Description: 线型PCM与A律和u律对数PCM相互转换的C++代码。-A linear PCM with the law and the law on several u PCM conversion of C code. Platform: |
Size: 9415 |
Author:周金喜 |
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Description: 线型PCM与A律和u律对数PCM相互转换的C++代码。-A linear PCM with the law and the law on several u PCM conversion of C code. Platform: |
Size: 9216 |
Author:周金喜 |
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Description: PCM的A律U律实现代码：16位线性量化的数据转换成8位非线性的PCM数据（13折线法）-U of A law PCM realize Legal Code: 16-bit linear quantization of the data into eight non-linear PCM data (13 line method) Platform: |
Size: 2048 |
Author:黄翠翠 |
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Description: pcm编码 有关线性、A律、mu律的pcm编码的matlab实现-pcm encoded on the linear, A law, mu law encoded pcm achieve matlab Platform: |
Size: 1603584 |
Author:jcl |
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Description: 完成pcm 16位线性与pcma律转换的几种源码！-Complete pcm 16-bit linear conversion with the law several pcma source! Platform: |
Size: 16384 |
Author:weiyabing |
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Description: G.711使用64Kbps的带宽，可将14bits转换成8bits。目前G.711有两个编码方式，A-law以及Mu-law。-G.711 defines two main compression algorithms, the µ -law algorithm (used in North America & Japan) and A-law algorithm (used in Europe and the rest of the world). Both are logarithmic, but A-law was specifically designed to be simpler for a computer to process. The standard also defines a sequence of repeating code values which defines the power level of 0 dB.
The µ -law and A-law algorithms encode 14-bit and 13-bit signed linear PCM samples (respectively) to logarithmic 8-bit samples. Thus, the G.711 encoder will create a 64 kbit/s bitstream for a signal sampled at 8 kHz.[5]
G.711 μ-law tends to give more resolution to higher range signals while G.711 A-law provides more quantization levels at lower signal levels. When using μ-law G.711 in networks where suppression of the all 0 character signal is required, the character signal corresponding to negative input values between decision values numbers 127 and 128 should be 00000010 and the value at the decoder output is- Platform: |
Size: 915456 |
Author:coco |
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Description: 一。产生长度为500的零均值，单位方差的高斯随机变量序列，用均匀pcm的方法用16电平进行量化：1）求所得的SQNR，该序列的前5个值，相应的量化值和相应的码字。2）画出量化误差（定义为输入值和量化值之间的差），同时 画出量化值作为输入值的函数的图。3）用128量化电平数重做2）题， 比较结果。
二。产生一个长度为500，按N(O,1)分布的随机变量序列，分别用16，128量化电平数和u=255的u律非线性进行量化，画出每种情况下量化器的误差和输入-输出关系，并求SQNR.
三。长度为500的非平稳序列a由两部分组成：前20个样本是按照均值为零和方差为400的高斯随机变量产生的，其余480个样本是根据均值和方差为1的高斯随机变量产生的，对这个序列分别用均匀pcm和非均匀pcm方法进行128电平量化，试比较两种情况下所得到的SQNR。
-One. Have a length of 500 zero mean, unit variance Gaussian random variables with uniform pcm way to quantify the level with a 16: 1) Find the income SQNR, the sequence of the first 5 values, the corresponding quantization value and the corresponding codeword. 2) Draw the quantization error (defined as the input value and quantify the difference between the values), and draw quantitative values as a function of input graph. 3) redo 128 the number of quantization level 2) title, compare the results.
II. Produce a length of 500, according to N (O, 1) distributed random variables, respectively 16,128 and the number of quantization level of u u = 255 to quantify non-linear law, draw each case quantizer error and input- output relationships, and seek SQNR.
III. Length of the non-stationary series is a 500 consists of two parts: the first 20 samples in accordance with zero mean and variance of 400 generated Gaussian random variable, and the remaining 480 samples are based o Platform: |
Size: 4096 |
Author:sun |
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