signal,part 4

‫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.‬‬

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