Problem 5.6 Use Parseval's identity to evaluate the following integrals: sinc(2t - 3)sinc(3t+1)dt (a) (b) o sinc¹(5t)dt.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Use Parseval’s identity to evaluate the following integrals:

(a) R ∞ −∞ sinc(2t + 1)sinc(3t − 1)dt

(b) R ∞ 0 sinc4 (3t)dt. please have a step by step solution and explain. I need to know how to approach the problem

Problem 5.6 Use Parseval's identity to evaluate the following integrals:
(a) sinc(2t - 3)sinc(3t+1)dt
(b) sinc¹(5t)dt.
Transcribed Image Text:Problem 5.6 Use Parseval's identity to evaluate the following integrals: (a) sinc(2t - 3)sinc(3t+1)dt (b) sinc¹(5t)dt.
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Follow-up Question

you did part a wrong. shouldnt there be 2 shifts? what happened to the shift from sin(3t+1)? why is it not included in the integral you gave? 

a)
sinc(2t-3) sinc(3t+1)dt
Let | =
The Parseval's identity:
=
sinc(2t-3)sinc(3t+1)dt
-∞0
-∞
Where G(s) = Fourier transform's
F(s)G(s)ds
f(t)g (t) = F(s)G(s)ds
=
S
|=
Xπ
50 [17 (2)] 0-10/31 x 1
e-ls/31,
³1×××
1
[²²/1 × ² / ×
-1
2
3
e ds
1 = 1/361/60
S
+1
-3e
-|s/31
X xe
1=[1-e-³]
53-|ds
Transcribed Image Text:a) sinc(2t-3) sinc(3t+1)dt Let | = The Parseval's identity: = sinc(2t-3)sinc(3t+1)dt -∞0 -∞ Where G(s) = Fourier transform's F(s)G(s)ds f(t)g (t) = F(s)G(s)ds = S |= Xπ 50 [17 (2)] 0-10/31 x 1 e-ls/31, ³1××× 1 [²²/1 × ² / × -1 2 3 e ds 1 = 1/361/60 S +1 -3e -|s/31 X xe 1=[1-e-³] 53-|ds
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Follow-up Question

NO i need it to be the Fourier transform as X(jw). Dont use this as s tranfer functions 

I dont understand the triangle. where did you get it from? i did not learn it that way. also you didnt explain the change in bounds in the integral. i dont understand how that triangle becomes (s+5). so can you please have a better explanation to this problem? why is f(t) sinc^2 (5t) ? i thought the given function  was sinc^4(5t).

Set f(t) = sinc2(5t) and f (s) =
= | * IF(S)|ds
1=
0
امام
1=
-ds
- ( *(s+52ds)
2
25
5
25
0
) = (3)
A
5
2
[ * (2 + 25 - 10s)ds
+ |
15
25 0
2 [ 3
1= +255-102²
25 3
2125
Transcribed Image Text:Set f(t) = sinc2(5t) and f (s) = = | * IF(S)|ds 1= 0 امام 1= -ds - ( *(s+52ds) 2 25 5 25 0 ) = (3) A 5 2 [ * (2 + 25 - 10s)ds + | 15 25 0 2 [ 3 1= +255-102² 25 3 2125
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