e^{j\theta} = \cos(\theta) + j\sin(\theta) \qquad\qquad \mbox{(Euler's Identity)}
e^{j\theta} = \cos(\theta) + j\sin(\theta) \qquad\qquad \mbox{(Euler's Identity)}
e^{j\theta} = \cos(\theta) + j\sin(\theta) \qquad\qquad \mbox{(Euler's Identity)}
a^{n_1} a^{n_2} = a^{n_1 + n_2}
e^{j\theta} = \cos(\theta) + j\sin(\theta) \qquad\qquad \mbox{(Euler's Identity)}
a^{n_1} a^{n_2} = a^{n_1 + n_2}
\left(a^{n_1}\right)^{n_2} = a^{n_1 n_2}
a^0 a = a^0 a^1 = a^{0+1} = a^1 = a
a^0 a = a^0 a^1 = a^{0+1} = a^1 = a
a^0 a = a
a^0 a = a^0 a^1 = a^{0+1} = a^1 = a
a^0 a = a
{a^0 = 1.}
a^{-1} \cdot a = a^{-1} a^1 = a^{-1+1} = a^0 = 1.
a^{-1} \cdot a = a^{-1} a^1 = a^{-1+1} = a^0 = 1.
{a^{-1} = \frac{1}{a}.}
a^{-1} \cdot a = a^{-1} a^1 = a^{-1+1} = a^0 = 1.
{a^{-1} = \frac{1}{a}.}
{a^{-M} = \frac{1}{a^M}}
A rational number is a real number that can be expressed as a ratio of two finite integers:
A rational number is a real number that can be expressed as a ratio of two finite integers:
\displaystyle x = \frac{L}{M}, \quad L\in\mathbb{Z},\quad M\in\mathbb{Z}
\displaystyle a^x = a^{L/M} = \left(a^{\frac{1}{M}}\right)^L.
\displaystyle a^x = a^{L/M} = \left(a^{\frac{1}{M}}\right)^L. Thus, the only thing new is a^{1/M} . Since
\displaystyle a^x = a^{L/M} = \left(a^{\frac{1}{M}}\right)^L. Thus, the only thing new is a^{1/M} . Since
\displaystyle \left(a^{\frac{1}{M}}\right)^M = a^{\frac{M}{M}} = a
we see that a^{1/M} is the M-th root of a. This is sometimes written
we see that a^{1/M} is the M-th root of a. This is sometimes written
{a^{\frac{1}{M}} = \sqrt[M]{a}.}
f(x)=e^x is one of the simplest imaginable infinite series:
f(x)=e^x is one of the simplest imaginable infinite series: e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots
f(x)=e^x is one of the simplest imaginable infinite series: e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots
The simplicity comes about because f^{(n)}(0)=1 for all n
f(x)=e^x is one of the simplest imaginable infinite series: e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots
The simplicity comes about because f^{(n)}(0)=1 for all n
e^{j\theta} = \sum_{n=0}^\infty \frac{(j\theta)^n}{n!} = 1 + j\theta - \frac{\theta^2}{2} - j\frac{\theta^3}{3!} + \cdots \,.
f(x)=e^x is one of the simplest imaginable infinite series: e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots
The simplicity comes about because f^{(n)}(0)=1 for all n
e^{j\theta} = \sum_{n=0}^\infty \frac{(j\theta)^n}{n!} = 1 + j\theta - \frac{\theta^2}{2} - j\frac{\theta^3}{3!} + \cdots \,.
\text{Re} \left(e^{j\theta}\right) = 1 - \frac{\theta^2}{2} + \frac{\theta^4}{4!} - \cdots
f(x)=e^x is one of the simplest imaginable infinite series: e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots
The simplicity comes about because f^{(n)}(0)=1 for all n
e^{j\theta} = \sum_{n=0}^\infty \frac{(j\theta)^n}{n!} = 1 + j\theta - \frac{\theta^2}{2} - j\frac{\theta^3}{3!} + \cdots \,.
\text{Re} \left(e^{j\theta}\right) = 1 - \frac{\theta^2}{2} + \frac{\theta^4}{4!} - \cdots
\text{Im} \left(e^{j\theta}\right) = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\
\displaystyle z = r e^{j\theta}
From the trig identity \sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B) , we have
From the trig identity \sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B) , we have
\displaystyle x(t) = A_x e^{j\omega_x t}
t_n = nT = \mbox{$n$th sampling instant (sec)} \omega_k = k\Omega = \mbox{$k$th frequency sample (rad/sec)} T = 1/f_s = \mbox{time sampling interval (sec)} \Omega = 2\pi f_s/N = \mbox{frequency sampling interval (rad/sec)}
\displaystyle e^{-j\omega_k t_n} = \cos(\omega_k t_n) - j \sin(\omega_k t_n)
X(\omega_k) , the DFT at frequency \omega_k , is a measure of the amplitude and phase.
Numpy
arrays, just as vectors.DFT decompose singals up into complex exponentials.
DFT decompose singals up into complex exponentials.
DFT decompose singals up into complex exponentials.
The transformation matrix W can be defined as W = \left( \frac{\omega^{jk}}{ \sqrt{N} } \right )_{\left(j,k=0,\ldots ,N-1\right)} ,
where {\displaystyle \omega =e^{-2\pi i/N}}
\omega = e^{-{\frac{2\pi i}{8}}} = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}
(Note that \omega^{8 + n} = \omega^{n}.)
$ git clone https://github.com/sbme-tutorials/sbe309-week3-demo.git
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