Nyquist Encirclement Decision for (G(s)H(s)=1-(s+1)) Around ((-1+j0))
We are asked: for the Nyquist plot of the loop transfer
will the plot encircle the point ((-1+j0))?
First simplify:
So the Nyquist plot is the locus of: as (\omega) runs from (0) to (\infty) (and similarly the conjugate branch for negative frequencies).
That means:
- For (\omega>0), (-j\omega) lies on the negative imaginary axis (purely imaginary with negative imaginary part).
- There is no way for (-j\omega) to acquire a real part of (-1).
Therefore the Nyquist curve cannot pass through ((-1+j0)), nor can it form a clockwise/counterclockwise encirclement around it—because the entire plot lies on the imaginary axis. Hence the correct option is:
(ii) No.
Mermaid sketch of the locus:
Key Nyquist concepts used:
- Nyquist stability criterion
- Encirclement
- Open-loop poles
(Note: In standard Nyquist, encirclements are counted for the point (-1+j0) relative to the winding number; here the curve is confined to the imaginary axis so the winding number about (-1+j0) is zero.)
Nyquist Plot Encirclement of -1 (Concept + Stability Criterion)
Determine whether the Nyquist plot of \(L(s)=-s\) encircles \((-1+j0)\)
- 1Step 1
Given (G(s)H(s)=1-(s+1)), simplify to (L(s)=-s).
- 2Step 2
Substitute (s=j\omega): (L(j\omega)=-j\omega).
- 3Step 3
For (\omega>0), (L(j\omega)) is purely imaginary with real part 0 and imaginary part negative.
- 4Step 4
The point ((-1+j0)) has real part (-1), but the curve always has real part 0, so it cannot even intersect that point.
- 5Step 5
Since the plot lies entirely on the imaginary axis, it cannot wrap around a point whose real coordinate is (-1). Thus the encirclement count about ((-1+j0)) is zero.
- 6Step 6
Zero encirclements implies the answer is (ii) No.
Geometric reason (why encirclement is impossible here)
Because: the Nyquist curve is confined to the line ( \Re{z}=0 ) (the imaginary axis).
But the point ((-1+j0)) lies strictly to the left at ( \Re=-1 ). A closed loop that remains entirely on the imaginary axis cannot wrap around a point with different real coordinate.
A simple “where is the locus?” diagram:
Pro Tip: Fast check for encirclement
A quick encirclement test: if the Nyquist locus has a fixed real part (e.g., always ), it cannot encircle a point whose real part differs.
Don’t confuse intersection with encirclement
Even if the curve never intersects ((-1+j0)), it still might encircle it. Here, geometric confinement to the imaginary axis prevents both intersection and encirclement.
Common Nyquist-criterion reasoning traps
Workflow to answer Nyquist encirclement multiple-choice questions
Simplify \(L(s)\)
1Algebraically reduce (G(s)H(s))."
Compute \(L(j\omega)\)
2Substitute (s=j\omega) to get the locus equation."
Plot/locate the locus
3Identify fixed real/imaginary constraints."
Check winding about \((-1+j0)\)
4Use geometry: can it reach/wrap around the point?"
Choose option
5If winding number is 0: pick (ii) No."
Decision for Nyquist encirclement around ((-1+j0))
Based on the locus (L(j\omega)=-j\omega), real part is always 0, so encirclement count about ((-1+j0)) is zero.
Knowledge Check
For (G(s)H(s)=1-(s+1)), which Nyquist option is correct regarding encirclement of ((-1+j0))?