Choice of Fs: satisfy Nyquist condition. Optional Fs = 60Hz；;

Selection of sampling points n: In order to accurately measure the frequency in the problem, each frequency component must be an integer multiple of fs/n ... Therefore, in the case of Fs=60Hz, n can be 12, 24, 36, 48.

When Fs=60Hz, N= 12, Fs/N=5Hz, X 1, X2 and X3 should appear at k = 3, 4 and 6 respectively;

When Fs=60Hz, N=24: Fs/N=2.5Hz, X 1, X2 and X3 should appear at k = 6, 8 and12 respectively;

The function f(ByVal x is Single) is Single.

f = x + Cos(x)

End function

The function df(ByVal x is Single) is Single.

df = 1 - Sin(x)

End function

Private subcommand 1_Click ()

tmp =(f(2 * Atn( 1))-f(0))/(2 * Atn( 1))

Form 1. Line (Form 1. ScaleLeft，tmp)-(Form 1。 ScaleWidth，tmp)，vbRed

For i = 0 to 2 * Atn( 1) step size 0.00 1

Form 1. Line (i，df(i))-(i + 0.00 1，df(i + 0.00 1))，vbBlue

Extended data:

The bandwidth of a communication system indicates its ability to transmit these different frequency components. In a structured cabling system, the unit of bandwidth is usually expressed in MHz.

The nominal bandwidth value of the super Class 5 wiring is 100MHz. Suppose a simple binary transmission is applied. Code, then theoretically, the maximum information transmission rate can be calculated by Nyquist equation:

C= 2 W Log 2 M, where w is bandwidth (HZ). M is the number of signaling units.

This leads to the theoretical information capacity of 2x 108 bits per second, which is 200Mb. In fact, this value will be reduced due to the influence of crosstalk and attenuation.

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