Relative Sensitivity

Posted on 2023-07-12 Edited on 2023-07-10 In analog

Assuming Target $T$ ( for example, the total resistance) is function of $x_1,x_2,...,x_N$, then total variation can be expressed as

\[\begin{align} dT &= \sum_{n=1}^N\frac{\partial T}{\partial x_n}dx_n \\ &= \sum_{n=1}^N\frac{\partial T}{\partial x_n}x_n\cdot \frac{dx_n}{x_n} \end{align}\]

Then, we obtain relative variation \[\begin{align} \frac{dT}{T} &= \sum_{n=1}^N\frac{\partial T}{\partial x_n}\frac{x_n}{T}\cdot \frac{dx_n}{x_n} \\ &= \sum_{n=1}^N S_{x_n}^T \cdot \frac{dx_n}{x_n} \end{align}\]

⭐ where $S_{x_n}^T=\frac{\partial T}{\partial x_n}\frac{x_n}{T}$ is relative sensitivity

relative sensitivity connect $\frac{dx_n}{x_n}$ with total relative variation $\frac{dT}{T}$

And $dT$ can be expressed as \[ dT =\sum_{n=1}^N S_{x_n}^T T\cdot \frac{dx_n}{x_n} = \sum_{n=1}^N x_n'\cdot \frac{dx_n}{x_n} \] ⭐ where $x_n'= S_{x_n}^T T$ is the contribution of $x_n$ in $T$

⭐ For parallel or series resistors, it can prove $\sum_{n=1}^N S_{x_n}^T = 1$ and $ _{n=1}^N x_n'=T$

parallel_pgx.drawio

Here $T= R_1 \parallel R_2 = \frac{R_1R_2}{R_1+R_2}$, and $T|_{R_1=8000, R_2=2000} = 1600$

We obtain relative sensitivity: \[\begin{align} S_{R_1}^T & = \frac{R_2}{R_1+R_2} \\ S_{R_2}^T & = \frac{R_1}{R_1+R_2} \end{align}\]

The contribution of $R_1$ and $R_2$ to $T$ \[\begin{align} R_1' &= S_{R_1}^T T | _{R_1=8000, R_2=2000} = 320 \\ R_2' &= S_{R_2}^T T | _{R_1=8000, R_2=2000} = 1280 \end{align}\]

scholar

Normalized sensitivity captures relative sensitivity

change in objective per change in design variable

Normalized sensitivity

reference

Olivier de Weck, Karen Willcox. MIT, Gradient Calculation and Sensitivity Analysis [pdf]

Karti Mayaram, ECE 521 Fall 2016 Analog Circuit Simulation, Sensitivity and noise analyses [https://web.engr.oregonstate.edu/~karti/ece521/lec16_11_09.pdf]