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1allegroallegro   数式の入力方法

[tex:x]
[tex:x^2]
[tex:x^{xy}]
[tex:e_2]
[tex:e_{xy}]
[tex:(x+y)]
[tex:\{x+y\}^2]
[tex:x\gt y]
[tex:x\lt y]
[tex:\int_{t0}^{t1} adt=v]

x

x^2

x^{xy}

e_2

e_{xy}

(x+y)

\{x+y\}^2

x\gt y

x\lt y

\int_{t0}^{t1} adt=v

とか

id:allegroトラックバックできるのかな

返信2006/05/29 02:02:52

2allegroallegro  

- Total power
-- [tex:TP = \int_{0}^{\infty} P(f)df]
- Average frequency
-- [tex:\bar{f} = \int_{0}^{\infty} fP(f)df/TP]
- Normalized second order central moment
-- [tex:NU2 = \int_{0}^{\infty}(f-\bar{f})^2 P(f)df/TP]
- Normalized third order central moment
-- [tex:NU3 = \int_{0}^{\infty}(f-\bar{f})^3 P(f)df/TP]
  • Total power
    • TP = \int_{0}^{\infty} P(f)df
  • Average frequency
    • \bar{f} = \int_{0}^{\infty} fP(f)df/TP
  • Normalized second order central moment
    • NU2 = \int_{0}^{\infty}(f-\bar{f})^2 P(f)df/TP
  • Normalized third order central moment
    • NU3 = \int_{0}^{\infty}(f-\bar{f})^3 P(f)df/TP
返信2006/07/22 08:43:56

3allegroallegro   TeXコマンド。

no title

奥村先生の資料。

返信2006/10/13 17:20:44

4allegroallegro  

 \phi^m(r) = \frac{1}{L-m+1}\bigsum^{L-m+1}_{i}\ln\frac{N^m(i)}{L-m+1}

 ApEn = \phi^m(r) - \phi^{m+1}(r)

[tex: \phi^m(r) = \frac{1}{L-m+1}\bigsum^{L-m+1}_{i}\ln\frac{N^m(i)}{L-m+1}]

[tex: ApEn = \phi^m(r) - \phi^{m+1}(r)]
返信2006/07/27 10:04:57

5allegroallegro  

[tex:D(t,f) = \int ^{\infty} _{-\infty} \int ^{\infty} _{-\infty} \int ^{\infty} _{-\infty} 
s(t'+\tau / 2)s*(t'-\tau /2)g(\theta, \tau) 
e^{-j2\pi\theta(t'-t)}e^{-j 2 \pi f \tau} d\theta dt' d\tau]

D(t,f) = \int ^{\infty} _{-\infty} \int ^{\infty} _{-\infty} \int ^{\infty} _{-\infty} s(t'+\tau / 2)s*(t'-\tau /2)g(\theta, \tau) e^{-j2\pi\theta(t'-t)}e^{-j 2 \pi f \tau} d\theta dt' d\tau

返信2006/07/28 06:27:23

6allegroallegro  

[tex:g_{spwv}(\theta,\tau) = \pi e^{-T^2\pi^2\theta^2} e^{-B^2\pi^2\tau^2}]

[tex:g_{wv}(\theta,\tau) = 1]

[tex:A(\theta, \tau) = \int ^{\infty} _{-\infty} s(t+\tau/2) s^*(t-\tau/2)e^{^j2\pi\theta t}dt]

g_{spwv}(\theta,\tau) = \pi e^{-T^2\pi^2\theta^2} e^{-B^2\pi^2\tau^2}

g_{wv}(\theta,\tau) = 1

A(\theta, \tau) = \int ^{\infty} _{-\infty} s(t+\tau/2) s^*(t-\tau/2)e^{^j2\pi\theta t}dt

返信2006/08/08 10:35:50

7allegroallegro   便利サイト

web上で入力した数式をPNGで落とせます。

返信2007/08/31 13:49:45