ﻓﺼﻞ ﭼﻬﺎرم
ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﺴﺘﻢهﺎي زﻣﺎن ﭘﻴﻮﺳﺘﻪ
٨٤
-١-٤ﺗﻌﺮﻳﻒ ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ زﻣﺎن ﭘﻴﻮﺳﺘﻪ f
) f (t ) ↔ F ( jω −1
ω = 2πf
dω
;
jωt
∞+
∫ F ( jω )e
∞−
1 2π
= ) f (t
∞+
− jωt ∫ f (t )e dt
,
f
= ) F ( jω
∞−
-٢-٤ﺷﺮاﻳﻂ وﺟﻮد ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ∞<
-١ﺳﻴﮕﻨﺎل ﻣﻄﻠﻘﺎ اﻧﺘﮕﺮال ﺑﺎﺷﺪ.
∞+
∫ f (t ) dt
∞−
-٢در ﻃﻮل هﺮ ﺑﺎزﻩ ﻣﺤﺪود ﺳﻴﮕﻨﺎل ﺗﻌﺪاد ﻣﺤﺪودي Maxﻳﺎ Minداﺷﺘﻪ ﺑﺎﺷﺪ. -٣در ﻃﻮل هﺮ ﺑﺎزﻩ ﻣﺤﺪود ﺳﻴﮕﻨﺎل ﺗﻌﺪاد ﻣﺤﺪودي ﻧﺎﭘﻴﻮﺳﺘﮕﻲ داﺷﺘﻪ ﺑﺎﺷﺪ. −at ﻣﺜﺎل (١ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل ، x (t ) = e u (t ) , a > 0را ﺑﺪﺳﺖ ﺁوردﻩ و اﻧﺪازﻩ و ﻓﺎز ﺁن را ﺑﺪﺳﺖ ﺁوردﻩ و ﺗﺮﺳﻴﻢ
ﻧﻤﺎﺋﻴﺪ. 1 a + jω
=
∞+ 0
1 e −(a + jω )t a + jω
∞+
∞+
∞+
0
0
∞−
−at − jωt −at − jωt −(a + jω )t dt = − ∫ e u (t ) e dt = ∫ e e dt = ∫ e
= ) X ( jω
; ) X ( jω ) = X ( jω ) ⋅ e j∠X ( jω
ω a ω =0
,
∞ω = ±
,
∠X ( jω ) = − tan −1 0 ∠X ( jω ) = π m 2
1
,
a2 +ω2 ω =0 ∞ω = ±
ﻣﺜﺎل (٢ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل زﻳﺮ را ﺑﺪﺳﺖ ﺁورﻳﺪ. ٨٥
= )X ( j ω
1 1 a = a , X ( j ω) = 0 ,
x (t ) = rect ( 1
− T1
X ( j ω) =
T1
∫T 1 × e
−
− jωt
dt = −
1
⇒ X ( jω ) = 2T 1
1
jω
e − jωt
T1 −T1
=−
1
jω
(e − jωT1 − e jωT1 ) =
t ) 2T1
t
T1 2 sin(ωT 1 )
ω
=
2 sin(ωT 1 ) 1 ωT 1 ⋅
T1
ωT 1 ) π = 2T sinc( ω ⋅ T ) 1 1 ωT 1 π π⋅ π
sin(π ⋅
sinc(x) =
sin(π x) π x :ﻳﺎدﺁوري
: در ﺣﺎﻟﺖ آﻠﻲ اﮔﺮ ﺳﻴﮕﻨﺎل ﭘﺎﻟﺴﻲ ﺑﻪ ﻓﺮم زﻳﺮ ﺑﺎﺷﺪ ﺁﻧﮕﺎﻩ:ﺗﺬآﺮ
x (t ) = A rect ( A − T1
2T 1 A × sinc(
T1
t ) 2T 1
t
ω ω ⋅T1 ) × π = )ﻧﺼﻒ ﻋﺮض ﭘﺎﻟﺲπ sinc( × ﺳﻄﺢ ﭘﺎﻟﺲX ( j ω ) =
ﺧﻮاص ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ-٣-٤ ﺧﻄﻲ ﺑﻮدن-١-٣-٤ f 1 (t ) ↔ F 1 ( jω ) ⇒ af 1 (t ) + bf 2 (t ) ↔ a F 1 ( jω ) + b F 2 ( jω ) f 2 (t ) ↔ F 2 ( jω )
ﺗﻐﻴﻴﺮ ﻣﻘﻴﺎس زﻣﺎﻧﻲ-٢-٣-٤ f (t ) ↔ F ( jω ) ⇒ f (at ) ↔
٨٦
1
a
F (j
ω ) a
. ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل زﻳﺮ را ﺑﺪﺳﺖ ﺁورﻳﺪ،( ﺑﻪ آﻤﻚ ﺧﻮاص٣ ﻣﺜﺎل
x (t ) 1
f (t ) = rect ( 1 −a
a
t
1
−1
t ) 2a
F ( jω ) = 2a sinc(
t
x (t ) = f (at ) ↔ X ( jω ) =
1
a
F (j
ω ⋅a) π :ﻣﻲداﻧﻴﻢ
ω ω 2a ω )= sinc( a ⋅ a ) = 2sinc( ) a π π a
ﺷﻴﻔﺖ زﻣﺎﻧﻲ-٣-٣-٤ f (t ) ↔ F ( jω ) ⇒ f (t − t 0 ) ↔ e − jωt F ( jω ) 0
.را ﺑﻪ دﺳﺖ ﺁورﻳﺪ
x (t ) = rect (
t − 2.5 3
)
( ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺗﺎﺑﻊ٤ ﻣﺜﺎل
t ω 3 x 1 (t ) = rect ( ) ⇒ X 1 ( jω ) = 3sinc( ⋅ ) π 2 ١) 3 x 1 (t ) 1
−
3 2
3 2
x (t ) = x 1 (t − 2.5) ⇒ X ( j ω ) = e − j 2.5ω ⋅ X 1 ( j ω ) = e − j 2.5ω × 3 sin(
٨٧
t
ω 3 ⋅ ) π 2 ٢)
) x (t 1
t
1
4
-٤-٣-٤ﺷﻴﻔﺖ ﻓﺮآﺎﻧﺴﻲ )) f (t ) ↔ F ( jω ) ⇒ e jω t f (t ) ↔ F ( j (ω − ω 0 0
) F ( j ω ) = 2π δ (ω
ﻣﺜﺎل (٥ﻋﻜﺲ ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ را ﺑﺪﺳﺖ ﺁورﻳﺪ.
dω = 1
jωt
∞+
∫ 2π δ (ω ) e
∞−
1 = ) f (t 2π
δ (t ) ↔ 1 ) 1 ↔ 2π δ (ω ± jω 0t eرا ﻧﻤﻲﺗﻮان ﻣﺴﺘﻘﻴﻤﺎ ﻣﺤﺎﺳﺒﻪ ﻧﻤﻮد .ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ دو راﺑﻄﻪ ﺑﺪﺳﺖ ﺁﻣﺪﻩ در ﺑﺎﻻ و ﺑﺎ اﺳﺘﻔﺎدﻩ ﺗﺬآﺮ :ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﻧﻤﺎﻳﻲ ﻣﺨﺘﻠﻂ
از ﺧﺎﺻﻴﺖ ﺷﻴﻔﺖ ﻓﺮآﺎﻧﺴﻲ ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ اﻳﻦ ﺗﻮاﺑﻊ را ﻣﺤﺎﺳﺒﻪ ﻣﻲآﻨﻴﻢ: ) e jω0t × 1 ↔ 2π δ (ω − ω 0 1 j ω 0t 1 − j ω 0t + e ) ↔ π δ(ω − ω 0 ) + π δ(ω + ω 0 ⇒ − jω t ⇒ e 2 2 ) e 0 × 1 ↔ 2π δ (ω + ω 0 f
) cos(ω 0t ) → π δ(ω − ω 0 ) + π δ(ω + ω 0
ﺑﻨﺎﺑﺮاﻳﻦ ﺗﺎﺑﻊ cos ω 0tﺟﺰو ﺗﻮاﺑﻌﻲ اﺳﺖ آﻪ ﺑﻪ آﻤﻚ ﺧﻮاص ،ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪاش ﺑﺪﺳﺖ ﻣﻲﺁﻳﺪ .ﺳﻴﮕﻨﺎل x (t ) = cos ω 0t ﻣﻄﻠﻘﺎ ﺣﻘﻴﻘﻲ و زوج اﺳﺖ ،هﻤﺎﻧﻄﻮرﻳﻜﻪ در ﺷﻜﻞ ﻣﺸﺎهﺪﻩ ﻣﻲﺷﻮد ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺁن ﻧﻴﺰ ﻣﻄﻠﻘﺎ ﺣﻘﻴﻘﻲ و زوج ﻣﻲﺑﺎﺷﺪ.
)) X ( j ω ) = f (cos(ω 0t π
ω
ω0
π − ω0
ﺑﺎ هﻤﺎن ﻣﻨﻄﻖ دﻧﺒﺎل ﺷﺪﻩ در ﺑﺎﻻ ﻣﻲﺗﻮان ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺗﺎﺑﻊ x (t ) = sin ω 0tرا ﻧﻴﺰ ﺑﺪﺳﺖ ﺁورد. 1 j ω 0t 1 − j ω 0t π π − ) ↔ δ(ω − ω 0 ) − δ(ω + ω 0 e e 2j 2j j j ) = −πj δ(ω − ω 0 ) + πj δ(ω + ω 0
= sin ω 0t
٨٨
ﺳﻴﮕﻨﺎل x (t ) = sin ω 0tﻣﻄﻠﻘﺎ ﺣﻘﻴﻘﻲ و ﻓﺮد اﺳﺖ .ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ اﻳﻦ ﺳﻴﮕﻨﺎل هﻤﺎﻧﻄﻮرﻳﻜﻪ در ﺷﻜﻞ ﻧﻴﺰ ﻣﺸﺎهﺪﻩ ﻣﻲﺷﻮد ﻣﻄﻠﻘﺎ ﻣﻮهﻮﻣﻲ و ﻓﺮد ﻣﻲﺑﺎﺷﺪ.
)) X ( j ω ) = f (sin(ω 0t jπ ω
ω0
− ω0
− jπ
٨٩
ﻣﺪوﻻﺳﻴﻮن-٥-٣-٤ f (t ) ↔ F ( j ω ) ⇒ f (t ) cos(ω 0t ) ↔ f (t ) cos(ω 0t ) =
1 1 F ( j (ω − ω 0 )) + F ( j (ω + ω 0 )) 2 2
1 1 1 1 f (t )e j ω 0t + f (t )e − j ω0t ↔ F ( j (ω − ω 0 )) + F ( j (ω + ω 0 )) 2 2 2 2 : اﺛﺒﺎت
( ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل زﻳﺮ را ﺑﺪﺳﺖ ﺁورﻳﺪ؟١ ﻣﺜﺎل
t rect ( ) cos(ω 0t ) ↔ ? a t ω a rect ( ) ↔ a sinc ( ⋅ ) a π 2
1
−
t rect ( ) a
a
a
2
2
f
−
t a
ω a ⋅ ) π 2
0 π
2π
1
t
⇒ rect ( ) cos(ω 0t ) ↔
a sinc(
a 2
sinc(
2π
a
−
π a
a
a
ω
ω - ω0 a ω + ω0 a a ⋅ ) + sinc( ⋅ ) 2 2 2 π π
ﻣﺸﺘﻖ ﻓﺮآﺎﻧﺴﻲ-٦-٣-٤ f (t ) ↔ F ( j ω ) ⇒ (− jt ) n f (t ) ↔ ٩٠
d nF ( j ω) dω n
-٧-٣-٤ﻣﺸﺘﻖ زﻣﺎﻧﻲ ) d n f (t ) ↔ ( j ω )n F ( j ω dt n
⇒ ) f (t ) ↔ F ( j ω
-٨-٣-٤ﻣﺰدوجﮔﻴﺮي )ﺗﻘﺎرن ﻣﺰدوج( ) f (t ) ↔ F ( j ω ) f ∗ (t ) ↔ F (− j ω ﺗﺬآﺮ: ) f (tﺳﻴﮕﻨﺎل ﺣﻘﻴﻘﻲ و زوج ⇐ ) F ( j ωﻣﻄﻠﻘﺎ ﺣﻘﻴﻘﻲ و زوج ﺧﻮاهﺪ ﺑﻮد. ) f (tﺳﻴﮕﻨﺎل ﺣﻘﻴﻘﻲ و ﻓﺮد ⇐ ) F ( j ωﻣﻄﻠﻘﺎ ﻣﻮهﻮﻣﻲ و ﻓﺮد ﺧﻮاهﺪ ﺑﻮد.
-٩-٣-٤ﺧﺎﺻﻴﺖ دوﮔﺎﻧﻲ ) f (t ) ↔ F ( j ω ) F (t ) ↔ 2π f (−ω
) f (t ) ↔ F (ω ) F (t ) ↔ 2π f (−ω
ﻳﺎ
1 ﻣﺜﺎل (٢ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل 1 + t 2 ∞
dt + ∫ e −at ⋅ e − jωt dt
− jωt
⋅e
0
at
= ) x (t
0
dt = ∫ e ∞−
را ﺑﻪ آﻤﻚ ﺧﻮاص ﺑﺪﺳﺖ ﺁورﻳﺪ.
− jωt
⋅e
−a t
∞+
∫e
= ) ⇒ F (ω
at
∞−
2a = 2 a +ω2 دوﮔﺎﻧﻲ 2a 2a −a −ω −a ω → 2π e = 2π e ⇒ 2 2 2 2 a +ω a +t 2a −a −ω −a ω ↔ 2π e = 2π e ﺑﻨﺎﺑﺮاﻳﻦ 2 2 a +t
↔
ﭘﺎراﻣﺘﺮ a=1را در ﻧﻈﺮ ﻣﻲﮔﻴﺮﻳﻢ: 2 1 −ω −ω ↔ 2π e ↔ X ( j ω) = π e = ) ⇒ x (t 2 2 1+t 1+t
٩١
−a t
e
f (t ) = e
.( ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل زﻳﺮ را ﺑﻪ آﻤﻚ ﺧﻮاص ﺑﺪﺳﺖ ﺁورﻳﺪ٣ ﻣﺜﺎل
x (t ) = a sinc(
1
−
t rect ( ) a
a
a
2
2
f
a sinc(
ω a ⋅ ) π 2
0 π
2π
1
t
−
2π
−
a
π a
t a ⋅ )↔? π 2
a
ω
a
t a ω ω ⋅ ) ↔ 2π rect (− ) = 2π rect ( ) a a π 2 ω t a ω x (t ) = a sinc( ⋅ ) ↔ 2π rect (− ) = 2π rect ( ) π 2 a a
a sinc(
a sinc( 1
−
2π
a
−
π a
0 π
a
t a ⋅ ) π 2
2π
ω a
2π rect ( )
f
t
a
1
−
a
a
2
2
ω
ﺿﺮب و آﺎﻧﻮﻟﻮﺷﻦ-١٠-٣-٤ f 1 (t ) ∗ f 2 (t ) ↔ F1 ( jω ) ⋅ F 2 ( jω ) f 1 (t ) ⋅ f 2 (t ) ↔
1 F 1 ( jω ) ∗ F 2 ( jω ) 2π
t t rect ( ) ∗ rect ( ) ↔ ? a a ٩٢
(٤ ﻣﺜﺎل
t t rect ( ) ∗ rect ( ) a a t t ω a rect ( ) ∗ rect ( ) ↔ a 2 sinc 2 ( ⋅ ) a a π 2
a
−a
t
اﻧﺘﮕﺮال-١١-٣-٤ f (t ) ↔ F ( j ω ) t
∫ f (λ)dλ ↔
−∞
F ( j ω) + π F ( j 0)δ (ω ) jω :اﺛﺒﺎت
+∞
t
−∞
−∞
f (t ) ∗ u (t ) = ∫ f ( λ )u (t − λ )dλ = ∫ f ( λ )dλ ↔ F ( j ω ) ⋅ U ( j ω ) u (t ) ↔ U ( j ω ) =
1
jω
+ π δ (ω )
F ( j ω )U ( j ω ) = t
⇒
∫ f ( λ ) dλ ↔
−∞
1
jω
F ( j ω ) + π δ (ω ) ⋅ F ( j ω )
F ( j ω) + π δ (ω ) ⋅ F ( j 0) jω
. ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺁن را ﺑﺪﺳﺖ ﺁورﻳﺪ. در ذﻳﻞ ﺗﺮﺳﻴﻢ ﺷﺪﻩ اﺳﺖx(t) ( ﺳﻴﮕﻨﺎل٥ ﻣﺜﺎل
٩٣
x (t ) 1.5
1 1
2
3
4
t
t rect ( )
x 1 (t )
3
1
1
1
4
t
− 1.5
1.5
t
t ω 3 rect ( ) ↔ 3sinc ( ⋅ ) 3 π 2
t
t ω 1 rect ( ) ↔ 0.5 sinc ( ⋅ ) π 2 1
t rect ( )
x 2 (t )
1
0.5 2
0.5
3
t
− 0.5 0.5
x (t ) = x 1 (t ) + x 2 (t ) ↔ X ( j ω ) = X 1 ( j ω ) + X 2 ( j ω ) t − 2.5 ω 3 x 1 (t ) = rect ( ) ↔ e − jω ( 2.5 ) ⋅ 3sinc ( ⋅ ) 3 π 2 t − 2.5 ω 1 x 2 (t ) = rect ( ) ↔ e − jω ( 2.5 ) ⋅ 0.5sinc ( ⋅ ) 1 π 2 ω 3 1 ω 1 x (t ) ↔ X ( j ω ) = e − jω ( 2.5) (3sinc ( ⋅ ) + sinc ( ⋅ )) π 2 2 π 2
راﺑﻄﻪ ﭘﺎرﺳﻮال-١٢-٣-٤ ∞
* ∫ f 1 (t )f 2 (t )dt =
−∞
1 2π
∞
∫F
1
( j ω )F 2* ( j ω )dω
−∞
. راﺑﻄﻪ ﭘﺎرﺳﻮال ﺑﻪ ﻓﺮم ﺳﺎدﻩﺗﺮي ﺑﻴﺎن ﻣﻲﺷﻮد، ﻳﻜﺴﺎن در ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺷﻮﻧﺪf 2 (t ) وf 1 (t ) اﮔﺮ ﺗﻮاﺑﻊ
f (t ) = f 1 (t ) = f 2 (t ) ⇒
∞
∫
−∞
2
f (t ) dt =
1 2π
∞
∫ F ( jω )
2
dω
−∞
. ﺑﻪ ﺻﻮرت زﻳﺮ اﺳﺖx(t) ( ﺳﻴﮕﻨﺎل٦ ﻣﺜﺎل : ﺑﺪون اﻧﺠﺎم ﻣﺤﺎﺳﺒﺎت ﺻﺮﻳﺢ ﻣﻄﻠﻮﺑﺴﺖ، ﺑﺎﺷﺪX ( j ω ) اﮔﺮ ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺁن ٩٤
) x (t 2
1
t
2
3
−1
1
∞
ب( )X ( j 0
اﻟﻒ( ) ∠X ( j ω
ج(
∫ X ( j ω )dω
dω
2
د(
∞−
∞
) ∫ X ( jω
∞−
ﺣﻞ اﻟﻒ x 1 (t ) ↔ X 1 ( j ω ) :ﭼﻮن ) x 1 (tﺣﻘﻴﻘﻲ و زوج اﺳﺖ X 1 ( j ω ) ،ﻧﻴﺰ ﻣﻄﻠﻘﺎ ﺣﻘﻴﻘﻲ و زوج ﺧﻮاهﺪ ﺑﻮد .ﺑﻪ
ﻋﺒﺎرت دﻳﮕﺮ∠X 1 ( j ω ) = 0 :
) x 1 (t 2
1
)x (t ) = x 1 (t − 1
ﻳﺎ
)x 1 (t ) = x (t + 1
t
2
1
−2 −1
) x (t ) = x 1 (t − 1) ↔ X ( jω ) = e − jω ⋅ X 1 ( j ω ) ⇒ ∠X ( j ω ) = −ω + ∠X 1 ( j ω ⇒ ∠X ( j ω ) = −ω ﺣﻞ ب: ∞+
∞+
∞−
∞−
∫ x (t )dt
− jωt = )∫ x (t )e dt ⇒ X ( j 0
= )X ( j ω
ﭼﻮن ﺳﻄﺢ زﻳﺮ ﻣﻨﺤﻨﻲ ) x(tﺑﺎ ﺳﻄﺢ زﻳﺮ ﻣﻨﺤﻨﻲ ) x 1 (tﺑﺮاﺑﺮ اﺳﺖ و ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺗﻌﺮﻳﻒ ﺗﺎﺑﻊ زوج دارﻳﻢ: 2
1
1
0
) dt + ∫ 2dt ) = 7ﻣﻌﺎدﻟﻪ ﺧﻂ( ∫ (X ( j 0) = 2 ﺣﻞ ج: ∞
∞
∞−
∞−
∫ X ( j ω )dω = 2π x (0) = 4π
jωt ⇒ ∫ X ( j ω )e dω
1 2π
= ) x (t
ﺣﻞ د :راﺑﻄﻪ ﭘﺎرﺳﻮال ﻣﻮرد ﻧﻈﺮ اﺳﺖ.
dt
2
∞
) ∫ x (t
∞−
2
X ( jω ) dω = 2π
∞
∫
∞−
2
= X ( jω ) dω
∞
∫
∞−
٩٥
1 2π
2
= x (t ) dt
∞
∫
∞−
-٤-٤ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل ﻣﺘﻨﺎوب اﮔﺮ ﺳﻴﮕﻨﺎﻟﻲ ﻣﺘﻨﺎوب ﺑﺎﺷﺪ ﺑﻪ اﻳﻦ ﻣﻌﻨﻲ اﺳﺖ آﻪ داراي ﺳﺮي ﻓﻮرﻳﻪ اﺳﺖ .ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل ﻣﺘﻨﺎوب را ﺑﻪ ﻃﻮر ﻣﺴﺘﻘﻴﻢ ﻣﺤﺎﺳﺒﻪ ﻧﻤﻲآﻨﻴﻢ .اﺑﺘﺪا ﺳﺮي ﻓﻮرﻳﻪ ﺗﺎﺑﻊ را ﺑﻪ دﺳﺖ ﺁوردﻩ ،ﺳﭙﺲ از ﺳﺮي ﻓﻮرﻳﻪ ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ را ﺑﻪ دﺳﺖ ﻣﻲﺁورﻳﻢ.
) x (t ) = x (t + T 0 )
,
∑ ak e
= ) x (t
∞k = −
2π 0
2π )t T0
( jk
∞
∞
∑ 2π a k δ (ω − k T k
f
= )→ X (j ω
2π )t T0
( jk
∞= −
∞
∑ ak e k
= ) x (t
∞= −
) − mT
ﻣﻬﻤﺘﺮﻳﻦ ﺳﻴﮕﻨﺎل ﻣﺘﻨﺎوﺑﻲ آﻪ ﻣﻲﺷﻨﺎﺳﻴﻢ ﻗﻄﺎر ﺿﺮﺑﻪ اﺳﺖ.
∞
∑ δ (t
= ) p (t
∞m = −
اﺑﺘﺪا ﺳﺮي ﻓﻮرﻳﻪ ﺗﺎﺑﻊ را ﺑﻪ دﺳﺖ ﺁوردﻩ: )
2π
T
δ (ω − k
2π
∞
∑T k
= )↔ P ( j ω
)t
2π
T
∞= −
( jk
e
1
∞
∑T k
= ) ; p (t
∞= −
1
T
= ak
) P (ω
) p (t
2π
T
ω
6π
T
4π
T
2π
T
− 2Tπ
4π
T
t
− 6Tπ −
2T 3T
T
− 3T − 2T − T
-٥-٤ﺗﺌﻮري ﻧﻤﻮﻧﻪﺑﺮداري ﺳﻴﮕﻨﺎل در اﻳﻦ روش ﺳﻴﮕﻨﺎل ) x(tدر ﻗﻄﺎر ﺿﺮﺑﻪ ﺿﺮب ﻣﻲﺷﻮد آﻪ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺧﻮاص ﺑﻴﺎن ﺷﺪﻩ ﺑﺮاي ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ،ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺣﺎﺻﻠﻀﺮب دو ﺗﺎﺑﻊ ﺑﺮاﺑﺮ ﺑﺎ آﺎﻧﻮﻟﻮﺷﻦ ﺁﻧﻬﺎﺳﺖ.
) x (t
) x s (t ) = x (t ) ⋅ p (t
) p (t
٩٦
1 )X ( j ω) ∗ P ( j ω 2π )
2π
2π
∞
∑T k
δ (ω − k
= )↔ P ( j ω
t
2π
T
= ) x s (t ) = x (t ) ⋅ p (t ) ↔ X s ( j ω jk
e
1
∞
∑T k
∞
= ) ∑ δ (t − mT
= ) p (t
T ∞ ∞ 1 2π 2π 1 2π = )X s (j ω ∑ ( ∗ )X ( j ω δ (ω − k )) = ∑ X ( j (ω − k )) 2π T T k = −∞ T k = −∞T ∞= −
∞= −
∞m = −
راﺑﻄﻪ ﺑﺪﺳﺖ ﺁﻣﺪﻩ ﺑﻴﺎن ﻣﻲآﻨﺪ آﻪ در اﺛﺮ ﻧﻤﻮﻧﻪﺑﺮداري از ﺳﻴﮕﻨﺎل ،ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل ﻧﻤﻮﻧﻪﺑﺮداري ﺷﺪﻩ ) x s (t ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل اوﻟﻴﻪ ) X ( j ωﺧﻮاهﺪ ﺑﻮد آﻪ اﻟﺒﺘﻪ در ﻓﻮاﺻﻞ
2π
1
Tﻋﻴﻨﺎ ﺗﻜﺮار ﺷﺪﻩ و داﻣﻨﻪ ﺁن ﻧﻴﺰ در Tﺿﺮب ﺷﺪﻩ اﺳﺖ.
)X s ( j ω
)X ( j ω 1
1
T
ω
4π
2π
T
T
ω1
− ω1
2π
T
−
4π
T
ω
−
ω1
ﻣﺜﺎل (٧ﻣﻄﻠﻮب اﺳﺖ ) X ( j ω ، sاﮔﺮ ) X ( j ωﺑﻪ ﺻﻮرت ﻣﻘﺎﺑﻞ ﺑﺎﺷﺪ.
)X ( j ω 2
π اﻟﻒ( 4
ﺣﻞ اﻟﻒ:
π
= T
=8
ب( 2
2π ج( 3
= T
= T
ω
1 2
2π
T )X s ( j ω
8
π ω
10
8
6
2
=
4
π
ﺑﺮاﺑﺮ ﺗﺒﺪﻳﻞ
×2
−2
−6
٩٧
− 10 − 8
− 2 −1
− ω1
ﺣﻞ ب:
=4
2π
T )X s ( j ω
4 4 = 2π π
ω
10
ﺣﻞ ج:
=3
2
6
×2
−2
− 10
−6
2π
T )X s ( j ω
3 3 = 2π π
ω
8
4 5
1 2
×2
− 5 − 4 − 2 −1
−8
-٦-٤ﺧﻼﺻﻪ ﻗﺒﻞ از ﺑﺪﺳﺖ ﺁوردن ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺑﺎ اﺳﺘﻔﺎدﻩ از ﺑﺎﻳﺪ ﺷﺮاﻳﻂ وﺟﻮد ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ )دﻳﺮﻳﻜﻠﻪ( را ﺑﺮرﺳﻲ آﺮد. در ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﻧﻴﺰ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ اﻧﺪازﻩ و ﻓﺎز ﻣﻲﺗﻮان ﻧﻮع ﻓﻴﻠﺘﺮ را ﺗﺸﺨﻴﺺ داد. ﺑﺎ داﻧﺴﺘﻦ ﺧﻮاص ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ،ﺑﺮﺧﻲ ﻣﺴﺎﺋﻞ ﺑﻪ روش ﺳﺎدﻩﺗﺮي ﻗﺎﺑﻞ ﺣﻞ اﺳﺖ. ﺗﺎﺑﻊ
cos ω 0t
ﺟﺰو ﺗﻮاﺑﻌﻲ اﺳﺖ آﻪ ﺑﻪ آﻤﻚ ﺧﻮاص ،ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪاش ﺑﺪﺳﺖ ﻣﻲﺁﻳﺪ و ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷﻜﻞ ﺣﺎﺻﻞ ،ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺁن
ﻣﻄﻠﻘﺎ ﺣﻘﻴﻘﻲ و زوج اﺳﺖ .هﻤﭽﻨﻴﻦ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷﻜﻞ ﺑﺪﺳﺖ ﺁﻣﺪﻩ ﺑﺮاي ، sin ω 0tﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺁن ﻣﻄﻠﻘﺎ ﻣﻮهﻮﻣﻲ و ﻓﺮد اﺳﺖ. اﮔﺮ ) f(tﺳﻴﮕﻨﺎﻟﻲ زوج و ﺣﻘﻴﻘﻲ ﺑﺎﺷﺪ F ( j ω ) ،ﻣﻄﻠﻘﺎ ﺣﻘﻴﻘﻲ و زوج و هﻤﭽﻨﻴﻦ اﮔﺮ ) f(tﺳﻴﮕﻨﺎﻟﻲ ﻓﺮد و ﺣﻘﻴﻘﻲ ﺑﺎﺷﺪ، ) F ( j ωﻣﻄﻠﻘﺎ ﻣﻮهﻮﻣﻲ و ﻓﺮد اﺳﺖ. اﮔﺮ ﺳﻴﮕﻨﺎﻟﻲ ﻣﺘﻨﺎوب ﺑﺎﺷﺪ ﺑﻪ اﻳﻦ ﻣﻌﻨﻲ اﺳﺖ آﻪ داراي ﺳﺮي ﻓﻮرﻳﻪ اﺳﺖ .ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺳﻴﮕﻨﺎل ﻣﺘﻨﺎوب را ﺑﻪ ﻃﻮر ﻣﺴﺘﻘﻴﻢ ﻣﺤﺎﺳﺒﻪ ﻧﻤﻲآﻨﻴﻢ .اﺑﺘﺪا ﺳﺮي ﻓﻮرﻳﻪ ﺗﺎﺑﻊ را ﺑﻪ دﺳﺖ ﺁوردﻩ ،ﺳﭙﺲ از ﺳﺮي ﻓﻮرﻳﻪ ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ را ﺑﻪ دﺳﺖ ﻣﻲﺁورﻳﻢ. ٩٨
ﻣﻬﻤﺘﺮﻳﻦ ﺳﻴﮕﻨﺎل ﻣﺘﻨﺎوﺑﻲ آﻪ ﻣﻲﺷﻨﺎﺳﻴﻢ ﻗﻄﺎر ﺿﺮﺑﻪ اﺳﺖ. در ﻧﻤﻮﻧﻪﺑﺮداري ﺳﻴﮕﻨﺎل ،ﺗﺎﺑﻊ ) x(tدر ﻗﻄﺎر ﺿﺮﺑﻪ ﺿﺮب ﻣﻲﺷﻮد .ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺧﻮاص ﺑﻴﺎن ﺷﺪﻩ ﺑﺮاي ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ،ﺗﺒﺪﻳﻞ ﻓﻮرﻳﻪ ﺣﺎﺻﻠﻀﺮب دو ﺗﺎﺑﻊ ﺑﺮاﺑﺮ ﺑﺎ آﺎﻧﻮﻟﻮﺷﻦ ﺁﻧﻬﺎﺳﺖ.
٩٩