FOUNDATION · CHAPTER 1 OF 8

What one feedback loop does

Derives the gain of one negative-feedback loop from its two summing-junction equations. Then shows what the loop does to a drifting gain, a disturbance and sensor noise, and what high loop gain can cost. These results apply wherever an engineer closes a loop.

Updated 13 Oct 2026

What you will be able to do

You will derive the closed-loop gain of a feedback loop from its two summing-junction equations. You will see how loop gain shrinks the effect of a drifting gain and of a disturbance. You will see why the sensor's own error and noise reach the output almost untouched. You will see what high loop gain can cost: in some loops it brings instability.

Before you start

Solving two linear equations in two unknowns
The quotient rule for derivatives
Complex numbers: magnitude, angle and cube roots
The Laplace transform; a transfer function $G (s) = Y (s) / U (s)$ , output over input with zero initial conditions
Poles of a rational transfer function with real coefficients (complex poles come in conjugate pairs); BIBO stability, bounded input giving bounded output, which holds exactly when every pole is in the open left half-plane

The loop in two equations

A negative-feedback loop measures the output through a sensor and subtracts the sensor signal from the reference. What is left drives the forward path. The output is the value at which both statements hold. With transfer functions the algebra is the same: at each $s$ away from poles, every transfer function is just a complex number.

THE TWO SUMMING-JUNCTION EQUATIONS

E C = R - H C = G E

Negative feedback, zero initial conditions, inputs applied at $t = 0$ ; unilateral Laplace transform, lower limit $0^{-}$ (for a signal zero before $t = 0$ the bilateral form agrees); $s = σ + j ω$ , $j$ the imaginary unit. $R$ reference, $C$ output, $E$ actuating signal, $G$ forward gain, $H$ sensor gain. With $H = 1$ , $E (s)$ is the transform of the error $e (t) = r (t) - c (t)$ . Example signals are in volts; constant gains are dimensionless.

CLOSED-LOOP GAIN

C C + G H C T = \frac{C}{R} = G (R - H C) = GR - G H C = GR = \frac{G}{1 + G H}

Holds when $1 + G H \neq = 0$ . $T$ is the closed-loop gain. The product $L = G H$ is the loop gain: the gain met once round the loop, dimensionless.

GAIN WHEN THE LOOP GAIN IS LARGE

T = \frac{1}{H} \cdot \frac{G H}{1 + G H} \approx \frac{1}{H}

The first form is exact for $H \neq = 0$ ; the approximation needs $G H ≫ 1$ . With positive constant gains, $T$ falls short of $1/ H$ by the fraction $1/ (1 + G H)$ of $1/ H$ . $G$ has nearly dropped out: the sensor gain sets the output.

WATCH OUT FOR

Loop gain is not $G$ . The test: multiply every gain met once round the loop. With $G = 100$ and $H = 0.1$ , $L = G H = 10$ , not 100.

WORKED EXAMPLE

A loop with negative feedback has $G = 100$ and $H = 0.1$ . The reference is $R = 1 V$ . Find $T$ , $C$ and $E$ , and check the result against $C = GE$ .

Solution

Loop gain: $L = G H = 100 \times 0.1 = 10$ . This is the number that appears in the denominator.
Closed-loop gain: $T = G / (1 + G H) = 100/11 = 9.09$ . Feedback has cut the gain from 100 to 9.09.
Output: $C = T R = 9.09 V$ , since $T$ is output over reference.
Actuating signal: $E = R - H C = 1 - 0.1 \times 9.09 = 1 - 0.909 = 0.0909 V$ : the reference minus the sensor signal.
Check: $GE = 100 \times 0.0909 = 9.09 V = C$ . Both equations hold at once, as they must.
Compare with $1/ H = 10$ : $T$ is 9.1% below it, the fraction $1/ (1 + L) = 1/11$ .

Answer: $T = 9.09$ , $C = 9.09 V$ , $E = 0.0909 V$ . A small actuating signal, amplified by $G = 100$ , gives the full output.

How much a drifting gain matters

Forward gains drift with temperature, supply and age. Sensitivity asks: if $G$ changes by a small fraction, by what fraction does $T$ change? It is the ratio of the two fractional changes, found by differentiating $T$ with respect to $G$ .

SENSITIVITY OF THE CLOSED-LOOP GAIN TO G

\frac{d T}{d G} S = \frac{d T / T}{d G / G} = \frac{( 1 + G H ) - G H}{( 1 + G H ) ^{2}} = \frac{1}{( 1 + G H ) ^{2}} = \frac{d T}{d G} \cdot \frac{G}{T} = \frac{1}{( 1 + G H ) ^{2}} \cdot \frac{G ( 1 + G H )}{G} = \frac{1}{1 + G H}

Small changes of $G$ with $H$ fixed; constant gains, $G \neq = 0$ . $S$ is dimensionless. With no feedback ( $H = 0$ ), $S = 1$ : $T$ drifts one for one with $G$ .

SENSITIVITY OF THE CLOSED-LOOP GAIN TO H

\frac{d T / T}{d H / H} = \frac{d T}{d H} \cdot \frac{H}{T} = - \frac{G ^{2}}{( 1 + G H ) ^{2}} \cdot \frac{H ( 1 + G H )}{G} = - \frac{G H}{1 + G H}

Small changes of $H$ with $G$ fixed; constant gains, $G \neq = 0$ . For $G H ≫ 1$ this is close to $- 1$ : a 1% error in the sensor gain gives about a 1% error in $T$ , in the opposite direction. The loop cannot correct what it measures wrongly.

Forward gain $G$	Change in $G$	Closed-loop $T$ ( $H = 0.1$ )	Change in $T$
100	0	9.091	0
80	-20%	8.889	-2.2%
120	+20%	9.231	+1.5%

A 20% fall in $G$ moves $T$ by only 2.2%, and a 20% rise by 1.5%. The sensitivity at $G = 100$ , $S = 1/11$ , predicts 20% divided by 11, that is 1.8%, for both. The figures differ because $S$ is the slope at one point and $T$ curves.

WATCH OUT FOR

Using $S$ for a large change. The test: recompute $T$ at the new $G$ and compare, as the table does.

Disturbance, sensor noise and the price

A disturbance enters the loop after some forward gain, for instance a load acting on the plant. The loop sees its effect at the output and pushes back. Sensor noise is different. It enters at the sensor, so the loop takes it for a true reading of the output and obeys it. Both transfer functions follow from the same two equations, with the extra signal added at its own point.

DISTURBANCE TO OUTPUT

C C C (1 + G_{1} G_{2} H) \frac{C}{D} = G_{2} (G_{1} E + D), E = - H C = - G_{1} G_{2} H C + G_{2} D = G_{2} D = \frac{G _{2}}{1 + G _{1} G _{2} H}

$R = 0$ ; the loop is linear, so the effects of $R$ and $D$ add. Holds when $1 + G_{1} G_{2} H \neq = 0$ . $D$ (volts) is added at the input of the plant $G_{2}$ ; $G_{1}$ is the controller before it. Loop gain $L = G_{1} G_{2} H$ . With no feedback ( $H = 0$ ), $C / D = G_{2}$ : the loop divides the disturbance's effect by $1 + L$ .

SENSOR NOISE TO OUTPUT

C C (1 + G H) C = G (R - H C - N) = GR - G H C - GN = GR - GN = T (R - N)

Holds when $1 + G H \neq = 0$ . $N$ (volts) is added to the sensor output $H C$ . The output carries $- N$ with exactly the gain $T$ that it gives $R$ , so $C / N = - T$ . For $G H ≫ 1$ , $C \approx - N / H$ : the sensor's error, read in output units, reaches the output in full.

WORKED EXAMPLE

A loop with negative feedback has controller $G_{1} = 10$ , plant $G_{2} = 10$ and sensor $H = 0.1$ . A disturbance $D = 1 V$ is added at the plant input, between controller and plant. Separately, noise $N = 0.1 V$ is added to the sensor output. With $R = 0$ , find the output in each case.

Solution

Loop gain: $G_{1} G_{2} H = 10 \times 10 \times 0.1 = 10$ , so $1 + L = 11$ . Both transfer functions share this denominator.
Disturbance: $C / D = G_{2} / (1 + L) = 10/11 = 0.909$ . So $C = 0.909 V$ .
Without feedback the same disturbance would give $C = G_{2} D = 10 V$ , so the loop has cut its effect by a factor of 11.
Noise: the forward path is $G = G_{1} G_{2} = 100$ , so $C / N = - T = - 100/11 = - 9.09$ .
So $C = - 9.09 \times 0.1 = - 0.909 V$ . The noise is a tenth the size of the disturbance but moves the output just as far.

Answer: Disturbance: $C = 0.909 V$ . Noise: $C = - 0.909 V$ . The loop suppresses the disturbance but passes the noise at gain $T$ .

WATCH OUT FOR

Thinking more loop gain quiets sensor noise too. The test: write $C / N$ and compare it with $T$ . They are equal in size, and both stay near $1/ H$ for any $L ≫ 1$ .

Each benefit above is a division by $1 + G H$ , so it is tempting to raise the gain without limit. A real forward path has lags, so $G H$ depends on $s$ . Suppose $1 + G H = 0$ at some $s = j ω$ ( $σ = 0$ ) with $G$ and $H$ finite there: then $G H = - 1$ , so $G \neq = 0$ , nothing cancels, and $T = G / (1 + G H)$ has a pole on the $j ω$ axis. For $ω > 0$ its conjugate sits at $- j ω$ , and a simple pair puts an undamped sinusoid at $ω$ in the response.

CONDITION FOR A CLOSED-LOOP POLE ON THE IMAGINARY AXIS

1 + G H (j ω) = 0 ⟺ ∣ G H (j ω) ∣ = 1 and ∠ G H (j ω) = - 18 0^{\circ}

$G$ , $H$ finite at $s = j ω$ , $ω$ in rad/s; the angle is taken modulo $36 0^{\circ}$ and written as a lag. LHP and RHP are the open left and right half-planes; the $j ω$ axis is in neither. No closed-loop pole in the RHP and only simple poles on the axis: the loop is marginally stable, which is not BIBO stable (a pole lies outside the LHP).

THREE EQUAL LAGS WITH UNITY FEEDBACK

G H (j ω) ∠ G H ∣ G H ∣ = \frac{K}{( 1 + j ω ) ^{3}}, ∣1 + j ω ∣ = 1 + ω^{2}, ∠ (1 + j ω) = tan^{- 1} ω = - 3 tan^{- 1} ω = - 18 0^{\circ} \Rightarrow tan^{- 1} ω = 6 0^{\circ} \Rightarrow ω = 3 = \frac{K}{( 1 + ω ^{2} ) ^{3/2}} = \frac{K}{8} = 1 \Rightarrow K = 8

$H = 1$ , $G = K / (s + 1)^{3}$ with $K \geq 0$ the loop gain at $s = 0$ (dimensionless); $ω > 0$ in rad/s, so $tan^{- 1} ω$ lies between $0^{\circ}$ and $9 0^{\circ}$ . The closed-loop poles solve $(s + 1)^{3} = - K$ : $s + 1$ has size $K^{1/3}$ at angles $6 0^{\circ}$ , $18 0^{\circ}$ and $30 0^{\circ}$ .

$K$	Closed-loop poles	Stability
0	-1 (three times)	stable
2	-2.26 and -0.37 ± j1.09	stable
8	-3 and ± j1.73	marginally stable, not BIBO stable
16	-3.52 and 0.26 ± j2.18	unstable, two poles in the RHP

As $K$ rises, the complex pair moves right and crosses the $j ω$ axis at $K = 8$ , as the condition predicts. Not every loop does this. With $G H = K / (s + a)$ or $K / ((s + a) (s + b))$ , $a, b > 0$ , the angle of $G H$ stays above $- 18 0^{\circ}$ at every finite $ω$ and $G H (0) > 0$ , so the poles start in the LHP at $K = 0$ and never reach the axis, for any $K \geq 0$ . In the three-lag loop, whose lag reaches $18 0^{\circ}$ , high loop gain buys insensitivity to $G$ and to disturbances at the cost of stability.

Check yourself

CHECK YOURSELF 1

A loop with negative feedback has constant gains $G = 50$ and $H = 0.2$ . What is the closed-loop gain $T = C / R$ ?

(A)
4.55
(B)
5.00
(C)
10.0
(D)
50

Answer and solution

Answer: (A)

Loop gain: $G H = 50 \times 0.2 = 10$ .
$T = G / (1 + G H) = 50/11 = 4.545$ .
The value $1/ H = 5$ is only the large-loop-gain approximation; $T$ falls short of it by the fraction $1/11$ , so B is not exact.

CHECK YOURSELF 2

A loop with negative feedback has constant gains $G = 200$ and $H = 0.05$ . Find the sensitivity $S = (d T / T) / (d G / G)$ of the closed-loop gain to $G$ . Give a number.

Answer and solution

Answer: $0.0909$ (accepted range $\pm 0.001$ )

$G H = 200 \times 0.05 = 10$ .
$S = 1/ (1 + G H) = 1/11$ .
$S = 0.0909$ .

CHECK YOURSELF 3

A loop with negative feedback has controller $G_{1} = 20$ , plant $G_{2} = 5$ and sensor $H = 0.5$ . A disturbance $D$ is added at the plant input, between controller and plant. With $R = 0$ , find $C / D$ . Give a number.

Answer and solution

Answer: $0.098$ (accepted range $\pm 0.001$ )

Loop gain: $G_{1} G_{2} H = 20 \times 5 \times 0.5 = 50$ .
$C / D = G_{2} / (1 + G_{1} G_{2} H) = 5/51$ .
$C / D = 0.0980$ .

What one feedback loop does

What you will be able to do

Before you start

The loop in two equations

How much a drifting gain matters

Disturbance, sensor noise and the price

Check yourself

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