Embed Size (px)
Citation preview
Modelação Numérica 2017
Aula 4, 22/Fev
• Álgebracomplexa
• CircuitoRLC
• Filtros
h7p://modnum.ucs.ciencias.ulisboa.pt
Transformada de Fourier
h7ps://en.wikipedia.org/wiki/File:Fourier_transform_Ame_and_frequency_domains_(small).gif
Generalização a qualquer função (não senos)
• QualquerfunçãoperiódicapodeserrepresentadacomoumasériedeFourier,i.e.:
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆
1
Aulapassada
Série de Fourier
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(1)
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(2)
1
Primeiraharmónica:
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(1)
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(2)
a1 cos2⇡t
T+ b1 sin
2⇡t
T= A cos
✓2⇡t
T+ �
◆
1
• Φ:Faseinicial
• A:Amplitude
• T:Período;Frequênciaf=1/T;Frequênciaangular:ω=2πf=2π/T
Aulapassada
Série de Fourier
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(1)
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(2)
1
Primeiraharmónica:
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(1)
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(2)
a1 cos2⇡t
T+ b1 sin
2⇡t
T= A cos
✓2⇡t
T+ �
◆
1
Aulapassada
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(1)
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(2)
a1 cos2⇡t
T+ b1 sin
2⇡t
T= A cos
✓2⇡t
T+ �
◆
ei✓ = cos ✓ + i sin ✓
cos
2⇡kt
T=
ei2⇡kt/T + e�i2⇡kt/T
2
sin
2⇡kt
T=
ei2⇡kt/T + e�i2⇡kt/T
2i
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
1X
k=1
ak � ibk2
ei2⇡kt/T +
1X
k=1
ak + ibk2
ei2⇡kt/T =
=
1X
k=�1cke
i2⇡kt/T
(3)
1
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(1)
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(2)
a1 cos2⇡t
T+ b1 sin
2⇡t
T= A cos
✓2⇡t
T+ �
◆
ei✓ = cos ✓ + i sin ✓
cos
2⇡kt
T=
ei2⇡kt/T + e�i2⇡kt/T
2
sin
2⇡kt
T=
ei2⇡kt/T + e�i2⇡kt/T
2i
f(t) =a02
+
1X
k=1
ak � ibk2
ei2⇡kt/T +
1X
k=1
ak + ibk2
ei2⇡kt/T =
=
1X
k=�1cke
i2⇡kt/T
(3)
1
• Directa:
• Inversa:
Transformada Discreta de Fourier
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
2
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
2
F=F(f)f=F-1(F)=F-1(F(f))
Aulapassada
Aulapassada
Interpretação do espectro: Frequência
Coseno,10ciclosexactos,T=10s,f=0.1Hz
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
2
F (f) =
Z 1
�1f(t)e�i2⇡ftdt
f(t) =
Z 1
�1F (f)ei2⇡ftdf
3
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(1)
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
✓a1 cos
2⇡t
T+ b1 sin
2⇡t
T
◆+
✓a2 cos
2⇡t
T/2+ b2 sin
2⇡t
T/2
◆+ ...
(2)
a1 cos2⇡t
T+ b1 sin
2⇡t
T= A cos
✓2⇡t
T+ �
◆
ei✓ = cos ✓ + i sin ✓
cos ✓ =
ei✓ + e�i✓
2
sin ✓ =
ei✓ � e�i✓
2i
cos
2⇡kt
T=
ei2⇡kt/T + e�i2⇡kt/T
2
sin
2⇡kt
T=
ei2⇡kt/T � e�i2⇡kt/T
2i(
cos
2⇡ktT =
ei2⇡kt/T+e�i2⇡kt/T
2
sin
2⇡ktT =
ei2⇡kt/T�e�i2⇡kt/T
2i
(3)
f(t) =a02
+
1X
k=1
✓ak cos
2⇡kt
T+ bk sin
2⇡kt
T
◆=
=
a02
+
1X
k=1
ak � ibk2
ei2⇡kt/T +
1X
k=1
ak + ibk2
e�i2⇡kt/T=
=
1X
k=�1cke
i2⇡kt/T
(4)
c0 =a02
, cn =
an � ibn2
, c�n =
an + ibn2
ck =
1
2⇡
Z T/2
�T/2f(t)e�i2⇡kt/T , k = 0,±1,±2
1
Interpretação do espectro: Frequência
Coseno,10ciclosexactos,T=10s,f=0.1Hz# -*- coding: utf-8 -*-"""Created on Sat Feb 18 02:10:48 2017
@author: susana"""
N=100.; T=10.; dt=1.t=np.arange(0.,N,dt)f=np.cos(2*pi * t/T)
tf=TFD(f)
plt.close(); plt.subplot(3,1,1); plt.plot(t, f)plt.title('Function f'); plt.grid();
plt.subplot(3,1,2); plt.plot(np.arange(0,N), np.real(tf))plt.title('Real(TFD)'); plt.grid();
plt.subplot(3,1,3); plt.plot(np.arange(0,N), np.imag(tf))plt.title('Imag(TFD)'); plt.grid();
1
fout=np.concatenate([tf[N/2:], tf[:N/2]])fNyq=1/(2*dt); df=1/(N*dt); freq=np.arange(-fNyq, fNyq,df)
plt.close(); plt.subplot(3,1,1); plt.plot(t, f)plt.title('Function f'); plt.grid();
plt.subplot(3,1,2); plt.plot(freq,np.real(fout))plt.title('Real(TFD)'); plt.grid();
plt.subplot(3,1,3); plt.plot(freq,np.imag(fout))plt.title('Imag(TFD)'); plt.grid();
1
Re-ordenando,obtém-seoespectrousual
Aulapassada
Filtro Digital
Input(sinaloriginal)
Output(sinalprocessado)Filtro
• Filtrosanalógicos:ex:circuitoRLC.• Filtroslinearesdigitais:• Não recursivos: o output em cada instante é uma
combinaçãolineardoinputeminstantesanteriores(fácildeanalisarcomanálisedeFourier).
• Recursivos:ooutputéumacombinaçãolineardoinputedooutputeminstantesanteriores.
Circuito RLC (Projecto 0)
Fontedetensão
Resistência
Bobina
Condensador
Circuito RLC
Resistência:
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
VR = RI
= RI0 sin(!t)(6)
2
VeIemfase,respostadependedeR.
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
• Componentes:
• Correntealterna:
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
2
Circuito RLC
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
• Componentes:
• Correntealterna:
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
2
Condensador:
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
VR = RI
= RI0 sin(!t)(6)
VC =
1
C
ZIdt
=
1
C
ZI0 sin(!t)dt
=
1
C
I0!(� cos(!t))
(7)
2
VeIdesfasados,VatrasadoemrelaçãoaI,respostadependedafrequência(1/ω,passa-alto).
Circuito RLC
Bobina:
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
VR = RI
= RI0 sin(!t)(6)
VC =
1
C
ZIdt
=
1
C
ZI0 sin(!t)dt
=
1
C
I0!(� cos(!t))
(7)
VL = LdI
dt
= Ld
dtI0 sin(!t)
= LI0! cos(!t)
(8)
2
VeIdesfasados,VadiantadoemrelaçãoaI,respostadependedafrequência(ω,passa-baixo).
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
• Componentes:
• Correntealterna:
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
2
LeideKirchoffdasmalhas+tensãosinusoidal:Comotodososcomponentestêmumcomportamentolinear:
Circuito RLC — Em conjunto...
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
• Componenteslineares:
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
VR = RI
= RI0 sin(!t)(6)
VC =
1
C
ZIdt
=
1
C
ZI0 sin(!t)dt
=
1
C
I0!(� cos(!t))
(7)
VL = LdI
dt
= Ld
dtI0 sin(!t)
= LI0! cos(!t)
(8)
V = VL + VC + VR
V = VL + VC + VR = V0 cos(!t)
IL = IC = IR = I
2
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
VR = RI
= RI0 sin(!t)(6)
VC =
1
C
ZIdt
=
1
C
ZI0 sin(!t)dt
=
1
C
I0!(� cos(!t))
(7)
VL = LdI
dt
= Ld
dtI0 sin(!t)
= LI0! cos(!t)
(8)
V = VL + VC + VR
V = VL + VC + VR = V0 cos(!t)
IL = IC = IR = I
I = I0 cos(!t+ �)
2
Circuito RLC — UVlizando a álgebra complexa
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
• Componenteslineares:
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
VR = RI
= RI0 sin(!t)(6)
VC =
1
C
ZIdt
=
1
C
ZI0 sin(!t)dt
=
1
C
I0!(� cos(!t))
(7)
VL = LdI
dt
= Ld
dtI0 sin(!t)
= LI0! cos(!t)
(8)
V = VL + VC + VR
V = VL + VC + VR = V0 cos(!t)
IL = IC = IR = I
I = I0 cos(!t+ �)
⇢V = V0 cos(!t) = Re(V0ei!t)I = I0 cos(!t+ �) = Re(I0ei(!t+�)
)
(9)
2
F (f) =
Z 1
�1f(t)e�i2⇡ftdt
f(t) =
Z 1
�1F (f)ei2⇡ftdf
VR = RI = RI0ei(!t+�)
VC =
1
C
ZIdt =
1
C
ZI0e
i(!t+�)dt =I0C
1
i!ei(!t+�)
= � i
!CI0e
i(!t+�)
VL = LdI
dt= L
d
dt(I0e
i(!t+�)) = Li!I0e
i(!t+�))
8<
:
VR = RI = RI0ei(!t+�)
VC =
1C
RIdt = 1
C
RI0ei(!t+�)dt = I0
C1i!e
i(!t+�)= � i
!C I0ei(!t+�)
VL = LdIdt = L d
dt(I0ei(!t+�)
) = Li!I0ei(!t+�))
(10)
3
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
DiferenciaçãonodomíniodotempocorrespondeamulAplicar
porωnodomínioespectral
Circuito RLC — UVlizando a álgebra complexa
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
• Componenteslineares:
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
VR = RI
= RI0 sin(!t)(6)
VC =
1
C
ZIdt
=
1
C
ZI0 sin(!t)dt
=
1
C
I0!(� cos(!t))
(7)
VL = LdI
dt
= Ld
dtI0 sin(!t)
= LI0! cos(!t)
(8)
V = VL + VC + VR
V = VL + VC + VR = V0 cos(!t)
IL = IC = IR = I
I = I0 cos(!t+ �)
⇢V = V0 cos(!t) = Re(V0ei!t)I = I0 cos(!t+ �) = Re(I0ei(!t+�)
)
(9)
2
F (f) =
Z 1
�1f(t)e�i2⇡ftdt
f(t) =
Z 1
�1F (f)ei2⇡ftdf
VR = RI = RI0ei(!t+�)
VC =
1
C
ZIdt =
1
C
ZI0e
i(!t+�)dt =I0C
1
i!ei(!t+�)
= � i
!CI0e
i(!t+�)
VL = LdI
dt= L
d
dt(I0e
i(!t+�)) = Li!I0e
i(!t+�))
8<
:
VR = RI = RI0ei(!t+�)
VC =
1C
RIdt = 1
C
RI0ei(!t+�)dt = I0
C1i!e
i(!t+�)= � i
!C I0ei(!t+�)
VL = LdIdt = L d
dt(I0ei(!t+�)
) = Li!I0ei(!t+�))
(10)
3
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
Integraçãonodomíniodotempocorrespondeadividirporωno
domínioespectral
Circuito RLC — UVlizando a álgebra complexa
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
• Componenteslineares:
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
VR = RI
= RI0 sin(!t)(6)
VC =
1
C
ZIdt
=
1
C
ZI0 sin(!t)dt
=
1
C
I0!(� cos(!t))
(7)
VL = LdI
dt
= Ld
dtI0 sin(!t)
= LI0! cos(!t)
(8)
V = VL + VC + VR
V = VL + VC + VR = V0 cos(!t)
IL = IC = IR = I
I = I0 cos(!t+ �)
⇢V = V0 cos(!t) = Re(V0ei!t)I = I0 cos(!t+ �) = Re(I0ei(!t+�)
)
(9)
2
F (f) =
Z 1
�1f(t)e�i2⇡ftdt
f(t) =
Z 1
�1F (f)ei2⇡ftdf
VR = RI = RI0ei(!t+�)
VC =
1
C
ZIdt =
1
C
ZI0e
i(!t+�)dt =I0C
1
i!ei(!t+�)
= � i
!CI0e
i(!t+�)
VL = LdI
dt= L
d
dt(I0e
i(!t+�)) = Li!I0e
i(!t+�))
8<
:
VR = RI = RI0ei(!t+�)
VC =
1C
RIdt = 1
C
RI0ei(!t+�)dt = I0
C1i!e
i(!t+�)= � i
!C I0ei(!t+�)
VL = LdIdt = L d
dt(I0ei(!t+�)
) = Li!I0ei(!t+�))
(10)
3
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
• Linearidadeinput-output• Desfasamento• Respostadependedafrequência
Circuito RLC — Em conjunto...
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
VR = RI
= RI0 sin(!t)(6)
VC =
1
C
ZIdt
=
1
C
ZI0 sin(!t)dt
=
1
C
I0!(� cos(!t))
(7)
VL = LdI
dt
= Ld
dtI0 sin(!t)
= LI0! cos(!t)
(8)
V = VL + VC + VR
V = VL + VC + VR = V0 cos(!t)
IL = IC = IR = I
2
F (f) =
Z 1
�1f(t)e�i2⇡ftdt
f(t) =
Z 1
�1F (f)ei2⇡ftdf
VR = RI = RI0ei(!t+�)
VC =
1
C
ZIdt =
1
C
ZI0e
i(!t+�)dt =I0C
1
i!ei(!t+�)
= � i
!CI0e
i(!t+�)
VL = LdI
dt= L
d
dt(I0e
i(!t+�)) = Li!I0e
i(!t+�))
8<
:
VR = RI = RI0ei(!t+�)
VC =
1C
RIdt = 1
C
RI0ei(!t+�)dt = I0
C1i!e
i(!t+�)= � i
!C I0ei(!t+�)
VL = LdIdt = L d
dt(I0ei(!t+�)
) = Li!I0ei(!t+�))
(10)
V = V0ei!t
=
R� i
!C+ i!L
�I0e
i(!t+�)
3
ImpedânciacomplexaZ(respostaemamplitudeefase)
Circuito RLC — Em conjunto...
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
V = V0 cos(!t)
I = I0 sin(!t)
VR = RI
= RI0 sin(!t)(6)
VC =
1
C
ZIdt
=
1
C
ZI0 sin(!t)dt
=
1
C
I0!(� cos(!t))
(7)
VL = LdI
dt
= Ld
dtI0 sin(!t)
= LI0! cos(!t)
(8)
V = VL + VC + VR
V = VL + VC + VR = V0 cos(!t)
IL = IC = IR = I
2
F (f) =
Z 1
�1f(t)e�i2⇡ftdt
f(t) =
Z 1
�1F (f)ei2⇡ftdf
VR = RI = RI0ei(!t+�)
VC =
1
C
ZIdt =
1
C
ZI0e
i(!t+�)dt =I0C
1
i!ei(!t+�)
= � i
!CI0e
i(!t+�)
VL = LdI
dt= L
d
dt(I0e
i(!t+�)) = Li!I0e
i(!t+�))
8<
:
VR = RI = RI0ei(!t+�)
VC =
1C
RIdt = 1
C
RI0ei(!t+�)dt = I0
C1i!e
i(!t+�)= � i
!C I0ei(!t+�)
VL = LdIdt = L d
dt(I0ei(!t+�)
) = Li!I0ei(!t+�))
(10)
V = V0ei!t
=
R� i
!C+ i!L
�I0e
i(!t+�)
3
F (f) =
Z 1
�1f(t)e�i2⇡ftdt
f(t) =
Z 1
�1F (f)ei2⇡ftdf
VR = RI = RI0ei(!t+�)
VC =
1
C
ZIdt =
1
C
ZI0e
i(!t+�)dt =I0C
1
i!ei(!t+�)
= � i
!CI0e
i(!t+�)
VL = LdI
dt= L
d
dt(I0e
i(!t+�)) = Li!I0e
i(!t+�))
8<
:
VR = RI = RI0ei(!t+�)
VC =
1C
RIdt = 1
C
RI0ei(!t+�)dt = I0
C1i!e
i(!t+�)= � i
!C I0ei(!t+�)
VL = LdIdt = L d
dt(I0ei(!t+�)
) = Li!I0ei(!t+�))
(10)
V = V0ei!t
=
R� i
!C+ i!L
�I0e
i(!t+�)
V0 =
R� i
!C+ i!L
�I0e
i� , V0
R� i!C + i!L
= I0ei�
3
f(t) =1X
k=�1cke
i2⇡kt/T
fNyquist =1
2�t
ck =
N�1X
t=0
f(tn)e�i2⇡tnk/N
f(tn) =1
N
N�1X
k=0
ckei2⇡tnk/N
ck = c⇤�k
8<
:
VR = RIVC =
1C
RIdt
VL = LdIdt
(5)
2
Aequaçãodiferencialpoderesolver-secomo(maissimples)umaequaçãoaritméAca
Circuito RLC — Exemplo
F (f) =
Z 1
�1f(t)e�i2⇡ftdt
f(t) =
Z 1
�1F (f)ei2⇡ftdf
VR = RI = RI0ei(!t+�)
VC =
1
C
ZIdt =
1
C
ZI0e
i(!t+�)dt =I0C
1
i!ei(!t+�)
= � i
!CI0e
i(!t+�)
VL = LdI
dt= L
d
dt(I0e
i(!t+�)) = Li!I0e
i(!t+�))
8<
:
VR = RI = RI0ei(!t+�)
VC =
1C
RIdt = 1
C
RI0ei(!t+�)dt = I0
C1i!e
i(!t+�)= � i
!C I0ei(!t+�)
VL = LdIdt = L d
dt(I0ei(!t+�)
) = Li!I0ei(!t+�))
(10)
V = V0ei!t
=
R� i
!C+ i!L
�I0e
i(!t+�)
V0 =
R� i
!C+ i!L
�I0e
i� , V0
R� i!C + i!L
= I0ei�
3
# -*- coding: utf-8 -*-"""Created on Thu Feb 3 15:58:50 2017
@author: susana"""
############ Módulos e definições
from __future__ import print_function # Facilita a função print: print("Hello")import numpy as np # Importar NumPy - funções numéricasimport matplotlib.pyplot as plt # Importar MatplotLib - gráficosfrom numpy import pi as pi
plt.rcParams['figure.figsize'] = 6, 8
#%% Circuito RLC# Cálculo da relação Tensão-Corrente (amplitude e fase)
# Parâmetros:R=1000. # Resistência, ohmL=1.0e-2 # Impedância da Bobine, HC=1.0e-6 # Capacidade do condensador, FV0=1.0 # Amplitude do potencial aplicado, V
#%% Cálculo:
omega = 2*pi* np.arange(10., 1.e6, 10.) # Vector de frequências angulares
Z = R - 1j/(omega*C) + 1j*omega*L ; # Impedância do circuito
I=V0/Z # Corrente
phi=np.angle(I) # Diferença de faseI0=np.abs(I)
# Gráficos:plt.subplot(2,1,1)plt.semilogx(omega,I0)plt.title('Circuito RLC')plt.ylabel('Amplitude (I0)')plt.autoscale(enable=True, axis='x', tight=True)plt.grid()
plt.subplot(2,1,2)plt.semilogx(omega,phi/pi)plt.xlabel('Frequencia Angular (omega)')plt.ylabel('Fase (phi/pi)')
1
Atenuação
Desfasamento
Circuito RLC — Exemplo
Input
Output
• Linearidadeinput-output:Inputeoutputsãosenosdeiguaisfrequências.• Desfasamento:Ooutputestáligeiramenteatrasadoemrelaçãoaoinput.• Atenuação:Ooutputtemmenoramplitudequeoinput.
Circuito RLC — Variando L e C
Filtropassa-bandaFiltropassa-baixo Filtropassa-alto
