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OFDM 16 lecture 2-ofdm and the DFT

2022-07-27 05:04:00 【Bai Xiaosheng in Ming Dynasty】

Reference resources ：

Iain Explains Signals, Systems, and Digital Comms

Preface ：

Shows how Orthogonal Frequency Division Multiplexing (OFDM) is implemented with a Discrete Fourier Transform (DFT), and how it relates to single carrier digital communications.

Catalog

1： Single channel transmission model

2： OFDM Transmission model

3： DFT

One Single channel transmission model

Single signal transmission model , I was talking about modulation Yes, I have .

The sender ：

The original bit Flow to 0,1, after QPSK Equal modulation ,0->1,1->-1

Then it is modulated into high-frequency analog signal by cosine oscillator

Add Gaussian white noise , Send the data through the channel

The receiving party

Multiply the cosine signal of the same frequency , You can decode t The moment bit Of $\hat{x}$

Two OFDM Transmission model

2.1 Two ways

As above, a channel of quadrature signal is added x_2

x(t)=x_1s_1(t)+x_2s_2(t)

s: Cosine oscillator , Modulate digital signals into analog signals

$\hat{x_1}=\int_{0}^Tx(t)s_1(t)dt$

$=x_1\int _{0}^{T}s_1^2(t)dt+0$

Empathy

$\hat{x_2}=x_2\int s_2^2(t)dt$

Orthogonality is the integration of 0
example ：
, $f_1=\frac{1}{T}$
$s_1(t)=cos2\pi f_1t$
$s_2(t)=cos2\pi f_2t$
$\int_{0}^T cos 2\pi f_1t* cos2\pi f_2tdt$
$=\frac{1}{2}\int cos2\pi(f_1+f_2)tdt+\frac{1}{2}\int cos2\pi(f_2-f_1)tdt+$
$=\frac{1}{2}\int cos 6\pi f_1tdt+\frac{1}{2}\int cos 2\pi f_1tdt$
$=\frac{1}{2} \frac{sin 6\pi f_1 t}{6\pi}|_{0}^T+\frac{1}{2} \frac{sin 2\pi f_1 t}{6\pi}|_{0}^T$

2.2 multiple

In practice , There can be no useless big numbers .

It can be used IDFT（ Inverse discrete Fourier transform ）Inverse discrete Fourier transform

$x(t)=\sum_{k=0}^{N-1}x_k e^{j(\frac{2\pi}{T}k)t}$

Every time channel The corresponding signal frequency is $\frac{k}{T}$ , With others channel Signal quadrature of

Another form

The Fourier transform
$F(w)=\int_{-\infty}^{\infty}f(t)e^{-jwt}dt$
Inverse Fourier transform
$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} F(w)e^{jwt}dw$
Principle of inverse Fourier transform ：

$\int \int f(x)e^{-jwx}dx e^{jwt}dw$
$=\int \begin{bmatrix} \int e^{-jw(x-t)}dw \end{bmatrix}f(x)dx$
$=\int 2\pi \delta (x-t)f(x)dx$
$=2\pi f(t)$
be
$f(t)=\frac{1}{2\pi}\int F(w)e^{jwt}dw$