Download EECS 145M Microcomputer Interfacing Lab: Spring 1994 Final Exam and more Exams Microcomputers in PDF only on Docsity!
NAME (please print)
STUDENT (SID) NUMBER
UNIVERSITY OF CALIFORNIA
College of Engineering Electrical Engineering and Computer Sciences Berkeley
EECS 145M: Microcomputer Interfacing Lab
LAB REPORTS:
1 ____________ (100 max) 8 _____________ (100 max) 21 ____________ (100 max)
2 ____________ (100 max) 9 _____________ (100 max) 22 ____________ (100 max)
3 ____________ (100 max) 10 ____________ (100 max) 23 ____________ (100 max)
- lowest grade dropped 25 ____________ (100 max)
HIGHEST 8 LAB GRADES ×5/
LAB PARTICIPATION
MID-TERM
MID-TERM
FINAL EXAM
TOTAL COURSE GRADE
_______________ (500 max) _______________ (100 max) _______________ (100 max) _______________ (100 max) _______________ (200 max) _______________ (1000 max)
COURSE LETTER
GRADE
Spring 1994 FINAL EXAM (May 20)
Answer the questions on the following pages completely, but as concisely as possible. The exam is to be taken closed book. Use the reverse side of the exam sheets if you need more space. Calculators are OK but not needed. In answering the problems, you are not limited by the particular equipment you used in the laboratory exercises. Many formulae from the course have been provided for you on the last page.
Partial credit can only be given if you show your work.
FINAL EXAM GRADE :
1 __________ (42 max) 3 __________ (60 max) 5 _________ (16 max)
2 __________ (50 max) 4 __________ (32 max) TOTAL __________ (200 max)
Problem 1 (total 42 points):
Define the following terms (should take no more than 40 words)
a. (7 points) Comparator circuit
b. (7 points) Harmonic (of a periodic signal)
c. (7 points) Data bus (for connecting 2 or more parallel outputs)
Problem 2 (50 points) The block diagram below shows a system for measuring the properties of eight 12-bit A/D converters automatically under computer control. The output of a single 16-bit D/A converter is used for all analog inputs. All eight A/Ds are started simultaneously and their outputs are latched onto 12-bit registers when they are done. The A/D conversion time is 10±1 μs.
A/D #
A/D #1 Tristatebuffer
12
Select #
Start Done Parallel output port #
Parallel input port
Micro- computer Select #
Start Done Tristate buffer
12
12
12
12
Parallel output port #
D/A Converter
16
12 Register
12 Register
Latch
Latch
Vref^ +^ Vref^ –
Vref^ +^ Vref^ –
a. (40 points) Describe the steps (both hardware and software) necessary to measure (i) the maximum absolute accuracy error, (ii) the minimum step size, and (iii) the maximum step size of each of the eight A/D converters. Assume that you know the reference voltages Vref+^ and Vref–. (Note: there are over ten steps)
Problem 2 (continued)
b. (10 points) Assuming that the parallel port read and write operations each take 10 μs and that the time to execute other program steps is negligible, how many seconds will the entire procedure take for the 8 A/D converters? (show work)
a. (15 points) Describe (or sketch) the Fourier magnitudes Mn from the FFT in step 3 as a func- tion of the frequency index n.
b. (5 points) To what frequency does the first Fourier magnitude M 1 correspond?
c. (10 points) Describe the gain vs. frequency relation that the low-pass filter should have.
d. (10 points) Explain whether a Hanning window would improve the recovered waveform.
e. (20 points) Describe in detail how a computer program would implement steps 2, 3, 4, and 5. (Note: there are over seven program steps)
Equations, some of which you might find useful:
V ( n ) = V ref^ −^ + n V ref^ +^ − V ref^ − 2 N
= V min + n V max^ − V min 2 N^ − 1
V out V in
1 + ( f / fc )^2 n
n = V − V ref^ − ∆ V
INTEGER
V ( n −1, n ) = V ref^ −^ +( n − 0.5)∆ V ∆ V = V ref^ +^ − V ref^ − 2 N^ − 1
G ( a ) =
exp − 1 2
a −μ σ
2 πσ^2
μ ≈ a = (^) m^1 ai i = 1
m
∑ rms^ =^ m^1 ∑ Ri^2 Ri =^ a^ +^ bni −^ Vi
a = st − rq ms − r^2
and b = mq − rt ms − r^2
where r = ∑ ni s = ∑ n i^2 q = ∑ niVi t = ∑ Vi
σ^2 = Var( a ) =
m − 1
Ri^2 i = 1
m
∑ =^
m − 1
( a i −^ a )
2 i = 1
m
∑ Var ( a^ )^ =^ Var ( a )/^ m
H ( f ) = h ( t ) e −^ j^2 π ft^ dt −∞
∞
∫ h ( t )^ =^
A for | t |≤ T 0 / 2 0 for | t |> T 0 / 2
⇒ H ( f ) = AT 0 sin(π T 0 f ) π T 0 f
H (^) n = hk e −^ j^2 π nk^ /^ N k = 0
N − 1
∑ hk =^
H (^) n N
e + j^2 π nk^ /^ N n = 0
N − 1
M (^) n = H (^) n = Re( H (^) n )^2 + Im( H (^) n )^2 tanφ (^) n = Im( H (^) n ) Re( H (^) n )
For hk = ai cos(2π ik / N ) + bi sin(2π ik / N ) H 0 = Na 0 H (^) n = ( N / 2)( an − jbn ) i = 0
N − 1
f max = fs/2 T = 1/fs S = NT ∆ f = 1/S h ( t ) = 0.5 [1.0 – cos(2π t / S )]
yi = A 1 xi − 1 + A 2 xi − 2 +... + AM xi − M + B 1 yi − 1 +... + BN yi − N If a ( t ) = b ( t ) convolved with c ( t ), then FFT( a ) = FFT( b ) multiplied by FFT( c )
f max =
2 N^ +^1 π T
e j θ^ = cosθ + j sin θ V ( t ) = V (0) e − t^ /^ RC
N = 8 9 10 11 12 13 14 15 16 2 N^ = 256 512 1,024^ 2,048^ 4,096^ 8,192^ 16,384^ 32,768^ 65,
Have a pleasant summer!
