Python中的矩形脉冲序列 [英] rectangular pulse train in python

问题描述

我正在尝试在python中实现矩形脉冲序列.

I'm trying to implement a rectangular pulse train in python.

我搜索了scipy，没有实现的信号. http://docs.scipy.org/doc/scipy/reference/signal. html

I searched scipy and there is no signal that implements. http://docs.scipy.org/doc/scipy/reference/signal.html

在matlab中有一个名为pulstran的信号: http://es.mathworks.com/help/signal/ref/pulstran. html

In matlab there is a signal named pulstran: http://es.mathworks.com/help/signal/ref/pulstran.html

matlab中的代码示例如下:

An example of code in matlab would be like this:

T=10; %Period
D=5; %Duration
N=10; %Number of pulses

x=linspace(0,T*N,10000);
d=[0:T:T*N];
y=pulstran(x,d,'rectpuls',D);
plot(x,y);
ylim([-1,2]);

我如何在python中实现此信号?

How i could implement this signal in python?

谢谢.

推荐答案

如果您只是在寻找周期性脉冲序列，例如您所给的示例，那么这是一个脉冲序列，该脉冲序列打开5个周期，然后关闭5个周期:

If you're looking for just periodic pulse trains, like the example you gave - here's a pulse train that is on for 5 cycles then off for five cycles:

N = 100 # sample count
P = 10  # period
D = 5   # width of pulse
sig = np.arange(N) % P < D

给予

plot(sig)

您可以在此处用linspace替换np.arange(N).请注意，这与您的代码等效不同，因为脉冲未居中.

You can replace np.arange(N) with your linspace here. Note this is not equivalent to your code, as the pulses are not centered.

这是一个完全可配置的脉冲序列:

And here's a fully configurable pulse train:

def rect(T):
    """create a centered rectangular pulse of width $T"""
    return lambda t: (-T/2 <= t) & (t < T/2)

def pulse_train(t, at, shape):
    """create a train of pulses over $t at times $at and shape $shape"""
    return np.sum(shape(t - at[:,np.newaxis]), axis=0)

sig = pulse_train(
    t=np.arange(100),              # time domain
    at=np.array([0, 10, 40, 80]),  # times of pulses
    shape=rect(10)                 # shape of pulse
)

给予:

我认为这是matlab的pulsetran函数比在python中单行实现更令人困惑的情况之一，这可能就是scipy不提供它的原因.

I think this is one of those cases where matlab's pulsetran function is more confusing than the one-line implementation of it in python, which is possibly why scipy does not provide it.

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