Is the frequency response of a circuit only valid for steady state (phasors)?

Question

This is my first post. I just finished a course about advanced circuit analysis techniques (the content was taken from textbooks by Sadiku, Hayt and Irwin.) As you know, phasors are used to solve sinusoidal AC circuits (and also non-sinusoidal AC circuits, with the help of Fourier series), however, the answer you get from analyzing a circuit with that technique is only the steady state response, not the complete response (which also includes the transient response). When we're introduced to the frequency response analysis, the only difference is that now frequency is an independent variable just like time, and we still solve the circuits with phasors.

So my question is this: is the answer obtained by a frequency analysis (output voltage, ratio of output to input voltage, transfer function, etc.) only valid for steady state? I'm assuming this because in freq. anal. we still use phasors, which give only the s.s. response.

Answer 1

Yes, it assumes the transients have decayed to zero.

But note that the frequency response contains information on the transients, this is why it's so useful. It's a steady state measurement that gives transient information, so filtering etc can take place in a steady state environment which, potentially, gives more accurate results.

Furthermore, SS frequency response can detect transients that may not even be visible on, say, a step response.

Answer 2

'A circuit' has a response, a behaviour.

That response can be completely described in the time domain, or the frequency domain. It doesn't matter how it is described, it's the same response. You can get from one to the other with the Laplace Transform.

Often, we use just part of the response, and throw away data that's not relevant to our immediate need. When we throw away data, we are then unable to transform that incomplete description into the other one.

For instance, we might take a time domain description and just focus on the percent overshoot and time to 50% of the step response. Or we might take a frequency domain description and use only the magnitude, ignoring the phase. When we throw the rest of the data away, we cannot uniquely reconstruct the time from the frequency, or vice versa.

There are of course some clues in an incomplete response. A large overshoot in the time domain suggests a steep edge in the frequency domain. A narrow passband in the frequency domain would suggest a long time to 50% step response. If the circuits are constrained in topology (so we have some knowledge outside the measured response), then there are some techniques that allow us to do rather better than 'suggesting' a form of response from partial data in the other domain. For instance, for the classic lowpass RC filter (a very simple and constrained circuit), you will often see a relationship given between its 3dB frequency response and its 50% step response.

There may be a good reason for taking data in one domain, when we actually want the other domain. For instance, measuring the impedance versus distance of a long transmission line is called Time Domain Reflectometry. This is often used for diagnosing where incorrect impedances are, due to perhaps faulty connectors, crushing or water ingress. The classic way of doing this was to launch a fast pulse into the line, and see the reflections on a 'scope. However, it's hard work to generate a lot of power in a very short pulse. These days, we get the same information by taking the data in the frequency domain with a network analyser, then transforming the full (amplitude and phase) frequency response measurements into the time domain. Same results, but we can get better frequency resolution and SNR than with a pulse.

Answer 3

There is an important theorem in control theory that states that any Linear time-invariant system is entirely characterized by its steady-state frequency response.

So as long as your circuit is linear and time-invariant, you can calculate any transient behavior based on the frequency response. Often this is a good enough assumption to allow practical calculations.

However, there exist many non-linear components, such as diodes and transistors. For these, frequency response to 2 volt signal could be very different than frequency response to 1 volt signal. And it might be that the system is linear during steady state operation, but nonlinear during transients.

There are also some time-varying components, like resistors that heat up. If you keep measuring them, the frequency response will change as the circuit heats up, until it achieves steady state.

So, to summarize: for many systems, the frequency response characterizes transient behavior also. But there are also many systems for which the frequency response doesn't fully characterize the steady-state performance either.

Answer 4

The frequency response ( Y = H(X) ) of a circuit gives the steady state behaviour of a circuit due to a sinusoidal input X.

Its possible to write a fourier series approximation any transient input X over some time interval.

The fourier series of input X will be a weighted sum of sinusoidal waveforms X = X0 + X1 + X2... where each Xn = Cn * sin(Wn * t)

If the circuit is linear then Y = H(X) = H(X0) + H(X1) H(X2)... Meaning that the output is the sum of the responses to each of the individual sinewaves in the fourier series.

In this way you can use the frequency response to determine the circuit output for any type of input waveform weather it be a sinewave or not.