z-transform derived from Laplace transform Consider a discrete-time signal x(t) below sampled every T sec. Lecture 15 δ(t) ⇔1(L6S5) xt( )=+ xδδ ( t ) x ( t− T )+ x δ ( t− 2 T )+ x δ ( t− 3 T )+ ..... 01 2 3 δ(t − T) ⇔ e−sT (L6S13) Discrete-Time System Analysis The Laplace transform of x(t) is therefore (Time-shift prop. L6S13): using z-Transform −−sT s23 T − s T Xs( )=+ x01 xe + xe 2 + xe 3 + ..... (Lathi 5.1) sT()σω+ j T σ T σ T Now define ze== e = ecosωω T + je sin T −−−123 Xz[ ]=+ x01 xz + xz 2 + xz 3 + ..... Peter Cheung Department of Electrical & Electronic Engineering Imperial College London URL: www.ee.imperial.ac.uk/pcheung/teaching/ee2_signals E-mail: [email protected] L5.8 p560 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 1 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 2 z-1 the sample period delay operator Laplace, Fourier and z-Tranforms sT From Laplace time-shift property, we know that ze = is time advance Definition Purpose Suitable for .. by T second (T is the sampling period). −−1 sT ∞ Therefore ze = corresponds to UNIT SAMPLE PERIOD DELAY. Laplace Xs()= xte ()−st dt Converts integral- Continuous-time system ∫−∞ differential equations to & signal analysis; stable As a result, all sampled data (and discrete-time system) can be transform algebraic equations or unstable expressed in terms of the variable z. Fourier ∞ More formally, the unilateral z-transform of a causal sampled Xxtedt()ω = ()− jtω Converts finite time Continuous-time; stable sequence: x[n] = x[0] + x[1] + x[2] + x[3] + transform ∫−∞ signal to frequency system, convergent domain signals only; best for is given by: steady-state N0 −1 −−−123 Discrete jnω T Converts finite discrete- Xz[ ]=+ x xz + xz + xz + ..... Xn[] xne []0 01 2 3 ω0 = ∑ time signal to discrete Discrete time, otherwise Fourier n=0 ∞ frequency domain same as FT −n transform NTsamples,= sample period = xnz[] 0 ∑ ω = 2/π T n=0 0 Converts difference Discrete-time system & z ∞ signal analysis; stable The bilateral z-transform for a general sampled sequence is: Xz[]= xnz []−n equations into ∞ transform ∑ algebraic equations or unstable Xz[]= xnz []−n n=−∞ ∑ L5.8 p560 n=−∞ PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 3 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 4 Example of z-transform (1) Example of z-transform (2) Find the z-transform for the signal γnu[n], where γ is a constant. Observe that a simple equation in z-domain results in an infinite By definition sequence of samples. Observe also that exists only for . Since u[n] = 1 for all n ≥ 0 (step function), For X[z] may go to infinity. We call the region of z-plane where X[z] exists as Region-of-Convergence (ROC), and is shown below. z-plane Apply the geometric progression formula: Therefore: L5.1 p496 L5.1 p496 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 5 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 6 z-transforms of δ[n] and u[n] z-transforms of cosβn u[n] Remember that by definition: Since From slide 5, we know Since Hence Also, for Therefore Therefore L5.1 p499 L5.1 p500 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 7 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 8 z-transforms of 5 impulses z-transform Table (1) Find the z-tranform of: By definition, Now remember the equation for sum of a power series: n r n+1 −1 ∑r k = k=0 r −1 −1 Let rz== and n4 z−5 −1 Xz[]= z−1 −1 z = (1− z−5 ) z −1 L5.1 p500 L5.1 p498 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 9 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 10 z-transform Table (2) Inverse z-transform As with other transforms, inverse z-transform is used to derive x[n] from X[z], and is formally defined as: Here the symbol indicates an integration in counterclockwise direction around a closed path in the complex z-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 z-tranform. One such technique is to use the z-transform pair table shown in the last two slides with partial fraction. L5.1 p498 L5.1 p494 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 11 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 12 Find inverse z-transform – real unique poles Find inverse z-transform – repeat real poles (1) Find the inverse z-transform of: Find the inverse z-transform of: Divide both sides by z and expand: Step 1: Divide both sides by z: Step 2: Perform partial fraction: Use covering method to find k and a0: Step 3: Multiply both sides by z: We get: Step 4: Obtain inverse z-transform of each term from table (#1 & #6): To find a2, multiply both sides by z and let z→∞: L5.1-1 p501 L5.1-1 p502 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 13 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 14 Find inverse z-transform – repeat real poles (2) Find inverse z-transform – complex poles (1) To find a1, let z = 0: Find inverse z-tranform of: Therefore, we find: Whenever we encounter complex pole, we need to use a special partial fraction method (called quadratic factors): Use pairs #6 & #10 Now multiply both sides by z, and let z→∞: We get: L5.1-1 p502 L5.1-1 p503 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 15 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 16 Find inverse z-transform – complex poles (2) Find inverse z-transform – long division To find B, we let z=0: Consider this example: Now, we have X[z] in a convenient form: Perform long division: Use table pair #12c, we identify A = -2, B = 16, and a = -3. Thus: Therefore: Therefore L5.1-1 p504 L5.1-1 p505 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 17 PYKC 3-Mar-11 E2.5 Signals & Linear Systems Lecture 15 Slide 18 .