April 7, 2010 page 1 S. Derenzo
Solutions for Midterm #2 - EECS 145M Spring 2010
1.1 Frequency aliasing is caused by sampling at a frequency fs that is less than twice the highest
frequency fmax in the waveform.
It may be avoided by using an analog low-pass filter to eliminate frequencies above fs/2.
[4 points off for increasing the sampling frequency. When sampling an arbitrary waveform
you don’t know the maximum frequency.]
1.2 Spectral leakage is caused when frequency components are not sampled for a whole number
of cycles, which results in a discontinuity between the last sample and the “next sample”,
which is also the first sample.
It is avoided by multiplying the sampled values by a windowing function that has zero value
and zero slope at the ends of the sampling window.
[3 points off for a solution that samples for a longer time- this does not eliminate the
discontinuity and long-range spectral leakage will still occur]
[3 points off if the only answer is to sample for an integer number of cycles- nothing is
known about the frequency components present]
2.1 Filter gain >0.99 for frequencies <78,400 Hz
2.2 Filter gain <0.01 for frequencies >177,800 Hz
2.3 S = M ∆t = M/fs = 216/218 Hz = 0.25 s
2.4 H0 corresponds to 0 Hz (d.c.); H1 corresponds to 1/S = 4 Hz
2.5 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 218 Hz]
2.6 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
2.7 For a 4,000 Hz sinewave, the primary FFT coefficients are H1000 and HM -1000. Additional
neighboring coefficients H999, H1001, HM-999, and HM-1001 are non-zero (actually half the value
of the primary coefficients) due to the side lobes produced by the Hann window.
[2 points off for omitting side lobes] [2 points off for omitting HM-999, HM-1000, and HM-1001]
2.8 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
Hann side lobes would be at Hk1000-1, Hk1000+1, HM-k1000-1, and HM-k1000+1.
[1 point off for omitting side lobes] [3 points off for omitting harmonics]
2.9 For a 4,002 Hz sinewave, H1000, H1001, HM-1000, and HM-1001 would be non-zero and of equal
magnitude, and the Hann side lobes would appear at H999, H1002, HM-999 and HM-1001.
[1 point off for omitting side lobes] [2 points off for omitting HM-1000 and HM-1001]
[4 points off for stating that all coefficients are non-zero]
2.10 The primary 4,000 Hz sinewave would produce non-zero values at H999, H1000, and H1001. A
second smaller sinewave of slightly higher frequency 4,000 + 4m Hz would produce non-
zero values at H1000+m -1, H1000+m, and H1000+m+1 (there are also complex conjugate
coefficients at HM-1000, etc.). For the smaller sinewave to appear as a separate peak, there
must be a valley between the coefficient H1001 and the coefficient at H1000+m, which requires
1000 + m > 1002, or m >2. The smallest value of m we can have is 3, which corresponds to
a frequency 12 Hz above 4,000 Hz.
