ECE356 – 1999 Solutions to the Midterm Test

1. We use the property

\frac{d}{dt}e^{At} = Ae^{At}

Hence A = \frac{d}{dt} e^{At}|_{t=0}

\frac{d}{dt} [ \begin{matrix}\frac{1}{2}(3e^{−t} − e^{−3t}) & \frac{1}{2}(e^{−t} − e^{−3t}) \\ −\frac{3}{2}(e^{−t} − e^{−3t}) & \frac{1}{2}(3e^{−t} − e^{−3t}) \end{matrix} ] = [ \begin{matrix}\frac{1}{2}(−3e^{−t} + 3e^{−3t}) & \frac{1}{2}(−e^{−t} + 3e^{−3t}) \\ −\frac{3}{2}(−e^{−t} + 3e^{−3t}) & \frac{1}{2}(−9e^{−3t} + e^{−t}) \end{matrix} ]
∴ A = [ \begin{matrix}0 & 1 \\ −3 & −4 \end{matrix} ]

2. (i) The characteristic polynomial is given by

p(s) = (s + K)s(s + 4) + K = s^{3} + 4s^{2} + K(s^{2} + 4s + 1)

Thus the root locus of the closed loop system has the same behavior as the root locus of the system

−  K  \frac{s^{2} + 4s + 1}{s^{2}(s + 4)}

Roots of s^{2} + 4s + 1 are at

\frac{−4 ± \sqrt{16 − 4}}{2} = \frac{−4 ± 2\sqrt{3}}{2} = −2 ± \sqrt{3}

Note that these are not zeros of the original system. They serve only as limiting values for the root locus. Note also that p(s) is stable for all K > 0 since its Routh table is

s^{3}:  1  4K
s^{2}:  4 + K  K
s^{1}: 15K + 4K^{2} 0
s^{0}:  K

Thus the root locus sketch looks like

−4

(The real axis break in point, which you don’t need to determine, can be estimated from

(s^{2} + 4s + 1)(3s^{2} + 8s) − (s^{3} + 4s^{2})(2s + 4)
= 3s^{4} + 20s^{3} + 35s^{2} + 8s − (2s^{4} + 12s^{3} + 16s^{2})
= s^{4} + 8s^{3} + 19s^{2} + 8s
= s(s^{3} + 8s^{2} + 19s + 8) = 0

Since s^{3} + 8s^{2} + 19s + 8 changes signs on [−1, 0], there is a real root in [−1, 0].) (ii) The normal behaviour of relative degree 3 systems, being unstable for K suffi-

ciently large, is for a characteristic polynomial of the form D(s) + KN(s) with D, N independent of K, and degree of D = degree of N + 3. This is not the case for this system and there is no contradiction.

3. (i)
P (iω) = \frac{iω + 3}{(iω + 2)^{2}(iω − 1)} = \frac{(iω + 3)(−iω + 2)^{2}(−iω − 1)}{(4 + ω^{2})^{2}(1 + ω^{2})}
= −\frac{12 + 9ω^{2} + ω^{4}}{(4 + ω^{2})^{2}(1 + ω^{2})} − i\frac{4ω}{(4 + ω^{2})^{2}(1 + ω^{2})}

At ω = ±∞, P (iω) = 0. At ω = 0, P (iω = −\frac{3}{4}. For ω in (−∞, 0), P (iω) is in 2nd quadrant, and in (0, ∞), P (iω) is in the 3rd quadratnt.

ω = ±∞ ω = 0 −\frac{3}{4} Nyquist plot Nyquist path

(ii) For stability, we need 1 counterclockwise encirclement of −\frac{1}{K} since there is one unstable open loop pole at s = 1. Thus we require

−\frac{3}{4} < −\frac{1}{K} < 0 or K > \frac{4}{3}