What is a complex frequency?

Discussion in 'Physics & Math' started by arfa brane, Mar 4, 2020.

  1. arfa brane call me arf Valued Senior Member

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    Yep, they're notes from a handout on network analysis. In this case, electronic networks, although I understand the math applies to any kind of reactive network. You could make a network of say, springs, and determine a transfer function from an input to an output, of forced oscillations. Such a network should have a transient response too.

    Laplace and Fourier are mentioned in these notes, but it seems the so-called s-domain of complex frequency doesn't need Fourier, it all hangs off the Laplace transform.
    But I'm trying to focus on what the relation between \( s = \sigma + j\omega\), a complex frequency, and a real frequency are. The real part of s is just a real number, the imaginary part is what corresponds in physics to an angular frequency (in radians per second).

    So a complex frequency is a real angular frequency \( \omega\) which is imaginary in the complex plane (or s-domain), and has an additional exponential term \(\sigma\) which is either a growing or decaying exponential curve; in other words, the real frequency also has an exponential envelope.
     
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  3. arfa brane call me arf Valued Senior Member

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    Complex frequencies aren't physical, but your video shows a system of coupled pendulums. I bet there's a transfer function (I can practically guarantee it).
     
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  5. Write4U Valued Senior Member

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    I observed that each individual pendulum displayed a different wave frequency, due to the different string lenghts and that these frequencies would alternately form chaotic and harmonic patterns.

    I think its beautiful to see those patterns emerge naturally and then imagine similar wave frequencies behave this way without any observers, perhaps at the atomic scale.
    IMO, the emergence of harmonic orders is an expression of natural mathematical art.

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    This information is summarized in the table below.

    Harmonic## ofWavesin String# ofNodes# ofAnti-nodesLength-WavelengthRelationship
    11/221Wavelength = (2/1)*L
    21 or 2/232Wavelength = (2/2)*L
    33/243Wavelength = (2/3)*L
    42 or 4/254Wavelength = (2/4)*L
    55/265Wavelength = (2/5)*L

    https://www.physicsclassroom.com/class/sound/Lesson-4/Fundamental-Frequency-and-Harmonics

    Truly ; "To the symphony of life, no one has the score"
     
    Last edited: Mar 9, 2020
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  7. Michael 345 New year. PRESENT is 72 years oldl Valued Senior Member

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    British BBC program did much the same demonstration as this



    But not found it

    Found this



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  8. CptBork Valued Senior Member

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    When you originally asked about complex frequencies I was thinking about exponential decays, but the responses everyone gave referred to real-valued frequencies being plugged into complex exponentials for calculating asymptotic network responses. Indeed the Laplace transform and Fourier transform are closely related and only differ by whether the s parameter is imaginary or real, as well as the limits of integration (either the whole imaginary number line or half of the real one). One way of deriving the Bromwich integral which inverts general Laplace transforms, is by recasting it in terms of inverse Fourier transforms. Anyhow, if you're looking at the transient response and the solution is an exponential with both real and imaginary parts, that solution indeed corresponds to a sinusoidal response with an exponentially decaying amplitude envelope.
     
  9. arfa brane call me arf Valued Senior Member

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    And those coupled pendulum motions in the videos, are just the system response to a transient "disturbance at the input" or in electronics a step function. This is the physical displacement from equilibrium, so you have a passive network response for n pendulums; they're coupled so they interact with each other in a, erm, complex way. But each pendulum has an exponential decay in the sinusoid, you need to know what this damping term \(-\sigma\) is. In theory if you have the initial displacement of each pendulum and you know the period of each you can write a transfer (response) function.
     
    Last edited: Mar 9, 2020
  10. arfa brane call me arf Valued Senior Member

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    But if you have an imaginary frequency you also have its conjugate. Maybe that's why it's Laplace in classical network analysis courses.
     
  11. arfa brane call me arf Valued Senior Member

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    So just doing this transform from the time domain f(t) to the complex frequency domain F(s) (in which real frequencies are a subspace) gives you some extra structure, albeit it's imaginary.
     
  12. arfa brane call me arf Valued Senior Member

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    An observation about those videos. The system of pendulums is constructed according to a formula which (I'm guessing here) means the coupling between them is damped, and each oscillates independently; the amplitudes don't get larger for any pendulum and all the amplitudes appear to decay at the same rate (fairly slowly).

    So it's a system that depends (to first order) on the period of each pendulum, and the response depends on an initial 'synchronous' displacement. I would like to see the response to the displacement of a single pendulum or to some forced oscillations. Here's a video that demonstrates a response to forced oscillations; the coupling is 'stronger' than in the other systems (where each pendulum's string is attached to a rigid bar).



    The coupling has a much more noticeable effect--the system of pendulums moves to a resonant state, or tries to maximise the number in equilibrium, say. The input frequency is stable, notice. Changing it will move the resonant response. What you see near the end, is the region of a system pole in the s-domain, but transformed into the time domain (obviously), It begins to appear after about 2min, and is also stable.

    Engineers like to design circuits that are stable--go figure huh?
     
    Last edited: Mar 10, 2020
  13. CptBork Valued Senior Member

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    The Laplace transform only integrates over the positive reals, not the negatives, thereby taking initial conditions into account. Using an imaginary-valued frequency and integrating over the entire line as in the Fourier transform, you're implicitly assuming knowledge of the input at all times infinitely far into the past (or at least dealing with a periodic signal that's been active for a very long time in the discrete case).
     
  14. arfa brane call me arf Valued Senior Member

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    That explains why knowledge of the input conditions is a part of the design process, perhaps. This includes no input, the system is "relaxed"; also this is why it's important to formulate a response to a transient input.

    In the case of electronics, you generally get to choose the nature of an input, and there are handy lookup tables of Laplace transform pairs.
     
  15. arfa brane call me arf Valued Senior Member

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    And the answer to why voltages and currents are in the same phase space, is the equivalence between an open voltage source and a closed current source in linear networks (c.f. Thevenin-Norton theorems).
     
  16. CptBork Valued Senior Member

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    To be honest, in my undergrad experience with signal processing and electronics, the Laplace transform was never used as much of anything but a novel curiosity. I think perhaps it's more important in systems involving various kinds of feedback, and in which one can assume linear relationships between all the different circuit elements. After moving beyond basic resistors, capacitors and inductors into diodes, transistors, amplifiers and logic gates, we basically stopped doing almost any math whatsoever and started using intuitive approximations and mathematical shortcuts instead.

    As to voltages and currents being in phase (watch your terminology, phase space refers to something completely different), that's not necessarily the case when inductors and capacitors are included in your circuit.
     
  17. arfa brane call me arf Valued Senior Member

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    Indeed. Phase space does have a different definition; perhaps I should say the voltages and currents can be represented by the same kind of vectors, called phasors.

    It's interesting that, in SHM (and in the space of rotating vectors for the forces, displacements etc), there is a correspondence (why wouldn't there be?) between electronic oscillators and mechanical oscillators. Namely, mass and inductance, mechanical damping and resistance, velocity and current, a spring constant and inverse capacitance are pairwise equivalent. Obviously these are very different physical things, but SHM 'generates' equivalence relations (perhaps).

    One observation though, is that the initial displacements x of the row of n pendulums in those videos (not the forced displacement one), corresponds to an electric charge, q, to the inputs of a circuit with n inputs
    The velocity of each pendulum, dx/dt, has a corresponding current dq/dt in a circuit. The mass of each pendulum corresponds to an inductor, and the damping due to say, friction, corresponds to a resistor.

    Resistance in ohms has units Joule-sec/sq. Coulomb. We have to think about these units in Newtonian terms of kg, m, and s. What does the square of a charge q, represent? Resistors as components don't look like they have much to do with squared Coulombs. They usually warm up a bit, and that's her (?)
     
    Last edited: Mar 12, 2020
  18. Confused2 Registered Senior Member

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    From memory the real part was just there in the exponential to ensure everything went to zero at infinity (unlike the Fourier series). We weren't asked or invited to understand quite how this worked - only that it did.
     
  19. CptBork Valued Senior Member

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    There's a simple math theorem which shows that any physical system with a position-dependent energy potential will behave like a simple harmonic oscillator for sufficiently small disturbances near local minima of the potential. Vibrations of one type at a given frequency tend to lead to vibrations of other types at the same frequency, because many systems have a roughly linear response to a given input. That's why voices can transmit cleanly through walls, why microphones can accurately pick up sounds and speakers can pump them back out.

    Ohms involve a relationship between voltage and current, which is where the squared Coulomb units come in. The resistance can also be approximated theoretically in terms of the length, cross-sectional area and material properties of the resistor in question, so the squared Coulombs can also be interpreted in terms of those properties.
     
  20. arfa brane call me arf Valued Senior Member

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    Hokay. So in electronics, where we can construct arbitrary networks of passive and active elements, we also have arbitrary transfer functions in an s-domain which can be written as a ratio of polynomials in s (i.e. with complex roots). The roots are the poles and zeros of the transfer function; a Laplace transform from s to t projects (resp, rotates) the imaginary frequencies onto the real line (or the affine 1-plane).

    The poles of a transfer function represent resonant states of a network under forced oscillations, the zeros relate maximum reactance/impedance to the physical response.
    So what I think I'm now interested in at this point is the differences between the differential equations of "motion" for a mass, m, and a charge, q. On the one hand you have \( m d^2 x/dt^2\), on the other, \( Ld^2 I/dt^2, \; I = dq/dt \). And so there's an equivalence between velocity for a mass m, and current for an inductance L, in that mv is equivalent to LI (in the s-domain). We also have that a displacement, x which is a vector in mechanics, is equivalent to a charge in an inductor.
     
    Last edited: Mar 15, 2020
  21. arfa brane call me arf Valued Senior Member

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    7,832
    I think it's just mathematics, really.

    The reason complex frequencies appear is because, in a real sine wave, if it has an envelope (of changes in amplitude) it is 'pure imaginary'. But make this real and the sine wave imaginary, and you have a complex representation of a sine (or cosine) wave with an exponential envelope. The exponential function shows up when you solve the differential equations of motion. It seems that wave motion and this mathematical function are closely related, but you just take that in your stride so to speak, in engineering.

    I mean, you learn that a capacitor discharging through a resistor follows an exponential curve, for one. I guess they assume you know Euler's identities.
     
  22. arfa brane call me arf Valued Senior Member

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    The other thing I remember about passive circuit elements is that capacitors store energy in an electric field, but inductors store energy in a magnetic field. There's a correspondence, therefore, between kinetic and potential energy in mechanics, and magnetic and electric energy in electronics.

    An inductor reacts to an input current by opposing it, so there's a reverse voltage across it given by -LdI/dt, the induced emf.
     
  23. arfa brane call me arf Valued Senior Member

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    Since I really don't have much else to do, I've reviewed something and decided I should look at wave dynamics/mechanics all over again.

    Particularly, the notion of continuity at a boundary. I think one of the physics books I have is saying that one of the properties of waves and boundaries, in general, is that any wave 'transmitted' through/across a potential barrier always has a lower amplitude than its incident wave (with which it must be--continuous).

    The book is really telling me there's a way to connect say, visible light and a pane of glass, to surface waves in a liquid and a barrier of some kind. Actually the book tells me what the barrier is in the second case, it's a sudden decrease in the depth of the medium, then a sudden but equal increase.

    I know that the propagation of surface waves in a liquid like water depends on gravity, although there is another dependence which I can ignore if I assume an incident wave has a constant wavelength.

    So there is a way to connect the gravitational field to the electromagnetic field, which is the mechanics of oscillating potentials.
     

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