorted by linear differential equations.

To describe a situation where you flip a switch at t = 0, you need to use a step function. Step functions cannot be represented by a single frequency, so phasor analysis breaks down. To handle this, you need to use Fourier analysis.

Answer 3

I know this is a late answer, but I want to give a different insight as to why phasors only give the steady state response.

Consider the well known RC circuit, with a driving source \$v_s=\text{V}\cos(\omega t + \phi)\$, then you have the differential equation:

$$ \text{RC}\dfrac{dv_c}{dt} + v_c=\cos(\omega t + \phi)$$

From a mathematics point of view, you can solve this differential equation by finding the homogeneous solution and a particular solution and when you add them up you get the general solution. So far so good.

Phasors give you a particular solution only (it doesn't give you the homogeneous solution, which is the transient solution) and the particular solution is what we call the steady state response.

In other words, the homogeneous solution (transient or natural response) is the solution to

$$ \dfrac{dv_c}{dt} + \dfrac{1}{\text{RC}}v_c=0$$

which you can find by the integrating factor method.

And the particular solution, using the fact that you can write the input source as \$\Re\{\text{V}e^{j\omega t}e^{j\phi}\}\$, where \$\Re\$ means the real part of. For the particular solution, we make a 'guess', based on the forcing function:

$$\dfrac{dv_c}{dt} + \dfrac{1}{\text{RC}}v_c= \dfrac{\text{V}}{\text{RC}}e^{j\phi}e^{j\omega t}$$

If you guess that your particular solution has the form \$v_{c,p}=\text{A}e^{j\omega t}\$, where \$A\$ will be a phasor too (it will have a magnitude and a phase in the end), just like \$\text{V}e^{j\phi}\$ is, by the definition of a phasor. $$j\omega \text{A}e^{j\omega t}+ \dfrac{1}{\text{RC}}\text{A}e^{j\omega t}= \dfrac{\text{V}}{\text{RC}}e^{j\phi}e^{j\omega t}$$ which you can simplify by dividing by \$e^{j\omega t}\$ and factoring the \$\text{A}\$ terms

$$\text{A}\bigg(j\omega+\dfrac{1}{\text{RC}}\bigg)= \dfrac{\text{V}}{\text{RC}}e^{j\phi}$$

$$ \text{A}=\dfrac{\text{V}e^{j\phi}}{j\omega\text{RC}+1}$$

And in the end, \$\text{A}\$, will be a phasor of the form:

$$\text{A}=|\text{A}|\angle{\theta} $$

So when you find \$v_{c,p}\$, you only have the particular solution (forced response, steady-state). You'd still need to find the solution to aforementioned homogeneous equation to have a full response.

In short, phasors give you a particular solution to the differential equation.

Answer 4

Although I think that @Big6 answer is a good one, I'd like to answer the OP from the Laplace transform point of view, for the sake of completeness.

So, suppose you have a linear SISO system that has a generic transfer function \$H(s)\$ which satisfies:

1. Is of order \$N\$, so it has \$N\$ poles.
2. Is stable, so all its poles have negative real part (they lie on the left-hand side of the complex plane).
3. The system does not have initial conditions (this is not mandatory but will shorten the explanation).

Also, suppose you feed the system's input with a generic sinusoidal function \$e(t)=A\cdot\cos(\omega\;t+\phi)\$.

Our first goal will be to find the system's time response using the Laplace Transform technique. To do so, we use the relation,

\begin{equation} R(s) = H(s) \cdot E(s) \tag{1} \end{equation}

where \$E(s)\$ ans \$R(s)\$ are the Laplace transforms of input and output signals, respectively, and \$H(s)\$ is the transfer function.

We start using basic trigonometry relations to find the Laplace transform of equation \$(1)\$:

$$ A\cdot\cos(\omega\;t+\phi) = A\cdot[\cos(\omega\;t)\cdot\cos(\phi) - \sin(\omega\;t)\cdot\sin(\phi)] $$ so $$ E(s) = \mathcal{L}\left\lbrace e(t) \right\rbrace = \mathcal{L}\left\lbrace A\cdot\cos(\omega\;t+\phi) \right\rbrace = A\cdot\left[\frac{s\cdot\cos(\phi)}{s^2+\omega^2} - \frac{\omega\cdot\sin(\phi)}{s^2+\omega^2}\right]\tag{2} $$

Combining equations \$(1)\$ and \$(2)\$ we have:

$$ R(s) = H(s) \cdot A\cdot\left[\frac{s\cdot\cos(\phi) - \omega\cdot\sin(\phi)}{s^2+\omega^2}\right]\tag{3} $$

We don't know much of \$H(s)\$ but, as it has \$N\$ poles, we can write them explicitly. Also, we write explicitly the two poles of \$E(s)\$. after that, we write the partial-fraction expansion of R(s):

$$\begin{align} R(s) &= \frac{N(s)}{(s-p_1)\cdots(s-p_n)}\cdot A\cdot\left[\frac{s\cdot\cos(\phi) - \omega\cdot\sin(\phi)}{(s-j\omega)\cdot(s+j\omega)}\right]\\ &= \color{blue}{\frac{k_1}{(s-p_1)} + \cdots + \frac{k_n}{(s-p_n)}} + \color{red}{\frac{k}{(s-j\omega)} + \frac{k^*}{(s+j\omega)}} \end{align}\tag{4} $$

where \$k_1 \cdots k_n\$ are the residues associated with poles \$ p_1\cdots p_n\$ and \$k, k^*\$ are the (conjugate) residues associated with the conjugate poles of the input signal.

Now the key part: Note that the blue terms correspond to the free or natural response of the system (due to system poles) and the red terms correspond to the forced response (due to poles introduced by the external input). Note also that because the circuit is supposed to be stable, the natural response will decay to zero after some time, and only the forced (red) response will remain, which will be the steady-state response of the whole circuit to that input.

So, if we're NOT INTERESTED in calculating the transient, we can obviate the blue terms and calculate only the steady-state (red) response, which we will denote \$R_{ss}(s)\$:

$$ \color{red}{R_{ss}(s) = \frac{k}{(s-j\omega)} + \frac{k^*}{(s+j\omega)}}\tag{5} $$

We calculate now the residue \$k\$ using the Heaviside Cover-Up method on equation \$(3)\$:

$$ \begin{align} k &= (s-j\omega)\cdot H(s) \cdot A\cdot\left[\frac{s\cdot\cos(\phi) - \omega\cdot\sin(\phi)}{(s-j\omega)\cdot(s+j\omega)}\right]\Biggr\vert_{s=j\omega} \\ &= H(j\omega) \cdot A \cdot\left[\frac{j\omega\cdot\cos(\phi)-\omega\sin(\phi)}{2j\omega}\right]\\ &= H(j\omega) \cdot \frac{A}{2} \cdot \left(\cos(\phi) + j\sin(\phi)\right) \end{align} $$

or, in polar form:

$$ k = \bigl|H(j\omega)\bigr| \cdot \frac{A}{2}\cdot e^{j(\measuredangle{H(j\omega)}+\phi)} \tag{6} $$

If we substitute this residue into equation \$(5)\$ (the conjugate is straightforward), find the Inverse Laplace of the two fractions and use Euler's formula to eliminate complex exponentials we obtain:

$$ r_{ss}(t) = A \cdot \bigl|H(j\omega)\bigr| \cdot \cos\bigl(\omega\;t + \phi + \measuredangle{H(j\omega)}\bigr) \tag{7} $$

Bear in mind that \$(7)\$ is the steady-state response to the input signal \$e(t)=A\cdot\cos(\omega\;t+\phi)\$. So, what has to do this result with phasor analysis? The answer is that if we express the input and output in phasor form we have: $$ \begin{align} \overline{\mathbf{E}} &= A\cdot e^{j\phi} \\ \overline{\mathbf{R}}_{ss} &= A\cdot \bigl|H(j\omega)\bigr|e^{j(\measuredangle{H(j\omega)}+\phi)} = \bigl|H(j\omega)\bigr|e^{j\measuredangle{H(j\omega)}} \cdot A\cdot e^{j\phi} \end{align} $$ so

$$ \overline{\mathbf{R}}_{ss} = H(j\omega) \cdot \overline{\mathbf{E}} \tag{8} $$

Equation \$(8)\$ is the same you obtain by phasor analysis, which explains why you don't obtain the transient response with it: because the entire phasor analysis procedure is designed to ignore the free or natural response, and goes directly to the forced response.

Answer 5

Like Adam and Auston already pointed out transient information is never included in phasor analysis, at least not in 2D. Putting it in would require an additional dimension, like this 3D phasor envelope of an amplitude modulated sine wave.

A transient, like the decay of a damped oscillation would look like a cone-shaped rotating vector decaying in an exponential fashion. Complex numbers would also no longer be suitable.

Answer 6

Phasors assume a sinusoidal waveform, sinusoids go on forever, they don't have any transient behavior. The transient part is lost the same way you can assume a capacitor that has been charging for "a long time" is fully charged.