Corrections to the Text and the Lecture Notes
Corrections to Textbook
Corrections to Lecture Notes
Corrections to Assignment Solutions
Corrections are shown in the order they were
reported. Corrections to previous corrections (!) are also given
below. Please let me know if you find an error that is not
listed below.
- Page 58, last paragraph and Figure 2.3: the frequencies of
the DFT components are in kiloradians/second not kHz.
- Page 17, Figure 1.10 (b): change P(f) to X(f).
- Page 19, Figure 1.12 (b): change Fc and Fs to fc and fs
(lower case) to agree with the text.
- Page 23, Example 1.2. When computing the maximum
passband signal level the text uses Vfs/sqrt(2) as the maximum
passband signal level. However, for a sinusoid of amplitude A
the rms level is A/sqrt(2) = (Vfs/2)/sqrt(2).
This changes the factor of 2^(B+1) to 2^B and changes Amin to 74
dB.
This also affects Equation 1.7 and Table 1.2.
Note that the calculation of the SQNR in section 1.4.2.2 does use
A/sqrt(2) as the signal level (the power is A^2/2).
- Page 128, property (2), "Delays or shifts." The Z transform
of x(n-m) is z^{-m} X(z).
This section lists corrections previous posted that turned out to
be unnecessary.
- I had written:
Page 23, Example 1.2, solution to part (3): the aliasing level is
computed for a sampling frequency of 194 kHz. For a sampling
frequency of 173 kHz the aliasing level is 13.3E-6 (-98 dB). The
aliasing level relative to the signal level at 4 kHz is 9.4E-4
(0.00094) percent.
But:
in part (3) we are solving for aliasing level relative to the
signal level at 4 kHz while in part (2) we were solving for
aliasing level relative to the maximum passband signal level.
The passband signal level at 4 kHz is 3 dB less than the maximum
so the required Amin is 83 dB. For this value of Amin Fs/2 is 97
kHz and Fs is 194 kHz.]
- I had written:
Page 23, Example 1.2, last paragraph. The second-last
sentence should read: "A lower sampling frequency will result if
we take 4 kHz as the aliasing frequency (as in Example 1.1). In
this case Fs = 86.2 + 4 = 90.2 kHz and the aliasing level at 4
kHz is equal to the quantization noise level."
But:
The original wording makes sense if ``aliasing frequency'' means
the difference between Fs and the frequency at which aliasing
happens (which is the way the term is used in the text).
-
FFT - Part 2, page 1. If we consider the lower 4-point DFT
box to be a single operation the inputs are x1, x3, x5 and x7.
When we ``open up'' the 4-point DFT box and see how it is
decimated into two 2-point DFTs, the inputs to the top 2-point
DFT are x1 and x5 and to the lower DFT the inputs are x3 and x7.
If we were to label the inputs to the 4-point DFT as y0, y1,
y2 and y3 (i.e. y0 = x1, y1 = x3, etc) then the inputs to the top
2-point DFT are y0 and y2 and those to the lower 2-point DFT are
y1 and y3.
z-Transform, page 3. The diagram and signal flow graph show
the feedback coefficients as b1, b2 and b3. To agree with the
notation given in the notes these should be -b1, -b2 and -b3.
The exercise should start "If a0=1 and and the bk's are zero..."
and "If the only non-zero b coefficient is ...".
If b1=1 the impulse response is {1,-1,1,-1,...} if b1=-1 the
impulse response is {1,1,1,1,...}.
Solutions to Assignment 1, Question 3. The equation for the
step size is correct but the equation evaluates to 76 microvolts,
not 839. I computed 55/(2^16-1) rather than 5/(2^16-1) (the '5'
key on my calculator ``bounces'').
Solutions to Assignment 4, Question 4. The signal flow
graph does not correspond to the equation transfer function in
the solution. The correct flow graph
does not have a direct branch between x(n) and y(n) since this
would correspond to a term of '1' in the numerator.
Solutions to Mid-Term Review Assignment. Question 1: the
level of the impulses shown in the sketch of the spectrum should
be 5/2 and 3/2 rather than 5 and 3. Question 6: the twiddle
factors of j and -j shown in the signal flow graph should be -j
and j instead. Question 8(a): the transfer function also has a
pole at zero.
Solutions to Assignment 1, Question 3. The equation for the
transfer function of the zero-order hold is wrong. It should be
H(f) = T sinc(Tf) = T sin(pi T f)/(pi f T). The numerical values
of |H(f)| in the table will be different but the general shape of
the graph will remain the same.
Solutions to Assignment 2, Question 6. The values of the
twiddle factors were computed using W^-k rather than W^k so the
values of the imaginary components of the twiddle factors should
be negated.
Solutions to Mid-Term, Question 3. The value of |H(z)| for
f=omega=0 is |H(z=1)| = 4/0.2 = 20, not |H(z=0)|=1.25.
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