How Do You Apply the Nyquist Criterion When the Open Loop Has a Pole on the jω Axis?

Studyloaf Team•October 10, 2026•Last verified October 10, 2026•8 min read

Detour around the pole on a small semicircle that keeps it outside the contour, so it does not count in P, and then work out where that detour maps: for a simple pole it becomes a semicircle of infinite radius that closes the Nyquist plot clockwise through the right half of the L-plane. With that closing arc drawn, count encirclements of −1 as usual. For L(s) = K/(s(s + 1)(s + 2)) the result is a stable closed loop for 0 < K < 6.

What you'll learn

  • Why the Nyquist contour cannot run straight through a pole on the jω axis
  • How the small detour around the pole maps to an infinite arc, and which way it turns
  • A fully worked example, L(s) = 3/(s(s + 1)(s + 2)), including the real-axis crossing and the low-frequency asymptote
  • What changes at K = 10, and how the closed-loop poles confirm it
  • The sign conventions that make Z = N + P work

Why can't the contour pass through the pole?

The Nyquist criterion comes from the argument principle, which counts the zeros and poles of 1 + L(s) inside a closed contour. That only works if the contour passes through no pole or zero of 1 + L(s). A pole of L(s) on the jω axis, such as an integrator's pole at s = 0, sits right on the standard contour, and L is infinite there, so the map is undefined at that point. The fix is to step around it on a semicircle of small radius ε and then let ε shrink to zero.

The conventions used here

Unity negative feedback with open-loop transfer function L(s). The Nyquist contour runs up the jω axis and back around the right half-plane on a large semicircle, which is clockwise. The detour passes to the right of the pole at the origin, which leaves that pole outside the contour. Then Z = N + P, where N is the number of clockwise encirclements of −1 + j0 by the plot of L(s), P is the number of open-loop poles inside the contour (strictly in the right half-plane) and Z is the number of closed-loop poles in the right half-plane. Some textbooks count counterclockwise encirclements and write Z = P − N; the two agree once the directions are matched.

The Nyquist contour indented around a pole at the originThe s-plane with the open-loop poles of L(s) marked at 0, minus 1 and minus 2 on the real axis. The contour runs up the imaginary axis, steps around the pole at the origin on a small semicircle of radius epsilon to its right, continues up to plus j R, and returns along a large semicircle of radius R through the right half-plane. It is traversed clockwise. Epsilon tends to zero and R to infinity.σjω−1−2radius εR → ∞+jR−jRclockwise
The contour for L(s) = K/(s(s + 1)(s + 2)): up the jω axis, around the pole at the origin on a small semicircle of radius ε to its right, and back along a large semicircle of radius R. Open-loop poles at 0, −1 and −2 are marked ×.

Where does the detour map?

On the detour s = εe^(jφ), with φ running from −90° up to +90°. Near the origin the factors (s + 1) and (s + 2) are close to 1 and 2, so L(s) is approximately (K/2ε)e^(−jφ). Its magnitude K/2ε grows without bound as ε shrinks, and its angle is −φ, which runs from +90° down to −90°: a decreasing angle means clockwise. So the tiny detour becomes an infinitely large semicircle that starts at the top of the L-plane (the end of the ω = 0− branch), swings clockwise through the positive real axis and arrives at the bottom (the start of the ω = 0+ branch). For a double pole at the origin the angle is −2φ, so the arc sweeps a full 360° clockwise instead of 180°.

Worked example: L(s) = 3/(s(s + 1)(s + 2))

  • Count P. The poles are at 0, −1 and −2. The pole at 0 is outside the indented contour and the other two are in the left half-plane, so P = 0.
  • Find the real-axis crossing. The denominator at s = jω is jω(2 − ω² + 3jω) = −3ω² + jω(2 − ω²). The imaginary part vanishes at ω = √2, where L(j√2) = 3/(−6) = −0.5.
  • Find the low-frequency behaviour. As ω → 0+, Re L(jω) → −3K/4 = −2.25 while Im L(jω) → −∞, so the ω > 0 branch comes in from far below along the line Re = −2.25.
  • Draw the ω < 0 branch as the mirror image of the ω > 0 branch in the real axis, and close the two with the infinite clockwise arc through the right half-plane.
  • Count N. The plot crosses the negative real axis only at −0.5, to the right of −1, and the closing arc lies far out on the right, so −1 is not encircled: N = 0.
  • Conclude: Z = N + P = 0, so the closed loop is stable. The gain margin is 1/0.5 = 2, or 6.02 dB.
Computed Nyquist plot of L(s) equals 3 divided by s(s plus 1)(s plus 2)The solid curve is L(j omega) for positive omega and the dashed curve its mirror image for negative omega. As omega approaches zero the curves run off towards minus and plus infinity along the line real part equals minus 2.25. They cross the negative real axis at minus 0.5 when omega equals root 2, to the right of the critical point minus 1, so the critical point is not encircled and the closed loop is stable. The detour around the origin maps to an arc of infinite radius closing the plot through the right half-plane.-3-2-10-3-2-10123Re L(jω)Im L(jω)−1−0.5 at ω = √2Re → −2.25ω > 0 solid, ω < 0 dashedω = 0− to 0+: infinite arcclosing on the right
The computed Nyquist plot for K = 3 (220 frequencies from ω = 0.35 to 50 rad/s, mirrored for negative ω). It crosses the real axis at −0.5, so −1 lies outside every loop of the closed curve.

What changes when K = 10?

Every point of the plot scales with K, so the crossing moves to −10/6 = −1.667, now to the left of −1. Tracing the closed curve, including the infinite arc, the critical point is circled twice clockwise, so N = 2 and Z = 2: two closed-loop poles in the right half-plane. Computing the closed-loop poles directly, as the roots of s³ + 3s² + 2s + K = 0, confirms it.

Closed-loop poles, roots of s³ + 3s² + 2s + K = 0
KCrossing of L(jω)Closed-loop polesRight half-plane poles
3−0.5−2.672, −0.164 ± j1.0470 (stable)
6−1−3, ±j1.4140, but two on the jω axis
10−1.667−3.309, 0.155 ± j1.7322 (unstable)

Checking with Routh-Hurwitz

For the characteristic equation s³ + 3s² + 2s + K = 0 the first column of the Routh array is 1, 3, (6 − K)/3 and K. All four are positive only for 0 < K < 6, the same limit the Nyquist plot gives, since the crossing −K/6 reaches −1 exactly at K = 6. At K = 6 the s¹ row is zero, and the auxiliary equation 3s² + 6 = 0 gives poles at s = ±j√2, the same frequency as the crossing.

Common mistakes

  • Counting the pole at the origin in P. With the detour to its right, it is outside the contour, so it does not count.
  • Closing the plot with an arc through the left half-plane, or counterclockwise. For a simple pole the arc turns clockwise through the right side.
  • Forgetting the ω < 0 branch, which is the mirror image of the ω > 0 branch, and so miscounting N.
  • Mixing conventions: Z = N + P needs N counted clockwise. If you count counterclockwise, use Z = P − N.
  • Treating the low-frequency end of the plot as finite. It runs off to infinity along Re = −3K/4.

Check your understanding

  • Q1. For the same L(s) with K = 2, where does the plot cross the negative real axis, and what is the gain margin? Answer: at −2/6 = −0.333, so the gain margin is 3, or 9.54 dB.
  • Q2. For K = 4.5, along what vertical line does the low-frequency part of the plot run? Answer: Re L(jω) → −3K/4 = −3.375.
  • Q3. At what gain does the closed loop have poles exactly on the jω axis, and where? Answer: K = 6, with poles at s = ±j√2 ≈ ±j1.414.

Related

Official source

Last verified

The GATE 2027 EC syllabus lists the Routh-Hurwitz and Nyquist stability criteria under Section 6, Control Systems (checked against the official PDF from IIT Madras on 10 October 2026). The crossing, the asymptote, the encirclement counts (by numerically winding the full contour) and the closed-loop poles were all computed in code.

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