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Significance of standard form of 1st and 2nd order transfer functions?

  1. Aug 1, 2014 #1
    I foolishly skipped most of my analogue electronics classes, and inevitably failed the exam. I'm now trying to revise for the resit but I'm so far behind that I just cannot understand any of the lecture slides, and i'm getting very stressed.

    The part of the module I am revising at the moment contains: transfer functions, poles and zeros, impulse response, step response, sine response, Bode plots, and frequency analysis.

    I understand very little so I have many questions, but as I can't ask a whole modules worth of questions in a thread I thought I'd start with this:

    What is the significance of the standard form of 1st and 2nd order transfer functions?

    The standard form of a first order transfer function is:

    (1) [tex]\tau \frac{dy}{dt} + y = k * x(t)[/tex]

    The laplace transform of this:

    (2) [tex]G(s) = \frac{Y'(s)}{X'(s)} = \frac{k}{\tau s+1}[/tex]

    but sometimes it is given as

    (3) [tex]H(s) = \frac{1}{\tau s +1} = \frac{a}{s+a}[/tex]

    The standard form of a second order transfer function is:

    (4) [tex]\tau ^{2} \frac{d^{2}y}{dt^{2}}+2 \tau \zeta \frac{dy}{dt} + y = k * x(t)[/tex]

    The laplace transform of this:

    (5) [tex]G(s) = \frac{Y(s)}{X(s)} = \frac{k}{\tau^2s^2 + 2\tau\zeta s+1}[/tex]

    but sometimes this is given as

    (6) [tex]H(s) = \frac{\omega_n^2}{s^2+2\zeta \omega_n s + \omega_n^2}[/tex]

    Here are my questions:

    • What is the physical meaning of "first" and "second order"? (apart from the fact that the highest power of the differential in the first is 1 and in the second is 2). How do I know if a system is first or second order?

    • Where do equations (1) and (4) come from? Why were these decided to be the "standard form"? What is so special about this form and how were these equations derived?

    • When given a first order system, why is sometimes equation (2) given, and sometimes equation (3) as the transfer function for this system? Likewise, when given a second order system why is equation (6) usually given, when the laplace transform is actually equation (5)?

    Thanks for reading!
    Last edited: Aug 1, 2014
  2. jcsd
  3. Aug 1, 2014 #2
    To get some physical insight, you should have a look at what kinds of systems you can model using first and second order linear differential equations with constant coefficients. A transfer function is just a tool we use to analyze these systems, but they're definitely not required. You could analyze them perfectly well without ever having to develop a transfer function.

    Are we talking purely mathematically? Just have a look at what order of differential equation the transfer function maps to and vice versa.

    'Order' is a mathematical term. In practice, if you can approximately model your system as a first or second order system, then you might say something along the lines of "the system has first/second order dynamics", where it's understood that we're talking about a LTI approximation.

    You tell me. Your (4) is unusual (the time constant of a second order system is given by [itex]\tau = \frac{1}{\zeta \omega_n}[/itex]). Usually you'll see it in (unforced) standard form as:
    \frac{{\mathrm{d}^2 x}}{{\mathrm{d}t}^2} + 2 \zeta \omega_n \frac{{\mathrm{d}x}}{{\mathrm{d}t}} + \omega_n^2 x = 0
    They're special because the time constant, natural frequency and damping factor characterize the dynamics of these systems in a meaningful way. I can look at numbers for these quantities and immediately tell you something meaningful about how the system behaves. These direct forms are a convenience.

    In (2) you see the time constant directly. In (3) you see the pole location directly. Your (4) is a bit wonky. (6) shows the natural frequency and damping factor directly.

    They're all good for something. It's just a matter of convenience.
    Last edited: Aug 1, 2014
  4. Aug 2, 2014 #3


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    Staff: Mentor

    Looking at (2) you can see immediately that at low-frequency (DC) it has a gain of k. Looking at (3) you can tell at a glance that the system's gain has begun dropping off by 3dB (1/√2) when frequency = a radians/second (the constant in the denominator). These are two defining characteristics of any system.

    As far as electronics goes, a first order system contains one storage element (either C or L), together with one or more resistors. A second order system contains some resistances together with twoǂ storage elements, either two C's or two L's or one of each.

    ǂ capacitors in parallel count as only one element, the same applies to multiple L's

    The differential equations arise because current is related to voltage for both C and L by a first-order equation.
  5. Aug 2, 2014 #4

    jim hardy

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    Science Advisor
    Gold Member

    You have it right - in Laplace one is linear the other is quadratic

    1 and 4 are ordinary differential equations that you learned to solve in that math class you took, Differential Equations.. They describe something in the real world. Laplace transforms are just a tool that simplifies the solution of such equations.

    That is purely the author's choice.
    As Miles Young pointed out, one form tells him a lot about the system's behavior. The other form might tell an electronics guy just what time constants he will need in order to create that function with operational amplifiers.
    Both electronic and mechanical types recognize that a quadratic in the denominator makes a system potentially oscillatory - if the denominator can resolve to zero the system can have an output for no input. Hence the dreaded phrase "Real roots in denominator" ...

    So an electronic oriented author may prefer one form and a mechanical oriented author might prefer the other.
    An astute student will become fluent at transforming between the two forms so he is at home in both worlds, mechanical and electronic.

    Sure it's a bit of work, but -
    "When the going gets tough , the tough get going. "
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