Signals and systems-3
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Discussion on LTI systems, Time domain representation of LTI systems.
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Signals and systems-3
- 1. Signal and Systems-III Prof: Sarun Soman Manipal Institute of Technology Manipal
- 2. Interconnection of LTI Systems Parallel Connection of LTI systems Output ݕ ݐ = ݕଵ ݐ + ݕଶ()ݐ = ݔ ݐ ∗ ℎଵ ݐ + ݔ ݐ ∗ ℎଶ()ݐ = ݔ ݐ ∗ {ℎଵ ݐ + ℎଶ(})ݐ Proof Prof: Sarun Soman, MIT, Manipal 2
- 3. Interconnection of LTI Systems Distributive Property for CT case: Distributive property for DT case: Cascade Connection of LTI systems Prof: Sarun Soman, MIT, Manipal 3
- 4. Interconnection of LTI Systems Interconnection Properties of LTI Systems Prof: Sarun Soman, MIT, Manipal 4
- 5. Interconnection of LTI Systems Relation between LTI system properties and the impulse response. Memory less LTI system The o/p of a DT LTI system To be memoryless y[n] must depend only on present value of input The o/p cannot depend on x[n-k] for ݇ ≠ 0 ݕ ݊ = ℎ ݇ ݊[ݔ − ݇] ஶ ୀିஶ Prof: Sarun Soman, MIT, Manipal 5
- 6. Time Domain Representations of LTI Systems A DT LTI system is memoryless if and only if ℎ ݇ = ܿߜ[݇] ‘c’ is an arbitrary constant CT system Output A CT LTI system is memoryless if and only if ℎ ߬ = ܿߜ(߬) ‘c’ is an arbitrary constant ݕ ݐ = න ℎ ߬ ݔ ݐ − ߬ ݀߬ ஶ ିஶ Prof: Sarun Soman, MIT, Manipal 6
- 7. Time Domain Representations of LTI Systems Causal LTI systems The o/p of a causal LTI system depends only on past or present values of the input DT system ݕ ݊ = ℎ ݇ ݊[ݔ − ݇] ஶ ୀିஶ future value of inputs Prof: Sarun Soman, MIT, Manipal 7
- 8. Time Domain Representations of LTI Systems For a DT causal LTI system ℎ ݇ = 0 ݂݇ ݎ < 0 Convolution Sum in new form For CT system ℎ ߬ = 0 ݂߬ ݎ < 0 Convolution integral in new form ݕ ݊ = ℎ ݇ ݊[ݔ − ݇] ஶ ୀ ݕ ݐ = න ℎ ߬ ݔ ݐ − ߬ ݀߬ ஶ Prof: Sarun Soman, MIT, Manipal 8
- 9. Time Domain Representations of LTI Systems Stable LTI Systems A system is BIBO stable if the o/p is guaranteed to be bounded for every i/p. Discrete time case: Bounding convolution sum Prof: Sarun Soman, MIT, Manipal 9
- 10. Time Domain Representations of LTI Systems Assume that i/p is bounded Condition for impulse response of a stable DT LTI system Prof: Sarun Soman, MIT, Manipal 10
- 11. Time Domain Representations of LTI Systems Continuous time case Impulse response must be absolutely integrable. න ℎ ߬ ݀߬ < ∞ ஶ ିஶ impulse response must be absolutely summable. Prof: Sarun Soman, MIT, Manipal 11
- 12. Tutorial Determine a DT LTI system characterized by impulse response ℎ ݊ = ( ଵ ଶ ) ]݊[ݑ is a)causal b) stable. Ans: ℎ ݊ = 0 ݂݊ ݎ < 0 System is causal Stable ℎ ݊ = (0.99)݊(ݑ + 3) Ans: ℎ ݊ ≠ 0, ݊ < 0 Non-causal Stable ℎ[݇] ஶ ୀ = ( 1 2 ) ஶ ୀ = 1 1 − 1 2 = 2 < ∞ (0.99) ஶ ୀିଷ (0.99)ିଷ 1 1 − 0.99 < ∞ Prof: Sarun Soman, MIT, Manipal 12
- 13. Tutorial A CT LTI system is represented by the impulse response ℎ ݐ = ݁ିଷ௧ݐ(ݑ − 1).Check for stability and causality. Ans: ℎ ݐ = 0, ݐ < 0 Causal ℎ ݐ = ݁௧ݑ −1 − ݐ Ans: ℎ ݐ ≠ 0, ݐ < 0 Non-causal stable න |ℎ ߬ |݀߬ ஶ ିஶ = න |ℎ ߬ |݀߬ ஶ ଵ = න |݁ିଷఛ |݀߬ ஶ ଵ = ݁ିଷ 3 < ∞ stable න |݁ఛ |݀߬ ିଵ ିஶ = ݁ିଵ < ∞ Prof: Sarun Soman, MIT, Manipal 13
- 14. Tutorial ℎ()ݐ = ݁ିସ|௧| Ans: ℎ ݐ ≠ 0, ݂ݐ ݎ < 0 Non-causal න |ℎ(߬)|݀߬ ஶ ିஶ = න |݁ସఛ |݀߬ ିஶ + න |݁ିସఛ |݀߬ ஶ = ଵ ଶ < ∞ stable Check for memory ,stability and causality. ℎ ݊ = ݁ଶ݊[ݑ − 1] ℎ ݊ ≠ 0, ݂݊ ݎ ≠ 0 Memory system ℎ ݊ = 0, ݂݊ ݎ < 0 Causal ݁ଶ > 1 h[n] is not absolutely summable unstable |݁ଶ| ஶ ୀଵ Prof: Sarun Soman, MIT, Manipal 14
- 15. Time Domain Representations of LTI Systems Step Response Characterizes the response to sudden changes in input. ݏ ݊ = ℎ ݊ ∗ ]݊[ݑ u[n-k] exist from −∞ to n h[k] is the running sum of impulse response. = ℎ ݇ ݊[ݑ − ݇] ஶ ୀିஶ ]݊[ݏ = ℎ ݇ ୀିஶ Prof: Sarun Soman, MIT, Manipal 15
- 16. Time Domain Representations of LTI Systems CT system Relation b/w step response and impulse response. ℎ ݐ = ݀ ݀ݐ )ݐ(ݏ ℎ ݊ = ݏ ݊ − ݊[ݏ − 1] Eg. Step response of an RC ckt ݏ ݐ = න ℎ ߬ ݀߬ ௧ ିஶ Prof: Sarun Soman, MIT, Manipal 16
- 17. Time Domain Representations of LTI Systems Prof: Sarun Soman, MIT, Manipal 17
- 18. Example Find the step response for LTI system represented by impulse response ℎ ݊ = ( ଵ ଶ )ݑ ݊ . Ans: = ଵି( భ మ )శభ ଵି భ మ , ݊ ≥ 0 ]݊[ݏ = ℎ ݇ ୀିஶ ]݊[ݏ = ( 1 2 ) ୀ Prof: Sarun Soman, MIT, Manipal 18
- 19. Example ℎ ݐ = ݑݐ ݐ Ans: = ݐଶ 2 , ݐ ≥ 0 ݏ ݐ = න ℎ ߬ ݀߬ ௧ ିஶ ݏ ݐ = න ߬݀߬ ௧ Prof: Sarun Soman, MIT, Manipal 19
- 20. Differential and Difference equation Representations of LTI Systems Linear constant coefficient differential equation: Linear constant coefficient difference equation: Order of the equation is (N,M), representing the number of energy storage devices in the system. Often N>M and the order is described using only ‘N’. ܽ ݀ ݀ݐ ݕ ݐ = ܾ ݀ ݀ݐ )ݐ(ݔ ெ ୀ ே ୀ ܽ݊[ݕ − ݇] = ܾ݊[ݔ − ݇] ெ ୀ ே ୀ Prof: Sarun Soman, MIT, Manipal 20
- 21. Differential and Difference equation Representations of LTI Systems Prof: Sarun Soman, MIT, Manipal 21
- 22. Differential and Difference equation Representations of LTI Systems Second order difference equation: Difference equation are easily arranged to obtain recursive formulas for computing the current o/p of the system. (1) Shows how to obtain y[n] from present and past values of the input. ܽ݊[ݕ − ݇] = ܾ݊[ݔ − ݇] ெ ୀ ே ୀ ݕ ݊ = 1 ܽ ܾ݊[ݔ − ݇] ெ ୀ − 1 ܽ ܽݕ ݊ − ݇ ே ୀଵ (1) Prof: Sarun Soman, MIT, Manipal 22
- 23. Differential and Difference equation Representations of LTI Systems Recursive evaluation of a difference equation. Find the first two o/p values y[0] and y[1] for the system described by ݕ ݊ = ݔ ݊ + 2ݔ ݊ − 1 − ݕ ݊ − 1 − ଵ ସ ݊[ݕ − 2]. Assuming that the input is ݔ ݊ = ( ଵ ଶ )ݑ ݊ and the initial conditions are y[-1]=1 and y[-2]=-2. Prof: Sarun Soman, MIT, Manipal 23
- 24. Solving Differential and Difference Equations Output of LTI system described by differential or difference equation has two components ݕ - homogeneous solution ݕ- particular solution Complete solution ݕ = ݕ + ݕ Eg. Prof: Sarun Soman, MIT, Manipal 24
- 25. Solving Differential and Difference Equations Output due to non zero initial conditions with zero input Prof: Sarun Soman, MIT, Manipal 25
- 26. Solving Differential and Difference Equations Step response, system initially at rest. Prof: Sarun Soman, MIT, Manipal 26
- 27. Solving Differential and Difference Equations The Homogenous Solution CT system Set all terms involving input to zero Solution ݎare the N roots of the characteristic equation ܽ ݀ ݀ݐ ݕ ݐ = ܾ ݀ ݀ݐ )ݐ(ݔ ெ ୀ ே ୀ ܽ ݀ ݀ݐ ݕ ݐ = 0 ே ୀ ݕ()ݐ = ܿ݁௧ ே ୀଵ Prof: Sarun Soman, MIT, Manipal 27
- 28. Solving Differential and Difference Equations DT system Set all terms involving input to zero Solution ݎ are the N roots of the characteristic equation. ܽ݊[ݕ − ݇] = ܾ݊[ݔ − ݇] ெ ୀ ே ୀ ܽ݊[ݕ − ݇] = 0 ே ୀ ݕ[݊] = ܿݎ ே ୀଵ Prof: Sarun Soman, MIT, Manipal 28
- 29. Solving Differential and Difference Equations If the roots are repeating ‘p’ times then there are ‘p’ distinct terms ݁௧, ݁ݐ௧, … . ݐିଵ݁௧ and ݎ , ݊ݎ … . . ݊ିଵݎ Eg. RC ckt depicted in figure is described by the differential equation ݕ ݐ + ܴܥ ௗ ௗ௧ ݕ ݐ = )ݐ(ݔ . Determine the homogenous solution. Prof: Sarun Soman, MIT, Manipal 29
- 30. Solving Differential and Difference Equations The homogenous equation is ݕ ݐ + ܴܥ ௗ ௗ௧ ݕ ݐ = 0 (1) Solution ݕ ݐ = ܿଵ݁భ௧ To determine ݎଵsubstitute in (1) ݕ ݐ + ܴܥ ݀ ݀ݐ ݕ ݐ = 0 ܿଵ݁భ௧ 1 + ܴݎܥଵ = 0 ܿଵ݁భ௧ ≠ 0 Characteristic equation 1 + ܴݎܥଵ = 0 ݎଵ = − 1 ܴܥ Homogenous solution of the system is ݕ ݐ = ܿଵ݁ି ଵ ோ௧ Note: ܿଵ is determined later, in order that the complete solution satisfy the initial conditions. ݕ ()ݐ = ܿ݁௧ ே ୀଵ Prof: Sarun Soman, MIT, Manipal 30
- 31. Example Determine the homogenous solution. ௗమ ௗ௧మ ݕ ݐ + 5 ௗ ௗ௧ ݕ ݐ + 6ݕ ݐ = 2ݔ ݐ + ௗ ௗ௧ )ݐ(ݔ Ans: Homogenous equation ௗమ ௗ௧మ ݕ ݐ + 5 ௗ ௗ௧ ݕ ݐ + 6ݕ ݐ =0 (1) Homogenous solution ݕ ݐ = ܿଵ݁భ௧ + ܿଶ݁మ௧ To find ݎଵand ݎଶ solve CE. Put ௗ ௗ௧ ݕ ݐ = ݎ (1) becomes ݎଶ + 5ݎ + 6 = 0 ݎ = −3, −2 ݕ ݐ = ܿଵ݁ିଷ௧ + ܿଶ݁ିଶ௧ ݕ()ݐ = ܿ݁௧ ே ୀଵ Prof: Sarun Soman, MIT, Manipal 31
- 32. Example Find the homogenous solution of the first order recursive system. ݕ ݊ − ߩݕ ݊ − 1 = ]݊[ݔ Ans: Set i/p to zero ݕ ݊ − ߩݕ ݊ − 1 = 0 (1) Substitute in (1) ܿଵݎଵ − ߩܿଵݎଵ ିଵ = 0 ݕ[݊] = ܿݎ ே ୀଵ ݕ[݊] = ܿଵݎଵ ܿଵݎଵ 1 − ߩݎଵ ିଵ = 0 1 − ߩݎଵ ିଵ = 0 ݎଵ = ߩ ݕ[݊] = ܿଵߩ Prof: Sarun Soman, MIT, Manipal 32
- 33. Example Ans: Set input to zero (1) Becomes 1 − 1 4 ݎିଵ − 1 8 ݎିଶ = 0 ݎଶ − 1 4 ݎ − 1 8 = 0 ݎ = 1 2 , − 1 4 ݕ ݊ = ܿଵ( 1 2 ) + ܿଶ(− 1 4 ) ݕ ݊ − 1 4 ݕ ݊ − 1 − 1 8 ݕ ݊ − 2 = ݊[ݔ − 1] N=2 ݕ ݊ − 1 4 ݕ ݊ − 1 − 1 8 ݕ ݊ − 2 = 0 (1) ݕ [݊] = ܿଵݎଵ +ܿଶݎଶ Prof: Sarun Soman, MIT, Manipal 33
- 34. Example ݕ ݊ + 9 16 ݕ ݊ − 2 = ݊[ݔ − 1] Ans: ݎଶ + 9 16 = 0 ݎ = ±݆ 3 4 ݕ ݊ = ܿଵ(݆ 3 4 ) + ܿଶ(−݆ 3 4 ) Prof: Sarun Soman, MIT, Manipal 34
- 35. Solving Differential and Difference Equations The Particular Solution ݕ Solution of difference or differential equation for a given input. Assumption : output is of same general form as the input. CT System Input Particular solution 1 c t ܿଵݐ + ܿଶ ݁ି௧ ܿ݁ି௧ cos(⍵ݐ + ⏀) ܿଵ cos ߱ݐ + ܿଶ sin(߱)ݐ DT System Input Particular solution 1 c n ܿଵ݊ + ܿଶ ߙ ܿߙ cos(Ω݊ + ⏀) ܿଵ cos Ω݊ + ܿଶ sin(Ω݊) Prof: Sarun Soman, MIT, Manipal 35
- 36. Example Consider the RC ckt .Find a particular solution for this system with an input ݔ ݐ = cos(߱.)ݐ Ans: From previous example ݕ ݐ + ܴܥ ௗ ௗ௧ ݕ ݐ = )ݐ(ݔ (1) ݕ ݐ = ܿଵcos(⍵)ݐ + ܿଶ sin(⍵)ݐ Substitute in (1) ݕ ݐ + ܴܥ ݀ ݀ݐ ݕ ݐ = cos(߱)ݐ ܿଵ cos ⍵ݐ + ܿଶ sin(⍵)ݐ − RC⍵ܿଵ sin ⍵ݐ + RC⍵ܿଶ cos ⍵ݐ = cos(⍵)ݐ Equating the coefficients of ܿଵand ܿଶ ܿଵ + RC⍵ܿଶ = 1 −RC⍵ܿଵ + ܿଶ = 0 Solving for ܿଵand ܿଶ ܿଵ = 1 1 + (RC⍵)ଶ ܿଶ = RC⍵ 1 + (RC⍵)ଶ Prof: Sarun Soman, MIT, Manipal 36
- 37. Example ݕ ݐ = 1 1 + (RC⍵)ଶ cos ⍵ݐ + RC⍵ 1 + (RC⍵)ଶ sin ⍵ݐ Determine the particular solution for the system described by the following differential equations. ݕ ݐ = ܿ 0 + 10ܿ = 4 ܿ = 2 5 ݕ ݐ = 2 5 b)ݔ ݐ = cos(3)ݐ Ans: ݕ ݐ = ܿଵcos(3)ݐ + ܿଶ sin(3)ݐ ݀ ݀ݐ ݕ ݐ = −3ܿଵ sin(3)ݐ + 3 ܿଶcos(3)ݐ −15ܿଵ sin 3ݐ + 15ܿଶ cos 3ݐ + 10ܿଵ cos 3ݐ + 10ܿଶ sin 3ݐ = 2 cos(3)ݐ Equating coefficients −15ܿଵ + 10ܿଶ = 0 10ܿଵ + 15ܿଶ = 2 Prof: Sarun Soman, MIT, Manipal 37
- 38. Example ܿଵ = 4 65 , ܿଶ = 6 65 ݕ ݐ = 4 65 cos 3ݐ + 6 65 sin(3)ݐ Prof: Sarun Soman, MIT, Manipal 38
- 39. Solving Differential and Difference Equations The Complete Solution Procedure for calculating complete solution 1. Find the form of the homogeneous solution ݕfrom the roots of the CE equation. 2. Find a particular solution ݕby assuming that it is of the same form as the input, yet is independent of all terms in the homogeneous solution. 3. Determine the coefficients in the homogeneous solution so that the complete solution ݕ = ݕ + ݕ satisfies the initial conditions. Prof: Sarun Soman, MIT, Manipal 39
- 40. Example Find the solution for the first order recursive system described by the difference equation. ݕ ݊ − 1 4 ݕ ݊ − 1 = ]݊[ݔ If the input ݔ ݊ = ( ଵ ଶ )]݊[ݑ and the initial condition is ݕ −1 = 8. Ans: N=1 Homogenous solution ݕ ݊ − 1 4 ݕ ݊ − 1 = 0 ݕ ݊ = ܿଵݎ ݎ − 1 4 ݎିଵ = 0 ݎ − 1 4 = 0 ݕ ݊ = ܿଵ( 1 4 ) Particular solution ݕ ݊ = ܿଶ( 1 2 ) ܿଶ( 1 2 ) − 1 4 ܿଶ( 1 2 )ିଵ = ( 1 2 ) ܿଶ − 1 2 ܿଶ = 1 ܿଶ = 2 Prof: Sarun Soman, MIT, Manipal 40
- 41. Example ݕ ݊ = 2 ( 1 2 ) Complete solution ]݊[ݕ = ݕ ݊ + ݕ[݊] ]݊[ݕ = ܿଵ ଵ ସ + 2 ( ଵ ଶ ) (1) ܶܿ ݂݀݊݅ ଵ-use initial conditions Particular solution exist only for n≥ 0 Translate the initial condition to n≥ 0 Rewrite the system equation in recursive form ݕ ݊ = ݔ ݊ + 1 4 ݕ ݊ − 1 ݕ 0 = ݔ 0 + 1 4 ]1−[ݕ = 1 + 1 4 ∗ 8 = 3 Substitute y[0]in (1) ݕ 0 = ܿଵ( 1 4 ) +2 ( 1 2 ) 3 = ܿଵ + 2 ܿଵ = 1 ]݊[ݕ = 1 4 + 2 ( 1 2 ), ݊ ≥ 0 Prof: Sarun Soman, MIT, Manipal 41
- 42. Example ݕ ݊ − 1 4 ݕ ݊ − 1 − 1 8 ݕ ݊ − 2 = ݔ ݊ + ݔ ݊ − 1 Input ݔ ݊ = ( ଵ ଶ ) , Ans: Homogenous solution N=2 ݕ ݊ − 1 4 ݕ ݊ − 1 − 1 8 ݕ ݊ − 2 = 0 ݕ ݊ = ܿଵݎ + ܿଶݎ To find ‘r’ ݎ − 1 4 ݎିଵ − 1 8 ݎିଶ = 0 ݎ = 1 2 , − 1 4 ݕ ݊ = ܿଵ( 1 2 ) +ܿଶ(− 1 4 ) Particular Solution ݕ ݊ = ܿଷ( 1 2 ) Same form as homogeneous solution ݕ ݊ = ܿଷ݊( 1 2 ) ]݊[ݑ ܿଷ݊( 1 2 ) − 1 4 ܿଷ(݊ − 1)( 1 2 )ିଵ − 1 8 ܿଷ(݊ − 2)( 1 2 )ିଶ = ( 1 2 ) + ( 1 2 )ିଵ Prof: Sarun Soman, MIT, Manipal 42
- 43. Example ܿଷ݊ − 1 4 ܿଷ(݊ − 1)( 1 2 )ିଵ − 1 8 ܿଷ(݊ − 2)( 1 2 )ିଶ = 1 + ( 1 2 )ିଵ ܿଷ݊ − 1 2 ܿଷ(݊ − 1) − 1 2 ܿଷ(݊ − 2) = 3 1 2 ܿଷ + ܿଷ = 3 ܿଷ = 2 ݕ ݊ = 2݊( 1 2 ) ]݊[ݑ Complete solution ݕ = ܿଵ( 1 2 ) +ܿଶ(− 1 4 ) +2݊( 1 2 ) ]݊[ݑ Determine the o/p of the system ݕ ݊ − 1 9 ݕ ݊ − 2 = ݊[ݔ − 1] ݔ ݊ = ݑ ݊ , ݕ −1 = 1, ݕ −2 = 0 Ans: Homogenous solution N=2 ݕ ݊ − 1 9 ݕ ݊ − 2 = 0 ݕ ݊ = ܿଵݎଵ + ܿଶݎଶ Find ‘r’ ݎ − 1 9 ݎିଶ = 0 ݎଶ − 1 9 = 0 Prof: Sarun Soman, MIT, Manipal 43
- 44. Example ݎ = ± 1 3 ݕ ݊ = ܿଵ 1 3 + ܿଶ − 1 3 Particular solution ݕ ݊ = ܿଷ]݊[ݑ Substitute in system eqn. ܿଷ − 1 9 ܿଷ = 1 ܿଷ = 9 8 ݕ ݊ = ଽ ଼ ]݊[ݑ Complete solution ݕ = ܿଵ 1 3 + ܿଶ − 1 3 + 9 8 ]݊[ݑ To find ܿଵand ܿଶ translate initial conditions to ݊ ≥ 0 Rewrite system eqn. in recursive form. ݕ ݊ = ݔ ݊ − 1 + 1 9 ݕ ݊ − 2 ݕ 0 = ݔ −1 + 1 9 ]2−[ݕ ݕ 0 = 0 ݕ 1 = ݔ 0 + 1 9 ]1−[ݕ Prof: Sarun Soman, MIT, Manipal 44
- 45. Example ݕ 1 = 10 9 Substitute y[0] and y[1] in complete solution y[n]. ܿଵ = − 7 12 , ܿଶ = − 13 24 ݕ ݊ = 9 8 ݑ ݊ − 7 12 1 3 − 13 24 − 1 3 Prof: Sarun Soman, MIT, Manipal 45
- 46. Characteristics of Systems Described by Differential and Difference Equations It is informative to express the o/p of a system as sum of two components. 1) Only with initial conditions 2) Only with input ݕ = ݕ + ݕ Natural response ࢟ (zero i/p response) System o/p for zero i/p Describes the manner in which the system dissipates any stored energy or memory. Prof: Sarun Soman, MIT, Manipal 46
- 47. Characteristics of Systems Described by Differential and Difference Equations Procedure to calculate ݕ 1. Find the homogeneous solution. 2. From the homogeneous solution find the coefficients ܿ using initial conditions. Note: homogeneous solutions apply for all time, the natural response is determined without translating initial conditions. Eg. A system is described by the difference equation ݕ ݊ − ଵ ସ ݕ ݊ − 1 = ݔ ݊ , ݕ −1 = 8. Find the natural response of the system. Prof: Sarun Soman, MIT, Manipal 47
- 48. Characteristics of Systems Described by Differential and Difference Equations Homogenous solution ݕ ݊ = ܿଵ 1 4 Use initial conditions to find ܿଵ Translation not required ݕ −1 = ܿଵ 1 4 ିଵ ܿଵ = 2 ݕ = 2 1 4 , ݊ ≥ −1 Prof: Sarun Soman, MIT, Manipal 48
- 49. Characteristics of Systems Described by Differential and Difference Equations The Forced Response ݕ System o/p due to the i/p signal assuming zero initial conditions. The forced response is of the same form as the complete solution. Eg. ݕ ݊ − ଵ ସ ݕ ݊ − 1 = ݔ ݊ .Find the forced response of this system if the i/p is ݔ ݊ = ଵ ଶ ]݊[ݑ Ans: Homogenous solution Prof: Sarun Soman, MIT, Manipal 49
- 50. Characteristics of Systems Described by Differential and Difference Equations ݕ ݊ − 1 4 ݕ ݊ − 1 = 0 ݕ ݊ = ܿଵ 1 4 Particular solution ݕ ݊ = ܿଶ 1 2 ]݊[ݑ ݕ ݊ = 2 1 2 ]݊[ݑ Complete solution ]݊[ݕ = 2 ଵ ଶ ݑ ݊ + ܿଵ ଵ ସ (1) To find ܿଵassume zero initial conditions ݕ −1 = 0 Translate initial conditions ݕ ݊ = ݔ ݊ + 1 4 ݕ ݊ − 1 ݕ ݊ = 1 2 ]݊[ݑ + 1 4 ݕ ݊ − 1 ݕ 0 = 1 + 0 Substitute in (1) ݕ 0 = 2 1 2 ݑ 0 + ܿଵ 1 4 ܿଵ = −1 ݕ ݊ = 2 1 2 − 1 4 , ݊ ≥ 0 Prof: Sarun Soman, MIT, Manipal 50
- 51. Example Identify the natural and forced response. ݕ ݊ − 3 4 ݕ ݊ − 1 + 1 8 ݕ ݊ − 2 = 2ݔ ݊ ݕ −1 = 1, ݕ −2 = −1 , ݔ ݊ = 2]݊[ݑ Ans: Homogeneous solution N=2 ݕ ݊ = ܿଵݎଵ + ܿଶݎଶ ݎ = 1 2 , 1 4 ݕ ݊ = ܿଵ 1 2 + ܿଶ 1 4 Natural response Find ܿଵand ܿଶusing initial conditions No translation required. 2ܿଵ + 4ܿଶ = 1 4ܿଵ + 16ܿଶ = −1 ࢟ = − ૡ Forced response. Particular solution ݕ ݊ = ܿଷ]݊[ݑ Prof: Sarun Soman, MIT, Manipal 51
- 52. Example ݕ ݊ = 32 3 ]݊[ݑ Complete solution ݕ ݊ = ݕ ݊ + ݕ[݊] ]݊[ݕ = ܿଵ ଵ ଶ + ܿଶ ଵ ସ + ଷଶ ଷ ]݊[ݑ (1) Forced response Find the coefficients ܿଵand ܿଶ using zero initial conditions. ݕ −1 = 0, ݕ −2 = 0 Translate the initial conditions ݕ 0 = 4, ݕ 1 = 7 Substitute y[0] and y[1] in (1) ܿଵ + ܿଶ = − 20 3 1 2 ܿଵ + 1 4 ܿଶ = − 11 3 ࢟ࢌ = ࢛ − ૡ + , ≥ Prof: Sarun Soman, MIT, Manipal 52
- 53. Example Find the forced response ݕ ݊ = ܿଵ 1 2 ݕ ݊ = ܿଶ − 1 2 ]݊[ݑ ܿଶ = 1 Complete solution ݕ ݊ = ܿଵ 1 2 + − 1 2 ]݊[ݑ Use zero initial conditions Translate the initial conditions ݕ 0 = 2 ܿଵ = 1 Prof: Sarun Soman, MIT, Manipal 53
- 54. Example ݕ ݊ + ݕ ݊ − 1 + 1 4 ݕ ݊ − 2 = ݔ ݊ + ݊[ݔ − 1] I/p ݔ ݊ = − ଵ ଶ ݑ ݊ , ݕ −1 = − 1, ݕ −2 = 2. Find the forced response. Ans: Homogeneous solution ݕ ݊ + ݕ ݊ − 1 + 1 4 ݕ ݊ − 2 = 0 N=2 ݕ ݊ = ܿଵݎଵ + ܿଶݎଶ (1) ݎ + ݎିଵ + 1 4 ݎିଶ = 0 1 + ݎିଵ + 1 4 ݎିଶ = 0 ݎଶ + ݎ + 1 4 = 0 ݎ = − 1 2 , − 1 2 Roots are repeating If a root ݎ is repeated p times then there are p distinct terms ݎ , ݊ݎ , … … . ݊ିଵݎ Eqn (1) is modified ݕ ݊ = ܿଵݎଵ + ܿଶ݊ݎଶ Prof: Sarun Soman, MIT, Manipal 54
- 55. Example ݕ ݊ = ܿଵ − 1 2 + ܿଶ݊ − 1 2 Particular solution ݕ ݊ = ܿଷ − 1 2 ]݊[ݑ ݕ [݊] must be independent of ݕ [݊] ݕ ݊ = ܿଷ݊ଶ − 1 2 ]݊[ݑ Substitute ݕ [݊] in system equation ܿଷ = − 1 2 ݕ ݊ = − 1 2 ݊ଶ − 1 2 ]݊[ݑ Complete solution ݕ = ݕ ݊ + ݕ [݊] ݕ = ܿଵ − ଵ ଶ + ܿଶ݊ − ଵ ଶ − ଵ ݊ଶ − ଵ ଶ ]݊[ݑ (2) Forced response Use zero initial conditions ݕ −1 = 0, ݕ −2 = 0. Translate the initial conditions to ݊ ≥ 0 ݕ ݊ = −ݕ ݊ − 1 − 1 4 ݕ ݊ − 2 + ݔ ݊ + ݊[ݔ − 1] Prof: Sarun Soman, MIT, Manipal 55
- 56. Example ݕ 0 = −ݕ −1 − 1 4 ݕ −2 + ݔ 0 + ]1−[ݔ ݕ 0 = 1,ݕ 1 = − ଵ ଶ Substitute y[0] and y[1] in (2) ܿଵ = 1, ܿଶ = 1 2 ݕ ݊ = − 1 2 + 1 2 ݊ − 1 2 − 1 2 ݊ଶ − 1 2 ݑ ݊ , ݊ ≥ 0 Prof: Sarun Soman, MIT, Manipal 56