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Unformatted text preview: 115 16 11—7 11—8 11—9 '11—10 A 11—12 A ten volt—(maximum) signal is to be digitized to a resolution of at most 0.01 volts. .. How many bits are needed in an A/D converter to do this? — 10 bits ..._‘I_ How does a successive approximation A/D converter work?
It tests each bit successively starting at the most significant bit. This bit drives adigital—to—analog converter that generates a signal for one side of a comparator.
The comparator compares the input analog signal with the output of the D/A and
if the input is higher, the bit remains set, otherwise it is reset. Each successive
bit is tested in this way. How does a dual slope A/D converter work?
It integrates the input signal for a ﬁxed time. At the end of this time, it discharges the integrated value at a fixed rate and measures the time taken to reach zero.
This time is then related to the digital value. How does a ﬂash converter work?
A ﬂash converter consists of 2N—1 comparators. The output code is produce in a flash in this way.
TheA/D converter conversion time is 100 us. What is the maximum frequency that can
be digitized without aliasing occurring? 5 KHz
An A/D converter is required to digitize a 1 KHz sinusoidal waveform. What is the
maximum allowable conversion time for the A/D? Assume a sampleandhold circuit
is being used to give the correct aperture time. 0.5 ms
An A/D is to digitize a 10 volt full—scale signal to a resolution of 1 part in 1024. 0.1 How many bits are required? 10 bits
0.2 When a 9 volt signal is being digitized, what is the accuracy of the measurement? The resolution is 10V/ 1024 = 9.8 mV. The accuracy at 9 volts is within (9.8 mV/9 V) = 0.1%
What is the accuracy when a 1 volt signal is digitized? (9.8 mV/l V): 0.9%
Atmnsducer is to be used to find the temperature over a range of —100 to 100" C. We
are required to read and display the temperature to a resolution of :1" C. The
transducer produces a voltage from five to plus five volts over this temperature range
with 5 millivolts of noise. Specify the number of bits in the A/D converter (a) based
on the dynamic range and (b) based on the required resolution. 0.1 The dynamic range is (10 V)/(0.005 V) = 2000. Thus, a 11—bit A/D converter is
required if the noise is to be :1/2 LSB.
0.2 The required resolution is 200 "/1 .. = 200. An 8—bit converter will meet these specifications.
An A/D converter is to be specified for the following measurement: The signal is DC (it will not vary during the conversion time); the signal range
is zero to ten volts; there is 1 mV of noise; when a one volt signal is being
measured, the measurement is to be within :0.5% of the true value. How many bits are required, and how would you specify the conversion time and aperture time?
The accuracy specification determines the resolution, with 4.05% at one volt
equal to :1/2 LSB. Therefore, at one volt, the A/D resolution is to be 0.01 V
giving a dynamic range of (10 V/0.01 V) = 1000. The number of bits required
is 10. The conversion time specification is not driven by aliasing considerations
because the signal is DC. Therefore, the conversion time should be determined
by thefrequency at which the signal processing software needs new information.
Likewise, with a DC signal, there is no aperture time requirement and a sample—
and—hold will not be needed. \2 Spring Semester 200.6
Final Quiz Print Your First Name (2 pts.)  Open book/open notes, 65 minute examination to be done in pencil., Calculators are permitted. MM Point System (for instructor and TA use only) Page 1 18 points
Page 2 31 points
Page 3 23 points
Page 4 28 points
TOTAL 100 points 1.0 ND & D/A Conversion You are given the task of sampling audio signals from a Neumann U87 microphone and preampliﬁer circuit.
The mic and preamplifier speciﬁcations are as follows: . Mic dynamic range = 110 st. Preampliﬁer dynamic range = 140 st
Mic frequency bandwidth = 522 KHz Preampliﬁer bandwidth = 11OOKHz
Mic minimum Voltage detection = 1 uV Preamplifier Power Supply Rails = +/ 3V 1. What should the reference voltages of the AID be if we neglect the mic and only consider the maximum
signal output characteristics of the preamplifier? (4 pts.) +l 3V 2. What should the reference voltages of the AID be if we take into account both the specifications for the
preampliﬁer and the mic? (4 pts.) 110 st = 20 log (Vmax/1uV) Vmax = 0.3162V Vmax/2 = 0.158V answer = +l 0.158V 3. If the maximum output of the preamplifier is measured on an Oscope to be +/— 1.0V peak to peak and the
AID rails have been set to +/ 2.5V, what will be the accuracy for a 16 bit AID? (4 pts.) 0.0038% Resolution = 5V/2"16 Accuracy = (Resolution/2V)*100% 4. If we use this to sample voice only, which is band limited to 8KHz, and a student sets the AID sampling
period to 50 usec. We will not observe aliasing. Explain why? (4 pts.) Tsample = 50 usec Fsample = 20 KHz Fnyquist = 10 KHz Fmax = 8 KHz Fsample > 2*Fmax (Shannon) or Fnyquist > Fmax (Nyquist) i
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references set at +/ 2.5V what is expected decimal digital value when the voltage at the input of the AID is measure to be 0.258V? (4 pts.)
Resolution = 5/2A16 +2.5V => 7FFF 0.258V/Resolution = 3381 or —3382 0 => 0
2.5V => 8000 6. What is the expected Hex value for the solution in #5? (3 pts.) 3382 => 1101 0011 0110
001011001001 +1 => 1111001011001010 => FZCAHeX 7. if we assume there is 1mV of noise in the preampliﬁer circuit, what is the range of digital decimal values for
the AID in #5? (4 pts.) 0.258 +I (0.00112) 0.2585, 0.2575 3388, 3375 => 13 or 14 bits 8. If we again assume there is 1mV of noise in the preampliﬁer circuit, how many bits are corrupted in the AID
described in #5? (4 pts.) 4 bits 9. Assuming again we have 1mV of noise in the preampliﬁer what is the expected dynamic range of the AID
described in #5? (4 pts.) 12 good bits * 6 st = 72 st or 20 log (5/0001) ~ 74 st 10. What is the expected accuracy for the AID in #5 assuming again we have 1mV of noise and a perfect
matched input signal to the AID? (4 pts.)
Accuracy = (resolution * 100%)Isignal span = (0.001*100%)l5 = 0.02%
11. A student would like to sample 30 seconds of data at 44100 Hz using the 16 bit AID previously described in
#5, how many bytes are required to buffer the sampled sound? (4 pts.) 44100*30*2 = 2,646 Kbytes = 2.646 Mbytes 12. The same student mentioned in #11 would now like to play the sound out at 44100 Hz to an unsigned 16
bit BIA with references set at +/ 6V. What is the expected voltage for $F000? (4 pts.) FFFF > 6V F000 => ? resolution = 12/2"16 (resolution*61,440) — 6 = 5.25V
8000 => 0V 0 => 6V ‘/ i
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 Spring '09

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