Bode Sensitivity

Consider again DC motor model with no disturbance (left side is open-loop version, right side is closed-loop version):

In the “nominal” situation, we have the motor with DC gain of $A$, and the overall transfer function, either open-loop or closed-loop, has some other DC gain, which we call $T$.

Suppose that, due to modeling error, changes in operating conditions, etc, the motor gain changes so that we have

$$A⟶A+\underset{\text{small perturbation}}{\underbrace{\delta A}}$$

This will cause a perturbation in the overall DC gain:

$$T⟶T+\delta T$$

The Bode sensitivity is defined as

$$\begin{array}{rl}S& =\frac{\delta T/T}{\delta A/A}\\ & =\text{relative error}\\ & =\frac{\text{normalized (percentage) error in }T}{\text{normalized (percentage) error in }A}\end{array}$$

Motor Example

Open-loop

In the nominal case, we have $T_{\text{ol}}=K_{\text{ol}}A=\frac{1}{A}A=1$.

In the perturbed case, we have:

$$\begin{array}{rl}A& ⟶A+\delta A\\ T_{\text{ol}}& ⟶K_{\text{ol}}(A+\delta A)=\frac{1}{A}(A+\delta A)=\underset{T_{\text{ol}}}{\underbrace{1}}+\underset{\delta T_{\text{ol}}}{\underbrace{\frac{\delta A}{A}}}\end{array}$$

Thus, the sensitivity is

$$S_{\text{ol}}=\frac{\delta T_{\text{ol}}/T_{\text{ol}}}{\delta A_{\text{ol}}/A_{\text{ol}}}=\frac{\delta A/A}{\delta A/A}=1$$

which means that a 5% error in $A$ will cause a 5% error in $T_{\text{ol}}$.

Closed-loop:

In the nominal case, we have

$$T_{\text{cl}}=\frac{A{K}_{\text{cl}}}{1+A{K}_{\text{cl}}}$$

In the perturbed case, we have

$$\begin{array}{rl}A& ⟶A+\delta A\\ T& ⟶T_{\text{cl}}+\delta T_{\text{cl}}\end{array}$$

How do we compute $\delta T_{\text{cl}}$? We can use a Taylor expansion:

$$T(A+\delta A)=T(A)+\frac{dT}{dA}(A)\delta A+\text{higher order terms}$$

In our case:

$$\begin{array}{rl}\frac{d{T}_{\text{cl}}}{dA}& =\frac{K_{\text{cl}}}{1+A{K}_{\text{cl}}}-\frac{A{K}_{\text{cl}}^2}{(1+A{K}_{\text{cl}}{)}^2}\\ & =\frac{K_{\text{cl}}}{(1+A{K}_{\text{cl}}{)}^2}\end{array}$$

which gives

$$\delta T_{\text{cl}}=\frac{K_{\text{cl}}}{(1+A{K}_{\text{cl}})}\delta A$$

Therefore, we have

$$\delta T_{\text{cl}}/T_{\text{cl}}=\frac{\frac{K_{\text{cl}}}{(1+A{K}_{\text{cl}})}\delta A}{\frac{A{K}_{\text{cl}}}{1+A{K}_{\text{cl}}}}=\frac{1}{1+A{K}_{\text{cl}}}\delta A/A$$

which finally gives us the sensitivity:

$$S_{\text{cl}}=\frac{\delta T_{\text{cl}}/T_{\text{cl}}}{\delta A/A}=\frac{1}{1+A{K}_{\text{cl}}}$$

From this we can conclude that for high gains $K_{\text{cl}}$, we get smaller relative error due to parameter variations in the plant model, $S_{\text{cl}}\ll 1$.