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.
٥٤