The mobile channel is a typical random channel. When the mobile station moves at the constant rate in the path of length d and terminals X and Y, it receives signals from signal source S, as shown in Figure 1. The radio waves start from signal source S and they are respectively accepted by the mobile station at points X and Y. So the difference between the two paths traveled by these two radio waves will be:, is the time consumed by the mobile station during its movement, is angle between incident wave and X or Y. It is assumed that the X point and Y point share the same angle. The phase variation of the received signals

due to path difference will be:, where is the carrier wavelength. Thus Doppler

frequency shift can be drawn. There is a Doppler shift for each signal component arriving at antenna relative to the different angle of the movement speed [2] .

is the ith component’s angle. is the maximum Doppler frequency shift.

The multipath effect is the delayed interference effect caused by the multipath transmission phenomenon of radio propagation channel. In the actual radio waves propagation channel (including all band), there are a lot of transmission paths with different time-delays. In the case of N paths, the channel output can be represented as:

and represent the attenuation and propagation-delay associated with the Nth multipath component and they are both the function of time. The complex envelope of the receiver’s input signal caused by the large amount of scattered components is the complex Gauss process. The expect is zero and there is a direct path. The amplitude obeys the Ricean Distribution. It is assumed that the channel input is a modulated signal [3] .

The form of the complex envelop is, take (3) into (2).

However, Thus,

The complex path attenuation can be defined as, Thus,

So the complex envelop of receiver’s input is, which defines the input and output

relationship of channel as a linear time-varying systems. The impulse response is:

where expresses propagation-delayed.

The data file 1, 2 and 3 are the tested impulse function values of channels. They are a series of impulse response sequences with an increasement during each unit sampling time, and the values are plural. I separate the real and imaginary parts.

According to data file 1, the actually measured data can be obtained at the velocity of 90 km/h and in the channel sampling frequency of 200 KHz. I use the method of Fourier model tree. Five steps of the regression tree of Fourier model are composed of data preprocessing, Fourier series approximation, model tree, pre-pruning and segmentation step. These steps are introduced in detail in the article “Fast time-varying channel model research for data processing of wireless communication.” Regression tree belongs to the improved method of decision tree. Generally speaking, the decision tree requires that data segmentation should be cut into small data set until they can’t be subdivided. Decision tree is a greedy algorithm, and with the given eigenvalues to partition the data, it will make the best choice in a given time but don’t care about whether the global optimum is reached or not. Data can be cut through the tree structure for complex data model, how to carry out effective data segmentation depends on the modeling way of the leaf node.

Firstly, I make model solving on imaginary parts of data and use the Fourier model tree to get the coefficients of the segmented Fourier series [4] .

{‘spInd’: 0, ‘spVal’: matrix([[673.]]),

‘right’: [matrix([[−0.00037289]]), matrix([[−0.00074577]]), matrix([[−0.01062968]])],

‘left’: [matrix([[0.00123526]]), matrix([[0.00247053]]), matrix([[0.01351784]])]}

The piecewise impulse response time is 673, the Fourier formula of the right part is:

The Fourier formula of the left part is:

And then draw the comparison picture of between the segmented Fourier series and the actually measured imaginary part data as shown in Figure 2.

As shown in Figure 2, the red line represents the fitting data, the blue line represents the actually measured data. Though the change trend can be fitted by the Fourier series with the change of velocity, the extension becomes more around the actually measured data and the data also gradually diffuse to surrounding due to the change of velocity. The deviation rate is calculated according to formula of variance.

and represent the data and fitting data respectively.

I calculate the error of square difference, getting Dev_img = 0.0594. The large deviation data appear much less likely making the square errors decrease because the actually measured data have a number of 72000. The Fourier coefficient is solved by the model tree according to the real part data of data file 1.

[matrix([[−7.57798329e−05]]), matrix([[−0.00015156]]), matrix([[ 0.04335678]])]

The Fourier formula is:

Draw the comparison picture of between Fourier series and the actually measured real part data as shown in Figure 3.

The fitting data of Fourier series cover the actually measured data well as shown in Figure 3, and the extensions around are not many. According to formula (10), I calculate the error of square difference, obtaining Dev_real = 0.0541. According to data file 2, the actually measured data can be obtained at the velocity of 270 km/h and in the channel sampling frequency of 200 KHz. I use the method of Fourier model tree. Firstly, I make model solving on imaginary parts of data and use the Fourier model tree to get the coefficients of the segmented Fourier series.

{‘spInd’: 0, ‘spVal’: matrix([[520.]]),

‘right’: [matrix([[−0.00013017]]), matrix([[−0.00026034]]), matrix([[0.008543]])],

‘left’: [matrix([[0.00037687]]), matrix([[0.00075373]]), matrix([[0.06409131]])]}

The Fourier series are:

Draw the comparison picture of between the Fourier series and the actually measured imaginary part data as shown in Figure 4.

As shown in Figure 4, the red line represents the fitting data, the blue line represents the actually measured data. According to the formula (10), I calculate square difference error of the Fourier series fitting data, getting Dev_img = 0.0970. The square error is bigger because there are a few large deviation data points and the up and downs of curve movements do not coincide with each other also can be a reason. The Fourier coefficients are solved according to the real part data of data file 2.

[matrix([[−0.00016071]]), matrix([[−0.00032143]]), matrix([[0.03061008]])]

The Fourier series are:

Draw the comparison picture of between the Fourier series and the actually measured real part data as shown in Figure 5.

The red line represents the fitting data, the blue line represents the actually measured data. According to the formula (10), I calculate square difference error of the Fourier series fitted data, getting Dev_real = 0.0643. According to data file 3, the actually measured data can be obtained at the velocity of 450 km/h and in the channel sampling frequency of 200 KHz. I use the method of Fourier model tree. Firstly, I make model solving on imaginary parts of data and use the Fourier model tree to get the coefficients of the segmented Fourier series.

{‘spInd’: 0, ‘spVal’: matrix([[313.]]),

‘right’: [matrix([[−0.00054486]]), matrix([[−0.00108971]]), matrix([[0.00911897]])],

‘left’: [matrix([[0.00014312]]), matrix([[0.00028623]]), matrix([[0.09310624]])]}

The segmented Fourier series are:

Draw the comparison picture of between the Fourier series and the actually measured imaginary part data as shown in Figure 6.

The red line represents the fitting data, the blue line represents the actually measured data. According to the formula (10), I calculate square difference error of the Fourier series fitted data, getting Dev_img = 0.1424.

The Fourier coefficients are solved according to the real part data of data file 3.

[matrix([[−0.00035213]]), matrix([[−0.00070427]]), matrix([[0.03083286]])]

The Fourier series are:

Draw the comparison picture of between the Fourier series and the actually measured real part data as shown in Figure 7.

The red line represents the fitting data, the blue line represents the actually measured data. According to the formula (10), I calculate square difference error of the Fourier series fitted data, obtaining Dev_real = 0.0641.

Imaginary parts Dev_img and real parts Dev_real are shown in Table 1 at the different velocities.

Making an illustration of velocity and offset is shown in Figure 8.

The cyan line represents deviation rate of imaginary parts, the red line represents deviation rate of real parts. With the increase of the velocity, the deviation rate of imaginary parts obviously increases and the deviation rate of real parts acts in the same way. With accelerating of the velocity, the deviation rate between the Fourier series fitted by the Fourier model tree and the actually measured data increases and so does the error.

In the condition of the certain carrier frequency of 3 GHz, the channel sampling frequency is 200 KHz and velocity is 90 Km/h. The maximum Doppler frequency shift can be solved in the condition of the velocity 90 Km/h according to the formula of Doppler frequency shift.

By the same token, the Doppler frequency shifts can be got on the conditions that the mobile station with speeds of velocity 180 Km/h, 270 Km/h and 450 Km/h, as shown in Table 2.

The relevant parameters of channel including multipath numbers, multipath delay and multipath attenuation gain are chosen according to the typical urban model in the reference [5] ’s appendix B. Specific parameters are shown in Table 3.

Three kinds of parameters as shown in Table 3 are substituted into the time-varying channel model to be simulated

obtaining the impulse response sequence which depicts channel characteristics. The method of the Fourier model tree is used to fitting solution. Firstly, data preprocessing is made and the impulse response sequences are divided into real and imaginary parts. Then we deal with real parts and derive the coefficients of the segmented Fourier series from the model tree.

{‘spInd’: 0, ‘spVal’: matrix([[487.]]),

‘right’: [matrix([[0.04664147]]), matrix([[0.09328293]]), matrix([[−0.13159072]])],

‘left’: {‘spInd’: 0, ‘spVal’: matrix([[786.]]),

‘right’: [matrix([[0.01709217]]), matrix([[0.03418433]]), matrix([[0.01846343]])],

‘left’: [matrix([[−0.00485146]]), matrix([[−0.00970292]]), matrix([[0.14601979]])]}}

According to five steps of regression tree of Fourier model, the first demarcation point is 487, and the left branch branches again. The left second demarcation point is 786. The Fourier series in the interval [0, 487] are:

The Fourier series to the right side of the left branch’s second branching point in the interval (487, 786] are:

The Fourier series to the left side of the left branch’s second branching point in the interval (786, 1000] are:

Draw the comparison between the real part fitting and the actually measured curve as shown in Figure 9.

The blue line represents the actually measured data. The cyan line represents fitting data. The red line repre- sents fitting data. The yellow line represents fitting data. The two sections before have a certain deviation, but there is a good fitting cover when the impulse response time reaches 786. According to formula (10), getting Dev_real = 0.2947. For the processing of imaginary parts, the coefficients of the segmented Fourier series are:

{‘spInd’: 0, ’spVal’: matrix([[484.]]),

‘right’: {‘spInd’: 0, ’spVal’: matrix([[103.]]),

‘right’: [matrix([[0.07269677]]), matrix([[0.14539348]]), matrix([[−0.06412545]])],

‘left’: [matrix([[0.01438901]]), matrix([[0.02877801]]), matrix([[0.03532326]])]},

‘left’: {‘spInd’: 0, ’spVal’: matrix([[838.]]),

‘right’: [matrix([[−0.03206781]]), matrix([[−0.06413562]]), matrix([[0.08239755]])],

‘left’: [matrix([[−0.02822166]]), matrix([[−0.0564433]]), matrix([[−0.03375118]])]}}

The level-one demarcation point is 484, the secondary right demarcation point is 103, and the secondary left demarcation point is 838. The segmented Fourier series are:

Draw the comparison between the imaginary part fitting and the actually measured curve in Figure 10.

As shown in Figure 10, the blue line represents the actually measured data. The cyan line represents fitting data. The red line represents fitting data. The yellow line represents fitting data. The pink line represents fitting data. According to formula (10), the fitting square difference deviation of the imaginary parts is Dev_img = 0.1677.

The Fourier model tree of the simulation data is fitted, but the overall effect is general and the deviation value is large. There is the phenomenon of too many segmentations. The fitting of real parts is divided into three Fourier series sections. The fitting of imaginary parts is divided into five sections of the Fourier series. Too many segmentations are also the cause which introduce error.

I use the simulation model of multipath fading channel to perform the 16 QAM modulation and draw the relationship diagram of BER changing with SNR in a range from 0 to 40 db.

As shown in Figure 11, the channel is the random channel due to the characteristics of multipath time-vary- ing channel [6] . The influence of the signal transmission in it has a big fluctuation. The BER gains no much improvement at the high SNR. The BER more shows partially random features and the fluctuation is more severe. The BER value is lowered a little overall, but there are no obvious rules found in details.

I carry out the mathematical abstraction over the whole multipath time-varying channel model which makes it regarded as a common transmission channel in order to compare with the white Gaussian noise channel and compare with AGWN channel in the same condition. For the multipath time-varying channel, the abstract mathematical model usually employs the Jakes time-varying channel model.

In the formula (30),

The formula (30), (31) and (32) are Jakes model.

I use the Jakes model to generate flat Rayleigh fading channel and compare it with the AGWN channel in the same SNR condition [7] .

As shown in Figure 12, the BER of AGWN channel will obviously decrease with the increase of SNR compared with the Jakes channel, while the one of the Jakes model channel basically maintain at the same level and has no obvious change, which coincides with the characteristics of the random channel.