Frequency response from Pole-Zero plot

Discussion in 'Physics' started by Abbas_BrainAlive, Aug 3, 2018.

  1. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
    Hello all!

    I could not find any satisfactory explanation for why, when computing the frequency response of a pole-zero plot, σ is taken to be zero, and thus s=jω.

    I mean why is the frequency response computed along the imaginary (jω) axis, and not throughout the (σ, jω) plane?

    and why does a crossing point (a pole or a zero) represents a change in slope by 20dB/decade in the frequency response?


    Could anyone please help me with that?
    I would be really grateful.


    Warm Regards,
    Abbas.
     
  2. danadak

    Well-Known Member

    Mar 10, 2018
    1,752
    359
  3. bogosort

    Active Member

    Sep 24, 2011
    170
    83
    The unsatisfactory reason is because that's how it was set up. A more intuitive reason is that the Laplace transform -- which gives us the transfer functions used in pole-zero analysis -- is a generalization of the Fourier transform.

    As you're probably aware, for a certain class of functions, the Fourier transform lets us express functions of time as functions of frequency. The inverse Fourier transform goes the other way. So, for example, if we set up a Cartesian plane with time in seconds as the horizontal axis, then the graph of sin(2πt) looks like a sinusoid with period 1 s. If we change the label associated with the horizontal axis to frequency in Hz , and graph the Fourier transform of sin(2πt), it looks like two spikes at ±1 Hz. This makes sense, as the frequency of sin(2πt) is 1 Hz.

    Notice that we didn't do anything particularly earth-shattering to the plane -- we simply changed one of its labels -- yet the resulting graphs are correct. This is a big hint that the Fourier transform is not just a time/frequency converter; in fact, it converts between any two conjugate variables, i.e., variables that are duals of each other in certain function spaces. (Another famous example of Fourier duals are position and momentum, which leads to Heisenberg's uncertainty principle in quantum mechanics.) The key takeaway here is that the labels we give to our graphs -- time, frequency, energy, etc. -- don't have any intrinsic meaning in the formulas. As long as the variables involved have a specific mathematical connection, the Fourier transform will convert between them.

    Now, consider how the Fourier transform of a function ƒ(t) is defined:

    \mathscr{F}(f(t)) = F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt

    In other words, for each frequency ω of interest, we take the pointwise product of our function and a complex exponential, and sum over all values of time. As Euler showed, any complex exponential is equivalent to a complex sum of sinusoids:

    e^{\pm j \omega t} = cos(\omega t) \pm j sin(\omega t)

    Since these sinusoids represent frequencies (and phase) at ω, the Fourier transform "picks out" each frequency component that's present in ƒ(t).

    Now consider the definition of the Laplace transform:

    \mathscr{L}(f(t)) = F(s) = \int_{0}^{\infty} f(t) e^{-s t} dt \qquad \small{\textrm{with } s = \sigma + j \omega}

    (We've used the unilateral form of the transform here, but that's irrelevant for our purposes.) It looks the same as the Fourier transform, except for that extra σ term in the exponential. Note that s = σ + jω is just a regular old complex number, with real part σ and imaginary part ω. When we graph regular old complex numbers, we call the vertical axis of the plane the purely imaginary part and the horizontal axis the purely real part.

    So, with the ω in our s variable as the imaginary part, any purely imaginary complex number s = jω gets mapped to the vertical axis. But notice that any such number s = jω turns the Laplace transform precisely into the Fourier transform. And since the Fourier transform maps time to frequency, the jω axis in the Laplace transform also maps to frequency.

    That's the only reason we associate frequency with the vertical axis in pole-zero diagrams.

    Because each pole or zero represents an first-order root of the transfer function. Recall the definition of a transfer function:

    H(s) = \frac{N(s)}{D(s)}

    where N(s) and D(s) are polynomials in s. Why this is the case involves the solution spaces of differential equations, but the essential thing is that zeros are the roots of N(s), and poles are the roots of D(s). The poles tell us the values of s for which the transfer function goes to infinity, while the zeros tell us the values that takes H(s) to zero.

    Each root, whether a pole or zero, can be expressed as a polynomial. For example, the root s = -1 corresponds to the polynomial equation s + 1 = 0. Since each root corresponds to a first-order equation, they each contribute first-order responses -- i.e., 20 dB per decade -- to their respective N(s) or D(s) polynomial. The order of the total transfer function H(s) is then the order of the remaining roots after H(s) has been simplified.

    Hope that helped, but let me know if something wasn't clear.

     
  4. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
  5. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
    Excellent explanation for both of my questions. Thanks a lot, bogosort!

    One thing that I still don't get is the magnitude Bode plotting from the transfer function. To be precise, I am facing difficulty in understanding the changes in the response of a (transfer) function as it traverses from one crossing point to another, be it a pole or zero. Every crossing point changes the slope of the plot by 20dB/dec, what does this change in slope physically mean?

    Thanking you once again.
    Best Regards,
    Abbas.
     
  6. bogosort

    Active Member

    Sep 24, 2011
    170
    83
    There are several different but equivalent ways of thinking about the poles and zeros of a transfer function. If you're familiar with basic electronics, I think RC filter circuits provide a nice way to gain intuitive understanding. They provide something tangible to associate with the equations.

    Consider the humble low-pass RC filter:
    [​IMG]
    where we take the output across the capacitor. With a single resistor-capacitor pair, this is a first-order filter. In the frequency domain, it has a cutoff frequency of ω = 1/(R1*C1) rad / s, and a -20 dB / decade slope.

    A useful way of looking at this circuit is as a frequency-dependent voltage divider. The capacitor C1 has an impedance that varies with frequency: when the input frequency is very low the capacitor presents a high impedance, allowing most of the signal to reach the output; as the input frequency increases, the capacitor's impedance diminishes, shunting more and more signal to ground. This is the classic LPF behavior described in terms of a voltage divider. From the formula for a voltage divider, we can write the filter's output as a function of the input and the circuit impedances:

    v_o(t) = v_i(t) (\frac{Z_{C1}}{R_1 + Z_{C1}})
    In the Laplace domain, we have

    <br />
\begin{align}<br />
v_o(t) &= V_o(s)<br />
v_i(t) &= V_i(s)<br />
R &= R<br />
Z_{C1} &= \frac{1}{sC_1}<br />
\end{align}
    Thus,

    V_o(s) = V_i(s)(\frac{\frac{1}{sC_1}}{R_1 + \frac{1}{sC_1}})
    By definition, a system's transfer function is the Laplace transform of the ratio of the system's output to its input, therefore

    G(s) = \frac{V_o(s)}{V_i(s)} = \frac{\frac{1}{sC_1}}{R_1 + \frac{1}{sC_1}} = \frac{\frac{1}{R_1C_1}}{s + \frac{1}{R_1C_1}} = \frac{a}{s + a}
    where a = 1 / (R1C1). This system has a single (real-valued) pole at s = -a, which in the frequency domain corresponds to a cutoff frequency at +a rad / s.

    Next, consider a second-order RC low-pass filter:
    [​IMG]
    The unity-gain voltage buffer between the RC sections is there solely to prevent them from loading each other down. As a second-order filter, we'd expect the slope of its frequency response to be twice as steep as the first-order version, i.e., -40 dB per decade. This makes intuitive sense: with twice the number of RC sections, we expect the filter to do twice the work. Let's see how this plays out in terms of poles.

    The transfer function of this filter is simply the product of the transfer function of each RC section, i.e.:

    H(s) = (\frac{a}{s + a})(\frac{b}{s + b}) = \frac{ab}{s^2 + (a + b)s + ab}

    where b = 1 / (R2C2). With a quadratic in the denominator, H(s) has two poles.

    To make this concrete, let R1 = R2 = 10 kΩ and C1 = C2 = 1 μC. Then, a = b = 100 rad / s. Comparing the Bode plots of each transfer function:

    [​IMG]

    We see the expected shapes, but notice that the scale on the vertical axis of the second-order magnitude plot is double that of the first-order plot. At 1000 rad / s -- i.e., one decade above the cutoff frequency -- the first-order filter is down -20 dB while the second-order filter is down -40 dB, just as we'd expect.

    So, in very brief summary, you can think of each new pole in a transfer function as an added RC section in the corresponding filter circuit. Of course, there are a whole bunch of other details, but hopefully this helps you gain a bit of physical intuition.
     
  7. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,277
    1,136
    Hi,

    The short answer is because we are using a sine forcing function and that assumes that a long time has passed so that the exponential parts have had sufficient time to become zero, and the time domain coefficient of the remaining time function is the same as the coefficient of the frequency domain function so we get the same amplitude either way.

    Consider the similarities:
    time domain: f(t)=A(w)*sin(w*t)+B*e^(-a*t)
    frequency domain: F(w)=A(w)

    Note that the coefficient in f(t) is A(w) and also in F(w), and that it does not matter what B is because the exponential part dies out for long 't' and that is an assumption for the frequency domain response too. So we effectively are comparing:
    A(w)*sin(w*t)
    and:
    A(w)

    where the lone A(w) is considered a sine function coefficient anyway: A(w)*sin(w*t) with the sin(w*t) part implied, so they are both the same.

    The only thing to remember is that for a sine (or cosine) forcing function in the frequency domain we assume that the exponential part has died down because an infinite time has passed because it is assumed that the sine function has been running since the beginning of all time t=-infinity. This also means that we DO in fact loose part of the response when we calculate in the frequency domain and that is the start up part of the response, but we dont always care about that.

    Another thing you might want to keep in mind is that the response to a sinusoidal forcing function is very much different than the response to a step function, and that an assumption in the frequency domain is that we are using a sinusoidal forcing function that has been 'on' for all time and so had no startup point in time.
     
    Last edited: Aug 14, 2018
  8. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
    I really appreciate your willingness to help, MrAl. Thank you very much.
     
  9. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
    Thank you once again, bogosort.

    How does that happen? From the transfer function, we got s=-a, and we also know that s=σ+jω. Comparing the two, we get σ=-a and ω=0; pole lying on the horizontal axis. So, where am I going wrong?
     
  10. danadak

    Well-Known Member

    Mar 10, 2018
    1,752
    359
    Its = 1 / (s + 1/a)

    upload_2018-8-14_8-2-3.png

    Regards, Dana.
     
  11. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
    Thank you very much danadak, I understand that very well.

    At the moment, I am questioning bogosort's conclusion about deriving the cut-off frequency from the pole-zero plot.

     
  12. bogosort

    Active Member

    Sep 24, 2011
    170
    83
    You've got it right, the pole lies on the real axis at -a. Such poles correspond to decaying exponentials in the time domain. This is necessarily the case for any first-order system: by the fundamental theorem of algebra, an nth order polynomial has n roots. Complex roots -- which give us the complex poles necessary for oscillatory responses -- always come in conjugate pairs, so first-order systems can only have real poles and monotonic responses. In other words, ω = 0 for all first-order s.

    As for the time constant, recall the definition of cut-off frequency (in rad / s) for a first-order RC circuit:

    f_c = 1 / (RC) = 1 / \tau

    In our example, we had

    a = 1 / (R_1C_1) = 1 / \tau

    Therefore, the pole at -a corresponds to the cut-off frequency at a.

    Makes sense?
     
  13. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
    Wonderful correlation, thanks a lot bogosort.

    I get the mathematical equivalence of the two equations. But, paying more attention to the equation, how does 1/(Ω-F) translate to Hz, if we compare the units? When referring to the pole/zero itself as a quantity/entity on the s-plot, how is it signifying the frequency when the frequency component itself is zero?

    Also, please allow me to wander a step deeper.

    What is the physical significance of a pole/zero being purely real (i.e. no frequency component), complex, and purely imaginary?

    Best regards,
    Abbas.
     
  14. bogosort

    Active Member

    Sep 24, 2011
    170
    83
    The transfer function we've been looking at tells us how the system responds to an impulse at its input, i.e., how the system behaves over time after it's been energized for a brief instant. This is sometimes called its natural response. The jω term characterizes how oscillatory the system is, as in how much a bell will ring after being struck, while the σ term tells us how fast the ringing will die out. When jω = 0, the system does not oscillate -- no matter what we put at its input, as soon as the input is stopped the output will decay to zero. The larger the value of σ (pushing the pole farther to the left on the real axis), the faster the response will decay; this corresponds to a fast time constant. If σ = 0, the ringing will go on "forever".

    As for the cut-off frequency, it's just a figure of merit derived from the system's time constant. It tells us the frequency at which half the input's power would be attenuated. I'm not sure what 1/(Ω-F) refers to, but if you're asking about the relationship between Hz and radians per second, the conversion constant is 2π. With ω in rad/s and f in Hz: ω = 2πf.

    The physical significance is damping, i.e., how much the system resists oscillation (as in a mass-spring system). Let's look at the possibilities for a 2nd-order system (since 1st-order systems can only have real poles). With two roots, there are four possible response types:

    Case 1: Two real, distinct poles. Let's call them σ1 and σ2. If we do the math, we'll see that the system has a response of the form

    y(t) = K_1e^{-\sigma_1t} + K_2e^{-\sigma_2t}

    This is called an overdamped response. With two decaying (real-valued) exponentials, this system cannot oscillate; it's behavior in the time domain will be sluggish with respect to transient inputs. Over time, the larger of the poles will dominate.

    Case 2: Two real, repeated poles. These systems have a response of the form

    y(t) = K_1e^{-\sigma_1t} + K_2 t e^{-\sigma_1t}

    This is a critically damped response. That factor of t in the second term fights against the decaying exponential, making this response faster than the overdamped response. In fact, this is the fastest possible transient response without oscillation.

    Case 3: Two complex conjugate poles at σ ± jω. The response is

    y(t) = Ae^{-\sigma t}(B cos \omega t + C sin \omega t)

    This is an underdamped response. With sinusoids in the response it is clearly oscillatory at ω, but the decaying exponential dampens the response over time. In other words, the natural response of such a system is a damped sinusoid with a decaying exponential envelope.

    Case 4: Purely imaginary poles at ±jω, leading to the following response:

    y(t) = A cos(\omega t - \phi)

    Obviously, this is purely oscillatory -- we call it an undamped response,. Without any damping such systems are metastable: a slight change in the pole's value could push the system into instability.

    Here's an image of the four response types from an old textbook of mine:

    [​IMG]

    Anyway, the point of pole-zero analysis is really to quantify the stability of systems, which becomes a big deal when you start to include gain stages and feedback paths. Frequency response is just a tiny portion of it. Remember, this stuff applies to way more than just filter circuits: your air-conditioning system, the way your pupils respond to light, entire processing plants -- any control system can be analyzed using pole-zero analysis. Are you studying this stuff for a class or for personal education/enjoyment?
     
  15. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
    Thanks a lot, bogosort, I am really grateful.

    Well, I studied control systems a long time ago, and was fairly good at it, yet I could not develop a satisfactory understanding of the 's' variable, and its physical significance.
    But, thanks to you, I can say that I do understand it now.
    I just need to do some work on relating σ and ζ (the damping ratio), and some more work on their relation with Q (the quality factor).

    I was referring to the units (resistance and capacitance). Though, more precisely, it should be 1/(radians-Ω-F).

    We have ƒc=1/(2π*R*C), or ωc=1/(R*C).
    The units don't match on both sides!


    Thanks a lot once again.
     
    Last edited: Aug 17, 2018
  16. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,277
    1,136
    Hi,

    Here is a rundown of the significance of the real and imaginary parts in their relationship to the signal response. In the drawing the point in the plane for each waveform shown is approximately near the center of the waveform. For example, the straight horizontal response near the point (0,0) is the response at (0,0).

    Note that the plane is divided into two halves, the LHP (left half plane) and the RHP (right half plane).
    Responses that have negative real parts are exponentially decreasing (LHP) and those that have positive real parts are exponentially increasing (RHP). Thus responses in the RHP are said to be unstable.
    Also, as we move up from the point (0,0) the frequency increases, so responses that are higher up have a higher frequency (and they are matched in the lower half plane by their conjugates).
    Signals that lie directly on the jw axis are sinusoidal and do not increase or decrease in amplitude with time. These are oscillators, and in most oscillators great care is taken to make sure the response stays either on the jw axis or varies back and forth left to right and right to left just a little so that the oscillations appear almost perfectly stable.
    You can also see that responses that lie on the real axis are either increasing or decreasing (except at 0,0) and contain no sinusoidal part.
     
  17. bogosort

    Active Member

    Sep 24, 2011
    170
    83
    My pleasure, I'm glad it helped.

    One way to look at ζ is as the ratio between the system's exponential decay rate and its natural frequency of oscillation; this, of course, gives the damping factor:

    \zeta = \frac{\sigma}{\omega}

    Algebraically, for a 2nd-order system of the form

    H(s) = \frac{c}{s^2 + bs + c}

    we have the relation

    \zeta = \frac{b/2}{\sqrt{c}}

    The units do match; that extra 2π factor ensures it. Look at both sides dimensionally: the quantity 2π is a dimensionless number. The product of ohm and farad has dimension of time, so in both cases the reciprocal must be a frequency. The frequency on the right corresponds to radians per second, on the left cycles per second. Both radians and cycles are dimensionless, but a cycle is bigger than a radian (there are 2π radians per cycle) so the extra factor makes them equal.
     
  18. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
    Thanks a lot, MrAl.
     
  19. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
    Thanks a lot, again, bogosort. I understand that very well, at least mathematically. But still, it will take me some time to understand that in physical terms.

    Moreover, this is what I was missing:
    How?
    Can you please explain this?

    Thanks a lot, once again.
    Abbas
     
  20. Abbas_BrainAlive

    Thread Starter Member

    Feb 21, 2018
    45
    5
    OOPS!

    I got my answer!

    doing elementary dimensional analysis did the trick!

    Thanks a lot, all of you!

    With warmest regards,
    Abbas.
     
    bogosort likes this.
Loading...