PYKC 3-Mar-11. Lecture 15 Slide 1. Laplace transform. Fourier transform. Discrete Fourier transform. transform. L5.8 p560.

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1 - derived from Laplace Lecture 15 Discrete-Time System Analysis using -Transform (Lathi 5.1) Peter Cheung Department of Electrical & Electronic Engineering Imperial College London URL: Consider a discrete-time signal x(t) below sampled every T sec. xt ( ) = xδ( t) + xδ( t T) + xδ( t 2 T) + xδ( t 3 T) The Laplace of x(t) is therefore (Time-shift prop. L6S13): Now define X( s) = x + xe + x e + x e +... st s2t s3t st ( σ+ jω) T σt σt = e = e = e cosωt + je sinωt X[ ] = x + x + x + x δ(t) 1 (L6S5) δ(t T) e st (L6S13) L5.8 p56 Lecture 15 Slide 1 Lecture 15 Slide 2-1 the sample period delay operator Laplace, Fourier and -Tranforms From Laplace time-shift property, we know that st is time advance by T second (T is the sampling period). 1 st = e corresponds to UNIT SAMPLE PERIOD DELAY. As a result, all sampled data (and discrete-time system) can be expressed in terms of the variable. More formally, the unilateral - of a causal sampled sequence: x[n] = x[] + x[1] + x[2] + x[3] + is given by: n= The bilateral - for a general sampled sequence is: = e X[ ] = x + x + x + x +... = xn [ ] X[ ] = x[ n] L5.8 p56 n= Lecture 15 Slide 3 Laplace Fourier Discrete Fourier Definition Purpose Suitable for.. st X() s = x() t e dt jωt X( ω) = x( t) e dt N 1 jn T ω = xne ω n= N samples, T = sample period Xn [ ] [ ] ω = 2 π / T X[ ] = x[ n] n= Converts integraldifferential equations to algebraic equations Converts finite time signal to frequency domain Converts finite discretetime signal to discrete frequency domain Converts difference equations into algebraic equations Continuous-time system & signal analysis; stable or unstable Continuous-time; stable system, convergent signals only; best for steady-state Discrete time, otherwise same as FT Discrete-time system & signal analysis; stable or unstable Lecture 15 Slide 4

2 Example of - (1) Example of - (2) Find the - for the signal γ n u[n], where γ is a constant. By definition Since u[n] = 1 for all n (step function), Apply the geometric progression formula: Observe that a simple equation in -domain results in an infinite sequence of samples. Observe also that exists only for. For X[] may go to infinity. We call the region of -plane where X[] exists as Region-of-Convergence (ROC), and is shown below. -plane : L5.1 p496 L5.1 p496 Lecture 15 Slide 5 Lecture 15 Slide 6 -s of δ[n] and u[n] -s of cosβn u[n] Remember that by definition: Since Since From slide 5, we know Hence Also, for L5.1 p499 L5.1 p5 Lecture 15 Slide 7 Lecture 15 Slide 8

3 -s of 5 impulses - Table (1) Find the -tranform of: By definition, Now remember the equation for sum of a power series: Let n k= r k n+ 1 r 1 = r 1 r = 1 and n= X [] = = (1 ) 1 L5.1 p5 L5.1 p498 Lecture 15 Slide 9 Lecture 15 Slide 1 - Table (2) Inverse - As with other s, inverse - is used to derive x[n] from X[], and is formally defined as: Here the symbol indicates an integration in counterclockwise direction around a closed path in the complex -plane (known as contour integral). Such contour integral is difficult to evaluate (but could be done using Cauchy s residue theorem), therefore we often use other techniques to obtain the inverse -tranform. One such technique is to use the - pair table shown in the last two slides with partial fraction. L5.1 p498 L5.1 p494 Lecture 15 Slide 11 Lecture 15 Slide 12

4 Find inverse - real unique poles Find inverse - repeat real poles (1) Find the inverse - of: Step 1: Divide both sides by : Find the inverse - of: Divide both sides by and expand: Step 2: Perform partial fraction: Use covering method to find k and a : Step 3: Multiply both sides by : Step 4: Obtain inverse - of each term from table (#1 & #6): We get: To find a 2, multiply both sides by and let : L5.1-1 p51 L5.1-1 p52 Lecture 15 Slide 13 Lecture 15 Slide 14 Find inverse - repeat real poles (2) Find inverse - complex poles (1) To find a 1, let = : Find inverse -tranform of:, we find: Whenever we encounter complex pole, we need to use a special partial fraction method (called quadratic factors): Use pairs #6 & #1 Now multiply both sides by, and let : We get: L5.1-1 p52 L5.1-1 p53 Lecture 15 Slide 15 Lecture 15 Slide 16

5 Find inverse - complex poles (2) Find inverse - long division To find B, we let =: Consider this example: Now, we have X[] in a convenient form: Perform long division: Use table pair #12c, we identify A = -2, B = 16, and a = -3. Thus: : L5.1-1 p54 L5.1-1 p55 Lecture 15 Slide 17 Lecture 15 Slide 18