Similar documents
Verilog HDL による回路設計記述

PLDとFPGA

( ) : 1997

2.5. Verilog 19 Z= X + Y - Z A+B LD ADD SUB ST (X<<1)+(Y<<1) X 1 2 LD SL ST 2 10

untitled


論理設計の基礎

Unconventional HDL Programming ( version) 1

Microsoft PowerPoint - Lec pptx

VHDL

untitled

Microsoft PowerPoint - FPGA

main.dvi

VLD Kazutoshi Kobayashi

曲面のパラメタ表示と接線ベクトル

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

MOSFET 6-2 CMOS 6-2 TTL Transistor Transistor Logic ECL Emitter Coupled Logic I2L Integrated

1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =

LSI LSI

Microsoft PowerPoint - Lec pptx


DELPHINUS EQUULEUS 2019 NASA SLS FPGA ( ) DELPHINUS 2

プログラマブル論理デバイス

1: ITT-2 DDR2 1.8V,.V(F) Config. Mem. JTAG XCFPV048 LEDs SWs Clock (VariClock) DDR2 DDR2 DDR2 FPGA XC5VFX0T General-Purpose LEDs SWs XTAL (2.68kHz) MC

VelilogHDL 回路を「言語」で記述する

Microsoft PowerPoint - Chap1 [Compatibility Mode]



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

Gmech08.dvi

1.1 1 A

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π

VHDL VHDL VHDL i

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


No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%

デザインパフォーマンス向上のためのHDLコーディング法

9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P

( ) ( )

impulse_response.dvi

keisoku01.dvi

HardCopy IIIデバイスの外部メモリ・インタフェース

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

, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

2 1 x 1.1: v mg x (t) = v(t) mv (t) = mg 0 x(0) = x 0 v(0) = v 0 x(t) = x 0 + v 0 t 1 2 gt2 v(t) = v 0 gt t x = x 0 + v2 0 2g v2 2g 1.1 (x, v) θ

Łñ“’‘‚2004


プリント

XC9500 ISP CPLD JTAG Port 3 JTAG Controller In-System Programming Controller 8 36 Function Block Macrocells to 8 /GCK /GSR /GTS 3 2 or 4 Blocks FastCO

橡卒論.PDF


1 nakayama/print/ Def (Definition ) Thm (Theorem ) Prop (Proposition ) Lem (Lemma ) Cor (Corollary ) 1. (1) A, B (2) ABC

LSI LSI 2

計数工学実験/システム情報工学実験第一 「ディジタル回路の基礎」

meiji_resume_1.PDF

pdf

7. y fx, z gy z gfx dz dx dz dy dy dx. g f a g bf a b fa 7., chain ule Ω, D R n, R m a Ω, f : Ω R m, g : D R l, fω D, b fa, f a g b g f a g f a g bf a

sikepuri.dvi

untitled

1 4 1 ( ) ( ) ( ) ( ) () 1 4 2

Stratix IIIデバイスの外部メモリ・インタフェース

スライド 1

橡200006youbou.PDF

「FPGAを用いたプロセッサ検証システムの製作」

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

if clear = 1 then Q <= " "; elsif we = 1 then Q <= D; end rtl; regs.vhdl clk 0 1 rst clear we Write Enable we 1 we 0 if clk 1 Q if rst =

組込みシステムシンポジウム2011 Embedded Systems Symposium 2011 ESS /10/20 FPGA Android Android Java FPGA Java FPGA Dalvik VM Intel Atom FPGA PCI Express DM

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

main.dvi

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 +

1, Verilog-HDL, Verilog-HDL Verilog-HDL,, FPGA,, HDL, 11, 1 (a) (b) (c) FPGA (d) 2 10,, Verilog-HDL, FPGA, 12,,,, html % netscape file://home/users11/

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>

( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +

LTC ビット、200ksps シリアル・サンプリングADC

B1 Ver ( ), SPICE.,,,,. * : student : jikken. [ ] ( TarouOsaka). (, ) 1 SPICE ( SPICE. *1 OrCAD

TULを用いたVisual ScalerとTDCの開発

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

IPSJ SIG Technical Report Vol.2009-CVIM-167 No /6/10 Real AdaBoost HOG 1 1 1, 2 1 Real AdaBoost HOG HOG Real AdaBoost HOG A Method for Reducing

24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x

電卓の設計 1

p06.dvi

Microsoft Word - 信号処理3.doc

(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

Microsoft PowerPoint - slide

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)

,. 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,,.

arma dvi

, FPGA Verilog-HDL

A Study of Adaptive Array Implimentation for mobile comunication in cellular system GD133

Cyclone IIIデバイスのI/O機能

t (x(t), y(t)), a t b (x(a), y(a)) t ( ) ( ) dy s + dt dt dt [a, b] a a t < t 1 < < t n b {(x(t i ), y(t i ))} n i ( s(t) ds ) ( ) dy dt + dt dt ( ) d

数学の基礎訓練I

= π2 6, ( ) = π 4, ( ). 1 ( ( 5) ) ( 9 1 ( ( ) ) (

N88 BASIC 0.3 C: My Documents 0.6: 0.3: (R) (G) : enterreturn : (F) BA- SIC.bas 0.8: (V) 0.9: 0.5:

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

2011de.dvi

高速データ変換

Kazutoshi Kobayashi (kobayasi kit.ac.jp)

PRECISION DIGITAL PROCESSOR DC-101

Transcription:

? FPGA FPGA FPGA : : :

? ( ) (FFT) ( ) (Localization)

? : 0. 1 2 3 0. 4 5 6 7 3 8 6 1 5 4 9 2 0. 0 5 6 0 8 8 ( )

? : LU Ax = b LU : Ax = 211 410 221 x 1 x 2 x 3 = 1 0 0 21 1 2 1 0 0 1 2 x = LUx = b 1 31 00 4 : y Ux Ly = b Ux = y (Matrix) Local ( )

? : F = 1 1 1 1 1 W W 2 W N 1 1 W 2 W 4 W 2N 2......... 1 W N 1 W 2N 2 W (N 1)(N 1) O(n 2 ) O(n log n), W n = 1

? : : CPU : L, C, R OP,.. :

? Problem Structure Operation Algorithm Hardware Topology Dynamics : : CPU :

? CPU Memory Instruction/Data Bus : : (Field Programmable Gate Array) ( )

FPGA? FPGA FPGA FPGA FPGA

FPGA : CMOS CMOS MOS A A A A NAND(A,B) B NOR(A,B) B Inverter NAND NOR

PLA X FPGA PLA (Programmable Logic Arrays) AND OR A B C D X=AB+ACD

FPGA FPGA LUT(Look Up Table) A B LUT (SRAM) NAND(A,B) AB 1 A B AB AB 1 1 OUT CMOS NAND AB 0 LUT

FPGA FPGA Logic Cell[LUT( ), FF( )] Matrix Logic Cell Logic Cell Logic Cell Logic Cell SW Logic Cell Logic Cell Logic Cell SW SW Logic Cell Logic Cell Logic Cell Logic Cell SW

FPGA 1. (Hardware Description Language) 2. 3. 4. FPGA FPGA 5. FPGA

FPGA HDL (Hardware Description Language) HDL: Verilog-HDL, VHDL, SFL : 4 Verilog module count(out, ck); output [3:0] out; input ck; reg [3:0] q; always @(posedge ck) begin q <= q+1; end assign out = q; endmodule

FPGA HDL (VCS ) module testcount; wire [3:0] out; reg ck; initial begin ck <= 0; #200 $finish; end always #10 begin ck <= ck; end count inst0 (out, ck); endmodule ck out[3] out[2] out[1] out[0]

FPGA (FPGA Express ) FPGA FPGA JTAG ROM

FPGA PCI PCI FPGA CPU Memory PCI Bridge FPGA board

FPGA FPGA 1 1 ( )

FPGA 1 FPGA CPU FPGA CPU

FPGA Web

? FPGA

integers binary codes Gray codes 0 000 000 1 001 001 2 010 011 3 011 010 4 100 110 5 101 111 6 110 101 7 111 100

1 1 2

x B=0.0010... y =0.... B x G=0.0011... y G= 0. 100... bit 3 bit 2 bit 1 bit 3 bit 2 bit 1 0 1 bit 0 bit 0 0 1/4 1/2 3/4 1 Binary code 2 0 1/4 1/2 3/4 1 Gray code

2 2 1 1 0 0 + 0 0 1 1 0 1 1 1 1 1 1 0 1 + 0 0 1 1 1 0 0 0 0 ( )

a b G G 0.11111 0.01111 Gray Code Adder.1000 c G 0.1 [0.1, 0.2]

2 1 2

Why Gray codes? 1/2 [ 1 X = 2 ε 1, 1 ] 2 + ε 2, ε 1,ε 2 > 0 X B=0. X G= 0. 100 bit 3 bit 2 bit 1 bit 3 bit 2 bit 1 0 1 bit 0 bit 0 0 1/4 1/2 3/4 1 Binary code 1 0 1/4 1/2 3/4 1 Gray code 1

x G x G X = [X, X] = [x 1, x + 1] 2 n x x : (integer), n x : (integer) x = 0. G 1 2 0 1 left 0 1/4 1/2 3/4 1 right center x G= 0.0 x G= 0. 1 x G= 0.1 1 2 1 2 1 2 0 1/4 1/2 3/4 1 0 1/4 1/2 3/4 1 0 1/4 1/2 3/4 1

a G b G = c G a G A, b G B c G C A B C A B A B C a G, b G = A, B = A B C = c G

d, d ( ) d = 2 n c A B C ( ) d = 2 n c C A B C d2 -n c A B d2 -n c A B A B, C d, d C A B conditions c G d d d 1, left 1 := P c d := 2d d := 2(d 1) d 1, right 1 := P c d := 2(d 1) d := 2d d 1 / 2 d 1 / 2, center 2 := 1, 0 d := 2(d 1 / 2 ) d := 2(d 1 / 2 ) P c = i b 1 j= c j (i b = 1 ) C

? FPGA

: : xy = 1 y = 1/x : x 2 + y 2 2 = 0, xy 1 = 0 x 4 + 2x 2 1 = 0, x 3 2x + y = 0

Ax = y x y y x a 11 a 12 a 1n a 21 a 22 a 2n.... a n1 a n2 a nn x 1 x 2. x n = y 1 y 2. y n LU L 1 u 11 u 12 u 1n 0 a 22 u 2n........ 0 0 u nn x 1 x 2. x n = l 11 0 0 l 21 l 22 0........ l n1 l n2 l nn y 1 y 2. y n x y y x

y(t) = t 0 g(t τ)x(τ)dτ x(t) y(t) y(t) x(t) Y(s) = G(s)X(s), G(s) 1 Y(s) = X(s) X(s) Y(s) Y(s) X(s)?

F(s) = = 0 0 f (t)e st dt f (t)e at e jωt dt f (t) e at (s = a + jω) :?

f (t) = L 1 [F(s)](t) = 1 2πj a+j a j F(s)e st ds s:, f (t):, F(s): = Bromwich ε FFT

1/ s f N (t) 1.5 1.0 0.5 FFT-Based square wave f N (t) 2.0 1.5 1.0 0.5 FFT-Based 1/ πt 0.0 0.0-0.5 0.0 0.5 1.0 1.5 2.0 F(s) = t 1 s(1 + e 0.1s ) -0.5 0.0 0.5 1.0 1.5 2.0 F(s) = 1 s t

F(s) G(s) t F(s) G(s) FFT-Based g N(hn) T T -1 inversion ^ N f (hn) T s T 1 t

s f (t) = L 1 [F(s)] = L 1 [s i F(s) s i ] = di dt i L 1 [ F(s) s i ] = s i + t i s domain t domain Conventional method F(s) FFT-Based inversion f (hn) N Proposed method F(s) 1 si i-th integral G(s) FFT-Based inversion g N (hn) i d dt i-th differential i ^ N f (hn)

1/ s f N (t) 1.5 1.0 0.5 FFT-Based square wave Proposed (i=4) f N (t) 2.0 1.5 1.0 0.5 FFT-Based 1/ πt Proposed (i=4) 0.0 0.0-0.5 0.0 0.5 1.0 1.5 2.0 F(s) = t 1 s(1 + e 0.1s ) -0.5 0.0 0.5 1.0 1.5 2.0 F(s) = 1 s t

? FPGA

(SFQ)

A A B B A B A B A B

: z = f (w 1 x 1 + w 2 x 2 + w n x n θ) x x x 1 2 w w w 3 3 θ x n 2 w n 1 z 1 0 f

(CML) A OUT OUT B A B CML ExOR

(Exp ) O1 O2 X1 X2 Y1 Y2

: (TDM, FDM, CDM, SDM,...) : FPGA V(t) A B C D 0 90 V(t) 180 270 A B C D

? :? A NOT B A B AND C A B OR C NOT A B 0 1 1 0 AND A B C 0 0 0 0 1 0 1 0 0 1 1 1 OR A B C 0 0 0 0 1 1 1 0 1 1 1 1

? :? CMOS : : IN OUT IN / OUT OUT / IN CMOS inverter Pass-transistor

? a + b = c. a b, c = a + b b c, a = c b

2 1 2 1 1 A B A B 0 0 1 1 A B 0 1 1 0 1 1 2 2 1 A B C A B C 0 0 0 0 1 1 1 0 1 1 1 0 A B C 0 0 1 0 1 0 1 0 0 1 1 1

AND A B AND C Z (a) (b) (c) Input Input Output Input Input Output Input Input Output A B C A C B B C A Z Z Z Z Z Z Z Z Z Z 0 0 Z 0 Z Z 0 Z Z 1 Z Z 1 1 Z 1 1 0 Z 0 0 Z Z 0 Z Z 0 0 0 0 0 Z 0 0 Z 0 1 0 0 1-0 1-1 Z Z 1 Z Z 1 Z Z 1 0 0 1 0 0 1 0 0 1 1 1 1 1 1 1 1 1

2 : X=1 X=0 (X,X)=(1,0) (X,X)=(0,1). : (X,X) = (0,0). A B A B : IN / OUT OUT / IN

Expression of the unfixed bit Two-way circuit 0, 1, Z OUT OUT open pull down (V) OUT OUT 0 0 5 1 5 0 Z 0 0

ExOR A =0 B C A=1 B C A, B and C Gate control Signal flow B B B B A A BBCC A A A A C C C C B C C ExOR B (a) (b) Input Input Output A A B B C C 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 1

AND A = 1 B C A = 0 A C C=1, A, B A A B B C C A A B B C AND C

A, B, C, S and C +. A, C, S, B. C+ C+ carry out a n b n FA(n) s n+1 sn c2 A A B B Reversible ExOR C C Reversible ExOR carry in sum S S a 1 b1 a 0 b 0 FA(1) c 1 FA(0) c 0 = 0 s 1 s 0 n-bit

2 2bit c 0 = a 0 b 0, c 1 = a 1 b 0 a 0 b 1, c 2 = a 1 b 1 (a 1 b 0 a 0 b 1 ), c 3 = a 1 b 1 a 1 b 0 a 0 b 1. A a 1 a 0 B b 1 b 0 Two-way AND a 1 b 1 a 0 b 1 a 1 b 0 a 0 b 0 Reversible ExOR 2 2 bit Reversible ExOR buffer C c 3 c 2 c 1 c 0 carry out carry out augend FA sum augend HA sum carry in addend addend a 2 b 2 FA a 1 b 2 FA HA a 2 b 1 a 0 b 2 a 0 b 1 a 0 b 0 a 1 b HA 1 FA a 2 b 0 HA a 1 b 0 c 5 c 4 c 3 c 2 c 1 c 0 3 3 bit