M
murat_mc
Guest
lahko kdo pomaga priti Fourierjevo transformacijo
sinc (sq) (1000 * pi * t)??(sinc kvadrat) v "w" domeno .....
sinc (sq) (1000 * pi * t)??(sinc kvadrat) v "w" domeno .....
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E
elmolla
Guest
hi Murat,
Lahko uporabite Fourierjeva transformacija parov zadobiti to, ali boste morali uporabiti izrek ostankov iz kompleksne analize za njegovo rešitev z uporabo ostankov Izrek
Lahko uporabite Fourierjeva transformacija parov zadobiti to, ali boste morali uporabiti izrek ostankov iz kompleksne analize za njegovo rešitev z uporabo ostankov Izrek
S
steve10
Guest
Imate ga.
N
neils_arm_strong
Guest
Fourierjeva transformacija v trikotne funkcijo sinc function.Then s simetrijo parov FT, FT o sinc kvadratnih bo tristransko funkcijo.
E
eecs4ever
Guest
Fourierjevo transformacijo sinc funkcija ni tristransko funkcijo.
Fourierjevo transformacijo sinc je škatla v obliki funkcije.Rectangular.
http://en.wikipedia.org/wiki/Sinc_function
Hope that helps.
Fourierjevo transformacijo sinc je škatla v obliki funkcije.Rectangular.
http://en.wikipedia.org/wiki/Sinc_function
Hope that helps.
P
purnapragna
Guest
hi vemo, da
F
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin(Wt)}{\pi t}' title="3 $ \ frac (\ sin (Wt)) (\ pi) t" alt='3$\frac{\sin(Wt)}{\pi t}' align=absmiddle>
je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$rect(\frac{\omega}{2W})' title="3 $ rect (\ frac (\ omega) (2W))" alt='3$rect(\frac{\omega}{2W})' align=absmiddle>in tudi vemo, da množenje v času domena je konvolucija pogostnosti domian.tj<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t)\times h(t) \longarrow \frac{1}{2 \pi}X(j\omega)*H(j\omega)' title="3 $ x (t) \ times h (t) \ longarrow \ frac (1) (2 \ pi) X (j \ omega) * H (j \ omega)" alt='3$x(t)\times h(t) \longarrow \frac{1}{2 \pi}X(j\omega)*H(j\omega)' align=absmiddle>uporabi
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$h(t) = x(t)' title="3 $ h (t) = x (t)" alt='3$h(t) = x(t)' align=absmiddle>
dobimo<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x^2(t) \longarrow \frac{1}{2\pi}X(j\omega)*X(j\omega)' title="3 $ x ^ 2 (t) \ longarrow \ frac (1) (2 \ pi) X (j \ omega) * x (j \ omega)" alt='3$x^2(t) \longarrow \frac{1}{2\pi}X(j\omega)*X(j\omega)' align=absmiddle>Za nas
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' title="3 $ x (t) = \ frac (\ sin (1000 \ pit)) (\ pi) t" alt='3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' align=absmiddle>Tako za FT
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin^{2}(Wt)}{(\pi t)^{2}}' title="3 $ \ frac (\ sin ^ (2) (Wt)) ((\ pi t) ^ (2))" alt='3$\frac{\sin^{2}(Wt)}{(\pi t)^{2}}' align=absmiddle>
izkaže, da je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$1000 tri(\frac{\omega}{2000 pi})' title="3 $ 1000 tri (\ frac (\ omega) (2000) pi)" alt='3$1000 tri(\frac{\omega}{2000 pi})' align=absmiddle>
.Upanje to vam pomaga
thnx
purna!Dodano po 2 minutah:hi vemo, da
FT of
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin(Wt)}{\pi t}' title="3 $ \ frac (\ sin (Wt)) (\ pi) t" alt='3$\frac{\sin(Wt)}{\pi t}' align=absmiddle>
je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$rect(\frac{\omega}{2W})' title="3 $ rect (\ frac (\ omega) (2W))" alt='3$rect(\frac{\omega}{2W})' align=absmiddle>in tudi vemo, da množenje v času domena je konvolucija pogostnosti domian.tj<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' title="3 $ x (t) \ times h (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * H (j \ omega)" alt='3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' align=absmiddle>uporabi
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$h(t) = x(t)' title="3 $ h (t) = x (t)" alt='3$h(t) = x(t)' align=absmiddle>
dobimo<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' title="3 $ x ^ 2 (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * x (j \ omega)" alt='3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' align=absmiddle>Za nas
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' title="3 $ x (t) = \ frac (\ sin (1000 \ pit)) (\ pi) t" alt='3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' align=absmiddle>Tako za FT
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin^{2}(Wt)}{(\pi t)^{2}}' title="3 $ \ frac (\ sin ^ (2) (Wt)) ((\ pi t) ^ (2))" alt='3$\frac{\sin^{2}(Wt)}{(\pi t)^{2}}' align=absmiddle>
izkaže, da je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$1000 tri(\frac{\omega}{2000 \pi})' title="3 $ 1000 tri (\ frac (\ omega) (2000 \ pi))" alt='3$1000 tri(\frac{\omega}{2000 \pi})' align=absmiddle>
.Upanje to vam pomaga
thnx
purna!Dodano po 1 minutah:hi vemo, da
FT of
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin(Wt)}{\pi t}' title="3 $ \ frac (\ sin (Wt)) (\ pi) t" alt='3$\frac{\sin(Wt)}{\pi t}' align=absmiddle>
je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$rect(\frac{\omega}{2W})' title="3 $ rect (\ frac (\ omega) (2W))" alt='3$rect(\frac{\omega}{2W})' align=absmiddle>in tudi vemo, da množenje v času domena je konvolucija pogostnosti domian.tj<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' title="3 $ x (t) \ times h (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * H (j \ omega)" alt='3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' align=absmiddle>uporabi
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$h(t) = x(t)' title="3 $ h (t) = x (t)" alt='3$h(t) = x(t)' align=absmiddle>
dobimo<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' title="3 $ x ^ 2 (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * x (j \ omega)" alt='3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' align=absmiddle>Za nas
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' title="3 $ x (t) = \ frac (\ sin (1000 \ pit)) (\ pi) t" alt='3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' align=absmiddle>Tako za FT
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin^{2}(1000 \pi t)}{(\pi t)^{2}}' title="3 $ \ frac (\ sin ^ (2) (1000 \ pi t)) ((\ pi t) ^ (2))" alt='3$\frac{\sin^{2}(1000 \pi t)}{(\pi t)^{2}}' align=absmiddle>
izkaže, da je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$1000 tri(\frac{\omega}{2000 \pi})' title="3 $ 1000 tri (\ frac (\ omega) (2000 \ pi))" alt='3$1000 tri(\frac{\omega}{2000 \pi})' align=absmiddle>
.Upanje to vam pomaga
thnx
purna!Dodano po 30 sekundah:hi vemo, da
FT of
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin(Wt)}{\pi t}' title="3 $ \ frac (\ sin (Wt)) (\ pi) t" alt='3$\frac{\sin(Wt)}{\pi t}' align=absmiddle>
je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$rect(\frac{\omega}{2W})' title="3 $ rect (\ frac (\ omega) (2W))" alt='3$rect(\frac{\omega}{2W})' align=absmiddle>in tudi vemo, da množenje v času domena je konvolucija pogostnosti domian.tj<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' title="3 $ x (t) \ times h (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * H (j \ omega)" alt='3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' align=absmiddle>uporabi
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$h(t) = x(t)' title="3 $ h (t) = x (t)" alt='3$h(t) = x(t)' align=absmiddle>
dobimo<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' title="3 $ x ^ 2 (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * x (j \ omega)" alt='3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' align=absmiddle>Za nas
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t) = \frac{\sin(1000 \pi t)}{\pi t}' title="3 $ x (t) = \ frac (\ sin (1000 \ pi t)) (\ pi) t" alt='3$x(t) = \frac{\sin(1000 \pi t)}{\pi t}' align=absmiddle>Tako za FT
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin^{2}(1000 \pi t)}{(\pi t)^{2}}' title="3 $ \ frac (\ sin ^ (2) (1000 \ pi t)) ((\ pi t) ^ (2))" alt='3$\frac{\sin^{2}(1000 \pi t)}{(\pi t)^{2}}' align=absmiddle>
izkaže, da je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$1000 tri(\frac{\omega}{2000 \pi})' title="3 $ 1000 tri (\ frac (\ omega) (2000 \ pi))" alt='3$1000 tri(\frac{\omega}{2000 \pi})' align=absmiddle>
.Upanje to vam pomaga
thnx
purna!
F
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin(Wt)}{\pi t}' title="3 $ \ frac (\ sin (Wt)) (\ pi) t" alt='3$\frac{\sin(Wt)}{\pi t}' align=absmiddle>
je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$rect(\frac{\omega}{2W})' title="3 $ rect (\ frac (\ omega) (2W))" alt='3$rect(\frac{\omega}{2W})' align=absmiddle>in tudi vemo, da množenje v času domena je konvolucija pogostnosti domian.tj<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t)\times h(t) \longarrow \frac{1}{2 \pi}X(j\omega)*H(j\omega)' title="3 $ x (t) \ times h (t) \ longarrow \ frac (1) (2 \ pi) X (j \ omega) * H (j \ omega)" alt='3$x(t)\times h(t) \longarrow \frac{1}{2 \pi}X(j\omega)*H(j\omega)' align=absmiddle>uporabi
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$h(t) = x(t)' title="3 $ h (t) = x (t)" alt='3$h(t) = x(t)' align=absmiddle>
dobimo<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x^2(t) \longarrow \frac{1}{2\pi}X(j\omega)*X(j\omega)' title="3 $ x ^ 2 (t) \ longarrow \ frac (1) (2 \ pi) X (j \ omega) * x (j \ omega)" alt='3$x^2(t) \longarrow \frac{1}{2\pi}X(j\omega)*X(j\omega)' align=absmiddle>Za nas
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' title="3 $ x (t) = \ frac (\ sin (1000 \ pit)) (\ pi) t" alt='3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' align=absmiddle>Tako za FT
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin^{2}(Wt)}{(\pi t)^{2}}' title="3 $ \ frac (\ sin ^ (2) (Wt)) ((\ pi t) ^ (2))" alt='3$\frac{\sin^{2}(Wt)}{(\pi t)^{2}}' align=absmiddle>
izkaže, da je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$1000 tri(\frac{\omega}{2000 pi})' title="3 $ 1000 tri (\ frac (\ omega) (2000) pi)" alt='3$1000 tri(\frac{\omega}{2000 pi})' align=absmiddle>
.Upanje to vam pomaga
thnx
purna!Dodano po 2 minutah:hi vemo, da
FT of
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin(Wt)}{\pi t}' title="3 $ \ frac (\ sin (Wt)) (\ pi) t" alt='3$\frac{\sin(Wt)}{\pi t}' align=absmiddle>
je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$rect(\frac{\omega}{2W})' title="3 $ rect (\ frac (\ omega) (2W))" alt='3$rect(\frac{\omega}{2W})' align=absmiddle>in tudi vemo, da množenje v času domena je konvolucija pogostnosti domian.tj<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' title="3 $ x (t) \ times h (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * H (j \ omega)" alt='3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' align=absmiddle>uporabi
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$h(t) = x(t)' title="3 $ h (t) = x (t)" alt='3$h(t) = x(t)' align=absmiddle>
dobimo<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' title="3 $ x ^ 2 (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * x (j \ omega)" alt='3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' align=absmiddle>Za nas
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' title="3 $ x (t) = \ frac (\ sin (1000 \ pit)) (\ pi) t" alt='3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' align=absmiddle>Tako za FT
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin^{2}(Wt)}{(\pi t)^{2}}' title="3 $ \ frac (\ sin ^ (2) (Wt)) ((\ pi t) ^ (2))" alt='3$\frac{\sin^{2}(Wt)}{(\pi t)^{2}}' align=absmiddle>
izkaže, da je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$1000 tri(\frac{\omega}{2000 \pi})' title="3 $ 1000 tri (\ frac (\ omega) (2000 \ pi))" alt='3$1000 tri(\frac{\omega}{2000 \pi})' align=absmiddle>
.Upanje to vam pomaga
thnx
purna!Dodano po 1 minutah:hi vemo, da
FT of
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin(Wt)}{\pi t}' title="3 $ \ frac (\ sin (Wt)) (\ pi) t" alt='3$\frac{\sin(Wt)}{\pi t}' align=absmiddle>
je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$rect(\frac{\omega}{2W})' title="3 $ rect (\ frac (\ omega) (2W))" alt='3$rect(\frac{\omega}{2W})' align=absmiddle>in tudi vemo, da množenje v času domena je konvolucija pogostnosti domian.tj<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' title="3 $ x (t) \ times h (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * H (j \ omega)" alt='3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' align=absmiddle>uporabi
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$h(t) = x(t)' title="3 $ h (t) = x (t)" alt='3$h(t) = x(t)' align=absmiddle>
dobimo<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' title="3 $ x ^ 2 (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * x (j \ omega)" alt='3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' align=absmiddle>Za nas
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' title="3 $ x (t) = \ frac (\ sin (1000 \ pit)) (\ pi) t" alt='3$x(t) = \frac{\sin(1000 \pit)}{\pi t}' align=absmiddle>Tako za FT
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin^{2}(1000 \pi t)}{(\pi t)^{2}}' title="3 $ \ frac (\ sin ^ (2) (1000 \ pi t)) ((\ pi t) ^ (2))" alt='3$\frac{\sin^{2}(1000 \pi t)}{(\pi t)^{2}}' align=absmiddle>
izkaže, da je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$1000 tri(\frac{\omega}{2000 \pi})' title="3 $ 1000 tri (\ frac (\ omega) (2000 \ pi))" alt='3$1000 tri(\frac{\omega}{2000 \pi})' align=absmiddle>
.Upanje to vam pomaga
thnx
purna!Dodano po 30 sekundah:hi vemo, da
FT of
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin(Wt)}{\pi t}' title="3 $ \ frac (\ sin (Wt)) (\ pi) t" alt='3$\frac{\sin(Wt)}{\pi t}' align=absmiddle>
je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$rect(\frac{\omega}{2W})' title="3 $ rect (\ frac (\ omega) (2W))" alt='3$rect(\frac{\omega}{2W})' align=absmiddle>in tudi vemo, da množenje v času domena je konvolucija pogostnosti domian.tj<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' title="3 $ x (t) \ times h (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * H (j \ omega)" alt='3$x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)' align=absmiddle>uporabi
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$h(t) = x(t)' title="3 $ h (t) = x (t)" alt='3$h(t) = x(t)' align=absmiddle>
dobimo<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' title="3 $ x ^ 2 (t) ---> \ frac (1) (2 \ pi) X (j \ omega) * x (j \ omega)" alt='3$x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)' align=absmiddle>Za nas
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$x(t) = \frac{\sin(1000 \pi t)}{\pi t}' title="3 $ x (t) = \ frac (\ sin (1000 \ pi t)) (\ pi) t" alt='3$x(t) = \frac{\sin(1000 \pi t)}{\pi t}' align=absmiddle>Tako za FT
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\frac{\sin^{2}(1000 \pi t)}{(\pi t)^{2}}' title="3 $ \ frac (\ sin ^ (2) (1000 \ pi t)) ((\ pi t) ^ (2))" alt='3$\frac{\sin^{2}(1000 \pi t)}{(\pi t)^{2}}' align=absmiddle>
izkaže, da je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$1000 tri(\frac{\omega}{2000 \pi})' title="3 $ 1000 tri (\ frac (\ omega) (2000 \ pi))" alt='3$1000 tri(\frac{\omega}{2000 \pi})' align=absmiddle>
.Upanje to vam pomaga
thnx
purna!
M
maxwell_28
Guest
basically, Fourierjeva transformacija tristranske funkcija sinc (sq), če tr razpon fn je do B (čas fn) in amplitudo je C potem Fourierjeve transformacije bo (BA) * c * sinc (sq) (( BA) f)