# 数学代写|傅里叶分析代写Fourier analysis代考|AMTH246

Quadrature amplitude modulation (QAM) transmits 2 signals using carriers of the same frequency but in phase quadrature. Figure $5.17$ shows the quadrature amplitude modulation and demodulation. The carrier frequency is $\cos \left(\omega_c t\right)$. Phase shifters are used to delay the carrier frequency to get $\sin \left(\omega_c t\right)$. The in-phase component of the message signal is multiplied by $\cos \left(\omega_c t\right)$. The quadrature component is multiplied by $\sin \left(\omega_c(t)\right.$. The sum of the two product signals is the modulator output
$$r e(t) \cos \left(\omega_c t\right)+i m(t) \sin \left(\omega_c t\right)$$
The modulated signal is transmitted using double sideband transmission. A similar process, in the demodulator at the right side, governed by the following equations produces the sum of the message signal and other high-frequency components.
\begin{aligned} x_1(t) &=2 y_{\text {qam }} \cos \left(\omega_c t\right) \ &=2\left(r e(t) \cos \left(\omega_c t\right)+i m(t) \sin \left(\omega_c t\right)\right) \cos \left(\omega_c t\right) \ &=r e(t)+r e(t) \cos \left(2 \omega_c t\right)+i m(t) \sin \left(2 \omega_c t\right) \ x_2(t) &=2 y_{q a m} \sin \left(\omega_c t\right) \ &=2\left(r e(t) \cos \left(\omega_c t\right)+i m(t) \sin \left(\omega_c t\right)\right) \sin \left(\omega_c t\right) \ &=i m(t)-i m(t) \cos \left(2 \omega_c t\right)+r e(t) \sin \left(2 \omega_c t\right) \end{aligned}
The message signal is obtained by lowpass filtering.

## 数学代写|傅里叶分析代写Fourier analysis代考|Hilbert Transform

A complex signal whose imaginary part is the Hilbert transform of its real part is called the analytic signal. An analytic signal has no negative frequency components. Hilbert transformer shifts the phase of every spectral component by $\frac{\pi}{2}$ radians. Typical applications are single sideband modulation in communication systems and sampling of bandpass signals.
The frequency response of the Hilbert transformer, for an even $N$, is
H(k)=\left{\begin{aligned} -j \text { for } k &=1,2, \ldots, \frac{N}{2}-1 \ 0 \text { for } k &=0, \frac{N}{2} \ j \text { for } k &=\frac{N}{2}+1, \frac{N}{2}+2, \ldots, N-1 \end{aligned}\right.
Basically, it is an all-pass filter that imparts a $\pm 90^{\circ}$ phase shift on the input signal. However, in practice, it is designed over the required bandwidth of a given input signal.

The IDFT of $H(k)$ is the impulse response. As the frequency response is imaginary and odd-symmetric, the impulse response is real and odd.
$$h(n)=\frac{1}{N} \sum_{k=0}^{N-1} H(k) e^{j \frac{2 \pi}{N} k n}=\frac{j}{N} \sum_{k=0}^{N-1} H(k) \sin \left(\frac{2 \pi}{N} k n\right)$$
For example,
\begin{aligned} x(n) &=\cos \left(\frac{2 \pi}{4} n\right) \leftrightarrow X(k)={0,2,0,2}, \ H(k) &={0,-j, 0, j}, \quad X(k) H(k)={0,-j 2,0, j 2} \end{aligned}
The Hilbert transform $x_h(n)$ of $x(n)$ is the IDFT of $X(k) H(k)$.
$$x_h(n)=\sin \left(\frac{2 \pi}{4} n\right) \leftrightarrow{0,-j 2,0, j 2}$$
Now, the analytic signal $x_a(n)$ corresponding to $x(n)$ is the complex signal with $x(n)$ as its real part and $x_h(n)$ as its imaginary part.

## 数学代写|傅里叶分析代写傅里叶分析代考|正交振幅调制

.

$$r e(t) \cos \left(\omega_c t\right)+i m(t) \sin \left(\omega_c t\right)$$

\begin{aligned} x_1(t) &=2 y_{\text {qam }} \cos \left(\omega_c t\right) \ &=2\left(r e(t) \cos \left(\omega_c t\right)+i m(t) \sin \left(\omega_c t\right)\right) \cos \left(\omega_c t\right) \ &=r e(t)+r e(t) \cos \left(2 \omega_c t\right)+i m(t) \sin \left(2 \omega_c t\right) \ x_2(t) &=2 y_{q a m} \sin \left(\omega_c t\right) \ &=2\left(r e(t) \cos \left(\omega_c t\right)+i m(t) \sin \left(\omega_c t\right)\right) \sin \left(\omega_c t\right) \ &=i m(t)-i m(t) \cos \left(2 \omega_c t\right)+r e(t) \sin \left(2 \omega_c t\right) \end{aligned}

## 数学代写|傅里叶分析代写傅立叶分析代考|希尔伯特变换

H(k)=\left{\begin{aligned} -j \text { for } k &=1,2, \ldots, \frac{N}{2}-1 \ 0 \text { for } k &=0, \frac{N}{2} \ j \text { for } k &=\frac{N}{2}+1, \frac{N}{2}+2, \ldots, N-1 \end{aligned}\right.

\begin{aligned} x(n) &=\cos \left(\frac{2 \pi}{4} n\right) \leftrightarrow X(k)={0,2,0,2}, \ H(k) &={0,-j, 0, j}, \quad X(k) H(k)={0,-j 2,0, j 2} \end{aligned}
Hilbert变换 $x_h(n)$ 的 $x(n)$ IDFT是多少 $X(k) H(k)$.
$$x_h(n)=\sin \left(\frac{2 \pi}{4} n\right) \leftrightarrow{0,-j 2,0, j 2}$$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: