signal,part 2

‫ﻓﺼﻞ دوم‬

‫ﺳﻴﺴﺘﻢهﺎي ‪ LTI‬زﻣﺎن ﭘﻴﻮﺳﺘﻪ و زﻣﺎن ﮔﺴﺴﺘﻪ‬

‫‪٣٢‬‬

‫ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ راﺑﻄﻪ آﺎﻧﻮﻟﻮﺷﻦ‬LTI ‫ ﺧﻮاص ﺳﻴﺴﺘﻢهﺎي‬-١-٢

‫زﻣﺎن ﭘﻴﻮﺳﺘﻪ‬

x (t )

h (t )

‫زﻣﺎن ﮔﺴﺴﺘﻪ‬

y (t )

h [n ]

x [n ]

x (t ) = δ (t ) ⇒ y (t ) = h (t )

y [n ]

x [n ] = δ [n ] ⇒ y [n ] = h [n ]

‫ ﺟﺎﺑﻪﺟﺎﻳﻲﭘﺬﻳﺮي‬-١-١-٢ y (t ) = x (t ) ∗ h (t ) =

+∞

∫ x (λ) h (t − λ) dλ

−∞

= h (t ) ∗ x (t ) =

∞

y [n ] = x [n ] ∗ h [n ] =

∑ x [k ] h [n − k ]

k = −∞

,

= h [n ] ∗ x [n ] =

∫ h (λ) x (t − λ) dλ

∞

∞

∑ h [k ] x [n − k ] k = −∞

−∞

‫ ﺗﻮزﻳﻊﭘﺬﻳﺮي‬-٢-١-٢ y (t ) = x (t ) ∗ (h1 (t ) + h 2 (t )) = x (t ) ∗ h1 (t ) + x (t ) ∗ h 2 (t ) h

1

x (t )

+

h

y (t )

2

y (t ) = (x 1 (t ) + x 2 (t )) ∗ h (t ) = x 1 (t ) ∗ h (t ) + x 2 (t ) ∗ h (t ) x 1 (t )

x 2 (t )

+

h (t )

y (t )

‫ ﺷﺮآﺖﭘﺬﻳﺮي‬-٣-١-٢ y (t ) = x (t ) ∗ (h1 (t ) ∗ h 2 (t )) = x (t ) ∗ (h 2 (t ) ∗ h1 (t ))

٣٣

x (t )

h

1

h

2

x (t )

h

2

h

1

y (t ) y (t )

‫ و ﭘﺎﺳﺦ ﭘﻠﻪ‬h(t) ‫ راﺑﻄﻪ ﺑﻴﻦ ﭘﺎﺳﺦ ﺿﺮﺑﻪ‬-٢-٢ ‫زﻣﺎن ﭘﻴﻮﺳﺘﻪ‬

‫زﻣﺎن ﮔﺴﺴﺘﻪ‬

u (t )

h (t )

y (t )

u [n ]

h [n ]

y [n ]

δ (t )

h (t )

h (t )

δ [n ]

h [n ]

h [n ]

y (t ) =

t

∑ h [k ]

∫ h ( λ ) dλ

dy (t ) dt

h [n ] = y [n ] − y [n − 1]

k = −∞

−∞

h (t ) =

n

y [n ] =

.‫ زﻳﺮ را ﺑﺪﺳﺖ ﺁورﻳﺪ‬LTI ‫( ﭘﺎﺳﺦ ﭘﻠﻪ ﺳﻴﺴﺘﻢهﺎي‬١ ‫ﻣﺜﺎل‬

h (t ) =

y (t ) =

t

1

∫ RC e

−∞

−λ

RC

  0  u ( λ) dλ =  t −λ −t  1 e RC dλ = 1 − e RC ∫0 RC

,

,

t <0 t ≥0

;

;

1

RC

e

−t RC

u (t )

t

0

0

t

y (t ) = (1 − e

λ

λ −

t RC

h [n ] = (−a ) n u [n ] ٣۴

(‫اﻟﻒ‬

)u (t )

(‫ب‬

‫‪1 − (−a ) n +1‬‬ ‫] ‪u [n‬‬ ‫‪1+a‬‬

‫‪‬‬ ‫‪0‬‬ ‫‪, n <0‬‬ ‫‪n‬‬ ‫‪‬‬ ‫‪k‬‬ ‫‪y [n ] = ∑ (−a ) u [k ] = ‬‬ ‫∞‪k = −‬‬ ‫‪n‬‬ ‫‪n +1‬‬ ‫) ‪∑ (−a ) k = 1 − (−a‬‬ ‫‪, n ≥0‬‬ ‫‪k =0‬‬ ‫) ‪1 − (−a‬‬

‫=‬

‫‪ -٣-٢‬ﺧﻮاص ﺳﻴﺴﺘﻢهﺎي ‪ LTI‬ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺗﺎﺑﻊ ﺗﺒﺪﻳﻞ‬ ‫‪ -١-٣-٢‬ﺣﺎﻓﻈﻪ‬ ‫زﻣﺎن ﮔﺴﺴﺘﻪ‬

‫] ‪y [n‬‬

‫زﻣﺎن ﭘﻴﻮﺳﺘﻪ‬

‫] ‪x [n‬‬

‫] ‪h [n‬‬ ‫∞‬

‫] ‪∑ h [k ] x [n − k‬‬

‫) ‪y (t‬‬

‫= ] ‪y [n‬‬

‫‪− λ) dλ‬‬

‫∞‪k = −‬‬

‫) ‪x (t‬‬

‫) ‪h (t‬‬ ‫∞‪+‬‬

‫‪∫ h (λ) x (t‬‬

‫= ) ‪y (t‬‬

‫∞‪−‬‬

‫ﺷﺮط ﺑﺪون ﺣﺎﻓﻈﻪ ﺑﻮدن ﺳﻴﺴﺘﻢ‪:‬‬

‫‪h [n ] = 0‬‬

‫‪; n = ±1, ± 2, ± 3, K‬‬

‫‪;t ≠ 0‬‬

‫‪h (t ) = 0‬‬

‫اﺛﺒﺎت )ﺳﻴﺴﺘﻢ زﻣﺎن ﮔﺴﺴﺘﻪ(‪:‬‬

‫‪y [n ] = K + h [−1]x [n + 1] + h [0]x [n ] + h [1]x [n − 1] + K‬‬ ‫اﻳﻦ ﺳﻴﺴﺘﻢ ﺑﻪ ﺷﺮﻃﻲ ﺑﺪون ﺣﺎﻓﻈﻪ اﺳﺖ آﻪ ﺧﺮوﺟﻲ در هﺮ ﻟﺤﻈﻪ ﺑﻪ ورودي در هﻤﺎن ﻟﺤﻈﻪ واﺑﺴﺘﻪ ﺑﺎﺷﺪ آﻪ ﺑﺮاي اﻳﻦ ﻣﻨﻈﻮر‬ ‫ﺑﺎﻳﺴﺘﻲ‪:‬‬

‫‪; k = cte‬‬

‫] ‪h [n ] = k δ [n‬‬

‫ﻳﺎ‬

‫] ‪⇒ h [n ] = δ [n‬‬

‫‪ -٢-٣-٢‬ﻣﻌﻜﻮس ﭘﺬﻳﺮي‬ ‫زﻣﺎن ﭘﻴﻮﺳﺘﻪ‬

‫) ‪y (t ) = x (t‬‬

‫) ‪h I (t‬‬

‫) ‪h (t‬‬

‫) ‪x (t‬‬

‫) ‪x (t ) = (h (t ) ∗ h I (t )) ∗ x (t‬‬ ‫‪٣۵‬‬

‫‪n ≠0‬‬

‫‪,‬‬

‫‪h [n ] = 0‬‬

‫) ‪h (t ) ∗ h I (t ) = δ (t‬‬

‫⇒ ) ‪δ (t ) ∗ x (t ) = x (t‬‬

‫زﻣﺎن ﮔﺴﺴﺘﻪ‬

‫] ‪y [n ] = x [n‬‬

‫] ‪h I [n‬‬

‫] ‪h [n‬‬

‫] ‪x [n‬‬

‫] ‪h [n ] ∗ h I [n ] = δ [n‬‬

‫‪ -٣-٣-٢‬ﻋﻠﻴﺖ‬ ‫∞‬

‫در ﺳﻴﺴﺘﻢ ﻋﻠﻲ ﺧﺮوﺟﻲ ﺑﻪ ﺁﻳﻨﺪﻩ ورودي ﺑﺴﺘﮕﻲ ﻧﺪارد‪ .‬ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ راﺑﻄﻪ ] ‪∑ h [k ] x [n − k‬‬ ‫‪k‬‬ ‫∞‪= −‬‬

‫ﺗﺎﺑﻊ ﺗﺒﺪﻳﻞ )ﭘﺎﺳﺦ ﺿﺮﺑﻪ( ﺑﻴﺎﻧﮕﺮ ﺳﻴﺴﺘﻢ ﻋﻠﻲ ﺧﻮاهﺪ ﺑﻮد‪.‬‬ ‫زﻣﺎن ﮔﺴﺴﺘﻪ‬

‫‪n <0‬‬

‫;‬

‫زﻣﺎن ﭘﻴﻮﺳﺘﻪ‬

‫‪t <0‬‬

‫‪h [n ] = 0‬‬

‫;‬

‫‪h (t ) = 0‬‬

‫ﺗﺬآﺮ‪ :‬ﺗﺎﺑﻊ ﺗﺒﺪﻳﻞ ) ‪ ( h [n ] = u [n ] ) h (t ) = u (t‬ﺑﻴﺎﻧﮕﺮ ﺳﻴﺴﺘﻢ ﻋﻠﻲ اﺳﺖ‪.‬‬

‫‪ -۴-٣-٢‬ﭘﺎﻳﺪاري)*‪(BIBO‬‬ ‫زﻣﺎن ﭘﻴﻮﺳﺘﻪ‬

‫) ‪y (t‬‬

‫) ‪h (t‬‬ ‫∞‬

‫∞‬

‫∞‬

‫∞‪−‬‬

‫∞‪−‬‬

‫∞‪−‬‬

‫∞ < ‪∫ h (λ) dλ‬‬ ‫∞<‬

‫) ‪x (t‬‬

‫= ‪∫ h (λ) x (t − λ) dλ‬‬

‫≤ ‪∫ h (λ) x (t − λ) dλ‬‬

‫∞‬

‫‪∫ h (t ) dt‬‬

‫‪:‬ﺷﺮط ﭘﺎﻳﺪاري‬

‫∞‪−‬‬

‫زﻣﺎن ﮔﺴﺴﺘﻪ‬

‫] ‪y [n‬‬

‫] ‪h [n‬‬

‫] ‪x [n‬‬

‫* ‪Bounded Input Bounded Output‬‬

‫‪٣۶‬‬

‫= ) ‪y (t‬‬

‫= ] ‪ y [n‬ﺷﺮط زﻳﺮ ﺑﺮاي‬

‫∞‬

‫∞‬

‫∞‪= −‬‬

‫∞‪= −‬‬

‫∞ < ] ‪∑ h [k ]x [n − k ] ≤ k∑ h [k‬‬ ‫‪k‬‬

‫= ] ‪y [n‬‬

‫∞‬

‫∞ < ] ‪∑ h [n‬‬

‫ﺷﺮط ﭘﺎﻳﺪاري‪:‬‬

‫∞‪k = −‬‬

‫ﻣﺜﺎل ‪ (٢‬ﺑﺎ اﺳﺘﻔﺎدﻩ از ﺗﺎﺑﻊ ﺗﺒﺪﻳﻞ ﺧﻮاص ﺳﻴﺴﺘﻢهﺎي ‪ LTI‬را ﺑﺮرﺳﻲ آﻨﻴﺪ‪.‬‬

‫) ‪y (t ) = x (t − t 0‬‬

‫) ‪x (t‬‬

‫) ‪h (t‬‬

‫اﻟﻒ( ﻋﻠﻲ ﺑﻮدن‬

‫) ‪h (t‬‬

‫ﺳﻴﺴﺘﻢ ﻋﻠﻲ‬

‫‪t‬‬

‫‪t0‬‬

‫) ‪h (t‬‬ ‫‪t‬‬

‫‪−t0‬‬

‫ﺳﻴﺴﺘﻢ ﻏﻴﺮ ﻋﻠﻲ‬

‫; ‪t 0 ≥ 0‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪h (t ) = y (t ) x (t ) =δ (t ) ⇒ h (t ) = δ (t − t 0 ) = ‬‬ ‫‪‬‬ ‫‪‬‬ ‫; ‪t 0 < 0‬‬

‫ب( ﺳﻴﺴﺘﻢ ﻣﻌﻜﻮس‬ ‫‪t −t =t ′‬‬ ‫‪h I (t − t 0 ) = δ (t ) ‬‬ ‫) ‪≤ h I (t ′) = δ (t ′ + t 0‬‬ ‫‪0‬‬

‫) ‪h (t ) ∗ h I (t ) = δ (t ) , δ (t − t 0 ) ∗ h I (t ) = δ (t‬‬

‫‪,‬‬

‫ﻣﻌﻜﻮسﭘﺬﻳﺮ‬ ‫ج( ﭘﺎﻳﺪاري‬

‫∞ < ‪− t 0 ) dt‬‬

‫ﭘﺎﻳﺪار‬

‫∞‪+‬‬

‫‪∫ δ (t‬‬

‫∞‪−‬‬

‫ﻣﺜﺎل ‪(٣‬‬ ‫‪١‬اﻟﻒ(‬

‫) ‪h (t ) = e at u (t‬‬

‫ﻋﻠﻲ و ﺣﺎﻓﻈﻪدار‬ ‫‪٢‬ب(‬

‫‪,‬‬

‫∞‬ ‫‪a < 0‬‬ ‫‪at‬‬ ‫‪at‬‬ ‫‪e‬‬ ‫‪u‬‬ ‫(‬ ‫‪t‬‬ ‫)‬ ‫‪dt‬‬ ‫‪e‬‬ ‫‪dt‬‬ ‫=‬ ‫=‬ ‫‪‬‬ ‫∫‬ ‫‪∫0‬‬ ‫‪a ≥ 0‬‬ ‫∞‪−‬‬

‫ﭘﺎﻳﺪار‬ ‫ﻧﺎﭘﺎﻳﺪار‬

‫∞‪+‬‬

‫]‪h [n ] = a n u [n + 2‬‬

‫‪n ≥ −2 ,‬‬

‫ﻏﻴﺮﻋﻠﻲ و ﺣﺎﻓﻈﻪدار‬

‫ﭘﺎﻳﺪار‬

‫‪,‬‬

‫ﻧﺎﭘﺎﻳﺪار‬

‫‪,‬‬

‫‪ a < 1‬‬ ‫‪⇒‬‬ ‫‪ a > 1‬‬

‫∞‬

‫‪ak‬‬ ‫∑‬ ‫‪k‬‬ ‫‪= −2‬‬

‫∞‬

‫= ]‪∑ a k u [k + 2‬‬

‫∞‪k = −‬‬

‫ﺗﻤﺮﻳﻦ‪ :١‬ﻣﻄﻠﻮب اﺳﺖ ﺗﺮﺳﻴﻢ )‪ y(t‬؟‬ ‫‪٣۴‬‬

‫) ‪y (t‬‬

‫) ‪x (t‬‬

‫) ‪h (t‬‬ ‫∞‬

‫) ‪∑ δ (t − 2kT‬‬

‫= ) ‪x (t‬‬

‫∞‪k = −‬‬

‫) ‪h (t‬‬ ‫‪1‬‬

‫‪t‬‬

‫‪T‬‬

‫‪−T‬‬

‫ﺗﻤﺮﻳﻦ ‪ :٢‬ﻣﻄﻠﻮب اﺳﺖ ﭘﺎﺳﺦ ﺿﺮﺑﻪ ﺳﻴﺴﺘﻢ ذﻳﻞ ?=)‪h(t‬‬

‫‪T‬‬

‫‪T‬‬

‫‪× a n +1‬‬ ‫‪+‬‬

‫) ‪y (t‬‬

‫‪× a2‬‬ ‫‪+‬‬

‫‪+‬‬

‫‪T‬‬

‫‪× a0‬‬

‫‪× a1‬‬ ‫‪+‬‬

‫‪+‬‬

‫) ‪x (t‬‬

‫‪+‬‬

‫ﻣﺜﺎل ‪ (٣‬راﺑﻄﻪ ﺑﻴﻦ ورودي دﻟﺨﻮاﻩ ]‪ x[n‬و ﺧﺮوﺟﻲ ]‪ y[n‬را ﺑﺮاي ﭘﺎﺳﺦ ﺿﺮﺑﻪ ﺳﻴﺴﺘﻢ ‪ LTI‬دادﻩ ﺷﺪﻩ ﺑﺪﺳﺖ ﺁورﻳﺪ‪.‬‬

‫‪n‬‬

‫] ‪h [n‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪3‬‬

‫‪0≤n ≤3‬‬

‫‪,‬‬

‫‪o .w‬‬

‫‪,‬‬

‫‪1‬‬

‫‪1‬‬ ‫‪‬‬ ‫‪h [n ] =  4‬‬ ‫‪0‬‬

‫‪1‬‬ ‫‪1‬‬ ‫)]‪(u [n ] − u [n − 4]) = (δ [n ] + δ [n − 1] + δ [n − 2] + δ [n − 3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪1‬‬ ‫)]‪y [n ] = x [n ] ∗ h [n ] = (x [n ] + x [n − 1] + x [n − 2] + x [n − 3‬‬ ‫‪4‬‬

‫= ] ‪h [n‬‬

‫ﻣﺜﺎل ‪ (۴‬اﮔﺮ ]‪ h 2 [n ] = u [n ] − u [n − 2‬اﻟﻒ( ﻣﻘﺪار ] ‪ h1 [n‬را ﺑﺪﺳﺖ ﺁورﻳﺪ‪ ،‬ﺗﺎﺑﻊ ﺗﺒﺪﻳﻞ آﻞ ﺳﻴﺴﺘﻢ ]‪ h[n‬دادﻩ ﺷﺪﻩ‬ ‫اﺳﺖ‪.‬‬

‫] ‪y [n‬‬

‫‪2‬‬

‫‪h‬‬

‫‪2‬‬

‫‪h‬‬

‫‪1‬‬

‫] ‪x [n‬‬

‫‪h‬‬

‫‪٣۵‬‬

‫‪11‬‬

‫] ‪h [n‬‬

‫‪10‬‬

‫‪8‬‬

‫‪5‬‬

‫‪4‬‬

‫‪1‬‬

‫‪n‬‬

‫‪1‬‬ ‫‪3‬‬

‫‪7‬‬

‫‪−1‬‬

‫)] ‪h [n ] = h1 [n ] ∗ (h 2 [n ] ∗ h 2 [n‬‬

‫‪,‬‬

‫]‪h 2 [n ] = δ [n ] + δ [n − 1‬‬

‫]‪h 2 [n ] ∗ h 2 [n ] = (δ [n ] + δ [n − 1]) ∗ (h 2 [n ]) = h 2 [n ] + h 2 [n − 1‬‬ ‫]‪= δ [n ] + δ [n − 1] + δ [n − 1] + δ [n − 2] = δ [n ] + 2δ [n − 1] + δ [n − 2‬‬ ‫]‪⇒ h [n ] = h1 [n ] + 2h1 [n − 1] + h1 [n − 2‬‬ ‫اوﻟﻴﻦ ﺟﺎﻳﻲ آﻪ ]‪ h[n‬ﻣﻘﺪار دارد ‪ n = 0‬اﺳﺖ ﭘﺲ در ﻧﻤﻮدارهﺎي ﺑﺎﻻ ﺑﻪ ازاي ‪ n = 0‬دارﻳﻢ‪:‬‬

‫‪n‬‬

‫‪3‬‬

‫‪2‬‬

‫‪1‬‬ ‫‪6‬‬

‫] ‪h 1[n‬‬

‫‪3‬‬

‫‪−1‬‬

‫‪1‬‬

‫‪a0 = 1‬‬

‫⇒‬

‫‪a1 = 3‬‬ ‫‪⇒ a2 = 3‬‬ ‫‪⇒ a3 = 2‬‬ ‫‪⇒ a4 = 1‬‬ ‫‪⇒ a5 = 0‬‬ ‫‪⇒ a6 = 0 K‬‬

‫‪+ 2a −1 + a −2 = 1‬‬ ‫⇒‬

‫‪+ 2a 0 = 5‬‬

‫‪+ 2a 1 + a 0 = 10‬‬ ‫‪+ 2a 2 + a 1 = 11‬‬ ‫‪+ 2a 3 + a 2 + 8‬‬ ‫‪+ 2a 4 + a 3 = 4‬‬ ‫‪+ 2a 5 + a 4 + 1‬‬

‫ب( اﮔﺮ ]‪ x[n‬ﺑﺮاﺑﺮ ﺑﺎﺷﺪ ﺑﺎ ]‪ x [n ] = δ [n ] − δ [n − 1‬ﻣﻄﻠﻮب اﺳﺖ ﻣﻘﺪار ﺧﺮوﺟﻲ ]‪ y[n‬؟‬ ‫اﻳﻦ ﻗﺴﻤﺖ آﺎﻣﻼ ﻣﺴﺘﻘﻞ از ﻗﺴﻤﺖ ﻗﺒﻞ اﺳﺖ ﭼﻮن ]‪ x[n‬دادﻩ ﺷﺪﻩ اﺳﺖ‪.‬‬

‫]‪y [n ] = h [n ] ∗ x [n ] = h [n ] − h [n − 1‬‬ ‫‪5‬‬

‫‪n‬‬

‫‪7‬‬

‫‪2‬‬ ‫‪−4‬‬

‫‪4‬‬

‫‪1‬‬

‫‪4‬‬

‫‪−1‬‬

‫] ‪y [n‬‬

‫‪0‬‬

‫‪−3‬‬

‫‪٣۶‬‬

‫‪a0‬‬ ‫‪a1‬‬ ‫‪a2‬‬ ‫‪a3‬‬ ‫‪a4‬‬ ‫‪a5‬‬ ‫‪a6‬‬

‫‪ -۴-٢‬آﺎﻧﻮﻟﻮﺷﻦ )زﻣﺎن ﭘﻴﻮﺳﺘﻪ(‬ ‫‪ -١-۴-٢‬روش ﺗﺮﺳﻴﻤﻲ‬ ‫ﺗﻮﺻﻴﻪ ﻣﻲﺷﻮد اﮔﺮ )‪ f(t‬ﺑﻪ ﻟﺤﺎظ رﻳﺎﺿﻲ ﺳﺎدﻩﺗﺮ اﺳﺖ از ﻓﺮﻣﻮل ‪ ٢‬و اﮔﺮ )‪ g(t‬ﺳﺎدﻩﺗﺮ اﺳﺖ از ﻓﺮﻣﻮل ‪ ١‬اﺳﺘﻔﺎدﻩ ﻣﻲآﻨﻴﻢ‪.‬‬ ‫دراﻳﻦ ﻣﺜﺎل از ﻓﺮﻣﻮل ‪ ٢‬اﺳﺘﻔﺎدﻩ ﻣﻲ آﻨﻴﻢ‪.‬‬ ‫∞‪+‬‬

‫∞‪+‬‬

‫= ‪y (t ) = f (t ) ∗ g (t ) = ∫ f ( λ ) g (t − λ) dλ‬‬

‫‪∫ g (λ) f (t‬‬

‫‪− λ ) dλ‬‬ ‫‪144424443‬‬

‫∞‪−‬‬

‫∞‪−‬‬

‫‪14442444‬‬ ‫‪3‬‬ ‫‪1‬‬

‫‪2‬‬

‫) ‪ f ( λ ) , g ( λ‬را ﻣﻲﺳﺎزﻳﻢ‪.‬‬ ‫ﺑﺮاي ﺳﺎﺧﺖ ) ‪ f (t − λ‬اول ﺑﺎﻳﺪ )‪ f (t + λ‬را ﺳﺎﺧﺖ ﭼﻮن ﻧﻤﻲداﻧﻴﻢ آﻪ ‪ t‬ﻣﺜﺒﺖ اﺳﺖ ﻳﺎ ﻣﻨﻔﻲ اﺳﺖ ﭘﺲ ﺑﻪ ﺻﻮرت ﻗﺮاردادي‬ ‫‪ t‬واﺣﺪ ﺑﻪ ﺳﻤﺖ ﭼﭗ ﺷﻴﻔﺖ ﻣﻲدهﻴﻢ‪ .‬ﺳﭙﺲ ) ‪ f (t − λ‬را ﻣﻲﺳﺎزﻳﻢ از اﻳﻦ ﻣﺮﺣﻠﻪ ﺑﻪ ﺑﻌﺪ ﺑﺎﻳﺪ در هﺮ ﻣﺮﺣﻠﻪ ﻣﻌﺎدﻟﻪ ﺧﻂ آﻨﺎر‬ ‫ﻧﻤﻮدار ﻧﻮﺷﺘﻪ ﺷﻮد‪.‬‬ ‫ﺣﺎل ) ‪ g ( λ‬را ﺛﺎﺑﺖ ﻧﮕﻪ داﺷﺘﻪ و ) ‪ f (t − λ‬را از ∞ ‪ −‬ﺑﻪ ﺳﻤﺖ ∞ ‪ +‬ﺷﻴﻔﺖ ﻣﻲدهﻴﻢ‪ .‬ﺑﺪﻳﻬﻲ اﺳﺖ در ﺟﺎﻳﻲ آﻪ دو ﺗﺎﺑﻊ‬ ‫هﻤﭙﻮﺷﺎﻧﻲ ﻧﺪاﺷﺘﻪ ﺑﺎﺷﻨﺪ ﺣﺎﺻﻠﻀﺮب ﺻﻔﺮ اﺳﺖ‪ .‬ﺳﭙﺲ ﺑﻪ ﻧﻘﻄﻪ اي ﻣﻲرﺳﺪ آﻪ ‪ max‬هﻤﭙﻮﺷﺎﻧﻲ را دارد و ﺑﻌﺪ دوﺑﺎرﻩ ﺑﻪ ﺟﺎﻳﻲ‬ ‫ﻣﻲرﺳﺪ آﻪ هﻴﭻ هﻤﭙﻮﺷﺎﻧﻲ ﻧﺪاﺷﺘﻪ ﺑﺎﺷﻨﺪ‪:‬‬ ‫ﻣﺜﺎل ‪(۵‬‬

‫) ‪g (t‬‬

‫) ‪f (t‬‬

‫‪3‬‬

‫‪2‬‬

‫‪t‬‬

‫‪t‬‬

‫‪−2‬‬

‫‪2‬‬ ‫)‪g (λ‬‬

‫‪2‬‬

‫‪3‬‬

‫‪λ‬‬

‫‪−2‬‬

‫‪2‬‬

‫) ‪f (− λ + t‬‬ ‫‪λ −t + 2‬‬

‫‪λ‬‬

‫‪t‬‬

‫‪3‬‬

‫‪2‬‬ ‫) ‪f (λ + t‬‬ ‫‪λ‬‬

‫‪t −2‬‬

‫‪٣٧‬‬

‫‪2‬‬ ‫) ‪2 − (λ + t‬‬

‫)‪f (λ‬‬

‫‪2‬‬

‫‪−t 2 −t‬‬

‫‪−λ+2‬‬

‫‪λ‬‬

‫‪2‬‬

‫‪2‬‬

‫‪1‬‬

‫‪∗=0‬‬ ‫‪3‬‬ ‫) ‪(4 − t 2‬‬ ‫‪4‬‬

‫= ‪+ 2)(3) dλ‬‬

‫‪+ 2)(3) dλ = 6‬‬

‫‪t‬‬

‫‪∫ (λ − t‬‬

‫=∗‬

‫‪−2‬‬

‫‪t‬‬

‫‪∫ (λ − t‬‬

‫=∗‬

‫‪t −2‬‬

‫‪3 2‬‬ ‫)‪(t − 8t + 16‬‬ ‫‪2‬‬

‫‪2‬‬

‫= ‪( λ − t + 2)(3) dλ‬‬

‫∫‬ ‫‪t‬‬

‫=∗‬

‫‪−2‬‬

‫‪∗=0‬‬

‫‪t < −2 ,‬‬ ‫‪‬‬ ‫‪− 2 ≤ t < 0 ,‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪0 ≤ t < 2 ,‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2 ≤ t ≤ 4 ,‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪,‬‬ ‫‪t ≥ 4‬‬

‫‪ -٢-۴-٢‬اﺳﺘﻔﺎدﻩ از ﻓﺮﻣﻮل‬ ‫اﮔﺮ ﺑﺎ اﺳﺘﻔﺎدﻩ از ﻓﺮﻣﻮل ﺑﺨﻮاهﻴﻢ آﺎﻧﻮﻟﻮﺷﻦ دو ﺳﻴﮕﻨﺎل را ﺑﺪﺳﺖ ﺑﻴﺎورﻳﻢ‪ .‬ﺑﺎﻳﺪ ﻣﻌﺎدﻟﻪ دو ﺳﻴﮕﻨﺎل ﺑﺮ ﺣﺴﺐ ﺗﻮاﺑﻊ رﻳﺎﺿﻲ دادﻩ‬ ‫ﺷﺪﻩ ﺑﺎﺷﻨﺪ ﻳﺎ اﻳﻨﻜﻪ ﺑﺘﻮان ﺁﻧﻬﺎ را ﺑﺮ ﺣﺴﺐ ﺗﻮاﺑﻊ وﻳﮋﻩ ﻓﺮﻣﻮﻟﻪ ﻧﻤﻮد‪.‬‬ ‫ﺑﺮاي ﻣﺜﺎل ﻗﺒﻞ )‪ f(t) , g(t‬را ﺑﺎﻳﺪ ﺑﻪ ﻓﺮم زﻳﺮ ﻧﻮﺷﺖ و ﺳﭙﺲ در ﻓﺮﻣﻮل آﺎﻧﻮﻟﻮﺷﻦ ﺟﺎﻳﮕﺬاري آﺮد‪:‬‬

‫))‪g (t ) = 3(u (t + 2) − u (t − 2‬‬

‫‪,‬‬

‫)‪f (t ) = 2u (t ) − r (t ) + r (t − 2‬‬

‫‪٣٨‬‬

(١ ‫ﻣﺜﺎل‬

x (t ) = 2u (t − 1) − 2u (t − 3) x (t )

h (t ) = u (t + 1) − 2u (t − 1) + u (t − 3)

,

h2 (t )

h1 (t )

h (t )

2

1 3

1

t

−1

1

t

1

1 3 t

1

−1

1

3

t −1

h (t ) = h1 (t ) + h2 (t )

y (t ) = x (t ) ∗ h (t ) =

∞

∫ h (τ )x (t

− τ )dτ

−∞

x (τ )

x (−τ + t )

x (τ + t ) 2

2

h (τ )

2

1 1

3

τ

1−t

3 −t

τ

t −3

t −1

τ

−1

1

3

τ −1

∗=0 , t − 1 < 1  t −1 − 1 ≤ t − 1 < 1 , ∗ = 2dτ = 2t ∫  −1  1 t −1  ≤ − < ∗ = + τ 1 t 1 2 , 2 d  ∫t −3 ∫1 − 2dτ = 12 − 4t   1 t −1 2 ≤ t − 1 < 3 , ∗ = 2dτ + − 2dτ = 12 − 4t ∫ ∫1  t −3  3  ≤ − < ∗ = 3 t 1 4 ,  ∫ − 2dτ = −2 (6 − t ) t −3  t − 1 ≥ 5 , ∗=0 

٣٩

‫) ‪y (t‬‬ ‫‪4‬‬

‫‪t‬‬

‫‪6‬‬

‫‪4‬‬

‫‪2‬‬

‫‪−4‬‬ ‫ﺗﺬآﺮ‪ :‬وﻗﺘﻲ دو ﺗﺎﺑﻊ ﭘﺎﻟﺴﻲ ﺑﺎهﻢ آﺎﻧﻮاﻟﻮ ﻣﻲﺷﻮﻧﺪ ﺣﺎﺻﻞ ﻳﻚ ﺗﺎﺑﻊ ذوزﻧﻘﻪ ﻣﻲﺷﻮد و اﮔﺮ اﻳﻦ دو ﺗﺎﺑﻊ ﭘﺎﻟﺴﻲ هﻢ ﻋﺮض ﺑﺎﺷﺪ ﺷﻜﻞ‬ ‫ﺣﺎﺻﻞ ﻳﻚ ﺗﺎﺑﻊ ﻣﺜﻠﺜﻲ ﺧﻮاهﺪ ﺑﻮد‪.‬‬

‫ﻣﺜﺎل ‪x (t ) = u (t ) − 2u (t − 2) + u (t − 5) (٢‬‬

‫‪,‬‬

‫) ‪h (t ) = e 2t u (1 − t‬‬

‫) ‪x (t‬‬

‫) ‪h (t‬‬

‫‪e2‬‬

‫‪1‬‬

‫‪1‬‬

‫‪t‬‬

‫‪5‬‬

‫‪2‬‬

‫‪t‬‬

‫‪1‬‬

‫‪−1‬‬ ‫‪− τ )dτ‬‬

‫∞‬

‫‪∫ x (τ )h (t‬‬

‫= ) ‪y (t‬‬

‫∞‪−‬‬

‫‪e2‬‬

‫) ‪h (τ + t‬‬

‫) ‪h (−τ + t‬‬

‫‪e2‬‬

‫‪e2‬‬ ‫) ‪e 2(τ +t‬‬

‫) ‪e 2( −τ +t‬‬ ‫‪τ‬‬

‫) ‪h (τ‬‬

‫‪τ‬‬

‫‪t −1‬‬

‫‪τ‬‬

‫‪1 −t‬‬

‫) ‪x (τ‬‬ ‫‪1‬‬

‫‪τ‬‬

‫‪5‬‬

‫‪2‬‬ ‫‪−1‬‬

‫‪٤٠‬‬

‫‪1‬‬ ‫‪1‬‬

‫‪e 2τ‬‬

‫‪∗=0‬‬ ‫)‪1 2 (t −5‬‬ ‫‪e‬‬ ‫‪−e2‬‬ ‫‪2‬‬ ‫‪1 2‬‬ ‫) ) ‪(e − e 2(t −2) + e 2(t −5) − e 2(t −2‬‬ ‫‪2‬‬ ‫‪1 2t‬‬ ‫) ) ‪(e − e 2(t −2 ) + e 2(t −5 ) − e 2 (t −2‬‬ ‫‪2‬‬

‫= ‪× (−1)dτ‬‬

‫) ‪2 ( −τ +t‬‬

‫‪5‬‬

‫‪∫t e‬‬

‫=∗‬

‫‪−1‬‬

‫‪5‬‬

‫= ‪dτ + ∫ e 2( −τ +t )dτ‬‬

‫) ‪2 ( −τ +t‬‬

‫‪2‬‬

‫‪2‬‬

‫‪∫e‬‬ ‫‪t‬‬

‫=∗‬

‫‪−1‬‬

‫‪5‬‬

‫‪2‬‬

‫‪2‬‬

‫‪0‬‬

‫= ‪∗ = ∫ e 2( −τ +t )dτ + ∫ − e 2 ( −τ +t )dτ‬‬

‫‪t − 1 > 5 ,‬‬ ‫‪‬‬ ‫‪2 < t − 1 ≤ 5 ,‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪0 < t − 1 ≤ 2 ,‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪,‬‬ ‫‪t − 1 ≤ 0‬‬ ‫‪‬‬

‫ﺗﻤﺮﻳﻦ‪ :‬ﺳﻴﺴﺘﻢ ‪ LTI‬ﺑﺎ ﭘﺎﺳﺦ ﺿﺮﺑﻪ )‪ h(t‬دادﻩ ﺷﺪﻩ‪ .‬ﭘﺎﺳﺦ ﺳﻴﺴﺘﻢ ﺑﻪ ورودي )‪ x(t‬را ﺑﺪﺳﺖ ﺁورﻳﺪ و ﻣﻘﺪار ﺧﺮوﺟﻲ را در‬ ‫‪1‬‬ ‫ﻟﺤﻈﺎت ‪4‬‬

‫‪t =−‬‬

‫‪3‬‬ ‫‪4‬‬

‫= ‪t‬‬

‫‪,‬‬

‫) ‪y (t ) = x (t ) ∗ h (t‬‬

‫‪,‬‬

‫‪,‬‬

‫‪3‬‬ ‫‪2‬‬

‫= ‪t‬‬

‫‪,‬‬

‫∞‪t = +‬‬

‫ﻣﺤﺎﺳﺒﻪ آﻨﻴﺪ‪.‬‬

‫) ‪x (t‬‬ ‫‪1‬‬ ‫?‬

‫‪−t‬‬ ‫) ‪t h (t ) = e δ (t ) + u (t‬‬

‫‪2‬‬

‫‪−1‬‬

‫‪−1‬‬

‫‪ -٥-٢‬آﺎﻧﻮﻟﻮﺷﻦ زﻣﺎن ﮔﺴﺴﺘﻪ‬ ‫‪ -١-٥-٢‬روش ﺗﺮﺳﻴﻤﻲ‬ ‫∞‪+‬‬

‫∞‪+‬‬

‫] ‪∑ x [k ] h [n − k ] = k∑ h [k ]x [n − k‬‬ ‫‪k‬‬ ‫∞‪= −‬‬ ‫‪1‬‬ ‫‪44‬‬ ‫‪42444‬‬ ‫‪3‬‬ ‫‪2‬‬

‫= ] ‪y [n‬‬

‫∞‪= −‬‬ ‫‪1‬‬ ‫‪44‬‬ ‫‪42444‬‬ ‫‪3‬‬ ‫‪1‬‬

‫ﭘﺲ از ﺗﺼﻤﻴﻢﮔﻴﺮي راﺟﻊ ﺑﻪ اﻳﻨﻜﻪ از راﺑﻄﻪ ‪ ١‬ﻳﺎ راﺑﻄﻪ ‪ ٢‬ﺧﺮوﺟﻲ ]‪ y[n‬را ﺣﺴﺎب آﻨﻴﻢ‪ ،‬ﻣﺮاﺣﻞ اﻧﺠﺎم ﻋﻤﻠﻴﺎت ﻋﻴﻨﺎ ﺷﺒﻴﻪ زﻣﺎن‬ ‫ﭘﻴﻮﺳﺘﻪ اﺳﺖ‪.‬‬ ‫روش ﺗﺮﺳﻴﻤﻲ را ﺑﺮاي ﺗﻌﻴﻴﻦ ﺧﺮوﺟﻲ ﺳﻴﺴﺘﻢ زﻣﺎن ﮔﺴﺴﺘﻪ ﻣﻄﺮح ﻣﻲآﻨﻴﻢ‪:‬‬ ‫ﻣﺜﺎل ‪(٣‬‬

‫‪n =0‬‬ ‫‪n =1‬‬ ‫‪n =2‬‬

‫‪2‬‬ ‫‪‬‬ ‫‪x [n ] = 3‬‬ ‫‪− 2‬‬ ‫‪‬‬

‫‪n = ±1‬‬ ‫‪n =0‬‬ ‫‪o .w‬‬

‫‪1‬‬ ‫‪‬‬ ‫‪h [n ] = 2‬‬ ‫‪0‬‬ ‫‪‬‬

‫‪٤١‬‬

‫] ‪x [n‬‬

‫‪3‬‬

‫] ‪h [n‬‬ ‫‪2‬‬

‫‪2‬‬

‫‪1‬‬

‫‪2‬‬

‫‪n‬‬

‫‪n‬‬

‫‪1‬‬

‫‪1‬‬

‫‪−1‬‬

‫‪−2‬‬

‫] ‪x [k‬‬

‫‪3‬‬

‫] ‪h [−k + n‬‬

‫‪k‬‬

‫‪1‬‬

‫‪2‬‬

‫‪2‬‬

‫‪1‬‬

‫‪2‬‬

‫‪k‬‬

‫] ‪h [k + n‬‬

‫‪2‬‬

‫‪2‬‬

‫] ‪h [k‬‬ ‫‪1‬‬

‫‪1‬‬

‫‪k‬‬

‫‪n −1 n n+1‬‬

‫‪−1−n − n n −1‬‬

‫‪k‬‬

‫‪−1‬‬

‫‪1‬‬

‫‪−2‬‬

‫] ‪y [n‬‬ ‫‪7‬‬

‫‪∗=0‬‬

‫‪6‬‬

‫‪2‬‬

‫‪n‬‬

‫‪,‬‬

‫‪+1 < 0‬‬

‫‪, ∗ = 1× 2 = 2‬‬ ‫‪, ∗ = 2× 2 +1× 3 = 7‬‬ ‫‪, ∗ = 1 × 2 + 2 × 3 − 2 ×1 = 6‬‬

‫‪+1 = 0‬‬ ‫‪+1 = 1‬‬ ‫‪+1 = 2‬‬

‫‪∗ = 1 × 3 + 2 × (−2) = −1‬‬

‫‪,‬‬

‫‪+1 = 3‬‬

‫‪∗ = 1 × (−2) = −2‬‬ ‫‪∗=0‬‬

‫‪,‬‬

‫‪+1 = 4‬‬ ‫‪+1 ≥ 5‬‬

‫‪2 3‬‬ ‫‪1‬‬

‫‪4 5‬‬

‫‪,‬‬

‫‪n‬‬ ‫‪‬‬ ‫‪n‬‬ ‫‪n‬‬ ‫‪‬‬ ‫‪n‬‬ ‫‪n‬‬ ‫‪‬‬ ‫‪n‬‬ ‫‪n‬‬ ‫‪‬‬

‫‪− 2 −1‬‬

‫‪−1‬‬ ‫‪−2‬‬

‫‪ -٢-٥-٢‬اﺳﺘﻔﺎدﻩ از ﻓﺮﻣﻮل‬ ‫اﺑﺘﺪا ﺑﺎﻳﺪ ﺗﻮاﺑﻊ را ﺑﺮ ﺣﺴﺐ ﺗﻮاﺑﻊ وﻳﮋﻩ و ﻳﺎ ﺗﻮاﺑﻊ رﻳﺎﺿﻲ ﺑﻴﺎن آﺮد‪ .‬اﮔﺮ ورودي و ﻳﺎ ﭘﺎﺳﺦ ﺿﺮﺑﻪ ﺑﺮ ﺣﺴﺐ ﺗﺎﺑﻊ ﺿﺮﺑﻪ ﺑﻴﺎن‬ ‫ﺷﺪﻩ ﺑﺎﺷﻨﺪ‪ ،‬ﺑﺪﻟﻴﻞ ﺧﺎﺻﻴﺖ ﺗﺎﺑﻊ ﺿﺮﺑﻪ ﻣﺤﺎﺳﺒﻪ آﺎﻧﻮﻟﻮﺷﻦ ﺑﺴﻴﺎر راﺣﺖ ﺧﻮاهﺪ ﺑﻮد‪.‬‬ ‫ﻣﺜﺎل ‪(٤‬‬

‫]‪x [n ] = 2δ [n ] + 3δ [n − 1] − 2δ [n − 2‬‬

‫]‪h [n ] = δ [n + 1] + 2δ [n ] + δ [n − 1‬‬

‫‪,‬‬

‫‪٤٢‬‬

y [n ]

y [n ] = x [n ] ∗ h [n ] = (2δ [n ] + 3δ [n − 1] − 2δ [n − 2]) ∗ h [n ] = 2h [n ] + 3h [n − 1] − 2h [n − 2] ‫ﻳﺎ‬

7

6

y [n ] = x [n ] ∗ h [n ] = x [n ] ∗ (δ [n + 1] + 2δ [n ] + δ [n − 1]) = x [n + 1] + 2x [n ] + x [n − 1]

2 2 3 − 2 −1

1

n

4 5

−1

−2

:‫ﺑﺪﻳﻬﻲ اﺳﺖ ﭘﺎﺳﺦ ﻣﺤﺎﺳﺒﻪ ﺷﺪﻩ در هﺮ دو ﺣﺎﻟﺖ ﻳﻜﺴﺎن ﺧﻮاهﺪ ﺑﻮد‬ 1 3

x [n ] = ( ) −n u [−n − 1] y [n ] =

∞

∑ x [n ] h [n − k ] =

k = −∞

1 u [−K − 1] =  0

1

∞

h [n ] = u [n − 1]

,

( ) k u [−k ∑ 3 k

(٥ ‫ﻣﺜﺎل‬

− 1] ×u [n − k − 1]

−

= −∞

1

; k ≤ −1 ; k >1

− 3 − 2 −1

k ⇒

0 u [n − K − 1] =  1

n − 1 < −1 ⇒ y [n ] =

n −1

1

k → −k

+∞

1

; k ≤ n −1 ; k > n −1

n −1

k

3 1 1 1 1 3 1 ( ) k = ( ) −n +1 × = ( ) −n +1 = (3) n −1 = (3) n 1 2 3 3 2 2 +1 3 1− 3

( )k = ∑ ∑ 3 k n k = −∞

−

=−

n − 1 ≥ −1 ⇒ y [n ] =

1 ( ) −k = ∑ k = −∞ 3 −1

∞

1

( )k ∑ 3 k

=

=1

1 1 1 × = 1 2 3 1− 3

y [n ] 1 8

1 6

1 2

−3 −2 −1 0 ٤٣

1 2

1 2

n

x [n ] = α n u [n ]

0 <α <1

h [n ] = u [n ]

,

h [n ] = u [n ] − u [n − 10]

(٦‫ﻣﺜﺎل‬

x [n ]

1

1

0

9

x [k ]

n

n

x [−k + n ] 1

1

1

n

6

2

x [n + k ]

6

y [n ] = ?

x [n ] = u [n − 2] − u [n − 7] (٧ ‫ﻣﺜﺎل‬

,

h [n ]

2

,

2−n

6 −n

k

n −6

n −2

k

h [k ] 1 0

9

n −2<0 , n −2 = 0, n −2 =2, n − 2 = 3, n − 2 = 4 ,

k

n − 2 = 10, n − 2 = 11, n − 2 = 12, n − 2 = 13, n − 2 ≥ 14 ,

∗=0 ∗=1 ∗=2 ∗=4

∗=5  M n − 2 = 9, ∗ = 5 

y [n ] 4 3

5

5

∗=4 ∗=3 ∗=2 ∗=1 ∗=0

4 3 2

2

1

1 16

1

٤٤

n

‫‪ -٦-٢‬ﺳﻴﺴﺘﻢهﺎي ‪ LTI‬ﺗﻮﺻﻴﻒ ﺷﺪﻩ ﺑﺎ ﻣﻌﺎدﻻت دﻳﻔﺮاﻧﺴﻴﻞ)زﻣﺎن ﭘﻴﻮﺳﺘﻪ(‬ ‫ﺑﻪ ﻃﻮر آﻠﻲ ﻣﻌﺎدﻟﻪ دﻳﻔﺮاﻧﺴﻴﻞ ﻳﻚ ﺳﻴﺴﺘﻢ ﺑﻪ ﻓﺮم روﺑﻪرو ﻧﻮﺷﺘﻪ ﻣﻲﺷﻮد‪:‬‬

‫) ‪d k x (t‬‬ ‫‪d k y (t ) M‬‬ ‫‪b‬‬ ‫=‬ ‫‪∑ k dt k‬‬ ‫‪dt k‬‬ ‫‪k =0‬‬

‫) ‪y (t‬‬

‫‪sys‬‬

‫‪N‬‬

‫‪∑ ak‬‬

‫‪k =0‬‬

‫) ‪x (t‬‬

‫ﺑﺮاي ﺣﻞ ﻣﻌﺎدﻟﻪ دﻳﻔﺮاﻧﺴﻴﻞ ﺑﺎﻳﺪ‬ ‫‪ (١‬ﻣﻌﺎدﻟﻪ هﻤﮕﻦ ﺣﻞ ﺷﻮد‪.‬‬ ‫‪(٢‬ﺟﻮاب ﺧﺼﻮﺻﻲ ﺑﻪ ازاي ورودي ﺧﺎص ﺗﻌﻴﻴﻦ ﮔﺮدد‪.‬‬ ‫‪ (٣‬ﺟﻮاب آﻠﻲ ﺳﻴﺴﺘﻢ ﻋﺒﺎرت اﺳﺖ از ﭘﺎﺳﺦ ﻋﻤﻮﻣﻲ ‪ +‬ﭘﺎﺳﺦ ﺧﺼﻮﺻﻲ و ﺑﺎ ﺗﻮﺟﻪ ﺑﺎ اﻳﻦ آﻪ ﭘﺎﺳﺦ ﻋﻤﻮﻣﻲ داراي ﺗﻌﺪادي‬ ‫ﺿﺮاﻳﺐ ﺛﺎﺑﺖ اﺳﺖ اﻳﻦ ﺿﺮاﻳﺐ از روي ﺷﺮاﻳﻂ ﺳﻜﻮن ﻳﺎ ﺷﺮاﻳﻂ اوﻟﻴﻪ ﺑﻪ دﺳﺖ ﻣﻲﺁﻳﻨﺪ‪.‬‬

‫) ‪d k y (t‬‬ ‫‪= 0 ⇒ y h (t ) = K‬‬ ‫‪d kt‬‬

‫‪N‬‬

‫‪∑ ak‬‬

‫‪k =0‬‬

‫‪y p (t ) = K‬‬

‫)‪١‬‬

‫)‪٢‬‬

‫) ‪y (t ) = y h (t ) + y p (t‬‬

‫)‪٣‬‬

‫ﺗﺬآﺮ‪ :‬ﺑﺮاي ﺑﻜﺎرﮔﻴﺮي ﺷﺮاﻳﻂ ﺳﻜﻮن ﺟﻬﺖ ﺗﻌﻴﻴﻦ ﭘﺎﺳﺦ ﺳﻴﺴﺘﻢ ‪ LTI‬ﺑﻪ ورودي دادﻩ ﺷﺪﻩ‪ ،‬ﻋﻨﻮان ﻋﻠﻲ ﺑﻮدن ﺳﻴﺴﺘﻢ ﺿﺮوري‬ ‫اﺳﺖ‪ .‬در ﻏﻴﺮاﻳﻨﺼﻮرت ﺑﺎﻳﺴﺘﻲ ﺷﺮاﻳﻂ اوﻟﻴﻪ دادﻩ ﺷﺪﻩ ﺑﺎﺷﺪ‪.‬‬ ‫ﻣﺜﺎل ‪(٨‬‬ ‫‪y (0) = 0‬‬ ‫‪t >0,‬‬ ‫‪y ′(0) = 0‬‬

‫‪,‬‬

‫‪x (t ) = t‬‬

‫‪,‬‬

‫) ‪d 2 y (t‬‬ ‫) ‪dy (t‬‬ ‫‪+2‬‬ ‫) ‪+ y (t ) = x (t‬‬ ‫‪2‬‬ ‫‪dt‬‬ ‫‪dt‬‬

‫ﺣﻞ‪:‬‬

‫) ‪d 2 y (t‬‬ ‫) ‪dy (t‬‬ ‫‪+2‬‬ ‫‪+ y (t ) = 0 ⇒ λ 2 + 2 λ + 1 = 0 ⇒ λ 1 = λ 2 = −1‬‬ ‫‪2‬‬ ‫‪dt‬‬ ‫‪dt‬‬ ‫‪⇒ y h (t ) = K 1 e −t + K 2 t e −t‬‬ ‫)‪t ≥ 0 ⇒ y p (t ) = At 2 + Bt + c ⇒ y p (t ) = (t − 2‬‬

‫‪t >0‬‬ ‫‪⇒ y (t ) = (2 + t ) e −t + (t − 2) , t > 0‬‬ ‫‪٤٥‬‬

‫‪,‬‬

‫‪,‬‬

‫‪y ′′(t ) + 2y ′(t ) + y (t ) = t‬‬

‫)‪y (t ) = K 1 e −t + K 2 t e −t + (t − 2‬‬ ‫‪‬‬ ‫‪y (0) = 0‬‬ ‫‪y ′(0) = 0‬‬ ‫‪‬‬

‫ﻣﺜﺎل ‪ (٩‬ﻣﻌﺎدﻟﻪ دﻳﻔﺮاﻧﺴﻴﻞ ﺳﻴﺴﺘﻢ ‪ LTI‬و ﻋﻠﻲ دادﻩ ﺷﺪﻩ اﺳﺖ‪ .‬ﻣﻄﻠﻮﺑﺴﺖ ﭘﺎﺳﺦ ﺳﻴﺴﺘﻢ ﺑﻪ ازاي ورودي دادﻩ ﺷﺪﻩ‪:‬‬

‫‪0 ; t ≤ −1‬‬ ‫‪1 ; t > −1‬‬

‫‪x (t ) = ‬‬

‫) ‪y ′(t ) + 2y (t ) = x (t‬‬

‫‪,‬‬

‫ﺣﻞ‪:‬‬

‫‪y h (t ) = K 1 e −2t (١‬‬ ‫‪1‬‬ ‫‪2 (٢‬‬

‫‪,‬‬

‫‪y ′(t ) + 2y (t ) = 0 ⇒ λ = −2‬‬

‫= ) ‪y ′(t ) + 2y (t ) = 1 ; t > −1 , y p (t‬‬

‫)‪ ٣‬اﻋﻤﺎل ﺷﺮط ﺳﻜﻮن ﺑﺮ اﺳﺎس ﻋﻠﻲ ﺑﻮدن ﺳﻴﺴﺘﻢ‬

‫‪t > −1‬‬

‫‪1‬‬ ‫‪1‬‬ ‫; ‪⇒ y (t ) = − e −2(t +1) +‬‬ ‫‪2‬‬ ‫‪2‬‬

‫‪t > −1‬‬

‫ﻳﺎ‬ ‫‪1‬‬ ‫‪2‬‬

‫‪1‬‬ ‫‪‬‬ ‫‪−2t‬‬ ‫; ‪y (t ) = K 1 e +‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪y (−1) = 0‬‬

‫‪1‬‬ ‫‪2‬‬

‫)‪y (t ) = (− e −2(t +1) + ) u (t + 1‬‬

‫‪ -٧-٢‬ﭘﺎﺳﺦ ﺑﻪ ورودي ﺿﺮﺑﻪ‪:‬‬ ‫‪ -١-٧-٢‬اﺳﺘﻔﺎدﻩ از راﺑﻄﻪ ﺑﻴﻦ ﭘﺎﺳﺦ ﺿﺮﺑﻪ و ﭘﺎﺳﺦ ﭘﻠﻪ در ﺳﻴﺴﺘﻢهﺎي ‪:LTI‬‬ ‫ﺑﺎ ﻳﻚ ﻣﺜﺎل اﻳﻦ راﺑﻄﻪ را ﺑﺮرﺳﻲ ﻣﻲآﻨﻴﻢ‪.‬‬

‫ﻣﺜﺎل ‪ (١٠‬ﺳﻴﺴﺘﻢ ‪ LTI‬و ﻋﻠﻲ اﺳﺖ‪.‬‬

‫) ‪x (t ) = δ (t‬‬

‫‪,‬‬

‫) ‪y ′′(t ) + y ′(t ) − 2y (t ) = x (t‬‬

‫ﺣﻞ‪z ′′(t ) + z ′(t ) − 2z (t ) = u (t ) :‬‬ ‫‪− 1 ± 3 − 2‬‬ ‫‪=‬‬ ‫‪2‬‬ ‫‪1‬‬

‫= ‪z ′′(t ) + z ′(t ) − 2z (t ) = 0 ⇒ λ 2 + λ − 2 = 0 ⇒ λ 1 , λ 2‬‬

‫‪z h (t ) = K 1 e −2t + K 2 e t‬‬ ‫‪1‬‬ ‫‪2‬‬

‫‪z p (t ) = −‬‬

‫‪٤٦‬‬

‫‪t >0‬‬

‫‪1 −2t 1 t 1‬‬ ‫; ‪e + e −‬‬ ‫‪6‬‬ ‫‪3‬‬ ‫‪2‬‬

‫= ) ‪⇒ z (t‬‬

‫‪1‬‬ ‫‪‬‬ ‫‪−2t‬‬ ‫‪+K2 et − ; t > 0‬‬ ‫‪z (t ) = K 1 e‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪z (0) = 0‬‬ ‫‪‬‬ ‫‪z ′(0) = 0‬‬ ‫‪‬‬ ‫‪‬‬

‫)ﻣﺸﺘﻖ ﭘﺎﺳﺦ ﭘﻠﻪ = ﭘﺎﺳﺦ ﺿﺮﺑﻪ(‬ ‫‪1 t 1‬‬ ‫) ‪dz (t‬‬ ‫= ) ‪e − ) u (t ) ; y (t‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪dt‬‬ ‫‪1 −2t 1 t‬‬ ‫‪1 −2t 1 t 1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪⇒ y (t ) = (− e‬‬ ‫‪+ e ) u (t ) + ( e‬‬ ‫) ‪+ e − ) δ (t ) ⇒ y (t ) = (− e −2t + e t ) u (t‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪6‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫‪6‬‬

‫‪z (t ) = ( e −2t +‬‬

‫‪ -٢-٧-٢‬ﻣﺤﺎﺳﺒﻪ ﭘﺎﺳﺦ ﺿﺮﺑﻪ ﺳﻴﺴﺘﻢ ﺑﻄﻮر ﻣﺴﺘﻘﻴﻢ‬ ‫ﺟﻬﺖ ﻣﺤﺎﺳﺒﻪ ﭘﺎﺳﺦ ﺿﺮﺑﻪ ﺳﻴﺴﺘﻢ ‪ LTI‬ﭘﺎﺳﺦ ﻣﻌﺎدﻟﻪ هﻤﮕﻦ را ﺑﺪﺳﺖ ﺁوردﻩ و ﺁن را ﺑﻌﻨﻮان ﭘﺎﺳﺦ آﺎﻣﻞ ﺳﻴﺴﺘﻢ در ﻧﻈﺮ ﻣﻲﮔﻴﺮﻳﻢ‪.‬‬ ‫ﭘﺎﺳﺦ آﺎﻣﻞ ﺑﺎﻳﺪ در ﻣﻌﺎدﻟﻪ دﻳﻔﺮاﻧﺴﻴﻞ ﺻﺪق آﻨﺪ آﻪ ﺑﺪﻳﻦ ﺗﺮﺗﻴﺐ ﺿﺮاﻳﺐ ﺛﺎﺑﺖ ﻣﺤﺎﺳﺒﻪ و ﭘﺎﺳﺦ آﺎﻣﻞ‪ ،‬آﻪ هﻤﺎن ﭘﺎﺳﺦ ﺿﺮﺑﻪ‬ ‫ﺳﻴﺴﺘﻢ ‪ LTI‬اﺳﺖ ﺗﻌﻴﻴﻦ ﺧﻮاهﺪ ﺷﺪ‪.‬‬ ‫ﺣﺎل ﻣﺜﺎل ‪ ١٠‬را ﺑﻪ اﻳﻦ روش ﻣﺠﺪدا ﺣﻞ ﻣﻲآﻨﻴﻢ‪:‬‬

‫) ‪x (t ) = δ (t‬‬

‫‪,‬‬

‫) ‪y ′′(t ) + y ′(t ) − 2y (t ) = x (t‬‬

‫‪y ′′(t ) + y ′(t ) − 2y (t ) = 0‬‬ ‫) ‪y h (t ) = K 1 e −2t + K 2 e t ⇒ y (t ) = (K 1 e −2t + K 2 e t )u (t‬‬

‫) ‪y ′(t ) = (−2K 1 e −2t + K 2 e t ) u (t ) + (K 1 e −2t + K 2 e t )δ (t‬‬ ‫) ‪y ′′(t ) = K = (4K 1 e −2t + K 2 e t ) u (t ) + (−4K 1 e −2t + 2K 2 e t ) δ (t ) + (K 1 e −2t + K 2 e t ) δ ′(t‬‬ ‫) ‪y ′′(t ) + y ′(t ) − 2y (t ) = δ (t‬‬ ‫) ‪⇒ ( −3K 1 e −2t + 3K 2 e t ) δ (t ) + (K 1 + K 2 ) δ ′(t ) − (−2K 1 e −2t + K 2 e t ) δ (t ) = δ (t‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫) ‪⇒ y (t ) = (− e −2t + e t ) u (t ) = h (t‬‬ ‫‪3‬‬ ‫‪3‬‬

‫‪٤٧‬‬

‫‪1‬‬ ‫‪‬‬ ‫‪K 1 = − 3‬‬ ‫‪⇒‬‬ ‫‪K = 1‬‬ ‫‪ 2 3‬‬

‫‪ -٨-٢‬ﺳﻴﺴﺘﻢهﺎی ‪ LTI‬ﺗﻮﺻﻴﻒ ﺷﺪﻩ ﺑﺎ ﻣﻌﺎدﻻت ﺗﻔﺎﺿﻠﻲ )زﻣﺎن ﮔﺴﺴﺘﻪ(‬ ‫هﺪف از ﺗﺤﻠﻴﻞ ﻣﻌﺎدﻟﻪ ﺗﻔﺎﺿﻠﻲ‪:‬‬ ‫‪ (١‬راﺑﻄﻪ ﻣﺴﺘﻘﻴﻤﻲ ﺑﻴﻦ ورودي و ﺧﺮوﺟﻲ ﺑﻪ دﺳﺖ ﺁورﻳﻢ ‪.‬‬ ‫‪M‬‬

‫] ‪∑ b k x [n − k‬‬ ‫‪k‬‬

‫= ] ‪y [n‬‬

‫‪=0‬‬

‫] ‪x [n ] = δ [n ] ⇒ y [n ] = h [n‬‬ ‫‪ (٢‬ﺧﺮوﺟﻲ را ﺑﻪ ازاي ورودي ﻣﺸﺨﺺ ﭘﻴﺪا آﻨﻴﻢ‪.‬‬

‫‪ -١-٨-٢‬روش ﺣﻞ ﻣﻌﺎدﻟﻪ ﺗﻔﺎﺿﻠﻲ‬ ‫] ‪x [n − k‬‬

‫‪M‬‬

‫‪N‬‬

‫‪=0‬‬

‫‪=0‬‬

‫‪∑ a k y [n − k ] = k∑ b k‬‬ ‫‪k‬‬

‫اﻟﻒ( روش ﻣﺴﺘﻘﻴﻢ‪:‬‬

‫] ‪y [n ] = y h [n ] + y p [n‬‬

‫ﭘﺎﺳﺦ آﺎﻣﻞ‪:‬‬

‫‪,‬‬

‫] ‪y p [n‬‬

‫ﭘﺎﺳﺦ ﺧﺎص‪:‬‬

‫‪,‬‬

‫] ‪ y h [n‬ﭘﺎﺳﺦ ﻋﻤﻮﻣﻲ‪:‬‬

‫ﺑﺎ ﻓﺮض ﺁﻧﻜﻪ ﻣﻌﺎدﻟﻪ ﺗﻔﺎﺿﻠﻲ ﻣﺮﺗﺒﻪ دوم اﺳﺖ ‪ ، N=2‬ﺑﺮاي ﺑﺪﺳﺖ ﺁوردن ﺟﻮاب ﻋﻤﻮﻣﻲ ﻣﺮاﺣﻞ زﻳﺮ را اﻧﺠﺎم ﻣﻲدهﻴﻢ‪:‬‬

‫‪a 0 y [n ] + a 1 y [n − 1] + a 2 y [n − 2] = 0‬‬ ‫‪y h [n ] = K r n ⇒ (a 0 + a 1r −1 + a 2 r −2 ) ⋅ kr n = 0‬‬

‫‪y h [n ] = k 1 (r1 ) n + k 2 (r 2 ) n‬‬

‫ﺣﻘﻴﻘﻲ و ﻣﺘﻤﺎﻳﺰ‬

‫‪y h [n ] = k 1 (r1 ) n + k 2 n (r 2 ) n‬‬

‫‪r1 , r 2‬‬ ‫‪r1 = r 2‬‬

‫‪2‬‬ ‫ﻣﺜﺎل ‪ (١‬ﻣﻄﻠﻮب اﺳﺖ ﺧﺮوﺟﻲ اﮔﺮ ‪ x [n ] = n‬و ‪ y [0] = 1‬ﺑﺎﺷﺪ‪.‬‬

‫]‪y [n ] + 2y [n − 1] = x [n ] − x [n − 1‬‬ ‫ﭘﺎﺳﺦ ﻋﻤﻮﻣﻲ‪:‬‬

‫‪y [n ] + 2y [n − 1] = 0 ⇒ y h [n ] = k r n ⇒ k r n + 2k r n −1 = 0‬‬ ‫‪k r n (1 + 2 r −1 ) = 0 ⇒ r = (−2) ⇒ y h [n ] = k (−2) n‬‬ ‫ﭘﺎﺳﺦ ﺧﺼﻮﺻﻲ‪:‬‬ ‫‪y[n] + 2y[n - 1] = n 2 − (n − 1) 2 = 2n − 1‬‬

‫‪٤٨‬‬

‫‪2‬‬ ‫‪‬‬ ‫‪A = 3‬‬ ‫‪‬‬ ‫‪B = 1‬‬ ‫‪‬‬ ‫‪9‬‬

‫‪3A = 2‬‬ ‫‪y p [n ] = An + B ⇒ An + B + 2( A (n − 1) + B ) = 2n − 1 ⇒ ‬‬ ‫⇒‬ ‫‪− 2 A + 3B = −1‬‬

‫ﺟﻮاب ﺧﺼﻮﺻﻲ‬

‫‪2‬‬ ‫‪1‬‬ ‫‪n+‬‬ ‫‪3‬‬ ‫‪9‬‬

‫= ] ‪⇒ y p [n‬‬

‫ﭘﺎﺳﺦ آﺎﻣﻞ‪:‬‬ ‫‪8‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪(−2) n + n +‬‬ ‫‪9‬‬ ‫‪3‬‬ ‫‪9‬‬

‫= ] ‪y [n‬‬

‫‪2‬‬ ‫‪1‬‬ ‫‪8‬‬ ‫‪‬‬ ‫‪n‬‬ ‫; = ‪y [n ] = k (−2) + n + ⇒ k‬‬ ‫‪3‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪‬‬ ‫‪y [0] = 1‬‬

‫ب( روش ﺑﺎزﮔﺸﺘﻲ‬ ‫ﺑﺎ ﻳﻚ ﻣﺜﺎل اﻳﻦ روش را ﻳﺮرﺳﻲ ﻣﻲآﻨﻴﻢ‪:‬‬ ‫ﻣﺜﺎل ‪ (٢‬ﺳﻴﺴﺘﻢ ‪ LTI‬و ﻋﻠﻲ ﺗﻮﺻﻴﻒ ﺷﺪﻩ ﺑﺎ ﻣﻌﺎدﻟﻪ ﺗﻔﺎﺿﻠﻲ زﻳﺮ را در ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ‪.‬‬ ‫‪1‬‬ ‫] ‪y [n − 1] + x [n‬‬ ‫‪4‬‬

‫= ] ‪y [n‬‬

‫ﻣﻄﻠﻮﺑﺴﺖ‪:‬‬ ‫اﻟﻒ( ﭘﺎﺳﺦ ﺿﺮﺑﻪ ]‪h[n‬‬

‫ب( ﭘﺎﺳﺦ ﺿﺮﺑﻪ ﻣﻌﻜﻮس ] ‪h I [n‬‬ ‫ج( ﺑﻪ ازاي ورودي دادﻩ ﺷﺪﻩ ]‪ x [n ] = δ [n − 1‬ﺧﺮوﺟﻲ را ﺑﺪﺳﺖ ﺁورﻳﺪ‪.‬‬

‫] ‪h [n ] = y [n ] x [n ]=δ [n‬‬

‫اﻟﻒ(‬

‫‪1‬‬ ‫] ‪h [n − 1] + δ [n‬‬ ‫‪4‬‬

‫= ] ‪⇒ h [n‬‬

‫‪:‬روش ﺑﺎزﮔﺸﺘﻲ‬ ‫‪1‬‬ ‫‪h [−1] + 1 = 1‬‬ ‫‪4‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫)‪h [1] = h [0] = (1‬‬ ‫‪4‬‬ ‫‪4‬‬

‫= ]‪h [0‬‬

‫‪1‬‬ ‫‪1‬‬ ‫‪h [1] = (1) 2‬‬ ‫‪4‬‬ ‫‪4‬‬

‫= ]‪h [2‬‬

‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫] ‪h [n − 1] = ( ) n ⇒ h [n ] = ( ) n u [n‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪٤٩‬‬

‫‪M‬‬

‫= ] ‪h [n‬‬

‫ب( ﺑﺮاي ﺑﺪﺳﺖ ﺁوردن ] ‪ h I [n‬ﺑﺎﻳﺪ ﻓﺮﻣﻮل زﻳﺮ ﺑﺮﻗﺮار ﺑﺎﺷﺪ‪:‬‬ ‫∞‬

‫] ‪∑ h [k ] h I [n − k ] = δ [n‬‬

‫⇒ ] ‪h [n ] ∗ h I [n ] = δ [n‬‬

‫∞‪k = −‬‬

‫‪1‬‬

‫∞‬

‫] ‪( ) k h I [n − k ] = δ [n‬‬ ‫∑‬ ‫‪4‬‬ ‫‪k‬‬ ‫‪=0‬‬

‫= ] ‪u [k ] h I [n − k‬‬

‫‪1‬‬

‫∞‬

‫‪( )k‬‬ ‫∑‬ ‫‪4‬‬ ‫‪k‬‬

‫⇒‬

‫∞‪= −‬‬

‫ﺳﺆال‪ :‬اﮔﺮ ﺳﻴﺴﺘﻤﻲ ‪ LTI‬و ﻋﻠﻲ ﺑﺎﺷﺪ ﺁﻳﺎ ﻣﻌﻜﻮس ﺳﻴﺴﺘﻢ ﻧﻴﺰ ‪ LTI‬و ﻋﻠﻲ ﺧﻮاهﺪ ﺑﻮد؟‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫] ‪h I [n − 1] + ( ) 2 h I [n − 2] + ( ) 3 h I [n − 3] + L = δ [n‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬

‫‪h I [n ] +‬‬

‫‪:‬روش ﺑﺎزﮔﺸﺘﻲ‬

‫‪hI [0] = 1‬‬ ‫] ‪h I [n‬‬ ‫‪1‬‬ ‫‪1‬‬

‫‪n‬‬

‫‪2 3‬‬

‫‪1‬‬ ‫‪4‬‬

‫‪−‬‬

‫‪1‬‬ ‫‪1‬‬ ‫‪hI [0] = 0 ⇒ hI [1] = −‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪hI [2] + hI [1] + ( )2 hI [0] = 0 ⇒ hI [2] = 0‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪1‬‬ ‫]‪hI [3] = 0 , L ⇒ hI [n ] = δ [n ] − δ [n − 1‬‬ ‫‪4‬‬

‫‪hI [1] +‬‬

‫ج( ﭘﺎﺳﺦ ﺿﺮﺑﻪ ﺳﻴﺴﺘﻢ ]‪ h[n‬در ﻗﺴﻤﺖ اﻟﻒ ﺑﺪﺳﺖ ﺁوردﻩ ﺷﺪ‪ ،‬ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺧﺎﺻﻴﺖ ‪) TI‬ﻧﺎﻣﺘﻐﻴﺮﺑﺎزﻣﺎن( ﺑﻮدن ﺳﻴﺴﺘﻢ دارﻳﻢ‪:‬‬ ‫‪1‬‬ ‫‪4‬‬

‫] ‪x [n ] = δ [n ] ⇒ h [n ] = ( ) n u [n‬‬ ‫‪1‬‬ ‫‪4‬‬

‫]‪x [n ] = δ [n − 1] ⇒ y [n ] = h [n − 1] = ( ) n −1 u [n − 1‬‬

‫ﻣﺜﺎل ‪ (٣‬ﺳﻴﺴﺘﻢ ‪ LTI‬ﺑﺎ ﻣﻌﺎدﻟﻪ ﺗﻔﺎﺿﻠﻲ ] ‪ y [n ] + 2y [n − 1] = x [n‬ﺗﻮﺻﻴﻒ ﺷﺪﻩ اﺳﺖ‪ .‬ﺑﺎ ﻓﺮض ﺳﻜﻮن اوﻟﻴﻪ )ﻋﻠﻲ(‬ ‫ﭘﺎﺳﺦ ﺿﺮﺑﻪ را ﺑﻪ دﺳﺖ ﺁورﻳﺪ‪.‬‬ ‫ﺗﺬآﺮ‪ :‬اﮔﺮ ورودي ﺿﺮﺑﻪ واﺣﺪ ﺑﻮد ﭘﻴﺸﻨﻬﺎد ﻣﻲﺷﻮد از روش ﺑﺎزﮔﺸﺘﻲ ﻣﻌﺎدﻟﻪ را ﺣﻞ آﻨﻴﺪ‪.‬‬ ‫روش اول‪ :‬ﺑﻜﺎرﮔﻴﺮي رواﺑﻂ ﺑﺎزﮔﺸﺘﻲ و ﺗﻌﻴﻴﻦ ﭘﺎﺳﺦ ﺑﻪ ازاء ورودي ﺧﺎص ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷﺮط ﺳﻜﻮن اوﻟﻴﻪ‬

‫] ‪x [n ] = δ [n ] ⇒ y [n ] = h [n‬‬ ‫] ‪h [n ] + 2h [n − 1] = δ [n ] ⇒ h [n ] = −2h [n − 1] + δ [n‬‬ ‫‪h [0] = −2h [−1] + 1 = 1‬‬ ‫‪h [1] = −2h [0] = −2‬‬ ‫‪h [2] = −2h [1] = (−2) 2‬‬ ‫‪h [3] = (−2) 3‬‬ ‫‪M‬‬ ‫‪٥٠‬‬

‫] ‪h [k ] = (−2) k ⇒ h [n ] = (−2) n u [n‬‬ ‫روش دوم‪ :‬ﺑﺪﺳﺖ ﺁوردن ﺧﺮوﺟﻲ ﺑﺮ ﺣﺴﺐ ورودي ﺑﻄﻮر آﻠﻲ ﺑﺎ اﺳﺘﻔﺎدﻩ از رواﺑﻂ ﺑﺎزﮔﺸﺘﻲ‪.‬‬

‫] ‪y [n ] + 2y [n − 1] = x [n‬‬ ‫]‪y [0] + 2y [−1] = x [0‬‬ ‫]‪y [1] + 2y [0] = x [1‬‬ ‫]‪y [2] + 2y [1] = x [2‬‬ ‫‪M‬‬

‫]‪y [n − 2] + 2y [n − 3] = x [n − 2‬‬ ‫]‪y [n − 1] + 2y [n − 2] = x [n − 1‬‬ ‫] ‪y [n ] + 2y [n − 1] = x [n‬‬

‫ﺑﺮاي ﺑﺪﺳﺖ ﺁوردن راﺑﻄﻪ ﺻﺮﻳﺢ ﺧﺮوﺟﻲ ﺑﺮ ﺣﺴﺐ ورودي )‪ (FIR‬ﺑﻪ ﻃﺮﻳﻖ ذﻳﻞ ﻋﻤﻞ ﻣﻲ آﻨﻴﻢ‪:‬‬

‫)]‪(−2) n (y [0] + 2y [−1] = x [0‬‬ ‫)]‪(−2) n −1 (y [1] + 2y [0] = x [1‬‬ ‫)]‪(−2) n −2 (y [2] + 2y [1] = x [2‬‬ ‫‪M‬‬ ‫)]‪(−2) 2 (y [n − 2] + 2y [n − 3] = x [n − 2‬‬ ‫)]‪(−2)1 (y [n − 1] + 2y [n − 2] = x [n − 1‬‬ ‫)] ‪(−2) 0 (y [n ] + 2y [n − 1] = x [n‬‬

‫∑‬

‫]‪⇒ y [n ] = (−2) 0 x [n ] + (−2)1 x [n − 1] + (−2) 2 x [n − 2] + L + (−2) n x [0‬‬ ‫‪n‬‬

‫‪n‬‬

‫‪k =0‬‬

‫‪k =0‬‬

‫] ‪∑ (−2) k x [n − k ] =∑ (−2) n −k x [k‬‬ ‫‪n‬‬

‫‪n‬‬

‫] ‪(−2) n k δ [k ] = (−2) n ∑ δ [k ] = (−2) n u [n‬‬ ‫∑‬ ‫‪k‬‬ ‫‪k‬‬ ‫‪−‬‬

‫‪=0‬‬

‫‪=0‬‬

‫روش ﺳﻮم‪ :‬اﺳﺘﻔﺎدﻩ از راﺑﻄﻪ ﺑﻴﻦ ﭘﺎﺳﺦ ﺿﺮﺑﻪ و ﭘﺎﺳﺦ ﭘﻠﻪ در ﺳﻴﺴﺘﻢهﺎي ‪LTI‬‬ ‫ﭘﺎﺳﺦ ﭘﻠﻪ‪:‬‬

‫] ‪z [n ] + 2z [n − 1] = u [n‬‬ ‫ﭘﺎﺳﺦ هﻤﮕﻦ‪:‬‬

‫‪z [n ] + 2z [n − 1] = 0‬‬ ‫‪z h [n ] = k (r ) n ⇒ k r n (1 + 2 r −1 ) = 0 ⇒ r = −2 ⇒ z h [n ] = k (−2) n‬‬ ‫‪٥١‬‬

‫=‬ ‫= ] ‪⇒ h [n‬‬

‫ﭘﺎﺳﺦ ﺣﺼﻮﺻﻲ‪:‬‬

‫‪1‬‬ ‫‪3‬‬

‫= ‪⇒ z p [n ] = A ⇒ A + 2A = 1 ⇒ A‬‬

‫‪z [n ] + 2z [n − 1] = 1 , n ≥ 0‬‬

‫ﭘﺎﺳﺦ آﺎﻣﻞ‪:‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫= ‪, n ≥ 0 ⇒ z [−1] = k (−2) −1 + = 0 ⇒ k‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫] ‪⇒ z [n ] = ( )(−2)n + , n ≥ 0 ⇒ z [n ] = [( )( −2)n + ]u [n‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬

‫‪z [n ] = k (r )n +‬‬

‫ﻣﺤﺎﺳﺒﻪ ﭘﺎﺳﺦ ﺿﺮﺑﻪ ﺑﺎ اﺳﺘﻔﺎدﻩ از ﭘﺎﺳﺦ ﭘﻠﻪ‪:‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫]‪⇒ h [n ] = z [n ] − z [n − 1] = ( )(−2) n u [n ] + u [n ] − (−2) n −1 u [n − 1] − u [n − 1‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫] ‪⇒ ( )(−2) n u [n ] − (−2) n −1 u [n ] = ( )( −2) n [ + 1]u [n ] = (−2) n u [n‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪2‬‬

‫‪ -٩-٢‬ﻧﻤﺎﻳﺶ ﺳﻴﺴﺘﻢ ‪ LTI‬ﺑﺎ اﺳﺘﻔﺎدﻩ از ﺑﻠﻮك دﻳﺎﮔﺮام )ﻣﺸﺘﻖﮔﻴﺮ و اﻧﺘﮕﺮالﮔﻴﺮ(‬ ‫زﻣﺎن ﭘﻴﻮﺳﺘﻪ‬ ‫‪t‬‬

‫‪∫ x (λ)dλ‬‬

‫‪t‬‬

‫∫‬

‫= ) ‪y (t‬‬

‫) ‪x (t‬‬

‫∞‪−‬‬

‫∞‪−‬‬

‫) ‪dx (t‬‬ ‫= ) ‪y (t‬‬ ‫‪dt‬‬

‫‪dt‬‬

‫‪d‬‬

‫) ‪x (t‬‬

‫زﻣﺎن ﮔﺴﺴﺘﻪ‬ ‫هﻢ ارز ﺑﺎ ﻣﺸﺘﻖ در ﭘﻴﻮﺳﺘﻪ ‪:‬‬

‫]‪y [n ] = x [n ] − x [n − 1‬‬

‫∑‬

‫] ‪x [n‬‬

‫‪−‬‬

‫‪D‬‬

‫هﻢ ارز ﺑﺎ اﻧﺘﮕﺮال در ﭘﻴﻮﺳﺘﻪ ‪:‬‬ ‫‪D‬‬ ‫‪n‬‬

‫] ‪∑ x [k‬‬ ‫‪k‬‬

‫= ] ‪y [n‬‬

‫∑‬

‫‪D‬‬

‫]‪x [n − 2‬‬

‫∑‬

‫∑‬

‫∞‪= −‬‬

‫‪٥٢‬‬

‫‪D‬‬

‫]‪x [n − 1‬‬

‫∑‬

‫‪D‬‬

‫] ‪x [n‬‬

y ′′(t ) + a 1 y ′(t ) + a 2 y (t ) = x (t ) + b1 x ′(t ) + b 2 x ′′(t )

(٤ ‫ﻣﺜﺎل‬

‫اﻟﻒ( ﻧﻤﺎﻳﺶ ﺳﻴﺴﺘﻢ ﺑﺎ اﺳﺘﻔﺎدﻩ از ﺑﻠﻮكهﺎي ﻣﺸﺘﻖﮔﻴﺮ‬

y (t ) = −

a1 b b 1 1 y ′(t ) − y ′′(t ) + x (t ) + 1 x ′(t ) + 2 x ′′(t ) a2 a2 a2 a2 a2 +

x (t ) d

d

dt

+

∑

1

∑

a2

y (t ) d

b1

∑

∑

dt

− a1 d

dt

b2

dt

−1 ‫ب( ﻧﻤﺎﻳﺶ ﺳﻴﺴﺘﻢ ﺑﺎ اﺳﺘﻔﺎدﻩ از ﺑﻠﻮكهﺎي اﻧﺘﮕﺮالﮔﻴﺮ‬

y ′′(t ) + a 1 y ′(t ) + a 2 y (t ) = x (t ) + b1 x ′(t ) + b 2 x ′′(t ) ⇒ y ′′(t ) = −a 1 y ′(t ) − a 2 y (t ) = x (t ) + b1 x ′(t ) + b 2 x ′′(t ) y (t ) = −a 1 ∫ y (t )dt − a 2 ∫ ∫ y (t )dt + ∫ ∫ x (t )dt + b1 ∫ x (t )dt +b 2 x (t ) b2

x (t )

∫

+

b1

+ ∑

∑

∑

∑

y (t ) − a1

∫ ∫

∫ −a2

y [n ] + 2y [n − 1] = x [n ] − x [n − 1]

(٥ ‫ﻣﺜﺎل‬

y [n ] = x [n ] − x [n − 1] − 2y [n − 1] + ∑

x [n ]

+ ∑

y [n ]

D

D −1

٥٣

−2

‫ﺗﻤﺮﻳﻦ‪ :‬رﺳﻢ ﺑﻠﻮك دﻳﺎﮔﺮام؟‬ ‫ﺳﻴﺴﺘﻢ زﻣﺎن ﭘﻴﻮﺳﺘﻪ‬

‫) ‪y ′′′(t ) + 3y ′′(t ) + 4y ′(t ) + 5y (t ) = x ′(t‬‬ ‫ﺳﻴﺴﺘﻢ زﻣﺎن ﮔﺴﺴﺘﻪ‬

‫] ‪y [n ] + 2y [n − 1] = x [n‬‬

‫‪ -١٠-٢‬ﺧﻼﺻﻪ‬ ‫ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ راﺑﻄﻪ آﺎﻧﻮﻟﻮﺷﻦ ﻣﻲﺗﻮان ﺧﻮاص ﺟﺎﺑﻪﺟﺎﻳﻲﭘﺬﻳﺮي‪ ،‬ﺗﻮزﻳﻊﭘﺬﻳﺮي و ﺷﺮآﺖﭘﺬﻳﺮي را ﺑﺮاي ﺳﻴﺴﺘﻢهﺎي ‪ LTI‬ﺗﻌﺮﻳﻒ‬ ‫آﺮد‪.‬‬ ‫ﺑﺎ ﻣﻌﺮﻓﻲ ﺗﺎﺑﻊ ﺗﺒﺪﻳﻞ ﻣﻲﺗﻮان ﺧﻮاص ﺟﺪﻳﺪي ﺑﺮاي ﺳﻴﺴﺘﻢهﺎي ‪ LTI‬ﺑﻴﺎن آﺮد‪.‬‬ ‫آﺎﻧﻮﻟﻮﺷﻦ ﺑﻪ دو روش ﺗﺮﺳﻴﻤﻲ و ﻓﺮﻣﻮل ﻗﺎﺑﻞ ﻣﺤﺎﺳﺒﻪ اﺳﺖ‪.‬‬ ‫در ﻣﻌﺎدﻻت دﻳﻔﺮاﻧﺴﻴﻞ‪ ،‬ﺑﺎ ﺣﻞ ﻣﻌﺎدﻟﻪ هﻤﮕﻦ و ﻳﺎﻓﺘﻦ ﺟﻮاب ﺧﺼﻮﺻﻲ ﺑﻪ ازاي ورودي ﺧﺎص‪ ،‬ﺟﻮاب آﻠﻲ ﺳﻴﺴﺘﻢ از ﺟﻤﻊ ﭘﺎﺳﺦ‬ ‫ﻋﻤﻮﻣﻲ و ﭘﺎﺳﺦ ﺧﺼﻮﺻﻲ ﺑﺪﺳﺖ ﻣﻲﺁﻳﺪ‪.‬‬ ‫ﺑﺮاي ﺑﻜﺎرﮔﻴﺮي ﺷﺮاﻳﻂ ﺳﻜﻮن ﺟﻬﺖ ﺗﻌﻴﻴﻦ ﭘﺎﺳﺦ ﺳﻴﺴﺘﻢ ‪ LTI‬ﺑﻪ ورودي دادﻩ ﺷﺪﻩ‪ ،‬ﻋﻨﻮان ﻋﻠﻲ ﺑﻮدن ﺳﻴﺴﺘﻢ ﺿﺮوري اﺳﺖ‪.‬‬ ‫ﺑﺎ اﺳﺘﻔﺎدﻩ از راﺑﻄﻪ ﺑﻴﻦ ﭘﺎﺳﺦ ﺿﺮﺑﻪ و ﭘﺎﺳﺦ ﭘﻠﻪ در ﺳﻴﺴﺘﻢهﺎي ‪ LTI‬و ﻳﺎ ﻣﺤﺎﺳﺒﻪ ﭘﺎﺳﺦ ﺿﺮﺑﻪ ﺳﻴﺴﺘﻢ ﺑﻪ ﻃﻮر ﻣﺴﺘﻘﻴﻢ ﻣﻲﺗﻮان‬ ‫ﭘﺎﺳﺦ ﺑﻪ ورودي ﺿﺮﺑﻪ را ﻳﺎﻓﺖ‪.‬‬ ‫ﺳﻴﺴﺘﻢهﺎي ‪ LTI‬را ﺑﺎ اﺳﺘﻔﺎدﻩ از ﺑﻠﻮك دﻳﺎﮔﺮام ﻧﻴﺰ ﻣﻲﺗﻮان ﻧﺸﺎن داد‪.‬‬

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