To get y[n] how do you IDTFT what rules did you use to get rid of the e^jnω
To get y[n] how do you IDTFT what rules did you use to get rid of the e^jnω
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.2P: Convert the following instantaneous currents to phasors, using cos(t) as the reference. Give your...
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To get y[n] how do you IDTFT what rules did you use to get rid of the e^jnω
![Now,
Y (ew) = B (w) + A (w) e³w,
⇒Y (ew) = {− sin (2w) — 4 sin (3w) + sin (4w) — 2 sin (5w)} + {6 cos w cos (2w) – 4 cos (5w) + 2 cos (3w) — 3 cos (4w)}e³w,
-
-
€13
-j3w
-j2w
-e
⇒Y (eviu) = {-(² 26 ¹ )
-
2j
⇒ Y (ejw) = {-
-j2w
In above equation (3),
3
y [3] = (- / + 3),
4
-
ej2w
+²2
2j
2j
⇒Y (ejw) = − 2ejów + ej5w
Taking IDTFT,
y[n] = { -2, (2-³), (¹+1), (−−), (3-2), 0, 3, (¹¹),
2j
2 j
2j
2€73w
j
-e
2j
-j4w
ej5w-e-j5w
pj2w te
-j2w
-
- (2₂) 2 (₂) } + {6 - ( + )
+
-
(²
2j
2j
2
ej4w
ej5w
-j5w
2e-j3w
-j4w
+
201² +22} + {3e +36-1-20_2e-jouw
2j
е
j
3
→y[3] = ( − − 2) = − + j(-2),
1/2
-
³/3
2
-
3
})
(1-1) +²² (1)
2
e
eju te jw
2
e
2j
j
ej²w
+ ³ (1 + 3) + √³ ( − 1 − ²) + ¹² ( ³3 - 4 )
1
ej³w
2
1
2j
2 j
+3
-
e jw
2
ejów
-j3w
4 (Now te stw ) + 2 (cft te shew) -
2
2
+e-j2w
(¹ + + 3) +
2j
(¹ + 1), (1 + ³), ( − ² + ¼ ) , } }
1
-2
2j
2j
+ e +e-j³w.
+e-j3w
3
2
( - 2 + 3) +
j
(3),
e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11149733-e604-46b1-89f7-e6cbe499df6d%2Fa05902a3-608a-464a-9ce6-cab87e01d2a7%2Fexifn9s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Now,
Y (ew) = B (w) + A (w) e³w,
⇒Y (ew) = {− sin (2w) — 4 sin (3w) + sin (4w) — 2 sin (5w)} + {6 cos w cos (2w) – 4 cos (5w) + 2 cos (3w) — 3 cos (4w)}e³w,
-
-
€13
-j3w
-j2w
-e
⇒Y (eviu) = {-(² 26 ¹ )
-
2j
⇒ Y (ejw) = {-
-j2w
In above equation (3),
3
y [3] = (- / + 3),
4
-
ej2w
+²2
2j
2j
⇒Y (ejw) = − 2ejów + ej5w
Taking IDTFT,
y[n] = { -2, (2-³), (¹+1), (−−), (3-2), 0, 3, (¹¹),
2j
2 j
2j
2€73w
j
-e
2j
-j4w
ej5w-e-j5w
pj2w te
-j2w
-
- (2₂) 2 (₂) } + {6 - ( + )
+
-
(²
2j
2j
2
ej4w
ej5w
-j5w
2e-j3w
-j4w
+
201² +22} + {3e +36-1-20_2e-jouw
2j
е
j
3
→y[3] = ( − − 2) = − + j(-2),
1/2
-
³/3
2
-
3
})
(1-1) +²² (1)
2
e
eju te jw
2
e
2j
j
ej²w
+ ³ (1 + 3) + √³ ( − 1 − ²) + ¹² ( ³3 - 4 )
1
ej³w
2
1
2j
2 j
+3
-
e jw
2
ejów
-j3w
4 (Now te stw ) + 2 (cft te shew) -
2
2
+e-j2w
(¹ + + 3) +
2j
(¹ + 1), (1 + ³), ( − ² + ¼ ) , } }
1
-2
2j
2j
+ e +e-j³w.
+e-j3w
3
2
( - 2 + 3) +
j
(3),
e
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