ece3101_Homework4_SOLNS_F19 (1).docx - B Olson 1 Determine the Fourier transform of the following function by viewing it as a series of shifted rect

# ece3101_Homework4_SOLNS_F19 (1).docx - B Olson 1 Determine...

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B. Olson 1) Determine the Fourier transform of the following function by viewing it as a series of shifted rect() functions and performing appropriate transform operations. You may use the provided tables. Define g ( t )= rect [ t / 2 ] { g ( t ) } = G ( ω )= 2 sinc ( ω )= 2 sin ( ω ) ω { g ( t 2 ) } = G ( ω ) e j 2 ω { g ( t )+ 2 g ( t 2 ) } = G ( ω ) ( 1 + 2 e j 2 ω ) = 2 sinc ( ω ) ( 1 + 2 e j 2 ω ) Working with Basic Fourier Transform Operations 2) Determine the Fourier Transform of the following. a) f ( t ) = e j 7 t sin ( 40 t ) b) f ( t ) = cos ( ω 0 t + π / 4 ) c) f ( t ) = e 4 t u ( t ) d) f ( t ) = e + 4 t u (− t ) e) f ( t ) = cos ( 40 t ) u ( t + 2 ) -3 -1 0 1 3 t f(t) 2 1 -1 0 1 t g(t) 1 -3 -1 0 1 2 3 t f(t) = g(t) + 2g(t-2) 2 1
a) f ( t ) = e j 7 t sin ( 40 t ) Root: g ( t ) = sin ( 40 t ) ↔G ( ω ) = π j [ δ ( ω 40 ) δ ( ω 40 ) ] f ( t ) = e j 7 t g ( t ) ↔F ( ω ) = G ( ω + 7 ) = π j [ δ ( ω + 7 40 ) δ ( ω + 7 40 ) ] b) Method 1 f ( t ) = cos ( ω 0 t + π 4 ) = 1 2 [ e j ( ω 0 t + π 4 ) + e j ( ω 0 t + π 4 ) ] = 1 2 e j π 4 e 0 t + 1 2 e j π 4 e 0 t = π e j π 4 δ ( ω ω 0 ) + π e j π 4 δ ( ω + ω 0 ) Method 2 f ( t ) = cos ( ω 0 t + π 4 ) = cos ( ω 0 t + ω 0 ω 0 π 4 ) = cos ( ω 0 ( t + π 4 ω 0 ) ) Root: g ( t ) = cos ( ω 0 t ) ↔G ( ω ) = π [ δ ( ω ω 0 ) + δ ( ω + ω 0 ) ] f ( t ) = g ( t + π 4 ω 0 ) ↔F ( ω ) = e + j π ω 4 ω 0 G ( ω ) = π [ δ ( ω ω 0 ) + δ ( ω + ω 0 ) ] e + j πω 4 ω 0 (This solution is equivalent to the previous one since the function e + j π ω 4 ω 0 would only be evaluated at ±ω o because of the delta function) ω + j π ( ¿¿ 0 ) 4 ω 0 δ ( ω ω 0 ) e + j π ω 0 4 ω 0 + δ ( ω + ω 0 ) e ¿ ¿ f ( t ) = π ¿ c) f ( t ) = e 4 t u ( t ) ↔F ( ω ) = 1 4 + (found on Table) d) f ( t ) = e + 4 t u (− t ) Root: g ( t ) = e 4 t u ( t ) ↔G ( ω ) = 1 4 + (found on Table)
f ( t ) = g ( t ) ↔ F ( ω ) = G ( ω ) = 1 4 + e) f ( t ) = cos ( 40 t ) u ( t + 2 ) = f ( t ) = 1 2 [ e j 40 t + e j 40 t ] u ( t + 2 ) Note the order of operations matters Method 1 Root: g ( t ) = u ( t ) ↔G ( ω ) = πδ ( ω ) + 1 j ω f 1 ( t ) = g ( t + 2 ) = u ( t + 2 ) ↔F 1 ( ω ) = G ( ω ) e + j 2 ω = e + j 2 ω [ πδ ( ω ) + 1 j ω ] f 2 ( t ) = f 1 ( t ) e j 40 t ↔F 2 ( ω ) = F 1 ( ω 40 ) = e + j 2 ( ω 40 ) [ πδ ( ω 40 ) + 1 j ( ω 40 ) ] f 3 ( t ) = f 1 ( t ) e j 40 t ↔ F 3 ( ω ) = F 1 ( ω + 40 ) = e + j 2 ( ω + 40 ) [ πδ ( ω + 40 ) + 1 j ( ω + 40 ) ] f ( t ) = 1 2 [ f 2 ( t ) + f 3 ( t ) ] ↔F ( ω ) = 1 2 [ F 2 ( ω ) + F 3 ( ω ) ] F ( ω ) = 1 2 { e + j 2 ( ω 40 ) [ πδ ( ω 40 ) + 1 j ( ω + 40 ) ] + e + j 2 ( ω + 40 ) [ πδ ( ω + 40 ) + 1 j ( ω + 40 ) ] } F ( ω ) = 1 2 { [ πδ ( ω 40 ) e + j 2 ( ω 40 ) + e + j 2 ( ω 40 ) j ( ω + 40 ) ] + [ πδ ( ω + 40 ) e + j 2 ( ω + 40 ) + e + j 2 ( ω + 40 ) j ( ω + 40 ) ] } F ( ω ) = 1 2 { [ πδ ( ω 40 ) e + j 2 ( 40 40 ) + e + j 2 ( ω 40 ) j ( ω + 40 ) ] + [ πδ ( ω + 40 ) e + j 2 ( 40 + 40 ) + e + j 2 ( ω + 40 ) j ( ω + 40 ) ] } F ( ω

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