NAME: ___Erik Lee________________________________________
Sensors and Sensor Systems: 615.747.31
michael.fitch@jhuapl.edu
Homework #1. Due 9/20/2012 by beginning of class. Can also be submitted on Blackboard, emailed,
etc. 100 pts total, problems as marked.
Problem I. (30pts total) Quick multiple Choice, fill-in-the-blank, and true/false. (2 pts each or as marked).
*Note on the unit conversions: Make sure you understand how to do this – you can use Google to check
your answers. Remember, you are the people that will develop the next generation of sensors and tools!
1. Sensors are specified in many different units. Let’s do some unit conversions: 0.002 MV/cm = ?
a. 2 kV/cm
b. 200 kV/m
c. 0.2 MV/m
d. All of the above.
e. None of the above.
2. Unit conversions: 500 atm = ?
a. 50.6 MPa
b. 380,000 torr
c. 5.06 x 107 N/m2
d. 7,348 psi (pounds per square inch)
e. All of the above
f. Only (a) and (c).
Pressure = Force / Area
1atm = 101324.99 pa
1atm = 760 torr
1atm = 101325 N/m^2
1atm = 14.69594 psi
3. Atmospheric pressure on Earth’s moon is about 10^-7Pa. Convert this to atm?
The restaurant on the moon has great food, but no atmosphere. (Dumb joke, I know.)
1 𝑎𝑡𝑚
101325 𝑃𝑎
=
𝑥 𝑎𝑡𝑚
1𝑒−7 𝑝𝑎
9.86923364e-13 atm
4. A good approximation for the number of seconds in a year is ( x 10^7). The percent error of using
this approximation is:?
0.447% error
Assuming 365.242374 days in a year.
5. The speed of light is approximately 3 x 10^8 m/s. Look up the exact value: _______________(2pts).
The percent error of using this approximation is: _______________? (2pts)
Speed of light = 299,792,458 m/s
Percent error = 0.0692%
6. How far does light travel in a year? (This is a light-year, or ly).___________________?
9.4605284 x 10^15 m
7. How far does light travel in a nanosecond (ns)? ________________(1 pt)? If a signal must travel
between two sensors in less than a nanosecond, how close together must the sensors be?
___________(1 pt)
Light travels 299.79 mm
Sensors must be within 299.79 mm
8. The speed of light in furlongs per fortnight is: _________________? (A furlong is 1/8 of a mile. A
fortnight is 2 weeks. Aren’t you glad we don’t use those units anymore?). Show your work.
299792458 𝑚𝑒𝑡𝑒𝑟
1 𝑚𝑖𝑙𝑒
8 𝑓𝑢𝑟𝑙𝑜𝑛𝑔 3600 𝑠𝑒𝑐𝑜𝑛𝑑 24 ℎ𝑜𝑢𝑟
14 𝑑𝑎𝑦
×
×
×
×
×
1 𝑠𝑒𝑐𝑜𝑛𝑑
1609.34 𝑚𝑒𝑡𝑒𝑟
1 𝑚𝑖𝑙𝑒
1 ℎ𝑜𝑢𝑟
1 𝑑𝑎𝑦
1 𝑓𝑜𝑡𝑛𝑖𝑔ℎ𝑡
1.8026𝑒12
𝑓𝑢𝑟𝑙𝑜𝑛𝑔𝑠
𝑓𝑜𝑟𝑡𝑛𝑖𝑔ℎ𝑡
9. Parallax is
a. A variant of Lacrosse played with two parallel sticks.
b. The apparent change in observation angle between two separated observation points.
c. In cards, a poker hand with the Lax of Hearts and the Lax of Spades.
10. The distance to nearby stars can be measured with stellar parallax: as the earth moves from one side
of its orbit to another, there is a small change in observation angle (with respect to distant galaxies). An
object with a parallax of one second of arc (1’’ or arcsecond) is said to be at a distance of one parsec.
There are 60 seconds of arc in one minute of arc (60’’ = 1’) and 60 minutes of arc in one degree. Convert
1” (one arcsecond) to radians: ______________.
1 arcseond = 4.848136e-5 radians
11. The mean distance from the Earth to the Sun is called an Astronomical Unit (AU). Look up the
value of an AU in km____________.
1AU = 149,597,870.691 kilo-meters
12. If the distance to the star is L=1 parsec, then L=1 AU where is the angle from 10. (Hint: draw the
triangle if you need to.) How many meters in a parsec? ______________ From the answer to 6, convert
1 parsec to light-years: 1 parsec = _____________ ly.
1 parsec = 3.08568e16 meters
1 parsec = 3.26163344 light years
13. The metric prefix atto- is 10^-18. How many cm in one atto-parsec? _________. A cube 1 attoparsec
on each side has a volume of how many cm^3? ________? A U.S. Fluid ounce is 29.5735295626 mL
(exactly). What is the percent difference between a cubic atto-parsec and a US Fluid Ounce?________
1 atto-parsec = 3.08568 centi-meters
1 cube atto-parsec = 29.38 cm^3
Percent difference =0.654%
14. You need a temperature sensor with an accuracy of ±1°K. You find a product with a specified
accuracy of ±2°F. True/False: this product meets your specification.
Problem II. Range Sensors (20 pts)
An optical distance sensor (or range-finder) was first developed by the Scottish firm of Barr and
Stroud in the 1880s. The same basic idea was in common military use until the 1940s
(supplanted by RADAR, and from 1965 onward: laser rangefinders). It consists of an
arrangement of lenses and prisms set at each end of a tube with a single eyepiece at the center.
The user measures the parallax angle by tilting a prism until the two images coincide. See the
picture below.
(a) Estimate the baseline of the two apertures based on the height of the men in the photo.
I estimate the base line between the apertures to be about 3 feet.
(b) The device shown below can measure parallax as large as 0.5 deg (30 minutes). Calculate
the range for an angle of 1 minute.
Given the angle of 1 minutes and assuming the baseline is 3 feet we can find the range
with the following equation.
B cot(theta) = R
3 cot(1/60) = 10313 ft
(c) Suppose the error on the parallax angle is 2 seconds of arc. What is the error on the range in
(b) derived from this angular error? (Assume that the error on the baseline is negligible.)
3 cot(1/60 + 2/3600) = 9980 ft
3 cot(1/60 - 2/3600) = 10669 ft
The range of error spans 688 ft.
(d) What is the maximum range at which it would be useful?
Using the difference of the previous equations and picking an arbitrary range error 10
feet we do not want to exceed we solve for x in the equation below.
3 cot(x/60 - 2/3600) - 3 cot(x/60 - 2/3600) <= 10
x >= 8.29 arc minutes
This translates into about 1244ft for the maximum range.
This is not desirable for an artillery unit in WWII. This is less than a quarter mile.
Gratefully by WWII the range finders were more accurate and there were other means
of measuring distance like radar.
Problem III. Some problems from Fraden Chapter 2. (20 pts)
1. Look at the figure inserted below from Fraden (3rd edition: Figure 2, p 23), (4th edition Figure
2.14b, p 39). Sometimes a sensor has a dead band, where the output is zero or close to zero
over a certain range. Give an example of how this could be a problem. Can you think of a
situation where this would be desirable? Discuss. (5 pts)
For example the dead band could be a problem in night vision goggles that just amplify
existing light. For complete darkness there would be no stimulus light to amplify and
would show the user nothing.
Switching to the example of a temperature sensing camera it might be useful to have
the dead band act as a filter for cool objects when searching for warm bodies. Though a
"dead-band filter" might potentially be useful, it would be more useful for the cool
objects to be sensed and provide context to the warm body. This would show the user
the landscape and surrounding objects.
2. Calibration error. Your bathroom scale (probably) has a zero-adjust. Is this all you need to
calibrate your scale? Why or why not? (5 pts)
NO, you need to be able to use known weights to make
sure the incremental measurements are accurate.
3. Suppose you are calibrating a set of mercury thermometers with a water-ice point (0 C) and a
water boiling point (100 C). You get the following data values for thermometers A, B, etc:
A: ice point +2 C, boiling point 102 C
B: ice point -1 C, boiling point 101 C
C: ice point +1, boiling point 98 C
D: ice point 0, boiling point 103 C

Generate calibration functions for each thermometer. (8 pts) Your calibration function
must be in the form of T = m*ta + b where T is the real (calibrated) temperature and ta
is the reading from thermometer A, and similarly for the other thermometers B, C, and
D.
A)
B)
C)
D)

T = ta – 2
T = 0.9803921 * tb + 1
T = 1.0309278 * tc - 1
T = 0.9708738 * td
You are lazy and just want to measure temperature change of a sample from a heating
pulse. Which one do you grab? (2 pts).
You grab thermometer A.
Problem IV. Optical Detectors: the Photo-electric effect. (10 pts)
Let’s take an example of an alkali photocathode such as CsTe (cesium telluride). You would expect that
the alkali metals (Na, K, Rb, Cs) to have both a low work-function and a high quantum efficiency because
the outer electron in these atoms is a single electron outside a closed shell, and is therefore weakly
bound. Chemists say these elements are strongly electro-negative. In pure form, these elements are
strongly reactive and pose some difficulties in handling, so they are incorporated into a matrix, tellurium
in this case, to stabilize the cesium. Such photo cathodes are said to have high Negative Electron Affinity
(NEA).
1. Show that Planck’s constant h and the speed of light c, together make: hc = 1240 eV nm (2
pts)
ℎ = 4.1356675 × 10−5 𝑒𝑉𝑠
𝑚
𝑐 = 299 792 458
𝑠
ℎ𝑐 = 0.0000012398419253 𝑒𝑉𝑚 = 1239.841𝑛𝑚
2. For ultraviolet light (above the work function), the quantum efficiency of CsTe is ~2%. If a UV
laser pulse of 1 µ J energy at 263 nm wavelength strikes the photocathode, how much charge
can be released? Express your answer in nano-Coulomb (nC). Show your work. (6 pts) [Note: in
this problem, the ‘quantum efficiency’ means the number of electrons emitted per incident
photon. Since this number is 2% or 0.02 you can think of it as the probability of electron
emission per incident photon. Hint: how many photons at 263nm are in a 1 µ j energy pulse?
How many electrons are in 1 nC?]
3. For copper metal, the quantum efficiency is ~1x10^-5 (or 0.001%). How much charge can be
released for the same laser pulse as (1)? (2 pts)
Problem V. Op-Amp Quick problems (20 pts)
a. Inverting Amplifier. (5pts) See figure below. Show that:
where the gain
𝑉𝑖𝑛 = 𝐼 ∗ 𝑅1
Vout = I ∗ R2
Gain =
Vout I ∗ R2 R2
=
=
Vin
I ∗ R1 R1
Gain =
b. Non-Inverting Amplifier. (5pts) See figure below. Show that:
R2
R1
where the gain
𝑉𝑖𝑛 = 𝐼 ∗ 𝑅1
Vout = I ∗ (R1 + R2)
Gain =
Vout I ∗ (R1 + R2)
R2
=
=1+
Vin
I ∗ R1
R1
Gain = 1 +
R2
R1
c. Op-Amp Integrator. (10pts) See figure below. The reset switch is opened at time t=0. Use the
capacitor relation (Q=CV) and the relation between charge and current to derive:
𝑡
𝑉𝑜𝑢𝑡 = ∫
0
−𝑉𝑖𝑛
𝑑𝑡 + 0
𝑅1 ∗ 𝐶1