MAIN : 2004/4/26(0:24)
|
|
|
- とき いいはた
- 5 years ago
- Views:
Transcription
1 Spice
2 Spice )
3 i v v = Ri, (R ) (1.1) R
4 2 1. i v R (1.1) v(t) = Ri(t) (1.2) (1.1) i = v R = Gv (1.3) G=1/R (conductance) (1.1) (1.1) 1. Georg Simon Ohm Georg Simon Ohm ( ) 2. Volta Alessandro Volta ( Alessandro Volta Die
5 galvanische Kette, mathematisch bearbeitet (1827) R = v/i 2 R a R b i a R a v b R b o 1.2 o, a, b v o, v a, v b v = v a v o,
6 4 1. v a = v a v b, v b = v b v o (1.4) v = v a v o = v a + ( v b + v b) v o = (v a v b) + (v b v o) = v a + v b (1.5) v = v a + v b (1.6) v a = R a i, v b = R b i (1.7) (1.6) v = (R a + R b )i (1.8) R = R a + R b (1.9) i = v R = v R a + R b (1.10) R a v a = R a i = v R a + R b v b = R b i = R b R a + R b v (1.11) v a v b v R a R b
7 i i a i b G a G b 1.3 i = i a + i b (1.12) G a = 1 R a, G b = 1 R b (1.13) i a = G a v, i b = G b v (1.14) v i i v R G G = G a + G b, G a i a = v, G a + G b G b i b = v (1.15) G a + G b
8 6 1. i a i b i G a G b A D V E N T U R E S in C Y B E R S O U N D Charles Wheatstone, Sir : Sir Charles Wheatstone, b. February 6, 1802, Barnwood, England, d. October 19, 1875, Paris, France 2. Wheatstone Bridges: Introduction 3. Wheatstone Bridge 4. Wheatstone Bridges v = E (1.16) 2 v 2 E R 1.5
9 i E v 1.4 v = R L R L + R E (1.17) R i R E v R L i = J (1.18) i R
10 8 1. i J v 1.6 i = R R + R L J (1.19) R i J R v R L R E = RJ J = E R (1.20)
11 Q V W W = QV (1.21) 2 V I 2 T Q W = QV (1.22) 2 P = W T = QV T = Q T V (1.23) T I = Q/T P = IV (1.24) 1. 5 (1) 1. Gustav Robert Kirchhoff (
12 Gustav Robert Kirchhoff, born March 12, 1824, Ko nigsberg, Prussia [now Kaliningrad, Russia], died Oct. 17, 1887, Berlin, Germany ( ) 3. Kirchhoff s Voltage Law (KVL) 4. Kirchhoff s Current Law (KCL) 1. 6 KVL KCL 2 )
13 LSI CG(Conjugate Gradient) 1.8 R a R c E a I a a R b I b b E b a b KVL E a = R a I a + R b (I a + I b ), E b = R c I b + R b (I a + I b ) (1.25) I a I b (R a + R b )I a + R b I b = E a R b I a + (R b + R c )I b = E b (1.26) A
14 12 1. A = R a + R b R b R b R b + R c, b = E a E b (1.27) Ax = b (1.28) x = (I a, I b ) T (7) R a, R b, R c (Computer Algebra) MAPLE MATHEMATICA MATLAB MATLAB LAPACK MATLAB R a = 1.1, R b = 2, R c = 3, E a = 7, E b = 12 MATLAB >> >> Ra=1.1; Rb=2; Rc=3; Ea=7; Eb=12; >> A=[Ra+Rb,Rb;Rb, Rb+Rc] A =
15 >> b=[ea;eb] b = 7 12 Ax=b >> x=a\b x = \ Ax = b MATLAB Spice Spice Linux Spice Windows Spice spice opus (licence zip Setup.exe 1. MATLAB SpiceOpus SpiceOpus (c) 1-> :
16 14 1. SpiceOpus (c) 1 -> edit C:\SpiceOpus\ex\ex2.cir TITLE ex2.cir V V V V Ra Rb Rc DC V DC V END Spice SpiceOpus (c) 2-> c:\spiceopus\ex\ex2.cir Circuit: TITLE test.cir Spice SpiceOpus (c) 3-> run SpiceOpus (c) 4-> show all Resistor: Simple linear resistor device rc rb ra model R R R resistance i p m 1 1 1
17 Vsource: Independent voltage source device v2 v1 model V V dc 12 7 acmag 0 0 m 1 1 i p MATLAB 2 V a = 50V, V b = 0V, R a = 100, R b = 200, R c = 50 (ex3.cir) DC circuit with R V V R R R R DC V end Spice SpiceOpus (c) 1 -> C:\SpiceOpus\ex\ex3.cir Circuit: DC circuit with R SpiceOpus (c) 2 -> run SpiceOpus (c) 3 -> print all v(1) = e+001 v(2) = e+001
18 16 1. v(3) = e+001 sweep = e+001 v#branch = e-001 Spice Spice3f5 SPICE SPICE SPICE (3) Johann Carl Friedrich Gauss Gauss, Karl Friedrich ( ) (4) Matrices and determinants (Don H. Johnson, Origins of the Equivalent Circuit Concept: The Voltage-Source Equivalent, Proceedings of the IEEE, APRIL, 2003, pp ) 1853 H.Helmholz Helmholz (1)
19 R a = 100, R b = 200, R c = R a R c E R b R L , R L = 200, E a = 75V R L R L Spice R L V T, E a R T Spice R T V T 1.10 R L ex3.cir) Thevenin Circuit V V R R R
20 18 1. R E12.DC V TF V(3) V.END Spice SpiceOpus (c) 1-> C:\SpiceOpus\ex\ex6.cir Circuit: Thevenin Circuit SpiceOpus (c) 2-> run SpiceOpus (c) 3-> show all Resistor: Simple linear resistor device r4 r3 r2 r1 model R R R R resistance 1e i 5e e p 2.5e e m Vsource: Independent voltage source device v model V dc 75 acmag 0 m 1 i p 18.8 SpiceOpus (c) 4-> print all
21 input_impedance = e+002 output_impedance = e+002 transfer_function = e-001 V T = = 50V, R T = output impedance = Lessons in Electric Circuits Volume I DC Linear Circuit Analysis, School of Computer Science and Engineering, Seoul National University 1. (pp.18-20) 2. ( ) Spice 1.6,1.7
22 C[ (farad)] 2.1 i v C q(t) 2.1 v(t) = 1 C i(t)dt (2.1)
23 i(t) = C dv(t) dt 2 (2.2) 2.1 q(t) q(t) = Cv(t) (2.3) i(t) = dq(t) dt (2.4) (Wikipedia) Capacitors 2. 3 L[ ] 2.2 i v L 2.2 v(t) = L di(t) dt (2.5)
24 22 2. i(t) = 1 L v(t)df (2.6) i(t) (linkage flux) φ(t) φ(t) = Li(t) (2.7) v(t) = dφ(t) dt φ (2) (pdf) Self-Inductance and Inductive Reactance (2.8) M ) M > 0 di 1 (t) v 1 (t) = L 1 dt v 2 (t) = M di 1(t) dt + M di 2(t) dt + L 2 di 2 (t) dt (2.9) 2.3 M > 0 (2-7) 2.4 M > 0 v 1 (t) = L 1 di 1 (t) dt v 2 (t) = M di 1(t) dt M di 2(t) dt + L 2 di 2 (t) dt (2.10)
25 i 1 M i 2 v 1 L 1 L 2 v i 1 M i 2 v 1 L 1 L 2 v p.27 p.28 Mutual Inductance
26 p(t) = v(t)i(t) = Ri 2 (t), W = t p(s)ds = t Ri 2 (s)ds (2.11) p(t) = v(t)i(t) = v(t)c dv(t) = d dt dt W = t { } 1 2 Cv2 (t), p(s)ds = 1 2 Cv2 (t) (2.12) p(t) = v(t)i(t) = L di(t) dt i(t) = d dt W = t { } 1 2 Li2 (t), p(s)ds = 1 2 Li2 (t) (2.13) p(t) = v 1 (t)i 1 (t) + v 2 (t)i 2 (t) = d dt t { 1 ( L1 i 2 2 1(t) + Mi 1 (t)i 2 (t) + L 2 i 2 2(t) )}, W = p(s)ds = 1 ( L1 i 2 2 1(t) + Mi 1 (t)i 2 (t) + L 2 i 2 2(t) ) (2.14) (2-12) W 0 k = M L1 L 2 1 (2.15)
27 ) (RL RC t=0 RL RC1
28 RC 2 R i(t) 1 V v(t) C 3.1 RC t = ( ) v(0) = u(t) 3.1 RC R i(t) V u(t) + v(t) C (t > = 0) u(t) = 0 (t < 0) RC ( ) (3.1)
29 v R (t) v R (t) + v(t) = V u(t) (3.2) v R (t) = Ri(t) i(t) = Cdv/dt (3-2) RC dv(t) dt + v(t) = V u(t) (3.3) (3.3) dx(t) dt + γx(t) = f(t) (3.4) (a) (3.4) f(t) dx(t) dt (3.5) + γx(t) = 0 (3.5) x(t) = Ae γt (3.6) A (t = 0 x (b) (3.4) (3.4) (3.5) (3.4) (3.7) (3.4) x(t) = e γt t e γs f(s)ds = t e γ(s t) f(s)ds (3.8) (3.4)
30 28 3. t ] x(t) = e [A γt + e γs f(s)ds (3.9) (c) (3.5) A (3.4) (1) t > 0 (3.3) v(t) = V (2) (3.3) RC dv(t) dt + v(t) = 0 (3.10) v(t) = A exp( t/rc) (3) t > 0 (3.3) v(t) = V + Ae t/rc (3.11) t = 0 v(t) = 0 A = V t > 0 (3.3) v(t) = V (1 e t/rc ) (3.12) Spice 3.3 Spice C:\SpiceOpus\ex\ex3-1.cir
31 R i(t) 2 V (t) + v(t) C RC Switch Closing in RC-series circuit V 1 0 PWL(0,0 10us,1V 10ms,1V) R k C uF.TRAN 1ms 10ms.PROBE.END SpiceOpus (c) 1-> C:\SpiceOpus\ex\ex3-1.cir Circuit: Switch Closing in RC-series circuit SpiceOpus (c) 2-> run SpiceOpus (c) 3-> plot v(2) (v(t) ) (pdf) 3. 2 RC 3. 3
32 V (t) L R i(t) V (t) + v(t) C 3.4 RLC L di(t) dt + Ri(t) + 1 C i(t)dt = V (t) (3.13) t L d2 i(t) dt 2 + R di(t) dt + 1 dv (t) i(t) = C dt (3.14) dv (t)/dt = 0 (3.14) L d2 i(t) dt 2 + R di(t) dt + 1 i(t) = 0 (3.15) C d 2 x(t) dt 2 + λ dx(t) dt (a) + ω 2 0x(t) = f(t) (3.16) (3.16) f(t)
33 d 2 x(t) dt 2 + λ dx(t) dt (3.17) ω 2 0x(t) = 0 (3.17) x(t) e pt (3.18) (3.18) (3.17) p 2 + λp + ω 2 0 = 0 (3.19) p ± = λ ± λ 2 4ω (3.17) x(t) = Ae p +t + Be p t (3.20) (3.21) (a-1) λ ((3.20) x(t) = e t/τ (A cos ωt + B sin ωt) (3.22) τ = 2/λ ω = ω 2 0 (λ/2)2 (b) f(t) = F cos ωt (3.23) (3.16) x(t) = F (A(ω) cos ωt + B(ω) sin ωt), λω A(ω) = (ω0 2 ω2 ) 2 + (λω) 2 B(ω) = ( ) ω 2 0 ω 2 (ω 2 0 ω2 ) 2 + (λω) 2 (3.24) (3.16)
34 32 3. (3.16) (3.17) (3.16) (3.25) (d) (3.17) (3.21) A, B (3.16) 1. ( 3.1 t = v(t) t > 0 v(0) = V (1) t > 0 v(t) (2) V = 1[V], R = 10(1+ /100)k[ ], C = 0.1µ[F] scilab MATLAB (3) 2 Spice
35 a(t) = A sin θ = A sin(ωt + φ) (4.1) θ = ωt + φ ω (angular frequency) A (amplitude) φ (phase angle) v(t) = Ri(t) (4.2) v(t) = L di(t) dt (4.3) i(t) = I sin ωt (4.4) v(t) = ωli cos ωt = ωli sin (ωt + π/2) = V sin (ωt + π/2) (4.5)
36 i(t) = C dv dt (4.6) v(t) = V sin ωt (4.7) i(t) = ωcv cos (ωt) = ωcv sin (ωt + π/2) = sin (ωt + π/2)(4.8) 4. 3 (1) p(t) = v(t)i(t) i(t) = I sin ωt p(t) = RI 2 sin 2 ωt = RI 2 (1 cos 2ωt)/2 (4.9) (2) (4.9) T P P = RI2 2 (4.10) I e = I 2 (4.11) P = RI 2 e (4.12) I e I e V e = V 2 (4.13) 100V 141V
37 Lessons In Electric Circuits Volume II-AC p.49-50
38 5 j e jθ = cos θ + j sinθ (5.1) (1) z x, y z = x + jy (5.2) x z x = Rez y z y = Imz (5.2) (Cartesian form) (2) (5.2) z = re θ (5.3) r = x 2 + y 2 (5.4) θ = arctan y x (5.5) (polar form)
39 r = z, θ = arg z (5.6) (3) z r θ z = r θ (5.7) z (1) a(t) = 2A e sin (ωt + φ) (5.8) A = A e e jφ (5.9) a(t) = Im[ 2A exp(jωt)] (5.10) A (5-4) phasor equivalent) (complex representation) a(t) (2) a 1 (t) = 2A e1 sin(ωt + φ 1 )
40 38 5. a 2 (t) = 2A e2 sin(ωt + φ 2 ) (5.11) A 1 = A e1 exp(jφ 1 ), A 2 = A e2 exp(jφ 2 ) (5.12) a(t) = a 1 (t) + a 2 (t) = 2Im[A 1 exp(jωt) + A 2 exp(jωt)] = 2Im[(A 1 + A 2 ) exp(jωt)] (5.13) a(t) = a 1 (t) + a 2 (t) A 1 + A 2 (3) a(t) = Im[ 2A exp(jωt)] (5.14) da dt (t) = Im[ 2jωA exp(jωt)] (5.15) a (t) jωa a(t)dt (5.16) A jω (5.17) e(t) = 2E e sin (ωt + φ) (5.18)
41 L R i(t) e(t) v(t) C 5.1 RLC E = E e e jφ (5.19) i(t) I 5-1 L di dt + Ri + 1 C idt = e (5.20) (jωli + R + 1 )I = E (5.21) jωc Z = jωli + R + 1 jωc = R + j(ωl 1 ωc ) (5.22) I = E Z (5.23) I = E, arg I = arg E arg Z (5.24) Z i(t) = 2I e sin (ωt + φ θ) (5.25) I e = E e, θ = arg Z (5.26) Z
42 40 5. Z = R 2 + (ωl 1 arg Z = arctan ωc )2 ωl 1 ωc R (5.27) 5. 2 (5.22) Z Z = E I (5.28) E I E/I Z (impedance) E = ZI (5.29) I = Y V (5.30) Y = 1/Z Y (admittance) Z (1) (2) John William Strutt Lord Rayleigh(Born: 12 Nov 1842
43 in Langford Grove (near Maldon), Essex, England, Died: 30 June 1919 in Terling Place, Witham, Essex, England) (3) ( ) Oliver Heaviside (Born: 18 May 1850 in Camden Town, London, England Died: 3 Feb 1925 in Torquay, Devon, England) (a) Maxwell (b) (c) 1887 induction coils (d) 1902 Kennelly-Heaviside Layer (e) Heaviside step function (4) Arthur Edwin Kennelly ( 1893 the American Institute of Electrical Engineers (AIEE) Impedance dc ac ) (5) Charles Proteus Steinmetz ( ) Breslau, Prussia 1889 General Electric in Schenectady 1902 New York city s Union College (1) (2) (Wechselstrom) C.P.Steinmetz, Theory and Calculation of Alternating Current Phenomena, McGraw-Hill Book Co., Inc., New York, Z = R + jx (5.31) R X R X Y = 1/Z
44 42 5. Y = G + jb (5.32) G B G B 5. 4 ) (1) (2) (diagram) (1) RC 5.2 RC 5-2 RC i(t) R = 5Ω e(t) = 10V 60Hz C = 100µF 5.2 RC E = 10[V] ω = 2πf = 120π Z = R + 1/(jωC) = 10 j/( ) = j[ ] Z arg Z MATLAB
45 >> k=1/(120*pi*0.0001) k = >> az=sqrt(5^2+k^2) az = >> s=-atan(1/(120*pi*0.0001*5)) s = >> s*180/pi ans = Z 5. 4 ) 43 Z = (5.33) I I = E/Z (5.34) I = 10 0 = 370.5mA (5.35) I θ E 5.3 RC e(t) = 210 sin 120πt (5.36)
46 44 5. i(t) = sin (120πt ) = sin (120πt )(5.37) Spice 5-2 ac r-c circuit v1 1 0 AC 10V r c u.AC lin 1 60Hz 60Hz.END c:\spiceopus\ex\ac10.cir Spice SpiceOpus (c) 1 -> c:\spiceopus\ex\ac10.cir Circuit: ac r-c circuit SpiceOpus (c) 2 -> run Warning: v1: has no value, DC 0 assumed SpiceOpus (c) 3 -> print i(v1) i(v1) = e-002, e-001 SpiceOpus (c) 4 -> print mag(i(v1)) mag(i(v1)) = e-001 SpiceOpus (c) 5-> print atan(imag(i(v1))/real(i(v1)))*180/pi atan(imag(i(v1))/real(i(v1)))*180/pi = e+001 (2) RC 5.4 RC ac r-c parallel circuit v1 1 0 AC 10V
47 5. 4 ) 45 i(t) e(t) = 10V 60Hz R = 5Ω C = 100µF 5.4 RC r c u.AC lin 1 60Hz 60Hz.END c:\spiceopus\ex\ac11.cir Spice SpiceOpus (c) 10 -> c:\spiceopus\ex\ac11.cir Circuit: ac r-c parallel circuit SpiceOpus (c) 11 -> run Warning: v1: has no value, DC 0 assumed SpiceOpus (c) 12 -> print i(v1) i(v1) = e+000, e-001 SpiceOpus (c) 13 -> print mag(i(v1)) mag(i(v1)) = e+000 SpiceOpus (c) 14 -> print atan(imag(i(v1))/real(i(v1)))*180/pi atan(imag(i(v1))/real(i(v1)))*180/pi = e+001 I I = deg (5.38)
48 46 5. Y = 1/R + jωc = j I = Y E = ( j)10 = j (5.39) (3) RL i(t) R = 5Ω e(t) = 10V 60Hz L = 10mH 5.5 RL 5.5 Z = R + jωl = π0.01 = 5 + j [ ] MATLAB >> abs(z) ans = >> a=angle(z) a = >> a*180/pi ans = Z = 6.262e 0.646j (5.40) I = E/Z = 10/Z = e 0.646j (5.41)
49 5. 4 ) 47 Spice ac r-l circuit v1 1 0 ac 10 r l m.ac lin end c:\spiceopus\ex\ac12.cir Spice SpiceOpus (c) 15 -> c:\spiceopus\ex\ac12.cir Circuit: ac r-l circuit SpiceOpus (c) 16 -> run Warning: v1: has no value, DC 0 assumed SpiceOpus (c) 17 -> print i(v1) i(v1) = e+000, e-001 SpiceOpus (c) 18 -> print mag(i(v1)) mag(i(v1)) = e+000 SpiceOpus (c) 19 -> print atan(imag(i(v1))/real(i(v1)))*180/pi atan(imag(i(v1))/real(i(v1)))*180/pi = e+001 (4) RL 5.6 Y = 1 R + 1 = j (5.42) jωl I = Y E = j (5.43) Spice ac r-l circuit
50 48 5. i(t) e(t) = 10V 60Hz R = 5Ω L = 10mH 5.6 RL v1 1 2 ac 10 r r l m.ac lin end \end{vernbatim} \begin{verbatim} SpiceOpus (c) 22 -> c:\spiceopus\ex\ac13.cir Circuit: ac r-l circuit SpiceOpus (c) 23 -> run Warning: v1: has no value, DC 0 assumed SpiceOpus (c) 24 -> print i(v1) i(v1) = e+000, e+000
2010 4 7 1 3 11 Electric source 3 111 Voltage source 3 112 Current source 3 113 3 12 Kirchhoff s law 4 121 Kirchhoff s voltage law 4 122 Kirchhoff s current law 4 2 5 21 Resistor 5 211 Ohm s law 5 212
c 2009 i
I 2009 c 2009 i 0 1 0.0................................... 1 0.1.............................. 3 0.2.............................. 5 1 7 1.1................................. 7 1.2..............................
128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds = 0 (3.4) S 1, S 2 { B( r) n( r)}ds
127 3 II 3.1 3.1.1 Φ(t) ϕ em = dφ dt (3.1) B( r) Φ = { B( r) n( r)}ds (3.2) S S n( r) Φ 128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds
I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )
I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17
LCR e ix LC AM m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x (k > 0) k x = x(t)
338 7 7.3 LCR 2.4.3 e ix LC AM 7.3.1 7.3.1.1 m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x k > 0 k 5.3.1.1 x = xt 7.3 339 m 2 x t 2 = k x 2 x t 2 = ω 2 0 x ω0 = k m ω 0 1.4.4.3 2 +α 14.9.3.1 5.3.2.1 2 x
0.1 I I : 0.2 I
1, 14 12 4 1 : 1 436 (445-6585), E-mail : sxiida@sci.toyama-u.ac.jp 0.1 I I 1. 2. 3. + 10 11 4. 12 1: 0.2 I + 0.3 2 1 109 1 14 3,4 0.6 ( 10 10, 2 11 10, 12/6( ) 3 12 4, 4 14 4 ) 0.6.1 I 1. 2. 3. 0.4 (1)
AC Modeling and Control of AC Motors Seiji Kondo, Member 1. q q (1) PM (a) N d q Dept. of E&E, Nagaoka Unive
AC Moeling an Control of AC Motors Seiji Kono, Member 1. (1) PM 33 54 64. 1 11 1(a) N 94 188 163 1 Dept. of E&E, Nagaoka University of Technology 163 1, Kamitomioka-cho, Nagaoka, Niigata 94 188 (a) 巻数
x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin
2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ
chap1.dvi
1 1 007 1 e iθ = cos θ + isin θ 1) θ = π e iπ + 1 = 0 1 ) 3 11 f 0 r 1 1 ) k f k = 1 + r) k f 0 f k k = 01) f k+1 = 1 + r)f k ) f k+1 f k = rf k 3) 1 ) ) ) 1+r/)f 0 1 1 + r/) f 0 = 1 + r + r /4)f 0 1 f
1 7 ω ω ω 7.1 0, ( ) Q, 7.2 ( Q ) 7.1 ω Z = R +jx Z 1/ Z 7.2 ω 7.2 Abs. admittance (x10-3 S) RLC Series Circuit Y R = 20 Ω L = 100
7 7., ) Q, 7. Q ) 7. Z = R +jx Z / Z 7. 7. Abs. admittance x -3 S) 5 4 3 R Series ircuit Y R = Ω = mh = uf Q = 5 5 5 V) Z = R + jx 7. Z 7. ) R = Ω = mh = µf ) 7 V) R Z s = R + j ) 7.3 R =. 7.4) ) f = π.
08-Note2-web
r(t) t r(t) O v(t) = dr(t) dt a(t) = dv(t) dt = d2 r(t) dt 2 r(t), v(t), a(t) t dr(t) dt r(t) =(x(t),y(t),z(t)) = d 2 r(t) dt 2 = ( dx(t) dt ( d 2 x(t) dt 2, dy(t), dz(t) dt dt ), d2 y(t) dt 2, d2 z(t)
ii 4 5 RLC 2 LC LC OTA, FDNR 6
There is nothing more practical than a good theory. James Clerk Maxwell ( 1831 1879) 1.1 1.1 2 3 MOSFET 3 MOSFET MOS CMOS CMOS CMOS ii 4 5 RLC 2 LC LC OTA, FDNR 6 iii 1 (90 30 ) 6 5 1 1 2013 3 1. 1.1...
) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)
4 4 ) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7
Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e
7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e z
(u(x)v(x)) = u (x)v(x) + u(x)v (x) ( ) u(x) = u (x)v(x) u(x)v (x) v(x) v(x) 2 y = g(t), t = f(x) y = g(f(x)) dy dx dy dx = dy dt dt dx., y, f, g y = f (g(x))g (x). ( (f(g(x)). ). [ ] y = e ax+b (a, b )
,, 2. Matlab Simulink 2018 PC Matlab Scilab 2
(2018 ) ( -1) TA Email : ohki@i.kyoto-u.ac.jp, ske.ta@bode.amp.i.kyoto-u.ac.jp : 411 : 10 308 1 1 2 2 2.1............................................ 2 2.2..................................................
<4D F736F F D B B BB2D834A836F815B82D082C88C602E646F63>
信号処理の基礎 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/081051 このサンプルページの内容は, 初版 1 刷発行時のものです. i AI ii z / 2 3 4 5 6 7 7 z 8 8 iii 2013 3 iv 1 1 1.1... 1 1.2... 2 2 4 2.1...
z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy
z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z
2016-08-18 3 1 9 1.1................................................... 9 1.2......................................... 9 1.3...................................................... 9 1.4......................................................
B1 Ver ( ), SPICE.,,,,. * : student : jikken. [ ] ( TarouOsaka). (, ) 1 SPICE ( SPICE. *1 OrCAD
![B1 Ver ( ), SPICE.,,,,. * : student : jikken. [ ] ( TarouOsaka). (, ) 1 SPICE ( SPICE. *1 OrCAD](/thumbs/101/152175529.jpg "B1 Ver ( ), SPICE.,,,,. * : student : jikken. [ ] ( TarouOsaka). (, ) 1 SPICE ( SPICE. *1 OrCAD")
B1 er. 3.05 (2019.03.27), SPICE.,,,,. * 1 1. 1. 1 1.. 2. : student : jikken. [ ] ( TarouOsaka). (, ) 1 SPICE ( SPICE. *1 OrCAD https://www.orcad.com/jp/resources/orcad-downloads.. 1 2. SPICE 1. SPICE Windows
214 March 31, 214, Rev.2.1 4........................ 4........................ 5............................. 7............................... 7 1 8 1.1............................... 8 1.2.......................
35
D: 0.BUN 7 8 4 B5 6 36 6....................................... 36 6.................................... 37 6.3................................... 38 6.3....................................... 38 6.4..........................................
Untitled
23 1 11 A 2 A.1..................................... 2 A.2.................................. 4 A.3............................... 5 A.4.................................... 6 A.5.......................
213 March 25, 213, Rev.1.5 4........................ 4........................ 6 1 8 1.1............................... 8 1.2....................... 9 2 14 2.1..................... 14 2.2............................
II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R
![II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R](/thumbs/95/125540821.jpg "II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R")
II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =
x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s
... x, y z = x + iy x z y z x = Rez, y = Imz z = x + iy x iy z z () z + z = (z + z )() z z = (z z )(3) z z = ( z z )(4)z z = z z = x + y z = x + iy ()Rez = (z + z), Imz = (z z) i () z z z + z z + z.. z
30 (11/04 )
30 (11/04 ) i, 1,, II I?,,,,,,,,, ( ),,, ϵ δ,,,,, (, ),,,,,, 5 : (1) ( ) () (,, ) (3) ( ) (4) (5) ( ) (1),, (),,, () (3), (),, (4), (1), (3), ( ), (5),,,,,,,, ii,,,,,,,, Richard P. Feynman, The best teaching
振動と波動
Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3
(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt
S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............
B 1 B.1.......................... 1 B.1.1................. 1 B.1.2................. 2 B.2........................... 5 B.2.1.......................... 5 B.2.2.................. 6 B.2.3..................
K E N Z U 01 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.................................... 4 1..1..................................... 4 1...................................... 5................................
Note.tex 2008/09/19( )
1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................
The Physics of Atmospheres CAPTER :
The Physics of Atmospheres CAPTER 4 1 4 2 41 : 2 42 14 43 17 44 25 45 27 46 3 47 31 48 32 49 34 41 35 411 36 maintex 23/11/28 The Physics of Atmospheres CAPTER 4 2 4 41 : 2 1 σ 2 (21) (22) k I = I exp(
85 4
85 4 86 Copright c 005 Kumanekosha 4.1 ( ) ( t ) t, t 4.1.1 t Step! (Step 1) (, 0) (Step ) ±V t (, t) I Check! P P V t π 54 t = 0 + V (, t) π θ : = θ : π ) θ = π ± sin ± cos t = 0 (, 0) = sin π V + t +V
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx
36 3 D f(z) D z f(z) z Taylor z D C f(z) z C C f (z) C f(z) f (z) f(z) D C D D z C C 3.: f(z) 3. f (z) f 2 (z) D D D D D f (z) f 2 (z) D D f (z) f 2 (
3 3. D f(z) D D D D D D D D f(z) D f (z) f (z) f(z) D (i) (ii) (iii) f(z) = ( ) n z n = z + z 2 z 3 + n= z < z < z > f (z) = e t(+z) dt Re z> Re z> [ ] f (z) = e t(+z) = (Rez> ) +z +z t= z < f(z) Taylor
sikepuri.dvi
2009 2 2 2. 2.. F(s) G(s) H(s) G(s) F(s) H(s) F(s),G(s) H(s) : V (s) Z(s)I(s) I(s) Y (s)v (s) Z(s): Y (s): 2: ( ( V V 2 I I 2 ) ( ) ( Z Z 2 Z 2 Z 22 ) ( ) ( Y Y 2 Y 2 Y 22 ( ) ( ) Z Z 2 Y Y 2 : : Z 2 Z
I
I 6 4 10 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2
filename=mathformula58.tex ax + bx + c =, x = b ± b 4ac, (.) a x + x = b a, x x = c a, (.) ax + b x + c =, x = b ± b ac. a (.3). sin(a ± B) = sin A cos B ± cos A sin B, (.) cos(a ± B) = cos A cos B sin
K E N Z U 2012 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.2................................... 4 1.2.1..................................... 4 1.2.2.................................... 5................................
1. ( ) 1.1 t + t [m]{ü(t + t)} + [c]{ u(t + t)} + [k]{u(t + t)} = {f(t + t)} (1) m ü f c u k u 1.2 Newmark β (1) (2) ( [m] + t ) 2 [c] + β( t)2
![1. ( ) 1.1 t + t [m]{ü(t + t)} + [c]{ u(t + t)} + [k]{u(t + t)} = {f(t + t)} (1) m ü f c u k u 1.2 Newmark β (1) (2) ( [m] + t ) 2 [c] + β( t)2](/thumbs/80/80934335.jpg "1. ( ) 1.1 t + t [m]{ü(t + t)} + [c]{ u(t + t)} + [k]{u(t + t)} = {f(t + t)} (1) m ü f c u k u 1.2 Newmark β (1) (2) ( [m] + t ) 2 [c] + β( t)2")
212 1 6 1. (212.8.14) 1 1.1............................................. 1 1.2 Newmark β....................... 1 1.3.................................... 2 1.4 (212.8.19)..................................
V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H
199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)
構造と連続体の力学基礎
II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton
ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +
2.6 2.6.1 ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.121) Z ω ω j γ j f j
<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
通信方式第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/072662 このサンプルページの内容は, 第 2 版発行当時のものです. i 2 2 2 2012 5 ii,.,,,,,,.,.,,,,,.,,.,,..,,,,.,,.,.,,.,,.. 1990 5 iii 1 1
<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
高速流体力学 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/067361 このサンプルページの内容は, 第 1 版発行時のものです. i 20 1999 3 2 2010 5 ii 1 1 1.1 1 1.2 4 9 2 10 2.1 10 2.2 12 2.3 13 2.4 13 2.5
Gmech08.dvi
145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2
2.2 ( y = y(x ( (x 0, y 0 y (x 0 (y 0 = y(x 0 y = y(x ( y (x 0 = F (x 0, y(x 0 = F (x 0, y 0 (x 0, y 0 ( (x 0, y 0 F (x 0, y 0 xy (x, y (, F (x, y ( (
(. x y y x f y = f(x y x y = y(x y x y dx = d dx y(x = y (x = f (x y = y(x x ( (differential equation ( + y 2 dx + xy = 0 dx = xy + y 2 2 2 x y 2 F (x, y = xy + y 2 y = y(x x x xy(x = F (x, y(x + y(x 2
LLG-R8.Nisus.pdf
d M d t = γ M H + α M d M d t M γ [ 1/ ( Oe sec) ] α γ γ = gµ B h g g µ B h / π γ g = γ = 1.76 10 [ 7 1/ ( Oe sec) ] α α = λ γ λ λ λ α γ α α H α = γ H ω ω H α α H K K H K / M 1 1 > 0 α 1 M > 0 γ α γ =
2 0.1 Introduction NMR 70% 1/2
Y. Kondo 2010 1 22 2 0.1 Introduction NMR 70% 1/2 3 0.1 Introduction......................... 2 1 7 1.1.................... 7 1.2............................ 11 1.3................... 12 1.4..........................
http://www.ike-dyn.ritsumei.ac.jp/ hyoo/wave.html 1 1, 5 3 1.1 1..................................... 3 1.2 5.1................................... 4 1.3.......................... 5 1.4 5.2, 5.3....................
4‐E ) キュリー温度を利用した消磁：熱消磁
( ) () x C x = T T c T T c 4D ) ) Fe Ni Fe Fe Ni (Fe Fe Fe Fe Fe 462 Fe76 Ni36 4E ) ) (Fe) 463 4F ) ) ( ) Fe HeNe 17 Fe Fe Fe HeNe 464 Ni Ni Ni HeNe 465 466 (2) Al PtO 2 (liq) 467 4G ) Al 468 Al ( 468
数学の基礎訓練I
I 9 6 13 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 3 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................
2 Chapter 4 (f4a). 2. (f4cone) ( θ) () g M. 2. (f4b) T M L P a θ (f4eki) ρ H A a g. v ( ) 2. H(t) ( )
http://astr-www.kj.yamagata-u.ac.jp/~shibata f4a f4b 2 f4cone f4eki f4end 4 f5meanfp f6coin () f6a f7a f7b f7d f8a f8b f9a f9b f9c f9kep f0a f0bt version feqmo fvec4 fvec fvec6 fvec2 fvec3 f3a (-D) f3b
(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0
1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45
. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n
003...............................3 Debye................. 3.4................ 3 3 3 3. Larmor Cyclotron... 3 3................ 4 3.3.......... 4 3.3............ 4 3.3...... 4 3.3.3............ 5 3.4.........
t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z
I 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)
D 1 l θ lsinθ y L D 2 2: D 1 y l sin θ (1.2) θ y (1.1) D 1 (1.2) (θ, y) π 0 π l sin θdθ π [0, π] 3 sin cos π l sin θdθ = l π 0 0 π Ldθ = L Ldθ sin θdθ
![D 1 l θ lsinθ y L D 2 2: D 1 y l sin θ (1.2) θ y (1.1) D 1 (1.2) (θ, y) π 0 π l sin θdθ π [0, π] 3 sin cos π l sin θdθ = l π 0 0 π Ldθ = L Ldθ sin θdθ](/thumbs/78/78633861.jpg "D 1 l θ lsinθ y L D 2 2: D 1 y l sin θ (1.2) θ y (1.1) D 1 (1.2) (θ, y) π 0 π l sin θdθ π [0, π] 3 sin cos π l sin θdθ = l π 0 0 π Ldθ = L Ldθ sin θdθ")
1 1 (Buffon) 1 l L l < L 1: 2 D 1 D 2 D 1 P 1 2 θ D 1 y θ y π θ π; 0 y L (1.1) 1 Georges Louis Leclerc Comte de Buffon Born: 7 Sept 1707 in Montbard, Cōte d Or, France Died: 16 April 1788 in Paris, France
m(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120)
2.6 2.6.1 mẍ + γẋ + ω 0 x) = ee 2.118) e iωt Pω) = χω)e = ex = e2 Eω) m ω0 2 ω2 iωγ 2.119) Z N ϵω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j 2.120) Z ω ω j γ j f j f j f j sum j f j = Z 2.120 ω ω j, γ ϵω) ϵ
第 章 交流回路素子とその性質 抵抗 コイル コンデンサ it) it) dit) vt) = L vt) = 1 it) 図. コイル インダクタ) [] 図.3 コンデンサ キャパシタ) [3] はインダクタである コイルの両端に印加された電圧 費電力が負である とは 電力がその回路素子から供給
![第 章 交流回路素子とその性質 抵抗 コイル コンデンサ it) it) dit) vt) = L vt) = 1 it) 図. コイル インダクタ) [] 図.3 コンデンサ キャパシタ) [3] はインダクタである コイルの両端に印加された電圧 費電力が負である とは 電力がその回路素子から供給](/thumbs/40/20581928.jpg "第 章 交流回路素子とその性質 抵抗 コイル コンデンサ it) it) dit) vt) = L vt) = 1 it) 図. コイル インダクタ) [] 図.3 コンデンサ キャパシタ) [3] はインダクタである コイルの両端に印加された電圧 費電力が負である とは 電力がその回路素子から供給")
1 ω *1 V m I m R V m = R I m ) L V m = ωl I m 9 V m = I m ω 9 *1 ω it) vt) = Rit).1 [1].1.1.1 resistor) resistor).1 [1] vt) it) vt) = Rit)..1) R resistance) Ω, Ohm conductance) S Siemens)).1. inductor)
211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,
() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y
![() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y](/thumbs/85/92979039.jpg "() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y")
5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =
( : December 27, 2015) CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x) f (x) y = f(x) x ϕ(r) (gradient) ϕ(r) (gradϕ(r) ) ( ) ϕ(r)
( : December 27, 215 CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x f (x y f(x x ϕ(r (gradient ϕ(r (gradϕ(r ( ϕ(r r ϕ r xi + yj + zk ϕ(r ϕ(r x i + ϕ(r y j + ϕ(r z k (1.1 ϕ(r ϕ(r i
18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb
![18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb](/thumbs/93/114079295.jpg "18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb")
r 1 r 2 r 1 r 2 2 Coulomb Gauss Coulomb 2.1 Coulomb 1 2 r 1 r 2 1 2 F 12 2 1 F 21 F 12 = F 21 = 1 4πε 0 1 2 r 1 r 2 2 r 1 r 2 r 1 r 2 (2.1) Coulomb ε 0 = 107 4πc 2 =8.854 187 817 10 12 C 2 N 1 m 2 (2.2)
64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k
63 3 Section 3.1 g 3.1 3.1: : 64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () 3 9.8 m/s 2 3.2 3.2: : a) b) 5 15 4 1 1. 1 3 14. 1 3 kg/m 3 2 3.3 1 3 5.8 1 3 kg/m 3 3 2.65 1 3 kg/m 3 4 6 m 3.1. 65 5
QMI_09.dvi
25 3 19 Erwin Schrödinger 1925 3.1 3.1.1 3.1.2 σ τ 2 2 ux, t) = ux, t) 3.1) 2 x2 ux, t) σ τ 2 u/ 2 m p E E = p2 3.2) E ν ω E = hν = hω. 3.3) k p k = p h. 3.4) 26 3 hω = E = p2 = h2 k 2 ψkx ωt) ψ 3.5) h
S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d
S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....
QMI_10.dvi
25 3 19 Erwin Schrödinger 1925 3.1 3.1.1 σ τ x u u x t ux, t) u 3.1 t x P ux, t) Q θ P Q Δx x + Δx Q P ux + Δx, t) Q θ P u+δu x u x σ τ P x) Q x+δx) P Q x 3.1: θ P θ Q P Q equation of motion P τ Q τ σδx
第1章 微分方程式と近似解法
April 12, 2018 1 / 52 1.1 ( ) 2 / 52 1.2 1.1 1.1: 3 / 52 1.3 Poisson Poisson Poisson 1 d {2, 3} 4 / 52 1 1.3.1 1 u,b b(t,x) u(t,x) x=0 1.1: 1 a x=l 1.1 1 (0, t T ) (0, l) 1 a b : (0, t T ) (0, l) R, u
Hanbury-Brown Twiss (ver. 2.0) van Cittert - Zernike mutual coherence
Hanbury-Brown Twiss (ver. 2.) 25 4 4 1 2 2 2 2.1 van Cittert - Zernike..................................... 2 2.2 mutual coherence................................. 4 3 Hanbury-Brown Twiss ( ) 5 3.1............................................
<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
MATLAB/Simulink による現代制御入門 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/9241 このサンプルページの内容は, 初版 1 刷発行当時のものです. i MATLAB/Simulink MATLAB/Simulink 1. 1 2. 3. MATLAB/Simulink
(1) (2) (3) (4) 1
8 3 4 3.................................... 3........................ 6.3 B [, ].......................... 8.4........................... 9........................................... 9.................................
QMII_10.dvi
65 1 1.1 1.1.1 1.1 H H () = E (), (1.1) H ν () = E ν () ν (). (1.) () () = δ, (1.3) μ () ν () = δ(μ ν). (1.4) E E ν () E () H 1.1: H α(t) = c (t) () + dνc ν (t) ν (), (1.5) H () () + dν ν () ν () = 1 (1.6)
z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy
f f x, y, u, v, r, θ r > = x + iy, f = u + iv C γ D f f D f f, Rm,. = x + iy = re iθ = r cos θ + i sin θ = x iy = re iθ = r cos θ i sin θ x = + = Re, y = = Im i r = = = x + y θ = arg = arctan y x e i =
Chap10.dvi
=0. f = 2 +3 { 2 +3 0 2 f = 1 =0 { sin 0 3 f = 1 =0 2 sin 1 0 4 f = 0 =0 { 1 0 5 f = 0 =0 f 3 2 lim = lim 0 0 0 =0 =0. f 0 = 0. 2 =0. 3 4 f 1 lim 0 0 = lim 0 sin 2 cos 1 = lim 0 2 sin = lim =0 0 2 =0.
n ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................
keisoku01.dvi
2.,, Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 4 Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 5 Mon, 2006, 401, SAGA, JAPAN Dept.
4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t
![4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t](/thumbs/70/62705042.jpg "4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t")
1 1.1 sin 2π [rad] 3 ft 3 sin 2t π 4 3.1 2 1.1: sin θ 2.2 sin θ ft t t [sec] t sin 2t π 4 [rad] sin 3.1 3 sin θ θ t θ 2t π 4 3.2 3.1 3.4 3.4: 2.2: sin θ θ θ [rad] 2.3 0 [rad] 4 sin θ sin 2t π 4 sin 1 1
L L L L C C C C (a) (b) (c) 4.4 (a) (b) (a) RG59/U 6.2mm ( ) 73Ω web page (c) 4 4 dx 4 J V dx dj Ydx Zdx dv Z,Y dv = JZdx, dj = VYdx (4.8) d 2 J dx 2
8 ( ) 2014 11 23 OP h g m r d 4 ( ) 4.2 + (lumped constant circuit) 4.2.1 3 (Poynting ) (strip line) (waveguide) (coaxial cable) GHz 4.4(a) dj dx dv Zdx Ydx ( ) (distributed constatn circuit) *1 4.4(c)
dynamics-solution2.dvi
1 1. (1) a + b = i +3i + k () a b =5i 5j +3k (3) a b =1 (4) a b = 7i j +1k. a = 14 l =/ 14, m=1/ 14, n=3/ 14 3. 4. 5. df (t) d [a(t)e(t)] =ti +9t j +4k, = d a(t) d[a(t)e(t)] e(t)+ da(t) d f (t) =i +18tj
mt_4.dvi
( ) 2006 1 PI 1 1 1.1................................. 1 1.2................................... 1 2 2 2.1...................................... 2 2.1.1.......................... 2 2.1.2..............................
zz + 3i(z z) + 5 = 0 + i z + i = z 2i z z z y zz + 3i (z z) + 5 = 0 (z 3i) (z + 3i) = 9 5 = 4 z 3i = 2 (3i) zz i (z z) + 1 = a 2 {
04 zz + iz z) + 5 = 0 + i z + i = z i z z z 970 0 y zz + i z z) + 5 = 0 z i) z + i) = 9 5 = 4 z i = i) zz i z z) + = a {zz + i z z) + 4} a ) zz + a + ) z z) + 4a = 0 4a a = 5 a = x i) i) : c Darumafactory
( ) Note (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e
![( ) Note (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e](/thumbs/98/136160482.jpg "( ) Note (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e")
( ) Note 3 19 12 13 8 8.1 (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e µ τ, e R, µ R, τ R (1a) L ( ) ) * 3) W Z 1/2 ( - )
,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.
9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,
DE-resume
- 2011, http://c-faculty.chuo-u.ac.jp/ nishioka/ 2 11 21131 : 4 1 x y(x, y (x,y (x,,y (n, (1.1 F (x, y, y,y,,y (n =0. (1.1 n. (1.1 y(x. y(x (1.1. 1 1 1 1.1... 2 1.2... 9 1.3 1... 26 2 2 34 2.1,... 35 2.2
e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,
01 10 18 ( ) 1 6 6 1 8 8 1 6 1 0 0 0 0 1 Table 1: 10 0 8 180 1 1 1. ( : 60 60 ) : 1. 1 e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1,
r d 2r d l d (a) (b) (c) 1: I(x,t) I(x+ x,t) I(0,t) I(l,t) V in V(x,t) V(x+ x,t) V(0,t) l V(l,t) 2: 0 x x+ x 3: V in 3 V in x V (x, t) I(x, t
1 1 2 2 2r d 2r d l d (a) (b) (c) 1: I(x,t) I(x+ x,t) I(0,t) I(l,t) V in V(x,t) V(x+ x,t) V(0,t) l V(l,t) 2: 0 x x+ x 3: V in 3 V in x V (x, t) I(x, t) V (x, t) I(x, t) V in x t 3 4 1 L R 2 C G L 0 R 0
[1] 1.1 x(t) t x(t + n ) = x(t) (n = 1,, 3, ) { x(t) : : 1 [ /, /] 1 x(t) = a + a 1 cos πt + a cos 4πt + + a n cos nπt + + b 1 sin πt + b sin 4πt = a
![[1] 1.1 x(t) t x(t + n ) = x(t) (n = 1,, 3, ) { x(t) : : 1 [ /, /] 1 x(t) = a + a 1 cos πt + a cos 4πt + + a n cos nπt + + b 1 sin πt + b sin 4πt = a](/thumbs/85/91634532.jpg "[1] 1.1 x(t) t x(t + n ) = x(t) (n = 1,, 3, ) { x(t) : : 1 [ /, /] 1 x(t) = a + a 1 cos πt + a cos 4πt + + a n cos nπt + + b 1 sin πt + b sin 4πt = a")
13/7/1 II ( / A: ) (1) 1 [] (, ) ( ) ( ) ( ) etc. etc. 1. 1 [1] 1.1 x(t) t x(t + n ) = x(t) (n = 1,, 3, ) { x(t) : : 1 [ /, /] 1 x(t) = a + a 1 cos πt + a cos 4πt + + a n cos nπt + + b 1 sin πt + b sin
120 9 I I 1 I 2 I 1 I 2 ( a) ( b) ( c ) I I 2 I 1 I ( d) ( e) ( f ) 9.1: Ampère (c) (d) (e) S I 1 I 2 B ds = µ 0 ( I 1 I 2 ) I 1 I 2 B ds =0. I 1 I 2
9 E B 9.1 9.1.1 Ampère Ampère Ampère s law B S µ 0 B ds = µ 0 j ds (9.1) S rot B = µ 0 j (9.2) S Ampère Biot-Savart oulomb Gauss Ampère rot B 0 Ampère µ 0 9.1 (a) (b) I B ds = µ 0 I. I 1 I 2 B ds = µ 0
磁性物理学 - 遷移金属化合物磁性のスピンゆらぎ理論
email: takahash@sci.u-hyogo.ac.jp May 14, 2009 Outline 1. 2. 3. 4. 5. 6. 2 / 262 Today s Lecture: Mode-mode Coupling Theory 100 / 262 Part I Effects of Non-linear Mode-Mode Coupling Effects of Non-linear
Quz Quz
http://www.ppl.app.keo.ac.jp/denjk III (1969). (1977). ( ) (1999). (1981). (199). Harry Lass Vector and Tensor Analyss, McGraw-Hll, (195).. Quz Quz II E B / t = Maxwell ρ e E = (1.1) ε E= (1.) B = (1.3)
I 1
I 1 1 1.1 1. 3 m = 3 1 7 µm. cm = 1 4 km 3. 1 m = 1 1 5 cm 4. 5 cm 3 = 5 1 15 km 3 5. 1 = 36 6. 1 = 8.64 1 4 7. 1 = 3.15 1 7 1 =3 1 7 1 3 π 1. 1. 1 m + 1 cm = 1.1 m. 1 hr + 64 sec = 1 4 sec 3. 3. 1 5 kg
.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,
[ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b
) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4
![) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4](/thumbs/92/107866707.jpg ") ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4")
1. k λ ν ω T v p v g k = π λ ω = πν = π T v p = λν = ω k v g = dω dk 1) ) 3) 4). p = hk = h λ 5) E = hν = hω 6) h = h π 7) h =6.6618 1 34 J sec) hc=197.3 MeV fm = 197.3 kev pm= 197.3 ev nm = 1.97 1 3 ev
IA
IA 31 4 11 1 1 4 1.1 Planck.............................. 4 1. Bohr.................................... 5 1.3..................................... 6 8.1................................... 8....................................
( ) ) ) ) 5) 1 J = σe 2 6) ) 9) 1955 Statistical-Mechanical Theory of Irreversible Processes )
( 3 7 4 ) 2 2 ) 8 2 954 2) 955 3) 5) J = σe 2 6) 955 7) 9) 955 Statistical-Mechanical Theory of Irreversible Processes 957 ) 3 4 2 A B H (t) = Ae iωt B(t) = B(ω)e iωt B(ω) = [ Φ R (ω) Φ R () ] iω Φ R (t)
1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1
sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω ω α 3 3 2 2V 3 33+.6T m T 5 34m Hz. 34 3.4m 2 36km 5Hz. 36km m 34 m 5 34 + m 5 33 5 =.66m 34m 34 x =.66 55Hz, 35 5 =.7 485.7Hz 2 V 5Hz.5V.5V V
JIS Z803: (substitution method) 3 LCR LCR GPIB
LCR NMIJ 003 Agilent 8A 500 ppm JIS Z803:000 50 (substitution method) 3 LCR LCR GPIB Taylor 5 LCR LCR meter (Agilent 8A: Basic accuracy 500 ppm) V D z o I V DUT Z 3 V 3 I A Z V = I V = 0 3 6 V, A LCR meter