Updated on: wto paĆș 05 00:16:51 CEST 2010

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  • What will all this be about?
  • slides
  • The schedule of lectures/homeworks/tests and labs
  • list of lab groups and scores
  • info on labs, example lab exercises
  • Current issues

  • Homework2 solutions If you want to take Exam2, contact me not later than Thu, 04 Feb.
  • Home page of EDISP
    (English) DIgital Signal Processing course
    winter 2009/10

    Schedule

    The lectures are on Tuesday, room 107, 14:15-16:00.

    There are lab exercises, 4 hours every second week, room 022 (basement).


    Labs will be on Mondays, 8-12 , in two subgroups of not more than 12 students.
    For the introductory lab (lab0) we met on 12 Oct, 9:15 room 022. You will be able to register for subgroups then.
    Next lab (lab1) will be

    The lab schedule is preliminary now (see schedule.pdf).

    After the Independence Day (11.11) "N" subgroup will have labs on weeks marked as "N" in the official elka calendar (and "P" on "P" weeks). First three labs are different....

    Books

    Book base

    The course is based on selected chapters of the book:

    A. V. Oppenheim, R. W. Schafer, Discrete-Time Signal Processing, Prentice-Hall 1989 (or II ed, 1999; also acceptable previous editions entitled Digital Signal Processing).

    Other books

    A free textbook covering some of the subjects can be found here: http://www.dspguide.com/pdfbook.htm
    The book is slightly superficial, but it can be valuable - at least as a quick reference.

    Additional books available in Poland:

    Please remember:

    Probably the best choice is to buy a used copy of O&S. It'll serve you for years, if you are interested in DSP. And it contains a lot of PROBLEMS to solve and learn!

    Or you may prefer to buy/borrow a laboratory scriptbook for CYPS, which is in Polish language (Cyfrowe Przetwarzanie Sygnałów, red. A Wojtkiewicz, Wydawnictwa PW).
     

    Lecture slides

    (You may always expect hand-made corrections and inserts at the lecture....)

    Lecture number = week number in schedule.


    Old slides below - this marker will be moved with slide update
    Exam 1 is scheduled for 2.02.2009, exam 2 for 9.02.2009.

    Examples of final tests (historical)

    Use them for study. Learn methods, not solutions.

    One test. Another test.

    There is no guarantee that the current test be identical ;-). It will be similar (the lecture was similar), but I might also put more focus on different subjects. The only base is the lecture content (live one, not only the published slides ....).

    The main rule: exam covers the whole course content (sampled), including the T1(H1)+T2(H2) area and also the lectures after the H2.

    Lab info: example lab exercises

    Disclaimer:

    These are called "examples" to underline the fact that they are not official. Some of them need review....

    Openly speaking, they are exercise sets current at the time of posting. I reserve the right to make some important modifications before the actual lab, to give different sets to different groups etc. (and I usually DO review the text before giving it....).


    Old instructions below - this marker will be moved with updates

    Homework solutions

    The solutions were prepared a year ago, the numbering of problems is different now.
    page 1
    page 2
    page 2A
    page 3
    page 4
    page 5
    page 6
    page 7
    page 8 - correction: ROC for u[n]-u[n-n0] is whole plane, because signal duration is finite; thus the denominator may be cancelled when you divide the polynomials (try this!)
    page 9 + some remarks:
    here a simple "manipulation type" solution for b) and c) is shown.
    Other way would be to calculate Inv. FT by integration over the passband - then a student usually forgets :-) to take BOTH passbands - for positive and negative frequencies:
    so, for HP, -π to π-θb and +θb to +π
    or, for BP, -θcb to -θcb and +θcb to +θcb
    page 10 - solution for the ex3 (old 2008/9 version, with unstable system due to c2=+0.81; with new version, c2=-0.81 and calculations are different)

    Test2 solutions

    page 1
    page 2
    page 3
    page 4
    page 5
    For Ex4 - just do it with matlab and you'll see the plot. The simplest way to do it by pencil was to use transformation from LP to HP, by shifting LP in frequency by π which results in multiplication in time by e-jπn=(-1)n. Be careful - this way passband in LP of {-θbb} is shifted to {π-θb π+θb}!
    Look at the homework!!.

    Exam 0 solutions

    Past things archive (Attic)

    Exam version A
    Exam version B
    1. In both cases the signal was sampled correctly (fs>2f)
      1. To calculate N0 it was enough to count no. of samples in period (or divide fs over f). Answer was 10(A) or 6(B). For N0 samples in period, θ0 was equal to 2π/N0.
      2. K-size DFT will have K discrete samples over <-π, +π) (we include -π, and exclude +π , but due to periodicity of spectrum it is only a convention)
        for a cosine, only two samples are non-zero: at k such that θk=±θsignal. Form the definition of θk you will see that this is for k=±4 (this is the result of taking K=4N0).
      3. You may label frequency axis with k=-K/2,....-1,0,1,...K/2 or with its periodic equivalent K-K/2,....K-1,0,1,...K/2
        to label with θ just use the expression for θk.
      1. H(z)=Y(z)/X(z) is easily obtainable from the time equation. It was
        0.2(1+z-1)/(1-0.8z-1) [A]
        -0.2(1-z-1)/(1+0.8z-1) [B]
      2. Zeros are roots of numerator: -1 [A] or +1 [B]
        Poles are roots of denominator: +0.8 [A] or -0.8 [B]. They are inside unit circle, so the system is stable (but I didn't ask...)
      3. Example graph: pic - please find a_0, b_0, b_1 by yourselves. If you are smart, you may save one multiplication by 0.2 (this is left as exercise to you).
      4. For x(n)= shifted delta, (a limited energy signal) you may take the impulse response and shift it appropriately. To find h(n) it is easiest to split H(z) into two fractions: (shown for [A], for [B] change some signs)
        0.2(1/(1-0.8z-1)+z-1/(1-0.8z-1))
        and lookup the inv.Z of 1/(1-0.8z-1) in the table. The final result is a sum of two identical exponentials shifted by 1 in time. Then, you shift h(n) to proper position....
      5. For x(n) = 1-(-1)n (a periodic signal) we see a DC component and a periodic component exp(jπn) with frequency of π. We find numerical values of H(0)=(2 or 0) and H(π)= (0 or -2) by substituting exp(0) and exp(π) for z, and finally
        y(n)=H(0)-H(π)·(-1)n
    2. The response was symmetrical around its midpoint (n= 2 or 4). Thus, it was a repsonse of a zero-phase filter delayed by 2 or 4.
      1. phase is linear φ=-(2 or 4)θ
        delay is constant and equal to (2 or 4)
      2. The response of filter is a rectangle modulated by exp(jπn). Thus, the characteristics is a sin(θL/2)/sin(θ), shifted to π. You may find the mainlobe width, you may plot exactly zeros of A(θ) etc.
    3. Time resolution is proportional to time duration of window, frequency resolution - to mainlobe width (which is prop. to 1/K ....).
      Rectangular window has narrowest mainlobe possible, but high sidelobes; so it is good for resolving signals close in frequency, but without large difference in amplitudes.
      Any other window will have wider mainlobe (so poorer resolution in f).
      1. There was nonlinearity introcuced by product of two samples (linear is multiplication by a constant only).
        Saying causal=yes because of BIBO was not enough; because FIR was enough; if you call BIBO, you have to prove it by finding relation between bound of input and output.
      2. LP filter with passband of π/4 (see the "lecture 17").
      3. Many shorter is better: by averaging we reduce the variance of estimate. (variance is huuuuuge with single FFT)
      4. β (some call it α) controls the shape of window - effectively the sidelobes level (high β - low sidelobes).
      5. Inv FT is calculated by summation when the spectrum is discrete ([B], periodic signal) and by integration when the spectrum is continuous ([A], limited energy signal).
      6. 3 buses are for opcode, data1 (signal), data2 (coefficient).
        Any instruction with dual move uses all three, e.g MAC instruction needs 2 data, so it is nice that we can load data in the same cycle in 56002 it can be coded as:
        mac a,b x:(r0+),x0 y:(r4+),y0
      7. Trivial
      8. def: order of n^2, FFT: order of n log2(n)
      9. y(n) length is, maximally, (length of h(n))+(length of x(n))-1. K=M+N-1. Here, we were asked to find M knowing K and N. Answer is, as you may guess, M=K-(N-1)
      10. The clue is in word "maximally". It may happen that for certain signal (e.g in the stopband....) the y(n) is shorter.....

    dr inż. Jacek Misiurewicz
    room 447 (GE)
    Office hours: Tue 16:30-17:00 (or by e-mail appointment)
    Institute of Electronic Systems

    email:jmisiure@elka.pw.edu.pl

     
     
     
     
     



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