Part 4 · Chapter 18

Closed-Loop Frequency Response and the Nichols Chart

Every frequency-domain result so far has been read off the open-loop function, yet what a user actually experiences is the closed loop — so this chapter builds the bridge, mapping contours of constant closed-loop magnitude and phase onto the open-loop plane as the M and N circles, folding them into the Nichols chart, and reading the resonant peak, the resonant frequency and the bandwidth straight off a single curve.

Control Systems Prof. Mithun Mondal Reading time ≈ 44 min
i What you'll learn
  • Why the closed-loop response \(T(j\omega)=G/(1+G)\) must be recovered from the open-loop plot rather than recomputed.
  • The geometric reading of \(M=|T|\) as a ratio of two distances in the \(G\)-plane.
  • The derivation of the constant-M circles and the constant-N circles, and what each family looks like.
  • The Nichols chart — log-magnitude against phase — and why a gain change becomes a rigid vertical shift on it.
  • The three closed-loop specifications: resonant peak \(M_r\), resonant frequency \(\omega_r\), and bandwidth \(\omega_b\).
  • The exact second-order correlation tying \(M_r\), \(\omega_b\) and phase margin back to \(\zeta\) and \(\omega_n\) from Chapter 8.
Section 18-1

Why the Closed Loop Needs Its Own Plot

Chapters 14 to 17 built an entire theory on one function: the open-loop transfer function \(G(j\omega)H(j\omega)\). There were good reasons for that. It factors into first- and second-order terms, so its Bode magnitude is a chain of straight-line asymptotes; its polar plot is drawn once and reused; and Nyquist's criterion turns closed-loop stability into a statement about how that open-loop curve encircles \(-1\). Gain margin and phase margin, from Chapter 17, are distances measured on the same open-loop curve.

But nobody buys an open loop. The quantity a servo user cares about is \(C(j\omega)/R(j\omega)\) — how faithfully the output follows a commanded sinusoid, at what frequency it starts to resonate, and beyond what frequency it simply stops responding. Gain and phase margins are indirect measures of those things: they say how far the loop is from breaking, not how the closed loop actually behaves. Chapter 17 ended by admitting exactly this. Here we close the gap.

The obvious approach — compute \(T(j\omega)=G/(1+G)\) numerically at a hundred frequencies — works but teaches nothing and destroys the structure that made the open-loop plot easy in the first place: \(1+G\) does not factor. So we do something cleverer. We leave the open-loop plot exactly as it is and lay over it a grid: a family of curves along which the closed-loop magnitude is constant, and a second family along which the closed-loop phase is constant. Once that grid is drawn, the closed-loop response is read off the open-loop curve by inspection, the way a contour map converts a walk across a hillside into a list of altitudes.

The whole chapter in one sentence. Nothing new is computed; the open-loop plane is simply re-labelled with closed-loop information. Every result that follows — M circles, N circles, the Nichols chart — is the same idea in a different coordinate system.
Section 18-2

Closed-Loop Magnitude and Phase

Take the unity-feedback loop of Chapter 9, whose closed-loop transfer function is \(T(s)=G(s)/(1+G(s))\). At a fixed frequency \(\omega\), the complex number \(G(j\omega)\) is one point in the plane. Write it in rectangular form,

The open-loop point, resolved
\[ G(j\omega) = X + jY, \qquad T(j\omega) = \frac{X+jY}{(1+X)+jY} \]

Define the two things we want as functions of \(\omega\): the closed-loop magnitude \(M(\omega)=|T(j\omega)|\) and the closed-loop phase \(\alpha(\omega)=\angle T(j\omega)\). Taking magnitudes and arguments of the fraction gives them directly.

Magnitude and phase in terms of X and Y
\[ M = \frac{\sqrt{X^2+Y^2}}{\sqrt{(1+X)^2+Y^2}}, \qquad \alpha = \tan^{-1}\!\frac{Y}{X} - \tan^{-1}\!\frac{Y}{1+X} \]

The magnitude expression deserves a second look, because it is not really algebra — it is geometry. The numerator \(\sqrt{X^2+Y^2}\) is the distance from the origin to the point \(G(j\omega)\). The denominator \(\sqrt{(1+X)^2+Y^2}\) is the distance from the point \(-1\) to that same point, since \(1+G = G-(-1)\). So the closed-loop magnitude at any frequency is simply a ratio of two lengths measured on the polar plot we already have.

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The geometric definition of M
\[ M(\omega) = \frac{\text{distance from the origin to } G(j\omega)}{\text{distance from } -1 \text{ to } G(j\omega)} \]

When the polar plot passes close to \(-1\) the denominator is small and \(M\) is large: the closed loop resonates. When the plot stays far from \(-1\), \(M\) never rises much above unity and the response is smooth. This is the quantitative version of the qualitative statement made in Chapter 17 — that proximity to \(-1\) means poor relative stability.

Similarly, \(\alpha\) is the difference of two angles: the angle the point subtends at the origin minus the angle it subtends at \(-1\). Both quantities are therefore determined by position alone. Two different systems whose polar plots happen to pass through the same point at some frequency have the same closed-loop magnitude and phase at that frequency, whatever else they do. That observation is what makes a universal overlay possible: the value of \(M\) is a property of the point, not of the system.

Section 18-3

Constant-M Loci: the M Circles

Fix a value of \(M\) and ask which points \(X+jY\) produce it. Squaring the magnitude relation and clearing the denominator turns the question into a quadratic in \(X\) and \(Y\).

Derivation of the constant-M locus
\[ M^2\left[(1+X)^2+Y^2\right] = X^2+Y^2 \]
\[ M^2 + 2M^2X + M^2X^2 + M^2Y^2 = X^2 + Y^2 \]
\[ (1-M^2)X^2 + (1-M^2)Y^2 - 2M^2X - M^2 = 0 \]

Two cases separate at once. If \(M=1\) the quadratic terms vanish and what remains is \(-2X-1=0\), a straight line. Otherwise divide by \((1-M^2)\) and complete the square in \(X\):

Completing the square for M ≠ 1
\[ X^2 + Y^2 - \frac{2M^2}{1-M^2}X - \frac{M^2}{1-M^2} = 0 \]
\[ \left(X - \frac{M^2}{1-M^2}\right)^{2} + Y^2 = \frac{M^2}{1-M^2} + \frac{M^4}{(1-M^2)^2} = \frac{M^2}{(1-M^2)^2} \]

That is the equation of a circle. Writing the centre with \(M^2-1\) in the denominator makes the sign convention easier to remember: for \(M>1\) the centre sits on the negative real axis.

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Constant-magnitude loci
\[ \text{centre} = \left(-\frac{M^2}{M^2-1},\; 0\right), \qquad \text{radius} = \left|\frac{M}{M^2-1}\right| \]

Every constant-M locus is a circle centred on the real axis, except \(M=1\), which degenerates into the vertical line \(X=-\tfrac12\). Circles with \(M>1\) lie to the left of that line and shrink onto the point \(-1\) as \(M\to\infty\); circles with \(M<1\) lie to the right, enclose the origin, and shrink onto it as \(M\to 0\).

The two limits are worth checking against intuition. As \(M\to\infty\) the centre tends to \(-1\) and the radius to zero: infinite closed-loop magnitude occurs only at the point \(G=-1\), which is exactly the condition \(1+G=0\) for a closed-loop pole on the imaginary axis. As \(M\to 0\) the circle collapses onto the origin, where \(G=0\) and the loop passes nothing. The family therefore interpolates smoothly between "no response" and "resonance without limit", and the \(M=1\) line is the watershed between them.

\(M\)\(M\) in dBCentre on real axisRadius
0.5−6.02+0.3330.667
0.707−3.01+1.0001.414
1.00.00straight line \(X=-0.5\)
1.2+1.58−3.2732.727
1.4+2.92−2.0421.458
2.0+6.02−1.3330.667
3.0+9.54−1.1250.375
−1.0000
Re Im M = 1.4 M = 2 M = 0.5 M = 1 −1 G(jω)
Constant-M circles collapse onto −1 as M grows; the polar plot grazes the largest M it reaches
Section 18-4

Constant-N Loci: the N Circles

The same treatment applied to the phase gives a second family. Let \(N=\tan\alpha\) and expand the tangent of the difference of the two arctangents that define \(\alpha\).

Derivation of the constant-phase locus
\[ N = \tan\alpha = \frac{\dfrac{Y}{X}-\dfrac{Y}{1+X}}{1+\dfrac{Y}{X}\cdot\dfrac{Y}{1+X}} = \frac{Y(1+X)-YX}{X(1+X)+Y^2} = \frac{Y}{X^2+X+Y^2} \]
\[ X^2 + X + Y^2 - \frac{Y}{N} = 0 \;\Longrightarrow\; \left(X+\tfrac12\right)^2 + \left(Y-\frac{1}{2N}\right)^2 = \frac14 + \frac{1}{4N^2} \]
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Constant-phase loci
\[ \text{centre} = \left(-\tfrac12,\; \frac{1}{2N}\right), \qquad \text{radius} = \tfrac12\sqrt{1+\frac{1}{N^2}}, \qquad N=\tan\alpha \]

Every N circle passes through both the origin and the point \(-1\) — substitute \(X=Y=0\) and \(X=-1,\,Y=0\) into \(X^2+X+Y^2-Y/N=0\) and both satisfy it identically, for every \(N\). The centres all lie on the vertical line \(X=-\tfrac12\), climbing as the phase lag decreases.

That both fixed points belong to every member of the family is not an accident. At \(G=0\) we have \(T=0\), and at \(G=-1\) the closed loop is singular; in both cases the phase is undefined, so every constant-phase curve is entitled to pass through them. The two special cases are equally quick: \(\alpha=90^{\circ}\) gives \(N=\infty\), a circle of radius \(\tfrac12\) centred at \(-\tfrac12\); \(\alpha\to 0\) sends the centre to infinity and the circle degenerates into the real axis itself.

One caution accompanies these curves. Since \(\tan\alpha=\tan(\alpha\pm 180^{\circ})\), a single N circle serves two phase angles differing by \(180^{\circ}\), and only an arc of it corresponds to each. In practice the correct branch is fixed by the quadrant in which \(G(j\omega)\) lies, and the charts printed in textbooks label the usable arcs directly.

\(\alpha\)\(N=\tan\alpha\)CentreRadius
15°0.268\((-0.5,\;1.866)\)1.932
30°0.577\((-0.5,\;0.866)\)1.000
45°1.000\((-0.5,\;0.500)\)0.707
60°1.732\((-0.5,\;0.289)\)0.577
90°\((-0.5,\;0)\)0.500
−30°−0.577\((-0.5,\;-0.866)\)1.000
Re Im α = 30° 60° 90° −60° −30° −1 every circle passes through 0 and −1
Constant-N circles: centres on the line X = −1/2, all threading the origin and the point −1
Section 18-5

Using the Circles on the Polar Plot

With both families printed as a background grid, the polar plot of \(G(j\omega)\) becomes a route across a contour map. Where the plot crosses an M circle, the closed-loop magnitude at that frequency equals the label on the circle; where it crosses an N circle, the closed-loop phase equals that label. Sweeping \(\omega\) along the plot and recording the labels reconstructs the entire closed-loop frequency response without a single complex division.

Three readings matter more than the rest. The largest M circle the plot reaches — the one it touches tangentially rather than cuts — gives the resonant peak \(M_r\), and the frequency at the point of tangency is the resonant frequency \(\omega_r\). The crossing of the \(M=0.707\) circle, on the far side of the peak, marks the bandwidth \(\omega_b\), because \(0.707\) is \(-3\) dB.

Tangency, not intersection. If the plot cuts an M circle it passes from one magnitude to another and back; the peak has not been found. Only where the curve just grazes a circle is the closed-loop magnitude stationary, and that is the definition of a resonance.

The construction also answers a design question: what loop gain \(K\) makes \(M_r\) equal a specified value? The geometry is pleasing. The M circle for a given \(M>1\) has its centre at distance \(d = M^2/(M^2-1)\) from the origin and radius \(r = M/(M^2-1)\). A straight line drawn through the origin tangent to that circle makes an angle \(\psi\) with the negative real axis given by \(\sin\psi = r/d\), and the algebra collapses beautifully.

The tangent-line angle
\[ \sin\psi = \frac{r}{d} = \frac{M/(M^2-1)}{M^2/(M^2-1)} = \frac{1}{M} \;\Longrightarrow\; \psi = \sin^{-1}\!\left(\frac{1}{M}\right) \]

The classical procedure follows. Draw the polar plot of \(G(j\omega)/K\), the shape with unit gain. Draw the line through the origin at angle \(\psi=\sin^{-1}(1/M_r)\) below the negative real axis. Construct the circle that is tangent both to that line and to the unit-gain plot, with its centre on the negative real axis; if that circle meets the negative real axis at \(-c\) on the side nearer the origin, then, because the true M circle meets the axis at \(-M/(M+1)\), the required gain is

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Gain for a specified resonant peak
\[ K = \frac{M_r}{(M_r+1)\,c} \]

The construction is exact but fiddly, and it has to be repeated from scratch whenever \(K\) changes, because multiplying by \(K\) rescales the polar plot radially — every point moves by a different amount. That single inconvenience is the reason the Nichols chart exists.

Section 18-6

The Nichols Chart

Nathaniel Nichols' contribution was a change of coordinates. Instead of plotting \(G(j\omega)\) in the complex plane, plot its log-magnitude in decibels as the ordinate against its phase in degrees as the abscissa, with frequency as the parameter running along the resulting curve. The point \(-1\) — magnitude 1, phase \(-180^{\circ}\) — becomes the single point \((-180^{\circ},\,0\text{ dB})\), and the M and N circles map into closed contours wrapped around it.

The payoff is immediate. Multiplying the loop gain by \(K\) multiplies every magnitude by \(K\), which adds \(20\log_{10}K\) decibels to every ordinate and leaves every phase untouched. On the Nichols chart a gain change is a rigid vertical translation of the whole curve — the shape is preserved, so one plot serves every gain.

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Why the chart is a design surface
\[ 20\log_{10}\!\left|KG(j\omega)\right| = 20\log_{10}K + 20\log_{10}\!\left|G(j\omega)\right|, \qquad \angle KG = \angle G \]

Slide the curve up or down until it is tangent to the desired \(M\) contour; the number of decibels you slid it is \(20\log_{10}K\). A phase-lead network, similarly, slides the curve to the right — which is exactly how the compensators of Part 5 are designed on this chart.

Reading the chart is a matter of four habits. The gain margin is the vertical distance, in decibels, from the curve's crossing of the \(-180^{\circ}\) vertical down to the \(0\) dB horizontal. The phase margin is the horizontal distance from the \(-180^{\circ}\) vertical to the point where the curve crosses \(0\) dB. The resonant peak is the highest M contour the curve touches, and the frequency marked there is \(\omega_r\). The bandwidth is the frequency at which the curve crosses the \(-3\) dB contour on its way down. All four come from one drawing.

phase −180° 0 dB +10 −10 −20 −270° −90° M = 3 dB 6 dB 0 dB G(jω) PM GM critical point
The Nichols chart: gain in dB against phase, with closed-loop M contours around (−180°, 0 dB)
One picture, both stability views. The gain margin and phase margin of Chapter 17 and the resonant peak of this chapter are legs of the same triangle on the Nichols chart: a curve that keeps its distance from \((-180^{\circ},0\text{ dB})\) has generous margins and a low \(M_r\). Relative stability measured by margins and relative stability measured by closed-loop peaking are two projections of one geometric fact.
Section 18-7

Resonant Peak, Resonant Frequency and Bandwidth

Three numbers summarise a closed-loop frequency response, and each has a plain engineering meaning.

The resonant peak \(M_r\) is the maximum of \(|T(j\omega)|\). It measures relative stability: a system with \(M_r=1.05\) is heavily damped and dull, one with \(M_r=3\) rings badly, and a design target of \(1.1\) to \(1.5\) — roughly \(0.8\) to \(3.5\) dB — is usual. The resonant frequency \(\omega_r\) is where that peak occurs and is a measure of speed. The bandwidth \(\omega_b\) is the frequency at which \(|T|\) has fallen to \(1/\sqrt2\) of its low-frequency value, that is by \(3\) dB; beyond it the system effectively stops responding. Bandwidth is the most useful single number in the set, because it governs both the speed of the transient and the amount of measurement noise the loop lets through.

For the standard second-order system of Chapter 8 all three follow in closed form. With \(u=\omega/\omega_n\),

Second-order closed-loop magnitude
\[ T(j\omega)=\frac{\omega_n^2}{(j\omega)^2+2\zeta\omega_n(j\omega)+\omega_n^2} \;\Longrightarrow\; |T| = \frac{1}{\sqrt{(1-u^2)^2+(2\zeta u)^2}} \]

The magnitude peaks where the quantity under the root is smallest. Differentiating that quantity with respect to \(u\) and setting the result to zero locates the peak.

Locating the peak
\[ \frac{d}{du}\left[(1-u^2)^2+4\zeta^2u^2\right] = -4u(1-u^2)+8\zeta^2u = 4u\left[u^2-1+2\zeta^2\right] = 0 \]
\[ u_r^2 = 1-2\zeta^2 \;\Longrightarrow\; \omega_r = \omega_n\sqrt{1-2\zeta^2} \]

The root \(u=0\) is the trivial one. The useful root exists only while \(1-2\zeta^2>0\), so a peak appears only for \(\zeta<1/\sqrt2\approx0.707\); above that damping the magnitude falls monotonically from unity and there is no resonance at all. Substituting \(u_r^2=1-2\zeta^2\) back into the denominator gives \((2\zeta^2)^2+4\zeta^2(1-2\zeta^2)=4\zeta^2(1-\zeta^2)\), so the peak height is

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Peak and its location
\[ M_r = \frac{1}{2\zeta\sqrt{1-\zeta^2}}\quad(\zeta \le 0.707), \qquad \omega_r=\omega_n\sqrt{1-2\zeta^2} \]

\(M_r\) depends on \(\zeta\) alone — it carries no information about speed. \(\omega_r\) carries the speed. The pair separates the two halves of the transient specification exactly as \(\zeta\) and \(\omega_n\) did in Chapter 8, which is no coincidence: they are the same two numbers wearing frequency-domain clothes.

Bandwidth follows from setting \(|T|=1/\sqrt2\), which requires the denominator squared to equal 2:

Bandwidth of the second-order system
\[ (1-u^2)^2+4\zeta^2u^2 = 2 \;\Longrightarrow\; u^4 - 2u^2\left(1-2\zeta^2\right) - 1 = 0 \]
\[ \omega_b = \omega_n\left[\left(1-2\zeta^2\right)+\sqrt{4\zeta^4-4\zeta^2+2}\right]^{1/2} \]

Only the positive root of the quadratic in \(u^2\) is physical. At \(\zeta=0.707\) the bracket evaluates to \(0+1=1\), so \(\omega_b=\omega_n\) exactly; at \(\zeta=0.5\) it gives \(\omega_b=1.272\,\omega_n\). Across the whole useful damping range \(\omega_b\) sits between about one and one-and-a-half times \(\omega_n\), which is why bandwidth is so often quoted as a proxy for natural frequency.

ω (log) |T| dB 0 dB −3 dB ωr ωb Mr cut-off slope
The three closed-loop specifications read from one magnitude curve

A fourth quantity, the cut-off rate, is the slope of the magnitude curve beyond \(\omega_b\). A steep cut-off is good for rejecting high-frequency noise, but steepness and peaking travel together: a sharp corner at the band edge almost always implies a large \(M_r\). The design compromise is to keep the cut-off as steep as the peak specification allows, and no steeper.

Section 18-8

Correlation with the Time Domain

The frequency-domain numbers are only worth having if they predict the step response, and for a dominant second-order pair they do so exactly. Two more relations complete the dictionary. The gain-crossover frequency of the open loop \(G(s)=\omega_n^2/[s(s+2\zeta\omega_n)]\) satisfies \(\omega_g=\omega_n\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}\), and the phase margin measured there is

Phase margin of the standard second-order loop
\[ \gamma = 90^{\circ} - \tan^{-1}\!\frac{\omega_g}{2\zeta\omega_n} = \tan^{-1}\frac{2\zeta}{\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}} \]
\(\zeta\)Phase margin \(\gamma\)\(M_r\)\(M_r\) (dB)\(\omega_r/\omega_n\)\(\omega_b/\omega_n\)Overshoot
0.222.6°2.5528.140.9591.51052.7 %
0.333.3°1.7474.850.9061.45437.2 %
0.443.1°1.3642.690.8251.37525.4 %
0.551.8°1.1551.250.7071.27216.3 %
0.659.2°1.0420.350.5291.1489.5 %
0.70765.5°1.0000.0001.0004.3 %

Read down the table and the story is one story. As damping rises the phase margin widens, the resonant peak falls towards unity, the resonance migrates to lower frequency and finally disappears, the bandwidth shrinks, and the overshoot subsides. Every column is a different instrument reading the same physical fact.

Two rules of thumb come out of the numbers and are worth carrying. For phase margins up to about \(60^{\circ}\), \(\zeta \approx \gamma/100\) with \(\gamma\) in degrees — at \(\gamma=51.8^{\circ}\) the true value is \(0.5\) against an estimate of \(0.518\). And a rough peak estimate \(M_r \approx 1/\sin\gamma\) is often quoted; it errs on the pessimistic side, giving \(1.27\) where the exact value is \(1.155\) at \(\zeta=0.5\), which makes it a safe design bound rather than a prediction.

Where the exactness ends. Every entry in that table is derived for an all-pole second-order system with no finite zeros. A real loop has extra poles and often a zero, and the numbers become estimates rather than identities. They stay trustworthy whenever a dominant complex pair exists — the condition Chapter 8 imposed for its own formulas — and they should be checked against a simulation whenever it does not. What survives in every case is the ordering: more peaking always means less damping, and more bandwidth always means a faster and noisier loop.

The practical attraction of these measures is that they can be obtained on hardware without a model. Drive the closed loop with a sinusoidal reference, sweep the frequency, and record the amplitude ratio. The peak of that record is \(M_r\), its location is \(\omega_r\), and the \(-3\) dB crossing is \(\omega_b\). No transfer function need ever be identified — which is precisely why frequency-domain specification survived into an era of computers, and why the Nichols chart is still the surface on which the lead and lag compensators of Part 5 are designed.

Section 18-9

Worked Examples

1 Locating a constant-M circle

Problem. Find the centre and radius of the constant-magnitude locus for \(M=2\), and state what it means for a polar plot to be tangent to it.

Solution. Substitute \(M=2\) into the centre and radius formulas of Section 18-3:

Working
\[ \text{centre} = -\frac{M^2}{M^2-1} = -\frac{4}{3} = -1.333, \qquad \text{radius} = \frac{M}{M^2-1} = \frac{2}{3} = 0.667 \]

The circle therefore runs from \(-2.000\) to \(-0.667\) on the real axis, wrapped around \(-1\). A polar plot tangent to it has \(M_r = 2\), or \(6.02\) dB — an overshoot of roughly 40 % if a dominant second-order pair applies, since \(M_r=2\) corresponds to \(\zeta\approx0.26\).

2 A constant-N circle

Problem. Find the centre and radius of the locus along which the closed-loop phase is \(\alpha = 30^{\circ}\), and verify that it passes through the origin.

Solution. Here \(N=\tan 30^{\circ}=0.5774\), so \(1/N^2 = 3\) exactly:

Working
\[ \text{centre} = \left(-\tfrac12,\; \frac{1}{2(0.5774)}\right) = (-0.5,\,0.866), \qquad r = \tfrac12\sqrt{1+3} = 1.000 \]

The distance from the centre to the origin is \(\sqrt{0.5^2+0.866^2}=\sqrt{0.25+0.75}=1.000\), equal to the radius, so the origin lies on the circle. The same computation with the point \(-1\) gives \(\sqrt{(-1+0.5)^2+0.866^2}=1.000\) as well, confirming the general property.

3 Closed-loop value from one open-loop point

Problem. For \(G(s)=\dfrac{4}{s(s+2)}\) with unity feedback, evaluate \(G(j2)\) and use it to obtain the closed-loop magnitude and phase at \(\omega=2\) rad/s.

Solution. Evaluate the open-loop point first, rationalising the denominator:

Working — the open-loop point
\[ G(j2)=\frac{4}{j2(j2+2)}=\frac{4}{-4+j4}=\frac{4(-4-j4)}{32}=-0.5-j0.5 \]

So \(X=-0.5\), \(Y=-0.5\). Because \(X=-\tfrac12\), the point lies exactly on the \(M=1\) line, and no further arithmetic is needed for the magnitude: \(M=1\). For the phase, use the N relation:

Working — the closed-loop phase
\[ N=\frac{Y}{X^2+X+Y^2}=\frac{-0.5}{0.25-0.5+0.25}=\frac{-0.5}{0}\to\infty \;\Longrightarrow\; \alpha=-90^{\circ} \]

Checking directly, \(T(s)=4/(s^2+2s+4)\) gives \(T(j2)=4/(-4+j4+4)=4/(j4)=-j\), of magnitude 1 and phase \(-90^{\circ}\). The two routes agree. This is the frequency at which the loop is driven at its natural frequency \(\omega_n=2\), where the closed-loop phase lag is always exactly a quarter cycle.

4 The three specifications for a second-order loop

Problem. For \(G(s)=\dfrac{25}{s(s+6)}\) with unity feedback, find \(\zeta\), \(M_r\), \(\omega_r\) and \(\omega_b\).

Solution. The closed loop is \(T(s)=25/(s^2+6s+25)\), so \(\omega_n=5\) rad/s and \(2\zeta\omega_n=6\), giving \(\zeta=0.6\).

Working
\[ M_r = \frac{1}{2(0.6)\sqrt{1-0.36}} = \frac{1}{1.2\times0.8}=1.042 \;\;(0.35\text{ dB}) \]
\[ \omega_r = 5\sqrt{1-2(0.36)} = 5\sqrt{0.28}=2.646\ \text{rad/s} \]
\[ \omega_b = 5\left[0.28+\sqrt{4(0.1296)-4(0.36)+2}\right]^{1/2} = 5\sqrt{1.3185}=5.741\ \text{rad/s} \]

A peak of only \(0.35\) dB confirms a well-damped loop; the phase margin, from the table of Section 18-8, is \(59.2^{\circ}\), and the step overshoot about 9.5 %.

5 Choosing the gain for a specified peak

Problem. For \(G(s)=\dfrac{K}{s(s+4)}\), find \(K\) so that \(M_r=1.3\). What tangent-line angle would the polar-plot construction of Section 18-5 use?

Solution. Invert the peak formula. Writing \(x=\zeta^2\), the condition \(2\zeta\sqrt{1-\zeta^2}=1/1.3=0.7692\) squares to \(4x(1-x)=0.5917\):

Working — the required damping
\[ 4x^2-4x+0.5917=0 \;\Longrightarrow\; x=\frac{4-\sqrt{16-9.468}}{8}=\frac{4-2.556}{8}=0.1805 \]
\[ \zeta=\sqrt{0.1805}=0.4249 \]

The smaller root is taken because a resonant peak requires \(\zeta<0.707\). The closed loop is \(T=K/(s^2+4s+K)\), so \(\omega_n=\sqrt K\) and \(2\zeta\omega_n=4\):

Working — the gain
\[ \omega_n = \frac{4}{2(0.4249)} = 4.707 \;\Longrightarrow\; K=\omega_n^2 = 22.2 \]

The resonance then sits at \(\omega_r=4.707\sqrt{1-2(0.1805)}=3.76\) rad/s. On a polar plot the same design would be done by drawing the tangent line at \(\psi=\sin^{-1}(1/1.3)=50.3^{\circ}\) below the negative real axis and scaling the unit-gain plot until it touched the circle tangent to that line.

6 Reading a third-order loop completely

Problem. For \(G(s)=\dfrac{10}{s(s+1)(s+5)}\) with unity feedback, find the gain margin, the phase margin, the resonant peak and the bandwidth.

Solution. The phase reaches \(-180^{\circ}\) when \(\tan^{-1}\omega+\tan^{-1}(\omega/5)=90^{\circ}\), that is when \(\omega\cdot(\omega/5)=1\), so \(\omega=\sqrt5=2.236\) rad/s. There

Working — gain margin
\[ |G| = \frac{10}{2.236\sqrt{6}\sqrt{30}} = \frac{10}{30}=0.333 \;\Longrightarrow\; \text{GM}=20\log_{10}3 = 9.54\ \text{dB} \]

Gain crossover occurs where \(\omega\sqrt{\omega^2+1}\sqrt{\omega^2+25}=10\), which solves numerically to \(\omega_g=1.227\) rad/s. The phase there is \(-90^{\circ}-50.8^{\circ}-13.8^{\circ}=-154.6^{\circ}\), so the phase margin is \(25.4^{\circ}\). For the closed loop, \(T=10/(s^3+6s^2+5s+10)\), whose magnitude is \(10/|(10-6\omega^2)+j(5\omega-\omega^3)|\):

Working — peak and bandwidth
\[ \omega=1.30:\quad |T| = \frac{10}{|-0.14+j4.303|}=\frac{10}{4.305}=2.32 \;\;(7.3\text{ dB}) \]
\[ \omega=2.00:\quad |T| = \frac{10}{|-14+j2|}=\frac{10}{\sqrt{200}}=0.7071 \;\;(-3.01\text{ dB}) \]

So \(M_r=2.32\) at \(\omega_r=1.30\) rad/s and \(\omega_b=2.00\) rad/s. The narrow phase margin of \(25^{\circ}\) and the large peak of \(7.3\) dB are the same warning stated twice: on the Nichols chart this curve pushes inside the \(6\) dB contour. Compare with the estimate \(M_r\approx1/\sin 25.4^{\circ}=2.33\), which here is almost exact.

Review

Chapter Summary

M as a ratio

\(M=|T|\) is the distance from the origin divided by the distance from \(-1\).

M circles

Centre \(-M^2/(M^2-1)\), radius \(|M/(M^2-1)|\); \(M=1\) is the line \(X=-0.5\).

N circles

Centre \((-0.5,\,1/2N)\), radius \(\tfrac12\sqrt{1+1/N^2}\); all pass through \(0\) and \(-1\).

Nichols chart

dB against phase; a gain change is a rigid vertical shift of \(20\log K\).

Peak and speed

\(M_r=1/(2\zeta\sqrt{1-\zeta^2})\), \(\omega_r=\omega_n\sqrt{1-2\zeta^2}\), peak only for \(\zeta<0.707\).

Bandwidth

The \(-3\) dB frequency; larger means faster and noisier, near \(\omega_n\) for usual \(\zeta\).

Practice

Practice Problems

Assume unity feedback and a stable closed loop throughout. For the second-order problems, identify \(\zeta\) and \(\omega_n\) before reaching for any formula; for the rest, sketch the geometry before computing.

  1. Find the centre and radius of the constant-magnitude circle for \(M=1.2\), and state on which side of the line \(X=-0.5\) it lies.
  2. Starting from \(M^2\left[(1+X)^2+Y^2\right]=X^2+Y^2\), show that \(M=1\) gives the straight line \(X=-\tfrac12\), and explain in words why the constant-magnitude locus for \(M=1\) must be the perpendicular bisector of the segment joining \(0\) and \(-1\).
  3. Find the centre and radius of the constant-phase circle for \(\alpha=45^{\circ}\), and verify by substitution that both \(0\) and \(-1\) lie on it.
  4. For \(G(s)=\dfrac{16}{s(s+4)}\), find \(\zeta\), \(M_r\), \(\omega_r\) and \(\omega_b\).
  5. A unity-feedback loop has a phase margin of \(45^{\circ}\). Estimate \(\zeta\), then \(M_r\) and the step overshoot, and say which of the three estimates you would trust least.
  6. For \(G(s)=\dfrac{K}{s(s+5)}\), choose \(K\) so that \(M_r=1.25\), and state the resulting \(\omega_r\) and \(\omega_b\).
  7. A Nichols plot crosses the \(-180^{\circ}\) line at \(-8\) dB and crosses \(0\) dB at a phase of \(-140^{\circ}\). State the gain margin and phase margin, and argue whether \(M_r\) is likely to exceed \(2\) dB.
  8. Explain why increasing \(K\) is a rigid vertical shift on the Nichols chart but a radial rescaling on the polar plot, and why that difference makes the Nichols chart the better surface for gain design.
Tip: when a problem gives you an open-loop transfer function and asks for a closed-loop quantity, decide first whether a dominant second-order pair exists. If it does, the formulas of Section 18-7 give \(M_r\), \(\omega_r\) and \(\omega_b\) in three lines and no chart is needed. If it does not — three poles of comparable magnitude, or a nearby zero — form \(T(j\omega)\) explicitly at a handful of frequencies around the suspected peak and read the maximum off the numbers, exactly as in Example 6. The charts are for design, where the gain is unknown; for analysis, where everything is known, arithmetic is faster.