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SGN-11007 Introduction to Signal Processing 8.1.2019

 

SGN-11007 Introduction to Signal Processing,
Final Exam, 8.1.2019,
Sari Peltonen

 

 

 

e Own calculators can be used in the exam.

e You may take the examination paper with you.
1. Explain briefly (in one or two sentences)

(a) Nyguist freguency, (1p)

(b) LDA, (1p)

(c) Causal filter, (1p)

(d) Passband ripple, (1p)

(e) Support vector, (1p)

(f) Multistage decimation. (1p)

2. (a) Calculate the DFT of the vector x(n) = (5,7,—1,—1)". (3p)
(b) The system shown below can be implemented as one LTI system. What is the impulse
response h(n) of that system expressed by using the impulse responses h; (n), h;(n)
and hs(n). (3p)

 

——— hz(n)

 

 

 

 

( J) yln)

 

 

ha(n)

 

 

 

3. Design using the window design method a filter (i.e. find out its impulse response) satis-
fying the following reguirements:

Stopband [12 kHz, 16 kHz]
Passband [0 kHz, 10 kHz]
Passband ripple 0.06 dB
Minimum stopband attenuation 48 dB
Sampling freguency 32 kHz

Use the tables below. (6p)

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SGN-11007 Introduction to Signal Processing 8.1.2019

4. (a) The pole-zero plot of a filter is shown below, and we know that its amplitude re-
sponse |H(e'")| € [0,1]. Sketch the amplitude response of the filter as accurately as it
is possible with the information provided. (2p)

(b) Is the filter stable? Why / why not? (2p)
(c) Is the filter FIR or IIR filter? Justify. (2p)

k

0.8t o ;

 

0.6+ o J
0.4+ | 4
0.2+ 4
of kanoaren 12 mei us 83 v ]

-0.2 J

Imaginary Part

-0.4- i
-0.6- . o d
0.8 o N

14 A nerous 4

 

 

Real Part

5. Filter j : j

pm + 7rn —1)+ yin —2)

is implemented in an eguipment having sampling freguency 16000 Hz. What is the am-
plitude response of the filter at the freguency 4000 Hz? (6p)

yln) =

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Tables

Ideal Impulse response when

filter type n yo n=0

Low-pass 2f.sinc(n -2nif.) 24.

High-pass —2fcsinc(n -2nf,) 1—2f.

Band-pass | 2f;sinc(n -27f2) — 2fisinc(n -2nf;) 2(170 — f1)

Band-stop | 2fisinc(n - 27f;) — 2fasinc(n - 27f2) | 1—2(f2—f)
Name of Transition Passband | Minimum Window expression w(n),
the window | bandwidth ripple stopband when |n| < (N — 1)/2
function (normalized) | (dB) attenuation (dB) |
Rectangular | 0.9/N 0.7416 21 1
Bartlett 3.05/N 0.4752 —|25 1-
Hanning 3.1/N 0.0546 44 0.540.5cos (57)
Hamming | 3.3/N 0.0194 53 0.54 + 0.46 cos (R)
Blackman —|5.5/N 0.0017 74 0.42 + 0.5 cos (27) + 0.08 cos (12)

Eguations
—b b? — 4ac

 

 

ax? + bx+c=0&x=
2a

Some Wikipedia pages that might be useful

 

Suppose two classes of observations have means Ko , h and covariances Yo, &1. Then the linear combination of features W « % will
have means % - hi; and variances WW fori= 0,1. Fisher defined the separation between these two distributions to be the ratio
of the variance between the classes to the variance within the classes
52 Famen — (R — 0-7 — (5-0 7)
iin — VY 0+V20 05 +2)5
This measure is, in some sense, a measure of the signal-to-noise ratio for the class labelling. It can be shown that the maximum
separation occurs when

5 (2o + 2) (fi, — Äy)
Wihen the assumptions of LDA are satisfied, the above eguation is eguivalent to LDA

Be sure to note that the vector 0 is the normal to the discriminant hyperplane. As an example, in a two dimensional problem, the line
that best divides the two groups is perpendicular to 15.

Generally, the data points to be discriminated are projected onto i; then the threshold that best separates the data is chosen from
analysis of the one-dimensional distribution. There is no general rule for the threshold. However, if projections of points from both
classes exhibit approximately the same distributions, a good choice would be the hyperplane between projections of the two means,
W- jin and W- fi, . In this case the parameter c in threshold condition 44 + % > c can be found esplicitly:

ad a Vatsa Lomaa
= 5: 5llo +) = 5Ä DR — 55 2 Pu

 

 

 

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SGN-11007 Introduction to Signal Processing 8.1.2019

 

A more condensed form of the difference eguation is:

1/2 a.
yln] = a (Din —4- 5 ajyin - 3)
120 ja

which, when rearranged. becomes:

o P
5 asuin -j] = Yy dialn —1]
0 10

To find the transfer tunction of the filter, we first take the Z-transform of each side of the above eguation, where we use the time-shift
property to obtain:

o
Yy ajz *Ylz) = S biz *X(2)
i=0

E

We define the transfer function tobe:
Y(2)

H(z)

 

Considering that in most IIR filter designs coefficient ag is 1. tne (IR filter transfer function takes the more traditional form.

Yiobit"
1+ Ei a;z”

 

H(2) =

 

 

Inversion of 2x2 matrices [edit]

The cofactor eguation listed above yields tne following result for 2 x 2 matrices. Inversion of these matrices can be done as tollows:1!

a =[ä 5] = 1 []- 1 [< 3]
— [e d =TtAl|-e a) ad-bel-e a)

 

 

 

Technigues sit] ————.

an analog continuous signal, then re-sampling at the new
rate, or calculating the values of the new samples directly from the old samples. The latter approach is more satisfactory, since it
introduces less noise and distortion 3! Two possible implementation methods are as follows

Conceptual approaches to sample-rate conversion include: converting to

4. 1f the ratio of the two sample rates is (or can be approximated byj'nd 1114] a fixed rational number (/M: generate an intermediate
signal by inserting L -10sbetween each of the original samples. Low-pass filter this signal at half of the lower of the two rates.
select every M-th sample from the filtered output, to obtain the result 15]

2. Treat the samples as geometric points and create any needed new points by interpolation. Choosing an interpolation method is
a trade-off between implementation complexity and conversion guality (according to application reguirements). Commonly used
are: ZOH (for film/video frames), cubic (for image processing) and windowed sinc function (for audio).

 

 

 

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