A Mathematical Approach to Classical Control by Andrew D. Lewis

By Andrew D. Lewis

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The following two exercises will recur as exercises in succeeding chapters. Since the computations can be a bit involved—certainly they ought to be done with a symbolic manipulation package—it is advisable to do the computations in an organised manner so that they may be used for future calculations. 3. Derive the full equations which govern the motion of the system using coordinates (x, θ) as in the figure. Here M is the mass of the cart and m is the mass of the pendulum. The length of the pendulum arm is .

V n } for Rn . 6. Write Ab,f v 1 = b11 v 1 + · · · + b 1 v .. Ab,f v = b1 v 1 + · · · + b v Ab,f v +1 = b1, +1 v 1 + · · · + b , +1 v + b +1, +1 v +1 + · · · + bn, +1 v n .. Ab,f v n = b1n v 1 + · · · + b n v + b +1,n v +1 + · · · + bnn v n . 46 2 State-space representations (the time-domain) 22/10/2004 7. Define an × matrix by   b11 b12 · · · b1 b21 b22 · · · b2    N Σ =  .. . ..  .  . .  . b 1 b 2 ··· b 8. The linear differential equation ˙ = N Σw w is called the zero dynamics for Σ.

EAt = 2(e−2t − e−t ) 2e−2t − e−t e2 (t) e3 (t) In this example, we do not use a step input, but rather a violent input: u(t) hΣ (t) et , t ≥ 0 hN,D (t) u(t) = 0, otherwise. 1Σ (t) 1N,D (t) Thus our inputω(t) blows up as time increases. The usual calculations, using zero initial condiφ (t) tions, give fj (t) 1 t e + 1 e−2t − 1 e−t −2t −t Re x(t) = 16 t 32 −2t 21 −t , y(t) = e − e . 10). Things are a bit more subtle with this example. The problem is that the large input is not being transmitted to the output.

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