how to interpret results from circuit sensitivity analysis

Question

If you do a search on circuit sensitivity, you come across a presentation from Ti that discusses it, Circuit Sensitivity.

In it, for example, they solve for the Q of a filter, and perform the sensitivity analysis that shows the sensitivity of Q to the C and L in the circuit. The analysis results in , for the sensitivity of Q to changes in C1, for example.

They go on to say that this result is the exponent in the equation for Q, . This is not directly obvious to me.

Although I can see that 1/2 is the exponent of (C1/L1), I don't understand the interpretation. If the change in Q due to a change in C1 is 1/2 (and -1/2 due to a change in L), I would expect, based on this result, that if C1 changes by 2, Q changes by 1, if it changes by 4, then Q doubles, etc.

I realize, however, that if I look a the equation for Q, that if C1 is doubled, Q increases by the sqrt(2) (and that it also decreases by the sqrt(x), where x is any multiple of L). However, it is not obvious to me why the result from the sensitivity analysis, , "tells" me this (for C1, for example). How would I make this connection, say, in more complex transfer functions or equations? In the case of this Q equation, it seems to me I'm better off just using the equation directly.

I would appreciate it if someone could explain why it is obvious that the result in this case applies an exponent, and not some other proportionality. How does one guarantee the proper interpretation of the result of the sensitivity analysis; does one need to consider the result in the context of the original equation?

Thank you!

Answer 1

Abstract

I don't remember when someone popularized the idea of sensitivity. But it is a really simple idea to gather. What you want to know is, "by what % will \$y\$ change if I change \$x\$ by some %?" Both of these are relative measures, but really important because parts, like resistors, will have specifications in percentages. So you may know the range of possible %-changes in something and wish to know how that may impact the %-change in something else.

Discussion

A %-change is defined in algebra as \$\frac{\Delta\,x}{x}\$. But that uses finite values and can only be an approximation, at best. Instead, calculus refines this and makes it perfectly accurate by instead saying \$\frac{\text{d}\,x}{x}\$. These are essentially similar, except that you don't need the limit statement saying, as \$\Delta\,x\to 0\$, anymore. (Here, we are saying, "take an infinitesimal change in \$x\$ and divide that by the current value of \$x\$." This is always a "tiny" percentage value -- in fact, less than any finite percent change but more than zero.)

So, if you want to find out how one thing changes (in %-terms) when something else changes (in %-terms), then you want to work out:

$$S^y_x=\frac{\frac{\text{d}\,y}{y}}{\frac{\text{d}\,x}{x}}=\frac{\text{d}\,y}{\text{d}\,x}\cdot\frac{x}{y}$$

You can later re-organize the above into \$\frac{\text{d}\,y}{y}=S^y_x\cdot \frac{\text{d}\,x}{x}\$, which is useful because once you have \$S^y_x\$ you are able to compute:

$$\%\text{-change}\:y=S^y_x\cdot \%\text{-change}\:x$$

A general example from your linked page

So, let's take their case where \$Q=R\sqrt{\frac{C}{L}}\$. You should be able to readily find that:

$$\begin{align*} \frac{\text{d}\, Q}{\text{d}\,C}&=\frac{R\,\sqrt{\frac{C}{L}}}{2\,C}=\frac12\,\frac{Q}{C}\\\\&\therefore\\\\S^{^\text{Q}}_{_\text{C}}&=\frac{\frac{\text{d}\,Q}{Q}}{\frac{\text{d}\,C}{C}}=\frac{\text{d}\,Q}{\text{d}\,C}\cdot\frac{C}{Q}=\frac12\,\frac{Q}{C}\cdot\frac{C}{Q}=\frac12 \end{align*}$$

This just means that:

$$\%\text{-change}\:Q=\frac12\cdot \%\text{-change}\:C$$

That's the interpretation.

This only works for relatively small %-change values. (If you are using large-scale changes then the above formula very well may not apply.)

Interpretation of your linked document

Let's take their case in point. From page 12, you have \$C=160\:\text{nF}\$, \$L=1.6\:\text{mH}\$, and \$R=70.7\:\Omega\$. Their transfer function on page 12 is correct, but it isn't in standard form. To get closer, first divide through by \$L\,C\$:

$$G_s=\frac1{s^2+\frac1{R\,C}s+\frac1{L\,C}}$$

Setting \$\omega_{_0}=\frac1{\sqrt{L\,C}}\$ and \$\alpha=\frac12\,\frac1{R\,C}\$, you find that the unitless damping factor is \$\zeta=\frac{\alpha}{\omega_{_0}}\$. Now you can write:

$$G_s=\frac1{s^2+2\zeta\,\omega_{_0}\,s+\omega_{_0}^2}$$

And that is now in standard form. Note also that \$Q=\frac1{2\,\zeta}\$.

Let's do some computations.

Find first that \$\omega_{_0}=\frac1{\sqrt{1.6\:\text{mH}\cdot 160\:\text{nF}}} = 62.5\times 10^3\:\frac{\text{rad}}{\text{s}}\$ (or \$f\approx 9.9472\:\text{kHz}\$.) Then find \$\zeta\approx 0.7072\$. (And find \$Q=0.707\$.)

Let's change the capacitance so that it is 5% more -- \$C=168\:\text{nF}\$.

Find \$\omega_{_0}=\frac1{\sqrt{1.6\:\text{mH}\cdot 168\:\text{nF}}} = 60.994\times 10^3\:\frac{\text{rad}}{\text{s}}\$ (or \$f\approx 9.707\:\text{kHz}\$.) Then find \$\zeta\approx 0.6902\$. Here now, the actual calculation results in \$Q\approx 0.7245\$.

From the sensitivity equation, we'd expect to see a %-change in \$Q\$ of about half of the percent change in \$C\$. So, a 5% increase in \$C\$ should mean about a 2.5% increase in Q; or an estimated result of \$Q=.707 \times 1.025 \approx 0.7247\$. Which is close enough.

Note: Adil Malik, in a comment below, correctly points out that this doesn't work for large changes. In that case, you have to look directly at the equation for \$Q\$, itself. That's why they used a square-root in the linked document. Not because of the sensitivity of \$\frac12\$. But instead because they looked directly at the large-scale equation. Thanks so much to Adil Malik for calling my attention to this.

The sensitivity equation only works well on small scale changes. In the linked document, they used the large-scale equation, instead.

Answer 2

@jrive - remember the definition of the sensitivity. This definition relates RELATIVE changes of Q (dQ/Q) with RELATIVE changes of one parameter (dC/C). More than that, because the differential quotient is involved, this definition applies to relatively small variations only.

You must not misuse these equation. For example, you must not change C by - for example - 100% and find the corresponding Q change. The sensitivity analysis is nothing else than a tool (or a method) to compare various filter topologies concerning their sensitivity to small parts tolerances. It gives a kind of "quality factor" .

Answer 3

I will present a slightly different interpretation to what the other answers have already said. You will see that the TI document is NOT wrong as far as I can see. The key to understanding this, is to realise that such sensitivity analysis is only a first order approximation of how a quantity will behave, if you change a parameter it depends on slightly.

Lets look at the actual equation without any approximations:

$$Q = R \sqrt{\frac{C}{L}}$$

Now lets perturb the capacitance ever so slightly and see the resulting change in quality factor: $$Q + \Delta Q = R \sqrt{\frac{C + \Delta C}{L}}$$ $$Q + \Delta Q = R \sqrt{\frac{C}{L}} \cdot\sqrt{1+\frac{\Delta C}{C}}$$ $$Q + \Delta Q = Q \cdot\sqrt{1+\frac{\Delta C}{C}}$$ Now here is the key, notice how the change in Q due to change in C is nonlinear. Okay so lets take a taylor series approximation and lets just keep the first linear term: $$Q + \Delta Q \approx Q \cdot (1+\frac{1}{2}\frac{\Delta C}{C})$$ So finally: $$ \frac{\Delta Q}{Q} \approx \frac{1}{2}\frac{\Delta C}{C} $$ Where $$S^Q_C = \frac{1}{2} $$

So relating dependence between two parameters using just a single sensitivity coefficient is an approximation, valid only for very small perturbations! To see the true change in the variable, you need to use the full expression, i.e. plug it into square root as the document does.

EDIT: You specifically ask why you cannot just say "the percentage change in Q varies with the 1/2 the percentage change in C1". As I try and explain in my answer, that is only really valid for very small changes in capacitance. You cannot expect the square root function to behave like that for large changes in capacitance.