[PPT].Lesson 04-Aliasing

[PPT].Lesson 04-Aliasing - Repeat Challenge 02 Challenge 02...

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Aliasing Challenge 03: Sampling Lesson 04 liasing Aliasing What is it? How do you measure it? How do you manage it? Challenge 04 Lesson 04 epeat Challenge 02 Repeat Challenge 02 1 st -Order Impulse Invariant System simple first- rder C ircuit is show. A simple first order RC circuit is show. The relationship between the input forcing function v ( t ) and voltage ld t h i tdt d i df i db th 1 t d developed across the capacitor, denoted v o ( t ), is defined by the 1 st order ordinary differential equation v hat is the equivalent discrete me model of the circuit’s impulse ) ( 1 = ) ( 1 ) ( 0 0 t v RC t v RC dt t dv What is the equivalent discrete-time model of the circuit s impulse response based on a sample rate of f s =1 k Sa/s, and RC=10 -2 ? Lesson 04 epeat Challenge 02 Repeat Challenge 02 It immediately follows that the system’s impulse response is given by: 1 = / - RC t If the sampling period is T s , the resulting discrete-time impulse esponse is: ) ( ) ( t u e RC t h response is: kT h T h or, for k 0, ) ( ] [ s s d k s RC kT s d RC T e RC T k h s / - = ] [ RC T s e / ; α Lesson 04 epeat Challenge 02 Repeat Challenge 02 or C 10 and 1000 z the following results For RC =10 -2 , and f s =1000 Hz, the following results. 0 0 0 1 . 0 10 / 10 / ; 904837 . 0 2 3 1 . 0 10 10 * 10 1 3 2 RC T e e e s s s d C kT h T k h ) ( ] [ k k s e RC T ) . ( . ) ( / . 904837 0 1 0 1 0 This represents the impulse response in the discrete time domain but does not reduce the impulse response to a difference equation. Lesson 04 epeat Challenge 02 Repeat Challenge 02 Where would you find a connection between a time-domain function nd difference equation? able of z ansforms (signals and systems and difference equation? – Table of z-transforms (signals and systems stuff). able 7 1 (Principles of Signals and System F Taylor McGraw ill Known, h d [k] (T s /RC) e -kT s /RC =(0.1) (0.904837) k for RC =10 -2 , and Table 7.1 (Principles of Signals and System, F. Taylor, McGraw-Hill 1 1 z kT u e k h s kT kT akT ) ( ] [ f s =1000 Hz, the following results. 1 1 904 0 1 1 z z e e z s s akT akT s d ) . ( Therefore: 1 1 904 0 1 1 0 1 1 0 1 0 1 0 z z e e z z kT u e k h s s s akT akT s akT d ) . ( . . . ) ( . ] [ Lesson 04 epeat Challenge 02 Repeat Challenge 02 rom 1 0 T akT . From 1 904 0 1 1 0 z kT u e k h s d ) . ( ) ( . ] [ the following difference equation arises h d [k] - 0.904837h d [k-1] = 0.1x[k] or h d k] = 0.904837h d [k-1] + 0.1x[k] x[k]= ] then y[k]: [k] y[k] If x[k]= [k], then y[k]: y[0] = 0.1 1]=(01)(09) x[k] T 0.904837 y[1] = (0.1)(0.9) y[2]= (0.1)(0.9)(0.9)= =(0.1)(0.9) 2 0.1 Lesson 04 epeat Challenge 02 >> den=[1 -0.9094837]; num=[0.1]; Repeat Challenge 02 >> x=[1 0 0 0 0 0 0 0 0 0 0 0 0 ] ; % impulse >> h=filter(num,den,x); % impulse response > plot(h) >> plot(h) h[k] Lesson 04 hallenge 03 Challenge 03 A continuous-time signal x(t) = cos( (t)), where (t) = 0 + sin( 0 t), is to be sampled at a rate f s . What is the system’s Nyquist sample rate if 0 = (2 ) 1000 rad/sec.?
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[PPT].Lesson 04-Aliasing - Repeat Challenge 02 Challenge 02...

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