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Lec_05

# Lec_05 - Lecture 5 Open-Loop Systems Relation between E(s...

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1 Lecture 5: Open-Loop Systems Relation between E * ( s ) and E ( z ) Pulse Transfer Function Open-Loop Systems with Digital Filters Reading: Chapter 4 of the textbook Relation Between E ( z ) and E * ( s ) e ( t ) e ¤ ( t ) e ( kT ) E ( z ) E ( s ) E ¤ ( s ) z = e Ts Sampler Data hold e ( t ) e ( kT ) ¹ e ( t ) sampler and hold Data hold e ( t ) e ¤ ( t ) ¹ e ( t ) T T Ideal sampler E ( z ) = P 1 k =0 e ( kT ) z ¡ k E ( z ) = P [ residues of E ( ¸ ) 1 1 ¡ z ¡ 1 e ] E ¤ ( s ) = P [ residues of E ( ¸ ) 1 1 ¡ e ¡ T ( s ¡ ¸ ) ] E ¤ ( s ) = 1 T P 1 k = ¡1 E ( s + jk! s ) + e (0) 2

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2 Pulse Transfer Function Adding a Plant Sampler Data hold e ( t ) e ( kT ) ¹ e ( t ) sampler and hold Data hold E ( s ) E ¤ ( s ) ¹ E ( s ) T Ideal sampler c ( t ) G p ( s ) Sampler c ( kT ) What is the relation of the input e ( t ) with the output c ( t ), or the relations of their sampled signals e ( kT ) and c ( kT ), k =0,1,2,…? 1 ¡ e ¡ Ts s In the second perspective: G p ( s ) C ¤ ( s ) C ( s ) Relation of E * ( s ) and C * ( s )? G ( s ) = 1 ¡ e ¡ Ts s G p ( s ) ( combined plant and data hold transfer function ) T Ideal sampler (optional)
3 After Simplification E ( s ) E ¤ ( s ) T G ( s ) C ( s ) C ( s ) = G ( s ) E ¤ ( s ) C ¤ ( s ) = 1 T P 1 k = ¡1 C ( s + jk! s ) C ¤ ( s ) T (assume c (0)=0 ) C ¤ ( s ) = G ¤ ( s ) E ¤ ( s ) C ( z ) = G ( z ) E ( z ) General Case In general, if we have A ( s ) = B ( s ) F ¤ ( s ) A ( s ) F ¤ ( s ) B ( s ) By following the same procedure, we have A ¤ ( s ) = B ¤ ( s ) F ¤ ( s ) F ¤ ( s ) B ¤ ( s ) A ¤ ( s ) Therefore, by replacing e sT with z , we have A ( z ) = B ( z ) F ( z ) F ( z ) B ( z ) A ( z )

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4 Interpretation of C ( z )= G ( z ) E ( z ) G ( z
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Lec_05 - Lecture 5 Open-Loop Systems Relation between E(s...

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