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ELEC 466 : DIGITAL SIGNAL PROCESSINGTERM 2 WINTER SESSION
1995/96
Solutions to Assignment 1Sampling and Convolution
(Revised Version)
Question 1 - Problem 1.2The signal has a bandwidth of 10 kHz and
the
minimum sampling frequency to avoidaliasing willbe twice this,
or 20 kHz.
If the signal is sampled at 16 kHz the portions ofthe signal
above 8 kHz (the folding frequency) willbe aliased. The original
and aliased components
will overlap in the frequency range of 6 to 10 kHz.In the
overlap region the positive and negative
frequency components will add up. The spectrumof a signal is
complex. Unfortunately, Figure 1.31does not indicate whether it is
the real, imaginary,magnitude, or phase components that are
shown.
If we assume Figure 1.31 shows the magnitude ofthe spectrum
(i.e. X ( f ) ) and that the phase is zeroat all frequencies then
the sum of the negative andpositive signal portions of the
components will be aconstant over the overlap range as shown
below
16 10 8 6 0 6 8 10 16
Xs(f)
Question 2 - Problem 1.9
An 8-bit ADC set up to quantize over the range + A
to A has an rms quantization error voltage of
( 2 A = 2
B
)
p
1 2
We will assume that we want to find the ratio ofsignalpower to
quantization power for an input sig-nal which is a sinusoid. Note a
signals power spec-trum does not define its probability
distributionand thus we cannot deduce either its peak value orthe
aliasing error of the signal in the passband sim-ply from a
knowledge of its spectrum.
A sinusoid of frequencyf
and amplitudeA
at the
input of the anti-aliasing filter will appear at theoutput with
an rms voltage of
A
p
2
1
p
1 + ( f = f
c
)
6
The filter passband extends from 0 to 3.4 kHz. Weneed to findthe
sampling frequency so that thelevel
ofa signalaliased to 3.4 kHz will have the samelevelas the
quantization noise.If we set the two levels equal and solve for f
,
f = f
c
1 5 2
2 B
1
( 1 = 6 )
WithB = 8
andf
c
= 3 4
we findf = 2 3 1
kHz. Sothe sampling frequency must be set to 26.5 kHz sothat a
signal component at 23 kHz, aliased down to3.4 kHz and have the
same level as the quantizationnoise.
If we wanted the maximum aliasing level over therange 0 to F
s
= 2 to be the same as the quantizationnoise level then we would
set F
s
= 4 6 kHz.If we follow the approach in the text in Example
1.2, we find that A min = 2 0 l o g (p
1 5 2
B
) =
5 0 dB (note the error in the text in the equation forA min).
SettingA min = 2 0 l o g 1 + ( f = f c )
6 1 = 2 wefindf = 2 3 1 kHz. Thus we would set the sampling
fre-quency to
2 3 1 + 3 4 = 2 6 5
kHz.
Question 3 - Problem 1.10
The quantization step size is
V
f s
= ( 2
B
1 ) = 5 = ( 2
1 6
1 ) = 8 3 9 V
Therms signal to quantizationnoiseratio(SQNR)
is given by equation 1.11:
6 0 2 B + 1 7 6 dB = 9 8 dB
Question 4 - Problem 1.15
The wording of this question leaves a bit to be de-sired. Assume
thespectrumin thefigure is thespec-trum that would result if the
samples at the outputof the DSP system were passed through an ideal
re-construction filter. That is, the digital signal con-sists of
samples of the sum of two sinusoids, oneat 1 kHz and one at 3 kHz.
Furthermore, assumethat y ( n ) is real so that the spectrum is
symmetricalabout the origin.
First, since the signal is sampled, the spectrumwill repeat at
multiples of the sampling frequency,15 kHz. Thus there will be
signal components atn F
s
1 , and n Fs
3 where n is all possible integersincluding zero. Only the
components at 0+1, 0+3,15-3, 15-1, 15+1, 15+3, 30-3, and 30-1 (1,
3, 12, 14,
1
