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 $X(k) H(k)$.
$$x_h(n)=\sin \left(\frac{2 \pi}{4} n\right) \leftrightarrow{0,-j 2,0, j 2}$$

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