Bode plots

Discussion in 'General Electronics Chat' started by sharanbr, Jun 27, 2015.

  1. sharanbr

    Thread Starter Active Member

    Apr 13, 2009
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    Hello All,

    Can I ask few basic questions related to Bode plots?

    1) Are Bode plots based on Laplace transform?
    2) What do Zeros and Poles signify?

    Do they signify frequencies?

    3) Further, if answer to 2) above is frequency, are Zeros frequency points where output is Zero

    If this is true then what does poles signify?

    4) I am also interested to find out how transfer functions for a given circuit can be determined (both in time domain and S domain)

    Thanks a lot ...
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    No. But the Laplace transform of a system and the Bode plot of it's response are related.

    Ambiguous -- the answer depends on context. But guessing at the context you are asking about, the key thing is that poles and zeros denote values of the domain variable at which the nature of the response changes.

    In circuit analysis, the poles and zeros are values of "complex frequencies", which is NOT the same as the frequency in the time domain. Remember that we are working in a transformed mathematical domain. The domain variable is complex and happens to have units of radian frequency, so we call it the complex frequency domain.

    They are values in the complex frequency domain at which the transfer function goes to zero. This is NOT the same thing as saying that these are values of real frequency at which the system output goes to zero. The reason is that the values of complex frequency that correspond to a zero are not real frequencies.

    Values in the complex frequency domain at which the transfer function goes singular (blows up to infinity).

    Transfer function is a frequency domain concept. Find the Laplace transform of the output and divide it by the Laplace transform of the input. In the time domain you can find the impulse response of the system and then take the Laplace transform of it -- again, the transfer function is a frequency domain concept.
     
  3. sharanbr

    Thread Starter Active Member

    Apr 13, 2009
    76
    1
    Dear Bahn,

    After looking at your response, I am taking one step backward and asking more fundamental question.

    If I take a linear system (e.g. y = 2*x - very simple system), would the output of such a system depend on frequency of inputs (frequency response). Intuitively, what could be the frequency where such a system would show zero output. Similarly, frequencies at which output would go to infinity.

    Unfortunately, this concept always confuses me.
    When I see such a system, for me, I always think that any input produces an output as per the equation y = 2*x.
    Now, where frequency aspect comes into picture, I am don't understand.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    The output of this system would have a flat Bode plot with a constant gain of 2 at all frequencies.

    But, again, do not confuse real frequencies with complex frequencies, they are not the same thing.

    In any system the input produces an output per the equations that describe the response of the system. In your example the output, at any frequency, is twice the input.

    But if you have a system that is governed by the first-order differential equation that describes an RC filter, the input still produces an output per the equation, but that equation produces a different response at different frequencies.
     
  5. sharanbr

    Thread Starter Active Member

    Apr 13, 2009
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    Thank you, Dear Bahn. One more fundamental hole plugged, I guess ...
     
  6. crutschow

    Expert

    Mar 14, 2008
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    3,244
    If the transfer function has no frequency dependent (S or jω) terms then the output will not change with frequency (in theory).
     
  7. sharanbr

    Thread Starter Active Member

    Apr 13, 2009
    76
    1
    Dear Bahn,

    Apologies. I am opening this thread again.
    The fundamental assumption we have made is that system is linear.
    Hence just impulse response of the system fully characterizes such a system.

    Now, with respect to your comment above (first order differential equation), would such a system be considered linear?
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    In general, no. The differential equation must be linear. It can be any order, but it must be linear.

    A non-linear system can still have a Bode plot (but not all systems can -- there ARE some constraints), but the system will not be fully characterized by the impulse response.
     
  9. sharanbr

    Thread Starter Active Member

    Apr 13, 2009
    76
    1
    Thanks. I will wait until I have better understanding of such systems. I am still going through basic notes on OP-AMP's and all my questions are coming from there ...
     
  10. sharanbr

    Thread Starter Active Member

    Apr 13, 2009
    76
    1
    I am reopening this thread after a little self study.

    If complex frequencies are not real frequencies then how to visualize impact of complex frequencies?

    How to co-relate complex with real frequency?

    I have a very basic question. Does the implication of poles and zeros mean that the system is unusable at these complex
    frequencies?
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    The domain variable for complex frequency is normally (almost universally) chosen to be 's' (which is really unfortunate given that 's' is also the unit for seconds, so the units of 's' are '1/s').

    The variable 's' is in a transformed mathematical space and to relate it to real world variable you have to transform the complex frequency domain function back into the time domain.

    We often write 's' in rectangular form as s = σ + jω. When transforming this back the solutions are of the general form e^-s so the real part becomes a dying exponential and the imaginary part becomes a sinusoid. But this is an oversimplification as the transformation depends on the function of 's', not just the variable 's'.

    We can put some limited meaning to it by looking at special cases. The most common is AC steady state in which all of the transients have died out. Under those conditions we set s = jω.

    Consider the simple transfer function

    <br />
H(s) \; = \; \frac{s}{s+\omega_0}<br />

    In steady state, this is

    <br />
H(j \omega) \; = \; \frac{j \omega}{j \omega + \omega_0}<br />
H(j \omega) \; = \; \frac{j}{j + \frac{\omega_0}{\omega}}<br />

    The root of the denominator has the real part, the σ, being -ωo and the imaginary part, the ω, being zero. That has no meaning in the real world.

    At a frequency of ω = ωo, the transfer function has an amplitude of 1/√2 and a phase angle of 45°.
     
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