- 2075 Asoj · 10 marks
Root locus of K/[(s²+2s+2)(s²+2s+5)]
Short answer
No real-axis locus. Asymptotes intersect at σ=-1 with angles ±45°,±135°. Departure angles: +90° from -1+j1, -90° from -1+j2. Complex breakaway at -1±j1.581 (K=2.25). Imaginary-axis crossing occurs at ω=1.871 rad/s (K=16.25). System stable for 0<K<16.25.
Sketch the root locus for the unity feedback system having the forward path transfer function G(s) = K/[(s²+2s+2)(s²+2s+5)].
Characteristic equation: .
1. Poles and zeros
- Poles: and (). No zeros.
- All 4 branches go to infinity.
2. Real-axis locus
There are no real poles or zeros, so no part of the real axis is on the locus.
3. Asymptotes
4. Angles of departure
- At : angles from : ; from : ; from : .
The branch leaves straight up.
- At : the three angles are all .
The branch leaves straight down.
So the two branches move toward each other along the line and meet.
5. Breakaway points
- : , so it is not on the locus.
- : at , and . Hence , a valid complex breakaway point.
At the poles meet at and then split horizontally (±90° from the vertical) toward the 45° and 135° asymptotes.
6. Imaginary-axis crossing (Routh)
| Row | Col 1 | Col 2 | Col 3 |
|---|---|---|---|
| 1 | 11 | ||
| 4 | 14 | ||
| 7.5 | |||
. Auxiliary equation: rad/s.
Sketch
|
|
.... | ....
...... | ......
...... .x ..X...
......*....... |
x. |
|
|
---------------------------------+----------------
|
|
x. |
......*....... |
...... .x ..X...
...... | ......
.... | ....
|
|
-3 -1 0
x pole * breakaway (-1 +- j1.58, K = 2.25)
X jw crossing (+-j1.87, K = 16.25)
Each pair moves along , meets at , and turns. One branch heads to the 135° asymptote; the other heads right, crossing the -axis at and following the 45° asymptote.
Answer: Complex breakaway at −1 ± j1.581 (K = 2.25). Departure angles are +90° from −1+j1 and −90° from −1+j2. The system is stable for 0 < K < 16.25 and oscillates at 1.871 rad/s when K = 16.25.
✓ Answer checked by Claude Opus 5.5 (high effort)
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