1 / 12

Lecture 2: System’s description (LTI, causal, finite dimensional)

Lecture 2: System’s description (LTI, causal, finite dimensional). Instructor: Dr. Gleb V. Tcheslavski Contact: gleb@ee.lamar.edu Office Hours: Room 2030 Class web site: http://ee.lamar.edu/gleb/dsp/index.htm. y n. x n. b 0. +. +. +. +. z -1. z -1. a 1.

libra
Download Presentation

Lecture 2: System’s description (LTI, causal, finite dimensional)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Lecture 2: System’s description (LTI, causal, finite dimensional) Instructor: Dr. Gleb V. Tcheslavski Contact:gleb@ee.lamar.edu Office Hours: Room 2030 Class web site:http://ee.lamar.edu/gleb/dsp/index.htm

  2. yn xn b0 + + + + z-1 z-1 a1 xn-1 * Δ yn-1 b1 memory z-1 z-1 b2 a2 yn-2 xn-2 z-1 z-1 … … The Signal Flow Graph for a BIBO stable LTI finite-dimensional system (2.2.1) * Δ yn is a function of previous (and current) inputs and outputs, therefore the system is causal. If ak and bm are constants → Linear Constant coefficient Difference Equation Signal Flow Graph (SFG): Remark: because we discretize the signal by amplitude, the system becomes non-linear!

  3. Difference equation (2.3.1) yn is a function of ak, bm, xn, memory = states. (2.3.2) If we know the initial conditions, we can solve the difference equation: If (2.3.1) is a linear constant coefficient difference equation, the output consists of a homogeneous solution and a particular solution: yn = yh,n + yp,n (2.3.3)

  4. Homogeneous (complementary) solution Homogeneous equation: (2.4.1) (2.4.2) (2.4.3) The characteristic equation has N characteristic roots:1, … N • If the system is real, xn, yn, and all the coefficients (a, b) are real • For LTI systems hn are real → roots are either real or complex conjugate pairs • is a form of solution, where ci depend on initial conditions. • if 1 = 2, solution in form of (2.4.4) (2.4.5)

  5. Particular solution “We apply an input and see what happens to the output for large n” Assumption: yp,n has the same form as xn: Ex.: (2.5.1) Characteristic equation: (2.5.2)

  6. xn z-1 xn-1n z-1 xn-1n … + + Alternative form Memory is empty (relaxed system) We can find a solution in form: yn = yzi,n + yzs,n (zero input + zero state) (2.6.1) The output will be linear iff both: zero input and zero state are linear. For a relaxed system: xn = n→ yn = hn- zero state response! (2.6.2)

  7. Example… Let’s find yzi,n for the example in (2.5.1). Assuming zero input, yp,n = k (a constant) and must satisfy the LCCDE (2.7.1) Difference Equation: (2.7.2) “large enough n” implies that all terms are “active”: in our case, n ≥ 2 In form of homogeneous solution (2.7.3) (2.7.4) (2.7.5) (2.7.7) (2.7.6)

  8. Example (cont) is LTI (2.8.1) Got so far: (2.8.2) (2.8.3) (2.8.4) By substituting (2.8.4) into (2.8.1) for large n, we get: (2.8.5) (2.8.6) (2.8.7) DE (2.8.8)

  9. Example (cont 2) What’s about the zero state solution yzs,n? (2.9.1) (2.9.2) (2.9.3) (2.9.4) (2.9.5)

  10. yss,n dominates For large n Example (cont 3) (2.10.1) Another representation: transient response steady state response (2.10.2)

  11. Example (cont 4) To evaluate a system’s unit pulse response hn when (2.11.1) Particular: (2.11.2) (2.11.3) DE: (2.11.4) (2.11.5) (2.11.6)

  12. Example (cont 5) (2.12.1) (2.12.2)

More Related