Name: ____Erik Lee____________________________________________________________
Sensors and Sensor Systems. 615.747.31 michael.fitch@jhuapl.edu
Homework #4 Due on 15 November, 2012 at the beginning of class.
1. (10 pts) Detector Deadtime.
𝑚1 𝑚2 − √𝑚1 𝑚2 (𝑚12 − 𝑚1 )(𝑚12 − 𝑚2 )
𝑚1 𝑚2 𝑚12
a. (1 pt) If all the m’s have units of counts per second (cps), what are the units of ?
𝜏=
𝑐𝑐
𝑐𝑐 𝑐 𝑐
−√ ( )( )
𝑠𝑠
𝑠𝑠 𝑠 𝑠
𝜏=
=
𝑐𝑐𝑐
𝑠𝑠𝑠
𝜏=
𝑐𝑐
𝑠𝑠
𝑐𝑐𝑐
𝑠𝑠𝑠
=
𝑐𝑐𝑠𝑠𝑠 𝑠
=
𝑠𝑠𝑐𝑐𝑐 𝑐
𝑠
𝑠𝑒𝑐𝑜𝑛𝑑𝑠 𝑝𝑒𝑟 𝑐𝑜𝑢𝑛𝑡
𝑐
b. (1 pt) What happens if m12 > (m1 + m2)? Is  real or imaginary? Does it make sense?
The time per count turns negative. This is imaginary and does not make sense. The dead time is a
measure of time that needs to be between two events for the sensor to detect two events. If the dead time
were negative one could argue that if two events happen at the same time the detector would sense the
first then go back in time and sense the second event.
c. (7 pts) Put this formula into a spreadsheet (or the program of your choice). Calculate  for these cases:
i.
ii.
iii.
iv.
v.
vi.
vii.
m1 = 1000 cps, m2 = 2000 cps, m12 = 2500 cps
𝜏 = 155.05 𝜇𝑠𝑝𝑐
m1 = 1000 cps, m2 = 2000 cps, m12 = 2900 cps
𝜏 = 25.98 𝜇𝑠𝑝𝑐
m1 = 1000 cps, m2 = 2000 cps, m12 = 2950 cps
𝜏 = 12.74 𝜇𝑠𝑝𝑐
m1 = 1000 cps, m2 = 2000 cps, m12 = 2990 cps
𝜏 = 2.51 𝜇𝑠𝑝𝑐
m1 = 1000 cps, m2 = 2000 cps, m12 = 2995 cps 𝜏 = 1.25 𝜇𝑠𝑝𝑐
m1 = 1000 cps, m2 = 2000 cps, m12 = 2999 cps
𝜏 = 0.25 𝜇𝑠𝑝𝑐
m1 = 1x106 cps, m2 = 2x106 cps, m12 = 2.999x106 cps
𝜏 = 250.10 𝑝𝑠𝑝𝑐
d. (1 pt) Write at least one sentence commenting on the usefulness or uselessness of this formula.
This formula is useful in determining how fast a sensor can detect events before multiple events blur into
the same event.
2. (30 pts) Gamma and Xray resolution: semiconductor detectors.
a. (5 pts) For Ge, the bandgap energy is 0.67 eV. For a gamma ray of 511 keV, how many electrons (N) would be
expected?
𝐸𝑝ℎ𝑜𝑡𝑜𝑛
511𝑘𝑒𝑉
𝑁=
=
= 762687 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
𝐸𝑔
0.67𝑒𝑉
b. (5 pts) If we repeated this process many times and had a statistical ensemble, the standard deviation in N would be
(for Poisson distributed errors: appropriate for counting singles) the square root of N. Usually the counts are
expressed as 𝑁 ± √𝑁. So in part a, the mean number of electrons would have a similar Poisson error. If the Poisson
error is the sole factor influencing the energy resolution, what is the E/E in percent that you expect for a
Germanium detector and a 511 keV gamma?
∆𝐸
√𝑁 √762687
=
=
= 0.1145%
𝐸
𝑁
762687
c. (5 pts) Cobalt-60 (Co60), a common calibration source, emits two gamma
rays, one at ~1.17 MeV and one at ~1.33 MeV. From part b, what percent
error would you expect for those energies? Look at Figure 1 below. What do
you estimate is the energy resolution in the figure? (Estimate the maximum
width of the peaks to give at least an upper limit on the energy resolution
from the figure – I realize the graphics quality is not that great.) Express
your answer as E/E in percent.
Figure 1 Energy spectrum of gamma rays
from Cobalt-60, as measured with a Ge
detector.
𝑁𝐶𝑜𝑏𝑎𝑙𝑡_1.17𝑀 =
𝐸𝑝ℎ𝑜𝑡𝑜𝑛
1.17𝑀𝑒𝑉
=
= 1746269 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
𝐸𝑔
0.67𝑒𝑉
∆𝐸
√𝑁 √1746269
=
=
= 0.07567%
𝐸
𝑁
1746269
∆𝐸
𝑤𝑖𝑑𝑡ℎ 1.18𝑀 − 1.16𝑀
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑
=
=
= 1.145%
𝐸
𝑁
1746269
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑
𝑁𝐶𝑜𝑏𝑎𝑙𝑡_1.33𝑀 =
𝐸𝑝ℎ𝑜𝑡𝑜𝑛
1.33𝑀𝑒𝑉
=
= 1985075 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
𝐸𝑔
0.67𝑒𝑉
∆𝐸
√𝑁 √1985075
=
=
= 0.070976%
𝐸
𝑁
1985075
∆𝐸
𝑤𝑖𝑑𝑡ℎ 1.34𝑀 − 1.32𝑀
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑
=
=
= 1.0075%
𝐸
𝑁
1985075
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑
d. (15pts) A detector system with a 10-bit digitizer (10 bits, or 210 = 1024, so the range is 0 to 1023) is calibrated with
60
Co and 137Cs sources. The 60Co lines at 1173 and 1332 keV are centered at channels 797 and 906 (the numbering
starts with 0). The 662 keV line from the Cs source is at channel 447. Determine the calibration function that relates
channels to energy (5pts). What are the minimum and maximum energies that can be measured without readjusting
the system (5pts)? Where would the 511keV and 1274.5 keV peaks from a 22Na source appear (5pts)?
𝐶ℎ𝑎𝑛𝑛𝑒𝑙 =
459
∗ 𝐸𝑛𝑒𝑟𝑔𝑦𝑘𝑒𝑉 − 6.5194
670
0=
459
∗ 𝐸𝑛𝑒𝑟𝑔𝑦𝑘𝑒𝑉 − 6.5194
670
1023 =
459
∗ 𝐸𝑛𝑒𝑟𝑔𝑦𝑘𝑒𝑉 − 6.5194
670
𝐶ℎ𝑎𝑛𝑛𝑒𝑙 =
𝐶ℎ𝑎𝑛𝑛𝑒𝑙 =
𝐸𝑚𝑖𝑛 = 9.509𝑘𝑒𝑉
𝐸𝑚𝑎𝑥 = 1502.777𝑘𝑒𝑉
459
∗ 511𝑘𝑒𝑉 − 6.5194
670
459
∗ 1274.5𝑘𝑒𝑉 − 6.5194
670
𝐶ℎ𝑎𝑛𝑛𝑒𝑙 = 344
𝐶ℎ𝑎𝑛𝑛𝑒𝑙 = 867
3. (10 pts) (10 pts) Consider the data size of a Hyperspectral Image. If the pixel array is N x M, and there are S spectral
channels, and each intensity value is B bits (typically 8 bits = 1 byte), A cube is NxMxS and the frame rate is F [for video
rates, F ~ 30 frames per second (= fps)]. The data rate for a hyperspectral cube generated F times for second is then:
NxMxSxFxB (in bits/sec).
a. Let N x M = 1024 x 1024, S=100, B=8 bits, F = 30 fps, what is the data rate?
Data rate = 1024 * 1024 * 100 * 30 * 8 = 25.165843 Gbits/sec
b. Let N x M = 2048 x 2048, S=1024, B= 12 bit, F = 1, what is the data rate?
Data rate = 2048 * 2048 * 1024 * 1 * 12 = 51.539607 Gbits/sec
c. What is the time required to transfer a single hyperspectral image (single frame) in cases (ba) and (cb) using a
wireless protocol (such as 802.11n) with a data rate of 11 Mbits/sec?
Data for one frame (case A) = 1024 * 1024 * 100 * 8 = 838.861 Mbits
Time for wireless transfer = 838.861 Mbits / 11 Mbits/s = 76.26 seconds
Data for one frame (case A) = 2048 * 2048 * 1024 * 12 = 51.539607 Mbits
Time for wireless transfer = 51.539607 Mbits / 11 Mbits/s = 4685.4188 seconds
d. What is the time required to transfer a single hyperspectral image (single frame) in cases (ba) and (cb) over a
dedicated fiber-optic communications line at 10 Gbits/sec?
Time for fiber optic transfer = 838.861 Mbits / 10 Gbits/s = 0.083886 seconds
Time for fiber optic transfer = 51.539607 Mbits / 10 Gbits/s = 5.15396 seconds
4. (10pts) Bio-Chem Sensors. Paragraph answer.
A. What are the advantages and disadvantages of PCR for bio-detection? (5 pts) PCR stands for the Polymerase
Chain Reaction. If this is not talked about in the bio-sensors class (ask!), you may need some internet searching to
help you with this question.
With PCR pathogens are detected by amplifying their unique DNA segments. If DNA amplification
occurs the pathogen has been identified to a certain degree of confidence. Typically these types of tests
required a stationary lab and expert knowledge. But companies are trying to make field units that can be
mobile and give faster results.
The major benefit from PCR is the ability to identify the very specific molecular DNA sequences of
known biological threats in a relatively short amount of time. Also units are being manufactured that can
be used out in the field to get the quickest results possible. On the down side the results from mobile units
are more prone to false-positives and are frequently double checked by other devices and even lab work.
Also the interpretation of the PCR results takes skilled knowledge and experience. Another disadvantage
is the PCR requires DNA in order to identify the substance; this cannot be used to identify a chemical
substance. This approach also requires a database of known sample to compare the results against.
B. What are the advantages and disadvantages of mass spectrometry for bio-detection? (5 pts)
A mass spectrometer is able to calculate the mass of a molecule. From this we are able to estimate what
atoms made up the molecule.
The advantage of a mass-spectrometer is the ability to detect any chemical or biological substance. Some
limitations include the size of the equipment. Designers will be hard pressed to make a mobile massspectrometer. This limits the mass spectrometer to labs dedicated to its purpose. Also because it is
measuring mass by calculating the time of flight the mass spectrometer needs to be located where there
are no vibrations that would interfere with the test results. This makes the idea of a mobile device
improbable. Also, because, molecular bonds can fall apart the analysis of the test results will have a
variable of inaccuracy.
5. (10 pts) Electron Microscope Images.
a. Here is an image of a spider. What artifacts do you see? What was done for sample preparation? Is this a life-like
pose? (see below)
Figure 2. Electron microscope: spider
One apparent artifact in this image is the placement of the detector. It’s creating the appearance of light
and dark or shaded areas. In preparation for the picture the spider had a thin layer of metal evaporated or
sputtered onto the sample so the electrons could bounce off the sample. No, this is not a life like pose;
Spiders are usually off the ground supported by their legs.
b. Here is an image of pollen grains. What artifacts do you see? What was done for sample preparation? Why do the
spikes in the center of the spherical grains look whiter or brighter?
Figure 3. Electron microsope: pollen grains
Again in this image is the artifact of the shading caused by the position of the detector, but also is the
artifact from the edges of the spikes. When the electrons collide with the spikes more electrons are
released, as compared to a flat surface. Because the spike is a narrow edge it exposes more electrons to be
released which translates into brighter pixel in the image.
6. (10 pts) Optical microscopy. Compare the following two images in Bright Field and Dark Field optical
microscopy. This is two copper lines with something(?) in between. What artifacts can you see? Is the damage a
hole or a blob on top (can you tell?)? Which image gives better information on edges? Which gives better
information on flat structures? Write a paragraph discussing what you can see and learn from these images. What
is your best guess as to what caused the damage here? Do you need other sensors or tests in order to figure that
out?
From the bright field image the center “blob” looks like it’s actually a pit from an explosion. It looks like
a small finger of copper (we can see a little piece sticking into the center from the left side) used to
connect the two larger traces. It’s possible this was some kind of fuse for a write-once programmable
chip. Where the write once chip literally has the design burned into it. We can also see in the yellowish
area some flecks left over from the explosion. The dark field image shows the obliterated pieces much
better. One could justify using another sensor like an electron microscope to look at this from another
angle and see the depth of the pit and the height of the copper wires. But these images are sufficient to
show there is a problem and suggest what might have gone wrong.
7. (20 pts) Mass Spectrometry. Learn and Do.
(a) (5 pts) Give a plausible guess for the number of Carbon atoms (x) and the number of Hydrogen atoms (y)
A possible number of 12 Carbon atoms could be attached to 26 Hydrogen atoms.
(b) (10 pts) Give a plausible explanation for each of the 5 major peaks (29, 43, 57, 71, 85) in the mass
spectrum.
The mentioned peaks are broken pieces of the C12H26 (170) molecule. When grouping the 43,
57, and 71 we get 171 which is close to the 171, just an extra hydrogen atom was picked up. We
can also see the same thing when grouping the 29, 57, and 85 together.
(c) (5 pts) What do you predict to be the ratio of peaks at (mass 170)/(mass 171) ?
The ratio will be 98.9 / (12 * 1.1) =͂ 7.49