Lecture 5: Test 1 review

Lecture 5: Test 1 review 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

Fundamental system’s properties (5.2.1) • Linearity: • Time-invariance: • Causality: • Stability (BIBO): (5.2.2) (5.2.3) (5.2.4) 1,2,3 Must hold for ANY inputs and constants! Most of properties are specified for the relaxed systems (empty states)!

Difference equations No driving terms – homogeneous solution. Whatever is stored (represents the memory of the system)  operations on the previous inputs  initial conditions  reflected in the homogeneous solution. Usually: (5.3.1) Remember and understand SFGs of an LTI system. states (5.3.2) xn yn zero-input transient zero-state steady-state memory (5.3.3) LTI The connection between (5.2.1), (5.2.2), and (5.2.3) is the same (in general) total response

Difference equations Example: find the total solution for the system specified by the following DE: For the following input and the initial conditions: Let’s find the homogeneous solution first: xn = 0. The characteristic eqn.: Characteristic roots are 1 = -3; 2 = 2. Therefore, the homogeneous solution is: For the particular solution, we assume: Therefore, combining with DE for large n:

Difference equations Which gives k = -2. Then, the total solution is: We find the constants by satisfying the initial conditions: Finally, the total solution is:

Difference equations Alternatively, the zero-input response can be found as a form of the homogeneous solution, where the constants are chosen to satisfy the initial conditions. From the equation for the homogeneous solution, we get: Therefore: C1 = -5.4; C2 = -1.6 The zero-input response is

Difference equations The zero-state response can be found as a form of the total solution when zero initial conditions are satisfied. We find the constants for the total solution as C1 = 3.6 and C2 = 6.4. Therefore, the zero-state response for n 0 and with i.c.s yzs,-2 = yzs,-1 = 0 is: Again, the total solution is:

LTI systems (5.4.1) If the system is linear: (5.4.2) since (5.4.3) Therefore: (5.4.4) Alternatively: (5.4.5) (5.4.6) (5.4.7)

SFG (5.5.1) (5.5.2) In general! (5.5.3) (5.5.4) - Relaxed system (5.5.5) If there is no input, the output must go to zero! What’s about unstable AND not-relaxed systems?

SFG Signal flow graph does not necessarily have to represent Fundamental of Direct 2 forms! They can be arbitrary… d1 z-1 A perfectly valid form of SFG: xn d2 c2 c1 d3 yn Find the LCCDE and state-space representation quantities! Regrouping an SFG might be a good idea! Unit step response: empty states, impulse at the input.

DTFT and DFT DTFT of a finite length sequence (sampled at ) is DFT{xn} generally, aliased result N “sampling” usually leads to “aliasing”! IDFT{XkHk} = - a circular convolution IDFT{XkHk} = - a linear convolution – a practical application after sufficient zero-padding “FFT” really means “Fast DFT”.

Suggestions You may need proofs, derivations, properties… Look at the examples covered in class. Think about “good questions” for the test! SFG  {A,b,c,d} Need to understand and be able to test for: Linearity, Time Invariance, BIBO, Causality. properties? What are and how to find: yss,n, ytr,n, yzi,n, yzs,n, yh,n, yp,n? Direct 1 and Direct 2 – implementation aspects Be able to interpret results of DFT, What are the frequency response (LTI), Impulse response, transient response? Purpose of scaling??

Lecture 5: Test 1 review