1) The signal x, it) is pestodfc ustth pertod T.=2. Thus at 2 -- dt + at -Tant 270 (- cos(an) %3D 2777 usten n=0 then xvo= actt %3D + 2 cosckn) cas(ant) Thus and x,(t)=1. It folboos then that L2,n=0, +nt0
1) The signal x, it) is pestodfc ustth pertod T.=2. Thus at 2 -- dt + at -Tant 270 (- cos(an) %3D 2777 usten n=0 then xvo= actt %3D + 2 cosckn) cas(ant) Thus and x,(t)=1. It folboos then that L2,n=0, +nt0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
In the image that I have attached, I have some questions about numbers 1 and 2. For number 1, how is the integral from -1 to 1 like in the formula, its -a/2 to a/2 so why did a =2 or is this not correct. Also for number 2, why does x_2(t) =1 and what does it mean, "It follows then that x_2,0 = 1 and x_2,n = 0, ∀n = 0.
Transcribed Image Text:) The signal x,lt) is pestodfe ustth pertad T.=2.
fant
dt
Thus
dt +
at
t
e
te
((- cos(an)
27377
ushen n=0
then xo= actdt
1-
+2E (- coscn) cas(ant)
Thus
,(E) =
x, (t)=1. It folbos then that
and
2,n=0, + n+o
3 The sigral is perfodie wofth pexfed To= 1.
To
ts'ete-faont ot = jef m
Thus
To
-ban + 1
4) The signal ct) is pexiodic with pertod T, = 27 ashereas cos (2:5)
is pestodic otth pesiod T, = 0.87. IH ollows that casct)+ coek.st5
is pesfodic oith pestod T=
The trigonometsic Fouriex sexies the even
cos (t)+ Ces (2-st)
=> T= 2T, =ST, = YT
%3D
Tz
087
sfgral co (t)+ Ces(2st) is
- d, cas (2) - ncos(3t)
DEI
Ey oqualiny the cofficlents a cas) bath sides we dseve
that an =0 fo8
all n
unleas
n= 2,5
o uohieh cose a=ag=|
Hence xy,2 = s =
fox n= 2,5
and yn= 0 fo8
all othes values of
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