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UNIVERSITY OF CALIFORNIA
College of Engineering
Electrical Engineering and Computer Sciences Department
145M Microcomputer Interfacing Lab
Final Exam Solutions May 15, 2009
1.1 Fourier Convolution Theorem: The Fourier transform of the convolution of two functions is
the product of their Fourier transforms (Textbook p 391, lecture slide #218).
[3 points off for stating the Fourier Frequency Convolution Theorem]
1.2 A periodic waveform is the convolution of a waveform with an infinite train of delta
functions. The Fourier transform of the latter is an infinite train of delta functions in
frequency. By the Fourier convolution theorem, the Fourier transform of a periodic
waveform is the product of its Fourier transform with an infinite train of delta functions in
frequency. The resulting Fourier transform has non-zero values only at discrete frequencies
(Textbook Figure 5.26, lecture slide #224).
r
t
Convolve
0
0
h ( t )
0
f
H ( f )
T r
2 T r
0 T r
2 T r
0 f
r
f
Multiply
0
T r
/ 2
h ( t ) h ( t ) h ( t ) h ( t ) h ( t )
-2 T r
-2 T r
/
6-bit flash A/D
6-bit D/A
6 6 MSB
Difference
amplifier
6-bit flash A/D
6 LSB
Analog input
S/H amplifier or holding capacitors were not required for full credit
1 Use one 6-bit flash A/D converter to determine the 6 most significant bits
2 Use the 6-bit D/A converter to convert the 6 most significant bits into an analog voltage
3 Use a difference amplifier to subtract the 6-bit D/A output from the analog input
4 Use the second 6-bit flash A/D converter to convert the analog difference from step 3 into
the 6 least significant bits
2. 3 The 6-bit A/D converter that determines the 6 most significant bits and the 6-bit D/A
converter that converts these into an analog voltage must be so accurate that the output of
the difference amplifier is accurate to 1/8 step size out of 2
12 = 4096 steps or one part in
32k. Since their accuracy is determined by the accuracy of their resistors, a resistor accuracy
of 1 part in 32k would be safe.
Here is an example: Imagine that the first stage has 64 steps each 64 mV wide for a total of
4096 mV and the second stage has 64 steps each 1 mV wide. Since the entire 12-bit
converter has 1 mV steps and an absolute accuracy of 1/8 LSB, the transition voltages of the
first stage must be accurate to 1/8 mV. As a worst case, imagine that the first 32 resistors of
the first stage have resistance R + e and the second 32 resistors have resistance R – e, where
e is a resistance error. The voltage at the center point of the resistor string of the first stage
would then be 4096 mV [32 (R + e) / 64R], which would be 2048 mV plus a error of 2048
mV (e/R) < 1/8 mV. This means that e/R < 1/16,000 and the resistors must be accurate to 1
part in 16k. Note that statistical analysis does not apply if the errors are systematic.
The 6-bit A/D converter that determines the 6 least significant bits only needs to have the
accuracy of a 6-bit converter, so the resistors need an accuracy of one part in 2
9 , or one part
in 512.
[Full credit for resistor accuracy 1 part in 32k or 1 part in 16k]
[3 points off for 1 part in 1,000] [4 points off for 1 part in 100]
[6 points off for expressing resistor accuracy in units of LSB- this will not be meaningful to
a resistor manufacturer]
9 if M=0, increase N by one and loop back to step 2
10 If M=1, save (D/A voltage step)(N-1/2) as the transition voltage
[1 points off for not starting the D/A input at zero]
[2 points off for not incrementing the D/A input until a A/D transition occurs]
[2 points off for not waiting for D/A to settle]
[2 points off for not waiting for A/D output data available signal; you do not know the
conversion time]
[2 points off for not enabling tri-state #1 while putting all others in high-impedance mode]
3.3 Send successive 16-bit numbers 0 to 2
- 1 to the D/A converter and convert the analog
output with the A/D. Whenever the A/D output value changes, store the corresponding D/A
value in a transition voltage table
Determine the D/A values corresponding to first and last A/D transition voltages, and the
equation of the line that passes through them. Linearity is a measure of how closely the
other measured transition values pass through the line.
[2 points off for determining maximum differential linearity or maximum absolute accuracy]
[3 points off for using the maximum deviation between the measured and ideal A/D output,
which can never be better than ½ LSB. The linearity error for the A/D is the difference
between the measured and ideal transition voltages, which can be much better than ½ LSB]
3.4 The method can determine the A/D accuracies to 1/16 LSB (±1/32 LSB was OK).
Note that 1 A/D LSB = 16 D/A LSBs.
[5 points off for an answer of 1/2 or 1 A/D LSB]
Buterworth filter
OE
OE
S / H
amplifiers
F E T
switches
A/D converter
Data available
Start conversion
Digital input port
Digital output port
Micro-computer
Buterworth filter
FET
FET
OE
OE
FET
FET
[3 points off if Butterworth is after the S/H- the Nyquist theorem requires LPF before
sampling]
[2 points off for using tri-states (which has digital input and output) rather than an FET
switches (which cannot be called a tri-state because it does not have three discrete
output states)]
4.2 Gain = 0.99 at f1/fc = 0.784 f1 = 0.784 * 25 kHz = 19.6 kHz
4.3 Gain = 0.001 at f2/fc = 2.371 f2 = 25 kHz * 2.371 = 59.3 kHz
4.4 minimum fs = f1 + f2 = 78.9 kHz
[2 points off for minimum fs = 2 f2]
1 Read the system timer to get the current tick count n in ms and set data storage index i = 0.
2 open both FET switches and set both S/H to sample mode
3 wait until tick count = n + 10i
4 bring both S/H to hold simultaneously
5 close FET switch 1
6 start conversion high
7 when DA high then read input port
8 start conversion low
9 open switch 1, close switch 2
10 start conversion high
11 when DA high then read input port
12 start conversion low
13 open switch 2
14 i = i + 1
15 loop back to step 3
[3 points off for not waiting for every 10
th
tick count to sample at an accurate 100 kHz]
[3 points off for not setting both S/Hs to hold mode simultaneously]
Standard Wi-Fi™
802.11b
Bluetooth™
ZigBee®
Battery Life (days) 0.5 – 5.
(short)
(medium)
(long)
from the central node to the host PC, we will need three types of wires for data, status and
control signal respectively.
[ 4 points off for not designing the appropriate interface between the central node and host PC]
[ 3 points off for using a system other than the experimental platform]
Solution 1:
Solution 2:
Note: The robotic arm has inertia and therefore won’t reverse the direction of stretching
immediately after you reverse the sign of A1. The absolute value of A2 will keep increasing until
the robotic arm starts rotating reversely. So, we need to wait several clock cycles before we
check absolute value of A2 again.
[ 3 points off if “several clock cycles waiting” was not included in solution2]
Our current experimental platform is not a real time system. If glitches happen when the torque
reaches the preset limit, the controller’s response (to reverse the direction of stretching) will be
delayed. The robotic device which is running in the velocity control mode will keep the original
angular velocity. That could lead to over-stretching or even hurt the patient.
[ 3 points off if you didn’t know the major potential risk is caused by the non-real-time controller]
Any real time system (DOS, RT Linux, Target PC, stand-alone micro-controller and etc.) will
reduce the risk.
[1 point off if giving solutions like “decreasing the maximum angular velocity”, or “adding
mechanical/electronic safety switches/stops”.]
145M Final Exam Grades:
Problem 1 2 3 4 5 6 Total
Average 12.6 33.1 38.5 41.8 24.6 17.1 167.
rms 4.7 5.4 4.5 2.9 4.8 2.9 16.
Maximum 15 40 45 45 30 25 200
145M Numerical Grades:
Short
labs
Long
labs
Lab Partic. Midterm #1 Midterm #2 Final Total
Average 88.9 373.4 100 83.9 77.1 167.8 891.
rms 14.2 50.3 0 9.2 10.5 16.1 81.
