Category: CDS 101/110 Fall 2003

Identifiers: FN H0 H1 H2 H3 H4 H5 H6 H7 H8 L0.0 L1.1 L1.2 L10.1 L2.1 L2.2 L3.1 L3.2 L4.1 L4.2 L5.1 L5.2 L6.2 L8.1 L9.1 L9.2

Questions

• What are the subscripts in L(s) = n_?(s)/d_?(s)

• What does 20dB/decade mean?

• Please elaborate on time delay.

• Can you explain the Final Value Theorem again?

• What are the subscripts in L(s) = n_?(s)/d_?(s)
  Submitted by: atiwari
  Submitted on: November 12, 2003
  Identifier: L6.2

  All the subscripts are L.
  
  L(s) = n_L(s)/d_L(s)
  
  

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• What does 20dB/decade mean?
  Submitted by: demetri
  Submitted on: November 13, 2003
  Identifier: L6.2

  Recall that the magnitude part of the Bode plot is shown on a special logarithmic scale called the deciBel scale. Specifically, we plot 20 times the base-10 logarithm of the function of interest.

  Now, on a base-10 logarithmic plot, each unit distance on the horizontal axis corresponds to a power of ten. I.e., the x-axis looks like 0.1, 1, 10, 100, etc. Each such unit is called a decade.

  So, when we say that a plot has a slope of 20dB/decade, we mean that in each decade unit the plot drops (or increases, as the case may be) 20 dB.

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• Please elaborate on time delay.
  Submitted by: demetri
  Submitted on: November 13, 2003
  Identifier: L6.2

  Time delay is a very common source of trouble in practical control engineering, and it is no small advantage that it can be incroporated directly into frequency-domain stability analysis. The stability implication of time-delay in a feedback system should be clear: attempting to use information from the past to control the present will certainly be harder than using current information.

  Time delays can arise in models from two main sources. First, one can actually have signal propagation delays. A traditonal example in which this occurs is flow-rate control in chemical processing. Because of the long lengths of pipe, the control action of modulating a valve takes time to propagate to the reservoir into which the flow dumps (this delay is roughly inversely proportional to the flow rate, so it is also a nonlinear effect, making it particularly nasty). Of course, the new wave of control applications in networking and networked systems offers a slew of problems in which time delay is important.

  A second, and perhaps more subtle source of time delay is in "fast unmodeled dynamics". Physical sensors and actuators are themselves dynamic elements, although high-quality devices will typically be designed so as to minimize their influence on overall system performance. Consider a simple DC motor: its inertia will cause some delay between the application of a voltage and the settling to the corresponding steady-state speed.

  Now, as control engineers, we certainly have the option of modeling the actuator and sensor dynamics explicitly, but we pay a (perhaps) heavy price in adding additional states to our system (equivalently, obtaining higher-order polynomials in the transfer function). A very common practical alternative to this is to model the sensor/actuator dynamics as being first-order with a known time constant which is significantly smaller than the dominant time-scales in the control process (for example, a small DC motor will typically respond much faster than the robot arm which it manipulates). In this case, we simply assume that signals propagate through the motor perfectly, but with some fixed delay. This is a _MODELING_ choice that we make to trade simplicity against accuracy, but it is often a good trade. In particular, having simple transfer functions allows us to use Bode/Nyquist arguments with much greater ease.

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• Can you explain the Final Value Theorem again?
  Submitted by: waydo
  Submitted on: November 17, 2003
  Identifier: L6.2 , H6

  For a signal f(t) with Laplace transform F(s), the Final Value Theorem says that lim(t->infinity) f(t) = lim(s->0) s F(s) Note that this is only true if F(s) has no poles in the right-half-plane, which makes sense - this is when f(t) will have a well-defined asymptotic value. Be careful - if the system is unstable (or marginally stable), the Final Value Theorem will still give you an answer, but it will be incorrect. With this in mind, if you have plant P(s) and controller C(s), then your closed-loop transfer function is P(s)C(s)/(1 + P(s)C(s)). The Laplace transform of a step input 1(t) is 1/s. Applying the Final Value Theorem (calling the output y(t)) we have lim(t->infinity) y(t) = lim(s->0) s (1/s) P(s)C(s)/(1 + P(s)C(s)).

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