Lesson 04-Aliasing

Lesson 04-Aliasing - Aliasing Challenge 03 Sampling Lesson...

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Lesson 04 Aliasing Challenge 03: Sampling Lesson 04 Aliasing What is it? How do you measure it? How do you manage it? Challenge 04
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Lesson 04 Repeat Challenge 02 1 st -Order Impulse Invariant System A simple first-order RC circuit is show. The relationship between the input forcing function v ( t ) and voltage developed across the capacitor, denoted v o ( t ), is defined by the 1 st order ordinary differential equation 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 ? ) ( 1 = ) ( 1 ) ( 0 0 t v RC t v RC dt t dv +
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Lesson 04 Repeat Challenge 02 It immediately follows that the system’s impulse response is given by: If the sampling period is  T s , the resulting discrete-time impulse response is: or, for  k 0, ) ( 1 = ) ( / - t u e RC t h RC t ) ( ] [ s s d kT h T k h k s RC kT s d RC T e RC T k h s α = / - = ] [ RC T s e / ; - = α
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Lesson 04 Repeat Challenge 02 For RC =10 -2 , and f s =1000 Hz, the following results. 1 . 0 10 / 10 / ; 904837 . 0 2 3 1 . 0 10 10 * 10 1 3 2 = = = = = = - - - - - - - RC T e e e s α ( 29 k k s s s d e RC T kT h T k h ) . ( . ) ( / . 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.
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Lesson 04 Repeat Challenge 02 1 1 904 0 1 1 1 1 - - - = - = - = z z e e z z kT u e k h s s s akT akT s akT d ) . ( ) ( ] [ Known, h d [k] (T s /RC) e -kT s /RC =(0.1) (0.904837) k for RC =10 -2 , and f s =1000 Hz, the following results. Where would you find a connection between a time-domain function and difference equation? – Table of z-transforms (signals and systems stuff). Table 7.1 (Principles of Signals and System, F. Taylor, McGraw-Hill 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 ) . ( . . . ) ( . ] [
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Lesson 04 Repeat Challenge 02 From 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] T 0.904837 y[k] If x[k]= δ [k], then y[k]: y[0] = 0.1 y[1] = (0.1)(0.9) y[2]= (0.1)(0.9)(0.9)= =(0.1)(0.9) 2 0.1 1 904 0 1 1 0 1 0 - - = z kT u e k h s akT d s ) . ( . ) ( . ] [ the following difference equation arises
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Lesson 04 >> den=[1 -0.9094837]; num=[0.1]; >> x=[1 0 0 0 0 0 0 0 0 0 0 0 0 ] ; % impulse >> h=filter(num,den,x); % impulse response >> plot(h) h[k] Repeat Challenge 02
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Lesson 04 Challenge 03   A continuous-time signal x(t) = cos( φ (t)), where  φ (t) = ϖ 0 + sin( ϖ 0 t), is to be sampled at a rate f s What is the system’s Nyquist sample rate if  ϖ 0 (2 π ) 1000 rad/sec.? Frequency = ϖ (t) = d φ (t)/dt = 0 + ϖ 0 cos( ϖ 0 t), f max = ϖ 0 /(2* π ) = 1000 Hz f Nyquist = 2 f max = 2000 Sa/s .
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Lesson 04 Sampling The classic DSP engineer would views the world through the veil of  periodic sampling (Shannon Theory) Two signals sampled at 1 Sa/s are to be compared on the basis of the  their time-series (sampled) signatures.     x 2 (t) = cos(2pt(9/10))+0.5*cos(2pt(8/10))    How would a plot of their discrete-time samples compare?
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Lesson 04-Aliasing - Aliasing Challenge 03 Sampling Lesson...

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