Lesson_22 - Lesson 22 Challenge 21 Lesson 22 Stability...

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Unformatted text preview: Lesson 22 Challenge 21 Lesson 22 Stability Challenge 22 Challenge 21 What is the transfer function of the filter shown in Figure 8-8? x[k] y[k] z-1 b b 1 a 1 Challenge 21 x[k] y[k] z-1 b b 1 a 1 Y(z)= a 1 z-1 Y(z) + b X(z) + b 1 z-1 X(z) or (1- a 1 z-1 )Y(z)= (b + b 1 z-1 )X(z) or H(z)=Y(z)/X(z) = (b + b 1 z-1 )/ (1- a 1 z-1 ) Lesson 22 For an LTI IIR (if a =1, called monic) Convolution theorem for at-rest system - +- = = = M m m N 1 m m m] x[k b m] y[k a a 1 y[k] ( 29 ( 29 ( 29 + = - +- = = =- = = M m-m m N 1 m m m M m m N 1 m m z X z b z a z Y m] x[k b m] y[k a y[k] z Y Z (difference equation) Lesson 22 Transfer function ( 29 ( 29 ( 29 = = =- =-- = - = = n m m M m m p z z z k N 1 m m m M m-m m ) ( ) ( z a 1 z b z X z Y z H Polynomial form Pole-zero form Lesson 22 Stability Motivational example: y[k]= 0.5 y[k-1] + 2x[k] which translates into: H(z)=2/(1-0.5z-1 ) = 2z/(z- 0.5 ) having a zero at the origin (z=0) and a pole at z=0.5 . According to the Table 1 (z-transforms), the impulse response is: h[k]=2(0.5) k u[k] - convergent Lesson 22 Magnitude frequency response lowpass filter. Maximum output expected for a DC input signal. H(1)=4 Lesson 22 Step (DC) Response X(z) = z/(z-1); H(z) = 2z/(z-0.5); Y(z) = H(z)X(z) = 2z 2 /((z-0.5) (z-1)) using methods that have not been formally introduced at this point, it can be...
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Lesson_22 - Lesson 22 Challenge 21 Lesson 22 Stability...

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