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Solutions for Midterm #2 - EECS 145M Spring 1999
1a The following are essential:
- Connect all 16 lines of one parallel output port to the input of the D/A converter
- Connect the analog output of the D/A to the analog input of the A/D
- Connect 12 lines of the output of the A/D to the parallel input port
- Connect 1 line of the other parallel output port to the A/D start conversion input
- Connect the data ready output of the A/D to one of the unused lines of the parallel input port.
- Start A/D conversion under computer program control
- Use the “Data ready” A/D output to signal the program that new data are available
Start Conversion
Parallel Output Port 1
Parallel Input Port
D/A Converter
A/D Converter
16
12
μ- computer
Data Ready
Analog data (^) Ready for input data
A/D output data
D/A input data
Parallel Output Port 2
1b 1 Set “Ready for input data” low, which makes “Data Ready” low 2 Start loop over all values of n from 0 to 2^16 – 1 3 Write n to the D/A converter 4 Write a low, then a high to “Ready for input data” to start A/D conversion 5 Read “Data ready” in a loop until it goes high 6 When the A/D converter finishes, it strobes the data onto the input port and sets the “Data Ready” line high 7 The program detects this and reads the input port 8 The program sets “Ready for input data” low, which causes the A/D converter to set “Data Ready” low 9 If the A/D output has changed from the last value read (say from m–1 to m), store the value of n, which corresponds to the Vm–1, m transition voltage. 10 Loop back to step 2 11 Tabulate the difference between the measured transition voltages Vm–1, m and the ideal transition voltages V(m–1,m) = (m–0.5)(4.095V/4095) = 0.001 V (m–0.5). The maximum value is the maximum absolute accuracy error.
Essential steps: (1) vary all 16 D/A bits; (2) read A/D only after Data Ready has gone high; (3) tabulate D/A input where A/D output changes; (4) compare transition voltages with ideal
[3 points off if only 12 D/A bits are varied. The transition voltages (or the center of the steps) cannot be determined accurately unless more than 12 bits of the accuracy of the 16-bit D/A is used.]
[4 points off if the method is not automatic] [3 points off if the transition voltages are not measured] [3 points off if handshaking steps not indicated]
1c As part 1b above, but compare the measured transition voltages Vm–1, m (as a function of m) with the straight line passing between the first measured transition voltage V 0 , 1 and the last measured transition voltage V4094, 4095. The largest deviation is the maximum linearity error. [3 points off if the straight line is defined in terms of Vref–^ and Vref+^ (which is required for the absolute accuracy error). The maximum linearity error requires a straight line that passed through the measured end points]
1d As part 1b above, but compare the A/D step sizes Vm, m–1 – Vm–1, m with their average value. The largest deviation is the maximum differential linearity error. Alternatively, the A/D step sizes could be determined as the number of successive D/A inputs that produce the same A/D output. Note: It was essential to use the concept of a “table of transition voltages” to answer parts b and c of this problem.
1e Since the D/A has an absolute accuracy of ±1LSB, and its step size is 16 times finer than the average step size of the A/D, this design can measure the A/D transition voltages to an accuracy of ±1/16 of the A/D LSB. Therefore, the accuracy is ±1/32 A/D LSB for the maximum absolute and linear errors and ±1.414/32 A/D LSB for the maximum differential linearity error (difference between two random errors). Note: Due to a typo on the 1997 midterm #2 solutions, ±1 was also accepted
2a Filter gain >0.99 for frequencies <78,400 Hz
[1 point off for giving a single frequency rather than a range]
2b Filter gain <0.01 for frequencies >177,800 Hz
[1 point off for giving a single frequency rather than a range]
2c S = M ∆t = M/fs = 2^16 /2^18 Hz = 0.25 s
2d H 0 corresponds to 0 Hz (d.c.); H 1 corresponds to 1/S = 4 Hz
2e The FFT produces coefficients Hn, where n = 0 to M–1. Therefore, the coefficient with the highest index is HM-1 or H65,535, which corresponds to 4 Hz. [2 points off for HM and 0 Hz] [3 points off for HM and 2^18 Hz]
2f The FFT coefficient that corresponds to the highest frequency is HM/2 or H32,768. The corresponding frequency is (M/2)/S = 131,072 Hz
2g For a 4,000 Hz sinewave, the primary FFT coefficients are H 1000 and HM-1000. Additional neighboring coefficients H 999 , H 1001 , HM-999 , and HM-1001 are non-zero (actually half the value of the primary coefficients) due to the side lobes produced by the Hanning window. [1 point off for omitting side lobes] [4 points off for omitting harmonics]
2h For a 4,000 Hz symmetric square wave, a sequence of harmonics will appear at odd multiples of the 4,000 Hz fundamental. So Hk1000 and HM-k1000 would be non-zero, and the Hanning side lobes would be at Hk1000-1, Hk1000+1, HM-k1000-1, and HM-k1000+1. [1 point off for omitting side lobes] [4 points off for omitting harmonics]
