利用mathematica画出一个周期为2Pi的三角波
时间: 2023-09-03 08:09:31 浏览: 71
可以使用 `Piecewise` 函数来实现三角波的绘制。具体代码如下:
```
Plot[Piecewise[{{x, 0 <= Mod[x, 2 Pi] < Pi}, {-x + 2 Pi, Pi <= Mod[x, 2 Pi] < 2 Pi}}], {x, 0, 4 Pi}, Exclusions -> None]
```
该代码会绘制一个周期为2Pi的三角波,其中 `Piecewise` 函数中的两个条件分别对应了三角波的上升部分和下降部分。 `Exclusions -> None` 参数可以避免在转折点处出现间断。
相关问题
利用mathematica画出一个周期三角波
可以使用Mathematica的内置函数TriangleWave来绘制一个周期三角波。下面是一个例子:
```mathematica
Plot[TriangleWave[x], {x, -1, 1}, PlotRange -> {-1.2, 1.2},
PlotStyle -> Thick, AxesLabel -> {"x", "y"},
LabelStyle -> Directive[Bold, Medium], ImageSize -> Medium]
```
这将绘制一个周期为2的三角波,其在x=0处取最大值1,x=1处取最小值-1,x=0.5处取值0。你可以通过更改函数的参数来调整三角波的周期和振幅。
用Mathematica画出一个椅子的模型
以下是使用Mathematica绘制椅子模型的代码:
```
chair = Graphics3D[
{
(* 座位部分 *)
Cuboid[{0, 0, 0}, {1.5, 0.25, 1}],
Cuboid[{0.25, 0.25, 0.25}, {1.25, 0.5, 0.75}],
Cuboid[{0.5, 0.5, 0.5}, {1, 0.75, 0.9}],
Cuboid[{0.5, 0.5, 0.1}, {1, 0.75, 0.25}],
Cuboid[{0.5, 0.5, 0.1}, {1, 0.6, 0.15}],
Cuboid[{0.5, 0.5, 0.4}, {1, 0.6, 0.35}],
Cuboid[{0.5, 0.5, 0.6}, {1, 0.6, 0.75}],
Cuboid[{0.5, 0.5, 0.6}, {0.7, 0.25, 0.75}],
Cuboid[{0.5, 0.25, 0.5}, {0.7, 0.4, 0.75}],
(* 背部支架 *)
Cylinder[{{1.25, 0.6, 0.75}, {1.25, 1.8, 0.75}}, 0.15],
Cylinder[{{1.25, 1.8, 0.75}, {0.8, 2.2, 0.75}}, 0.15],
Cylinder[{{1.25, 1.8, 0.75}, {1.5, 1.8, 1.2}}, 0.15],
(* 背部部分 *)
Cuboid[{0.8, 2.2, 0.6}, {1.5, 2.4, 1.2}],
Cuboid[{0.8, 2.2, 0.75}, {1.5, 2.4, 0.9}],
Cuboid[{0.8, 2.4, 0.6}, {1.5, 2.5, 1.2}],
Cuboid[{0.8, 2.4, 0.6}, {1.5, 2.7, 0.75}],
(* 扶手部分 *)
Cuboid[{0, 0.25, 0}, {0.25, 2.2, 1}],
Cuboid[{0.25, 0.25, 0.2}, {0.5, 1, 0.8}],
Cuboid[{0.25, 0.25, 0.2}, {0.5, 0.75, 0.6}],
Cuboid[{0.25, 0.75, 0.3}, {0.5, 1, 0.5}],
Cuboid[{0.25, 1.2, 0.65}, {0.5, 1.6, 0.75}],
Cuboid[{0.25, 1.6, 0.5}, {0.5, 2, 0.75}],
Cuboid[{0.25, 2, 0.25}, {0.5, 2.2, 0.5}]
}
]
Show[chair, ViewPoint -> {1.3, -2.4, 2}]
```
运行代码后,将会绘制出一个3D模型的椅子。