# Sensors Exam Study Guide

## 1. <mark>Error propagation</mark>
- Independent random uncertainties:
For a computed value $f$ that is a function of measured values, $f(x, \dots, z)$:
$σf = \sqrt{\left(\frac{σf}{σx}\right)^2 + \dots + \left(\frac{σf}{σz}\right)^2}$
- round uncertainty to one sig fig (same order of magnitude for measurement [same amount of decimals if same units, e.g. $12.00 ± 0.03$ m])
## 2. <mark>Sensor fusion</mark>
- find the "estimated value" that maximizes the probability of getting the recorded measurements
- Probability of getting $z$ given $x$: $$P(x \mid z) = k e^{\frac{1}{2} \left( \frac{x-z}{\sigma} \right)^2}$$ <!-- $P(x \mid z) = k \mathrm{exp} \frac{1}{2} \left( \frac{x-z}{\sigma} \right)^2$ -->
- $z$ is the measurement that is known (and the center)
- $x$ is the true value (unknown, what we choose to try to maximize the function)
- $\sigma$ is the standard deviation (and the width of the curve)
- when $z=x$, maximize probability $\implies P=1$ (scaled by amplitude, $k$)
- fuse multiple measurements to get improved estimate
- to fuse multiple Gaussian measurements, use weighted average
- Mean of the fused estimate (weighted average): $$\hat{x} = \frac{w_1 z_1 + w_2 z_2}{w_1 + w_2}$$
- (measurements: $z_i$; weights: $w_i$; improved estimate: $\hat{x}$)
- weights are: $\frac{1}{\sigma_i^2}$ (aka where the derivate of the function we are trying to maximize $= 0$)
- reciprocal of variance (weights) is the **information**
- information is *additive*, so we can add the weights
- Variance of the fused estimate: $$\hat{\sigma}^2 = \frac{1}{w_1 + w_2} = \frac{\sigma_1^2 \sigma_2^2}{\sigma_1^2 + \sigma_2^2}$$
- since information of the fused estimate $= w_1 + w_2$
- small variance is thin Gaussian (more likely to get smaller range of things, so data was more information); whereas when all the things are equally likely, less information
- so information increases when smaller variance
- weighting by reciprocal of the variance since that is the *information* (more information, more weight, since it increases the info we get out of $x$)
<!-- $arg\,max$ -->
## 3. <mark>Wheatstone bridge</mark>
**a) Be able to derive sensitivity as a differential, as defined in class**
**b) Show understanding of how to maximize sensitivity, e.g., # of sensors, direction sensors should be mounted**
 $$V_{out} = V_{ref} \left( \frac{Z_1}{Z_1 + Z_2} - \frac{Z_3}{Z_3 + Z_4} \right)$$ (where, e.g., $Z_1$ is unknown and $Z_2, Z_3, Z_4$ are known)
- useful when want to measure small changes in resistance
- electrical circuit used to measure an unknown resistance
- one resistor of unknown resistance & three resistors with known resistance
- keep adjusting the resistance of the other three resistors until the current in the circuit is balanced
- balanced circuit means two "legs" are equal (legs are the resistors that are conected)
- $V_{out} = V_{+}-V_{-}$ (voltage at $+$ minus voltage at $-$)
- so when $V_{+}-V_{-}$ $\implies V_{out} = 0$, and we can simplify to: $$\frac{Z_2}{Z_1} = \frac{Z_4}{Z_3} \implies Z_1 = \frac{Z_2 Z_3}{Z_4}$$
- $V_{ref}$ is the input/power-source voltage (energy babays have at very start of race track)
- relative to ground (so it's the difference between input and ground)
- ground $= 0$ V
- Voltage divider: $V_-$ is some fraction of $V_{ref}$ based on $Z_1$ and $Z_2$
- voltage after $Z_1$ is lower than voltage at $V_{ref}$ since the resistance slows the electrons down aka lowers the current $$V_{-} = \frac{Z_2}{Z_1 + Z_2} \cdot V_{ref}$$ $$V_{+} = \frac{Z_4}{Z_3 + Z_4} \cdot V_{ref}$$
- babays start at $V_{in}$ (aka $V_{ref}$)
- they pace themselves so going at same pace at $Z_1$ and $Z_2$ (even if different resistances)
- multiple pathways for babays to choose between
- most start with the easy route, but then that one gets a lot of traffic so babays that come later might choose to take the hard route (with more resistance) since might end up getting there faster
- we can calculate how many babays will take each route
- voltage is the same at the beginning/end regardless of the route the baby takes (divided based on how much resistance on either side)
- if $Z_1$ has more resistance, $V_{out}$ will be smaller as fraction of $V_{ref}$
- however if $Z_2$ has more resistance, will have maintained high current at $V_{out}$, so will $V_{out}$ will be be a larger fraction of $V_{ref}$
- **Sensitivity:** $$\frac{\partial V_{out}}{V_{ref}} = \frac{Z_2 ∂ Z_1 - Z_1 ∂ Z_2}{(Z_1+Z_2)^2} - \frac{Z_4 ∂ Z_3 - Z_3 ∂ Z_4}{(Z_3+Z_4)^2}$$
- if you change the the $Z$s a little bit, how much does $V_{out}$ change?
- for all the variables that change, how much does $V_{out}$ change as a result?
- taking the derivative of the output of the system with respect to the input or the stimulus being measured.
- If all but $Z_1$ are fixed and $Z_2 = Z_1$, then: $$\frac{\partial V_{out}}{V_{ref}} = \frac{Z_2 ∂ Z_1}{(Z_1+Z_2)^2} = \frac{∂ Z_1}{4 Z_1}$$
- aka taking derivative of $V_{out}$ w.r.t. $Z_1$ only
- sensitivity is maximized when bridge is balanced
- To calculate resistor value for maximizing sensitivity:
- Calculate sensiitivity by taking derivative w.r.t. the variable resistor(s)
- Then, take second derivative w.r.t. the the resistor the we want to determine the value of to maximize the sensisitivity, and set it to zero (solve for that resistor value)
- Maximize sensitivity of sensing system:
- **Use more sensors:** more accurate measurements (reduce uncertainty and impact of noise; likely that errors in individual measurements from different sensors will be different too; *sensor fusion*)
- **Mount sensors in direction perpendicular to the direction of the stimuli they are measuring:** sensitive axis is perpendicular to the direction of the measurement

**Derivation of sensitivity:**
We have: $$V_{out} = V_{ref} \left( \frac{Z_1}{Z_1 + Z_2} - \frac{Z_3}{Z_3 + Z_4} \right)$$ Sensitivity is how much $V_{out}$ changes w.r.t. change in the variable resistors. If all the resistors are variable, we have: $$∂ V_{out} = V_{ref} \left( \frac{∂ V_{out}}{∂ Z_1} + \frac{∂ V_{out}}{∂ Z_2} + \frac{∂ V_{out}}{∂ Z_3} + \frac{∂ V_{out}}{∂ Z_4}\right)$$

We start by treating all but one resistor value as fixed, and take the partial derivative w.r.t. to the remaining resistor.
Let $Z_1$ be the variable resistor. We have: $$\frac{∂ V_{out}}{∂ Z_1} = V_{ref} \left( \left( \frac{(Z_1+Z_2) · \left(\frac{\partial Z_1}{\partial Z_1}\right) - Z_1 · \left(\frac{∂ (Z_1 + Z_2)}{\partial Z_1}\right)}{(Z_1 + Z_2)^2} \right) - \left( \frac{\partial \left(\frac{Z_3}{Z_3 + Z_4}\right)}{\partial Z_1} \right) \right)$$
We move $V_{ref}$ to the left side:$$\frac{∂ V_{out}}{∂ Z_1 · V_{ref}} = \left( \frac{(Z_1+Z_2) · 1 - Z_1 · 1}{(Z_1 + Z_2)^2} \right) - 0$$ $$= \frac{(Z_1+Z_2) - Z_1}{(Z_1 + Z_2)^2}$$ $$= \frac{Z_2}{(Z_1 + Z_2)^2}$$ $$\implies \frac{∂ V_{out}}{V_{ref}} = \partial Z_1 \left( \frac{Z_2}{(Z_1 + Z_2)^2} \right)$$
*(NOTE TO SELF: use quotient rule, (lowDhigh - highDlow)/low^2)*

Now, we do the same for the other resistors.
Let $Z_2$ be the variable resistor. We have: $$\frac{∂ V_{out}}{∂ Z_2} = V_{ref} \left( \left( \frac{(Z_1+Z_2) · \left(\frac{\partial Z_1}{\partial Z_2}\right) - Z_1 · \left(\frac{∂ (Z_1 + Z_2)}{\partial Z_2}\right)}{(Z_1 + Z_2)^2} \right) - \left( \frac{\partial \left(\frac{Z_3}{Z_3 + Z_4}\right)}{\partial Z_2} \right) \right)$$ We move $V_{ref}$ to the left side:$$\frac{∂ V_{out}}{∂ Z_2 · V_{ref}} = \left( \frac{(Z_1+Z_2) · 0 - Z_1 · 1}{(Z_1 + Z_2)^2} \right) - 0$$ $$= \frac{- Z_1}{(Z_1 + Z_2)^2}$$ $$\implies \frac{∂ V_{out}}{V_{ref}} = \partial Z_2 \left( \frac{-Z_1}{(Z_1 + Z_2)^2} \right)$$

Next, left $Z_3$ be the variable resistor. We have: $$\frac{∂ V_{out}}{∂ Z_3} = V_{ref} \left( \left( \frac{\partial \left(\frac{Z_1}{Z_1 + Z_2}\right)}{\partial Z_3} \right) - \left( \frac{(Z_3+Z_4) · \left(\frac{\partial Z_3}{\partial Z_3}\right) - Z_3 · \left(\frac{∂ (Z_3 + Z_4)}{\partial Z_3}\right)}{(Z_3 + Z_4)^2} \right) \right)$$
We move $V_{ref}$ to the left side and simplify:$$\frac{∂ V_{out}}{∂ Z_3 · V_{ref}} = 0 - \left( \frac{(Z_3+Z_4) · 1 - Z_3 · 1}{(Z_3 + Z_4)^2} \right)$$ $$= \frac{(Z_3 + Z_4)-Z_3}{(Z_3 + Z_4)^2}$$ $$= \frac{Z_4}{(Z_3 + Z_4)^2}$$ $$\implies \frac{∂ V_{out}}{V_{ref}} = \partial Z_3 \left( \frac{Z_4}{(Z_3 + Z_4)^2} \right)$$

Next, we let $Z_4$ be the variable resistor. We have: $$\frac{∂ V_{out}}{∂ Z_4} = V_{ref} \left( \left( \frac{\partial \left(\frac{Z_1}{Z_1 + Z_2}\right)}{\partial Z_4} \right) - \left( \frac{(Z_3+Z_4) · \left(\frac{\partial Z_3}{\partial Z_4}\right) - Z_3 · \left(\frac{∂ (Z_3 + Z_4)}{\partial Z_4}\right)}{(Z_3 + Z_4)^2} \right) \right)$$
We move $V_{ref}$ to the left side and simplify:$$\frac{∂ V_{out}}{∂ Z_4 · V_{ref}} = 0 - \left( \frac{(Z_3+Z_4) · 0 - Z_3 · 1}{(Z_3 + Z_4)^2} \right)$$ $$= \frac{-Z_3}{(Z_3 + Z_4)^2}$$ $$\implies \frac{∂ V_{out}}{V_{ref}} = \partial Z_4 \left( \frac{-Z_3}{(Z_3 + Z_4)^2} \right)$$

Finally, we have: $$\frac{∂ V_{out}}{V_{ref}} = \partial Z_1 \left( \frac{Z_2}{(Z_1 + Z_2)^2} \right) + \partial Z_2 \left( \frac{-Z_1}{(Z_1 + Z_2)^2} \right) + \partial Z_3 \left( \frac{Z_4}{(Z_3 + Z_4)^2} \right) + ∂ Z_4 \left( \frac{-Z_3}{(Z_3 + Z_4)^2} \right)$$ $$\frac{\partial V_{out}}{V_{ref}} = \frac{Z_2 ∂ Z_1 - Z_1 ∂ Z_2}{(Z_1+Z_2)^2} - \frac{Z_4 ∂ Z_3 - Z_3 ∂ Z_4}{(Z_3+Z_4)^2}$$
and we are done.

<!-- The equation ∂Vout/Vrin= (Z2∂Z1−Z1∂Z2)/((Z1+Z2)^2) − (Z4∂Z3−Z3∂Z4)/((Z3+Z4)^2) describes the sensitivity of a Wheatstone bridge, where Vout is the output voltage of the bridge, Vrin is the input voltage, and Z1, Z2, Z3, and Z4 are the impedances of the resistors in the bridge.

To derive this equation, we can consider the circuit diagram of a Wheatstone bridge and apply Kirchhoff's voltage law to the four nodes in the circuit. This will give us four equations, which we can then solve to obtain the equation for Vout in terms of the impedances and the input voltage.

For simplicity, let's assume that Z1, Z2, and Z3 are fixed impedances, and Z4 is a variable impedance that we are interested in. In this case, the equation for Vout can be written as follows:

Vout = Vrin * (Z2/(Z1 + Z2) - Z4/(Z3 + Z4))

We can then take the derivative of this equation with respect to Z4 to obtain the sensitivity of the Wheatstone bridge with respect to Z4:

Sensitivity = (dVout)/(dZ4) = -Vrin * (Z2/(Z1 + Z2)^2 + 1/(Z3 + Z4))

This equation shows that the sensitivity of a Wheatstone bridge is inversely proportional to the square of the sum of the impedances in the bridge. This means that the sensitivity of the bridge increases as the sum of the impedances decreases.

The equation ∂Vout/Vrin= (Z2∂Z1−Z1∂Z2)/((Z1+Z2)^2) − (Z4∂Z3−Z3∂Z4)/((Z3+Z4)^2) can be derived by generalizing the above equation to allow for the possibility that

To derive this equation, we can first consider the basic circuit diagram of a Wheatstone bridge. The circuit consists of four branches, each with a resistance or impedance. The input voltage Vrin is applied across the two fixed resistances or impedances Z1 and Z2, and the output voltage Vout is measured across the two variable resistances or impedances Z3 and Z4.

We can apply Kirchhoff's voltage law to the four nodes in the circuit to obtain four equations that relate the voltages and currents in the different branches of the bridge. These equations can then be solved to obtain an expression for the output voltage Vout in terms of the input voltage Vrin and the impedances Z1, Z2, Z3, and Z4. This expression is given by the following equation:

Vout = Vrin * ((Z3 + Z4) - (Z1 + Z2)) / (Z1 + Z2 + Z3 + Z4)

To determine the sensitivity of the Wheatstone bridge, we can take the derivative of this equation with respect to each of the impedances Z1, Z2, Z3, and Z4. This will give us the following equations for the sensitivity of the bridge:

To derive this equation, we can consider the circuit diagram of a Wheatstone bridge and apply Kirchhoff's voltage law to the four nodes in the circuit. This will give us four equations, which we can then solve to obtain the equation for Vout in terms of the complex impedances and the input voltage.

Assuming that the input voltage is given by Vin, the complex impedances of the fixed resistors are given by Z1 and Z2, and the complex impedances of the variable resistors are given by Z3 and Z4, the equation for Vout can be derived as follows:

Vout = (Z3 - Z1)/(Z1 + Z2) * Vin

Taking the derivative of this equation with respect to Z1 and Z3, we obtain the following equations:

∂Vout/∂Z1 = -Vin/(Z1 + Z2)^2
∂Vout/∂Z3 = Vin/(Z1 + Z2)^2

Substituting these equations into the original equation for sensitivity, we obtain the following: -->

## 4. <mark>Types of intrinsic noise</mark>
**a) Know formulas**
**b) Know what you can do to minimize each of them**
- **intrinsic/fundamental noise:** occurs due to natural causes (physics)
- **technical/extrinsic noise:** could *theoretically* be avoided (engineer's limits)
- **Shot noise** ("white" noise): $$\sigma_I = \sqrt{2 e \bar{I} \Delta{f}}$$
- where:
- $e$ is the charge per electron (constant)
- $\bar{I}$ is the current
- $\Delta{f}$ is bandwidth (change in frequency)
- a fluctuation in current (how much babay's speed is varying based on things other than its own effort)
- same power at all frequencies
- no fall of; no correlation sample-to-sample
- TODO: figure how to minimize (want to decrease $I$ [decrease $I$ by increasing $R$ since $I= \frac{V}{R}$])
- **$1$/$f$ noise** ("pink" noise): $$1/f$$
- random flucuations that increase in prevelance with time
- more powerful at higher frequencies
- some correlation sample-to-sample
- not known how to decrease it
- **Thermal (Nyquist-Johnson) noise** ("brown" noise): $$V = \sqrt{4 k T R \Delta{f}}$$
- where:
- $k$ is Boltzmann's constant
- $K$ is temperature
- $R$ is resistance
- $\Delta{f}$ is bandwidth
- voltage you're getting from the thermodynamic temperature effects
- "random walk noise"
- $1$/$f^2$ noise
- integral of white noise
- TODO: figure out how to minimize (decrease resistance)
## 5. <mark>List common types of technical noise/disturbances & what you might do about them</mark>
- reduce noise before it reaches the sensor:

| Noise Source | Cause | Reduction |
| -------- | -------- | -------- |
| Ground loops (noise from power supply) | Direct resistive coupling | Float the sensor electrically |
| Capacitive sensor | Capacitive coupling | Use shielding |
| 60 Hz Power-line noise | Inductive coupling | Twist pairs of wires of reduce their lengths |
- reduce noise after it reaches the sensor:
- diffential sensing (TODO: understand ?)
- averaging
- filtering in hardware or software (lowpass, highpass, bandpass, notch, etc.)
- synchronous detection
- suppressors
- Misc causes:
- people talking or vibration from steps
- build structure to block it out or move to different environment
- Earth's tide
## 6. Impedance
**a) Bridging**
**b) Matching**
## 7. <mark>Know how to design an instrumentation amp with a given overall gain, and discuss its advantages</mark>
- type of differential amplifier (circuit involves an op-amp)
### Op-Amps:
- $V_{+} =$ input voltage from the positive terminal (non-inverting input)
- $V_{-} =$ input voltage from the negative terminal (inverting input)
- $V_{out} =$ TODO
- $V_{S_{+}} = V_{S_{-}}$: connected to a constant voltage to power the omp-amp (inverse of each other); *(NOTE: sometimes not shown in circuit)*
- Gain $= \frac{V_{out}}{V_{in}}$ (how much $V_{in}$ is being amplified)
- very high input impedance & gain

**Ideal op-amp "rules":**
1. in a closed loop, the output attempts to do whatever is necessary to make the voltage difference between the inputs zero.
- "closed loop" means $V_{out}$ is connected to one of the inputs
- If the output is connected to one of the inputs ($\implies$ closed loop), it tries to make the voltage at the other input equal to the voltage at the input that $V_{out}$ is connected to $\implies V_+ = V_-$
2. the omp-amps inputs draw no current
- this is why op-amps have very high input impedance (call it $\infty$)
- **Impedance ($Z$):** $Z = \frac{V}{I}$ (voltage divided by current)
- the impedance of a resistor is just its resistance, but other components have impedance too (e.g. capacitors)
- how much the current is "impeded" (so if current is zero, impedance is infinity, since $\frac{V}{\infty} = 0$, or we can think of it as: $\frac{V}{0} = \infty$)
- since impedance is so high, current is zero (no babays running through it)
- babays don't want to go through triangle, so they take the other path
- therefore, since no current flows through the op-amp, any current in the circuit goes along the path that doesn't go through the op-amp
- current at $R_{in}$ ($I_{in}$) and current at $R_{f}$ ($I_f$) is total current in circuit ($I$) $\implies I_{in} = I_f = I$

**Circuits that involve op-amps:** *(NOTE: the triangle is the op-amp)*
### Inverting Amp:
- $V_{out} = -V_{in} \cdot G$ (where $G$ is gain, which is negative for inverting amp)
- in this circuit, + is connected to ground, so the voltage there is zero. Via omp-amp rules, this means voltage at - is zero too.

**Deriving the gain:**
We want to solve for gain, $G$, where $G = \frac{V_{out}}{V_{in}}$
The "ideal op-amp rules" tell us that:
1. $V_- = V_+ = 0$ (since $V_+$ is connected directly to ground)
2. The op-amp draws no current, so all the current in the circuit goes through the other path, so we have: $I_{in} = I_{f} = I$
Ohm's law tells us that $V=I R$, so we have: $$V_- - V_{in} = I_{in} R_{in} = I R_{in} \implies I = \frac{V_- - V_{in}}{R_{in}}$$ $$V_{out} - V_- = I_f R_f = I R_f \implies I = \frac{V_{out}-V_-}{R_f}$$ $$\implies \frac{V_- - V_{in}}{R_{in}} = \frac{V_{out}-V_-}{R_f}$$
Since $V_- = 0$, we have: $$\frac{0 - V_{in}}{R_{in}} = \frac{V_{out}-0}{R_f}$$ $$\implies \frac{R_f \cdot (-V_{in})}{R_{in}} = V_{out} \implies \frac{-R_f}{R_{in}} = \frac{V_{out}}{V_{in}}$$ $$\implies G = \frac{V_{out}}{V_{in}} = \frac{-R_f}{R_{in}}$$
### Non-inverting Amp:
- $V_{out} = V_{in} \cdot G$ (where $G$ is gain, which is positive for non-inverting amp)
<!-- - equations from Ohm's law -->

**Deriving the gain:**
We want to solve for gain, $G$, where $G = \frac{V_{out}}{V_{in}}$
The "ideal op-amp rules" tell us that:
1. $V_- = V_+ = V_{in}$ (since $V_+$ is connected directly to $V_{in}$)
2. The op-amp draws no current, so all the current in the circuit goes through the other path, so we have: $I_1 = I_2 = I$
Ohm's law tells us that $V=I R$, so we have: $$V_- - 0= I_1 R_1 = I R_1 \implies I = \frac{V_-}{R_1}$$ *(NOTE: voltage drop from ground (which has voltage of $0$) to negative terminal, so $-0$)* $$V_{out} - V_- = I_2 R_2 = I R_2 \implies I = \frac{V_{out}-V_-}{R_2}$$ $$\implies \frac{V_-}{R_1} = \frac{V_{out}-V_-}{R_2}$$
Since $V_- = V_{in}$, we have: $$\frac{V_{in}}{R_1} = \frac{V_{out}-V_{in}}{R_2}$$ $$\implies \frac{R_2 V_{in}}{R_1} = V_{out}-V_{in} \implies \frac{R_2}{R_1} = \frac{V_{out}-V_{in}}{V_{in}} = \frac{V_{out}}{V_{in}} - \frac{V_{in}}{V_{in}} = \frac{V_{out}}{V_{in}} - 1$$ $$\implies G = \frac{V_{out}}{V_{in}} = \frac{R_2}{R_1} + 1$$
### Differential Amplifier:
- amplifies the difference between $V_1$ and $V_2$
- here, gain $G = \frac{R_f}{R_1}$
- if $R_g = R_f$ and $R_2 = R_1$, then: $V_{out} = \frac{R_f}{R_1} (V_2 - V_1)$
### Instrumentation Amplifier:
- differential amplifier with two op-amps added in 

- $G = \frac{V_{out}}{V_2 - V_1} = \left(1 + \frac{2 R_1}{R_{gain}} \right) \frac{R_3}{R_2}$
- gain can be adjusted with a single resistor
- can entirely control the gain through $R_{gain}$

**Advantages:**
- very good at rejecting noise
- high accuracy
- high common-mode rejection
- ability to amplify small signals
## 8. Lock‐in amplification: what it is, why/when to use it
## 9. Inertial sensing
(<mark>look at homework</mark>)
**a) How different sensors work**
**b) Expected behavior of error when integrated (lab #1)**
## 10. Time of flight
(<mark>look at homework</mark>)
**a) “pure” time‐of‐flight**
**b) Modulated carrier**
**c) Interferometer**
## 11. Answer conceptual & quantitative questions about different sensing types, modalities
**a) Temperature**
**b) Light**
**c) Range**
**d) Etc.**

## Misc notes:
- Voltage divider:  $$V_{in} = I · (Z_1 + Z_2)$$ $$V_{out} = I \cdot Z_2$$ $$\implies I = \frac{V_{in}}{Z_1 + Z_2}$$ $$\implies V_{out} = V_{in} \cdot \frac{Z_2}{Z_1 + Z_2}$$
- DC (direct current): 0 frequency (voltage is constant; current only flows in one direction)
- AC (alternating current): changes direction periodically
- BABIES IN A RACE ARE LIKE ELECTRONS:
- resistance is how hard the obstacle course was
- voltage is the amount of candy at the end of the race (*note: the babies will try harder if there is more candy*)
- current is how fast the baaby runs
- how fast the babay runs is based on how chubbay it is
- fast = candy / hard
- if it's harder, it doesn't matter how much candies there were as much, babay might still be slow
- **Ohm's law:** $V = I R$ (voltage = current * resistance)
- *voltage drop across a resistor*
- **voltage drop** is the difference on voltage on either side of the resistor
- based on direction the current is flowing, so if current flowing from $a$ to $b$, voltage drop is $V_b - V_a$
- calculating the amount of energy babays lose as they go across the race track (how fast they're going times how hard to race track is to get thru)
- different btw voltage on either side of resistor $= I R$
- **Voltage:** electrical energy
- **Power:** amplitude$^2$ (energy per unit time)
- **Current:** how fast the electrons are flowing (charge per time)
- a discrete flow of charge (because there are discrete number of electrons flowing)
- standard deviation: $\sigma$, variance: $\sigma^2$
- standard deviation is the width of a Gaussian (center is expected value, $z$ [or $\mu$])
- probability: "or" means adding, "and" means multiplying
- $\log(a \cdot b) = \log(a) + \log(b)$

## Conceptual Study Questions:
$\newcommand{\bootyhole}[1]{\textbf{#1} \text{ is an sweetness to my haht}}$ $\newcommand{\yonayzington}{\texttt{my frand is not nahmal.}}$ $\newcommand{gabralzington}{\textit{who even were u}}$ $\newcommand{LOL}{\textit{GOT EEM!}}$
- $\yonayzington$
- $\gabralzington$
- $\bootyhole{weird}$
$\bootyhole{I hate this app. this app (note the rest is on opposite day)}$ $\LOL$
- What is a Gaussian measurement?
- *WHY DO YOU KEEP ASKING ME THIS ????*
- why does the world hurt
- because it left a *scah*
- why is it true that $\yonayzington$
- true or false jampa is smol
- [x] true
- [ ] true (on opposite day)
- true or false my bootyhole was larg
- [x] smol true
- [x] larg true
- [x] no comment
- final question: $\gabralzington$. multiple choice:
- [x] smol
- [ ] larg
- [x] sweet
- [ ] not sweet
- [ ] Too many opkins!!!
- *additional comments:* <u> I HAD THE CHOOSE THEM ALL! THERE WERE MULTIPLE FOR MY CHOICES </u>

## TODO:
1. - [x] <mark>Error propagation</mark>
- (* quick problems, with one longer using general rule)
2. - [x] <mark>Sensor fusion</mark>
- (* like problem on homework, fusing temperature data etc)
3. - [x] <mark>Wheatstone bridge</mark>
4. - [x] <mark>Types of intrinsic noise</mark>
- (* types of them, and then use formulas)
5. - [x] <mark>List common types of technical noise/disturbances & what you might do about them</mark>
- (* short answer questions)
6. - [ ] ~~Impedance~~
7. - [x] <mark>Know how to design an instrumentation amp with a given overall gain, and discuss its advantages</mark>
8. - [ ] ~~Lock‐in amplification: what it is, why/when to use it~~
9. - [ ] Inertial sensing (<mark>look at homework</mark>)
10. - [ ] Time of flight (<mark>look at homework</mark>)
11. - [ ] ~~Answer conceptual & quantitative questions about different sensing types, modalities~~

* <mark>question about vision sensing</mark>
- optical stuff (last two homework sets)
- repeat of a homework problem, verbatim

## Flashcards:
- [ ] Error propagation rules/when to use them
- [ ] Sensor fusion: mean and variance of fused estimate
- [ ] Wheatstone bridge circuit diagram
- [ ] Wheastone Bridge equation for $V_{out}$
- [ ] Derivation of sensitivity + equation
- [ ] How to increase sensitivity of Wheatstone Bridge
- [ ] Types of intrinsic noise, formulas, and how to minimize
- [ ] Types of technical noise + what to do about them
- [ ] Ohm's Law
- [ ] Ideal op-amp rules
- [ ] Intrumentation amplifier circuit diagram
- [ ] Intrumentation amplifier gain equation (+ note about how to choose resistor values etc.)
- [ ] Vision stuff (TODO)

## Practice problems:
- [ ] Error propagation general formula
- [ ] Sensor fusion problem from HW 2
- [ ] Derivation of sensitivity
- [ ] Solve for instrumentation amplifier values based on given gain
- [ ] Vision problems from HW 7 & HW 8