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Szereg Fouriera sygnału trapezowego


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#1 piotrw86

piotrw86

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Napisano 24.02.2009 - 22:43

Witam!

Mam zadanie do wykonania, z którym nie jestem w stanie sobie poradzić :rolleyes:, mianowicie:

Sygnał_trapezowy800x600.JPG

Dla danego sygnału trapezowego (patrz rysunek/plik) muszę określić wzór, który ten sygnał opisuje, a następnie sprowadzić go do postaci trygonometrycznego szeregu fouriera...

Wg mnie funkcja jest opisana takim wzorem: (sorry za zapis, ale nie umialem w Tex'ie zrobić klamry do tego wzoru :/)

f(t) = (2A/(T-T1))*t dla <kT, (T-T1/2)+kT)
A dla <(T-T1/2)+kT, (T+T1)/2+kT) k nalezy do calkowitych
((-2A)/(T-T1))*t + (2AT)/(T-T1) dla <(T+T1)/2 + kT, T+kT)

Wspolczynnik a_0 do szeregu Fouriera wychodzi mi:

a_0={A}\over{2}+{AT_1}\over{2T}

Pozostają do obliczenia 2 współczynniki: a_k i b_k.
Będę wdzięczny jeżeli ktoś mi pomoże.
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Afroman

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Napisano 25.09.2011 - 17:55

#2 Kinia7

Kinia7

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Napisano 17.01.2016 - 15:47

f(t)=\left{\ \begin{array}{lcrcccl} \fr{2A}{T-T_1}\cd t -\fr{2kAT}{T-T_1}& \ dla\ & kT & \leq& t & < & \fr12(T-T_1)+kT\\ A & \ dla\ & \fr12(T-T_1)+kT & \leq & t & < & \fr12(T+T_1)+kT\\ \fr{-2A}{T-T_1}\cd t+\fr{2(k+1)AT}{T-T_1} & \ dla\ & \fr12(T+T_1)+kT & \leq & t & < & T+kT\end{array}
a_n=\fr2T\int_{-\fr T2}^{\fr T2}f(t)\cos\fr{2n\p t}{T}dt
a_o=\fr2T\int_{-\fr T2}^{\fr T2}f(t)dt=\fr4T\int_{0}^{\fr T2}f(t)dt=\fr4T\(\int_{0}^{\fr{T-T_1}{2}}\fr{2A}{T-T_1}\cd tdt+\int_{\fr{T-T_1}{2}}^{\fr T2}Adt\)=\fr4T\(\fr{A(T-T_1)}{4}+\fr{AT_1}{2}\)=\fr{A(T+T_1)}{T}
a_n=\fr2T\int_{-\fr T2}^{\fr T2}f(t)\cos\fr{2n\p t}{T}dt=        dla   n>0
=\fr2T\(A\int_{-\fr T2}^{\fr{T_1-T}{2}}\cos\fr{2n\p t}{T}dt+\fr{2A}{T_1-T}\int_{\fr{T_1-T}{2}}^0t\cos\fr{2n\p t}{T}dt+\fr{2A}{T-T_1}\int_0^{\fr{T-T_1}{2}}t\cos\fr{2n\p t}{T}dt+A\int_{\fr{T-T_1}{2}}^{\fr T2}\cos\fr{2n\p t}{T}dt\)=
=\fr{2A}{T}\(\int_{-\fr T2}^{\fr{T_1-T}{2}}\cos\fr{2n\p t}{T}dt+\fr{2}{T_1-T}\int_{\fr{T_1-T}{2}}^0t\cos\fr{2n\p t}{T}dt+\fr{2}{T-T_1}\int_0^{\fr{T-T_1}{2}}t\cos\fr{2n\p t}{T}dt+\int_{\fr{T-T_1}{2}}^{\fr T2}\cos\fr{2n\p t}{T}dt\)=
=\fr{2A}{T}\(I_1+I_2+I_3+I_4\)
I_1=\int_{-\fr T2}^{\fr{T_1-T}{2}}\cos\fr{2n\p t}{T}dt=\fr{T}{2n\p}\|\ \\\sin\fr{2n\p t}{T}\\\ \|_{-\fr T2}^{\fr{T_1-T}{2}}=\fr{T}{2n\p}\(\sin\fr{n(T_1-T)\p}{T}+\sin n\p \)=\fr{T}{2n\p}\sin\fr{n(T_1-T)\p}{T}
I_2=\fr{2}{T_1-T}\int_{\fr{T_1-T}{2}}^0t\cos\fr{2n\p t}{T}dt=\fr{T}{n(T_1-T)\p}\|\ \\\fr{\cos\fr{2n\p t}{T}}{\fr{2n\p}{T}}+t\sin\fr{2n\p t}{T}\\\ \|_{\fr{T_1-T}{2}}^0=
\ \ \ \ \ =\fr{T}{n(T_1-T)\p}\(\fr{T}{2n\p}+0-\fr{T\cos\fr{n(T_1-T)\p}{T}}{2n\p}-\fr{T_1-T}{2}\sin\fr{n(T_1-T)\p}{T}\)=
\ \ \ \ \ =\fr{T}{n(T-T_1)\p}\(-\fr{T}{2n\p}+\fr{T\cos\fr{n(T-T_1)\p}{T}}{2n\p}+\fr{T-T_1}{2}\sin\fr{n(T-T_1)\p}{T}\)
I_3=\fr{2}{T-T_1}\int_0^{\fr{T-T_1}{2}}t\cos\fr{2n\p t}{T}dt=\fr{T}{n(T-T_1)\p}\|\ \\\fr{\cos\fr{2n\p t}{T}}{\fr{2n\p}{T}}+t\sin\fr{2n\p t}{T}\\\ \|_0^{\fr{T-T_1}{2}}=
\ \ \ \ \ =\fr{T}{n(T-T_1)\p}\(\fr{T\cos\fr{n(T-T_1)\p}{T}}{2n\p}+\fr{T-T_1}{2}\sin\fr{n(T-T_1)\p}{T}-\fr{T}{2n\p}-0\)
I_4=\int_{\fr{T-T_1}{2}}^{\fr T2}\cos\fr{2n\p t}{T}dt=\fr{T}{2n\p}\|\ \\\sin\fr{2n\p t}{T}\\\ \|_{\fr{T-T_1}{2}}^{\fr T2}=\fr{T}{2n\p}\(\sin n\p-\sin\fr{n(T-T_1)\p}{T} \)=-\fr{T}{2n\p}\sin\fr{n(T-T_1)\p}{T}
a_n=\fr{2A}{T}\(-\fr{T}{n\p}\sin\fr{n(T-T_1)\p}{T}+\fr{T}{n(T-T_1)\p}\(\fr{T\cos\fr{n(T-T_1)\p}{T}}{n\p}+(T-T_1)\sin\fr{n(T-T_1)\p}{T}-\fr{T}{n\p}\)\)=
\ \ \ \ \ =\fr{2AT}{n^2(T-T_1)\p^2}\(\cos\fr{n(T-T_1)\p}{T}-1\)
b_n=\fr2T\int_{-\fr T2}^{\fr T2}f(t)\sin\fr{2n\p t}{T}dt=        dla   n>0
=\fr2T\(A\int_{-\fr T2}^{\fr{T_1-T}{2}}\sin\fr{2n\p t}{T}dt+\fr{2A}{T_1-T}\int_{\fr{T_1-T}{2}}^0t\sin\fr{2n\p t}{T}dt+\fr{2A}{T-T_1}\int_0^{\fr{T-T_1}{2}}t\sin\fr{2n\p t}{T}dt+A\int_{\fr{T-T_1}{2}}^{\fr T2}\sin\fr{2n\p t}{T}dt\)=
=\fr{2A}{T}\(\int_{-\fr T2}^{\fr{T_1-T}{2}}\sin\fr{2n\p t}{T}dt+\fr{2}{T_1-T}\int_{\fr{T_1-T}{2}}^0t\sin\fr{2n\p t}{T}dt+\fr{2}{T-T_1}\int_0^{\fr{T-T_1}{2}}t\sin\fr{2n\p t}{T}dt+\int_{\fr{T-T_1}{2}}^{\fr T2}\sin\fr{2n\p t}{T}dt\)=
=\fr{2A}{T}\(I_5+I_6+I_7+I_8\)
I_5=\int_{-\fr T2}^{\fr{T_1-T}{2}}\sin\fr{2n\p t}{T}dt=-\fr{T}{2n\p}\|\ \\\cos\fr{2n\p t}{T}\\\ \|_{-\fr T2}^{\fr{T_1-T}{2}}=-\fr{T}{2n\p}\(\cos\fr{n(T_1-T)\p}{T}-\cos n\p \)=
\ \ \ \ \ =\fr{T}{2n\p}\((-1)^{n}-\cos\fr{n(T-T_1)\p}{T}\)
I_6=\fr{2}{T_1-T}\int_{\fr{T_1-T}{2}}^0t\sin\fr{2n\p t}{T}dt=\fr{T}{n(T_1-T)\p}\|\ \\\fr{\sin\fr{2n\p t}{T}}{\fr{2n\p}{T}}-t\cos\fr{2n\p t}{T}\\\ \|_{\fr{T_1-T}{2}}^0=
\ \ \ \ \ =\fr{T}{n(T_1-T)\p}\(0-0-\fr{T\sin\fr{n(T_1-T)\p}{T}}{2n\p}+\fr{T_1-T}{2}\cos\fr{n(T_1-T)\p}{T}\)=
\ \ \ \ \ =\fr{T}{n(T-T_1)\p}\(\fr{T-T_1}{2}\cos\fr{n(T-T_1)\p}{T}-\fr{T\sin\fr{n(T-T_1)\p}{T}}{2n\p}\)
I_7=\fr{2}{T-T_1}\int_0^{\fr{T-T_1}{2}}t\sin\fr{2n\p t}{T}dt=\fr{T}{n(T-T_1)\p}\|\ \\\fr{\sin\fr{2n\p t}{T}}{\fr{2n\p}{T}}-t\cos\fr{2n\p t}{T}\\\ \|_0^{\fr{T-T_1}{2}}=
\ \ \ \ \ =\fr{T}{n(T-T_1)\p}\(\fr{T\sin\fr{n(T-T_1)\p}{T}}{2n\p}-\fr{T-T_1}{2}\cos\fr{n(T-T_1)\p}{T}-0+0\)
I_8=\int_{\fr{T-T_1}{2}}^{\fr T2}\sin\fr{2n\p t}{T}dt=-\fr{T}{2n\p}\|\ \\\cos\fr{2n\p t}{T}\\\ \|_{\fr{T-T_1}{2}}^{\fr T2}=-\fr{T}{2n\p}\(\cos n\p-\cos\fr{n(T-T_1)\p}{T}\)=
\ \ \ \ \ =\fr{T}{2n\p}\(\cos\fr{n(T-T_1)\p}{T}-(-1)^n\)
b_n=\fr{2A}{T}(I_5+I_6+I_7+I_8)=0
f(x)=\fr{a_o}{2}+\sum_{n=1}^\infty\(a_n\cos\fr{2n\p t}{T}+b_n\sin\fr{2n\p t}{T}\)=
\ \ \ \ \ \ \ \ =\fr{A(T+T_1)}{2T}+\fr{2AT}{(T-T_1)\p^2}\sum_{n=1}^\infty \fr{1}{n^2}\(\cos\fr{n(T-T_1)\p}{T}-1\)\cos\fr{2n\p t}{T}
pre_1453041841__trapez_szereg_fouriera.j
pre_1453041912__trapez_szereg_fouriera1.
pre_1453041964__trapez_szereg_fouriera2.
 

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