## 电子工程代写|电子系统工程代写Digital Systems Engineering代考|Distortionless LTI Systems

Let us take sinusoidal radiated wave at frequency $\omega$ being received at a distance $d$ from the transmitter. The electric field function between the two transmitting and receiving ends will be of the form
$$E_{T}=(A / z) \sin (\omega t-\beta z)$$

where $A$ is a constant, $\beta=2 \pi / \lambda$ and $z$ is the direction of propagation. The $A / z$ term says that the field decays as $1 / z$ as the wave moves away from the transmitting station, while the argument $\omega t-\beta z$ ensures that the wave is a travelling wave.
On receiving the signal, at the receiving end, the electric field (and therefore the voltage received) is of the form
$$E_{R}=B \sin (\omega t-\theta)$$
where $B=A / d$ and $\theta=\beta d$. We notice that the amplitude is not a function of frequency but let us examine the nature of $\theta$ (let $f$ be the frequency in Hertz):
\begin{aligned} \theta &=\beta d \ &=(2 \pi / \lambda) d \ &=(2 \pi f / c) d \quad(c \text { is the velocity of the wave; } c=f \lambda) \ &=(\omega / c) d \ &=\omega(d / c) \ &=\omega t_{0} \quad\left(\text { where } t_{0}=d / c\right) \end{aligned}
And we know that the wave is undistorted. So what are the conditions of distortionless transmission?

1. The amplitude of the wave is not a function of frequency. In this case, it is $A / d$.
2. The phase of the wave is a linear function of frequency. In this case, it is $\omega(d / c)$.
These results are shown in Fig. 2.40. The figure shows the amplitude and phase characteristics of the signal with frequency $\omega$. In fact, this is the FT of the impulse response of the ideal distortionless channel. If we compare with Fig. 2.31, then we find that both figures are identical and that Fig. 2.40 represents the Fourier transform of
$$B \delta\left(t-t_{0}\right)$$

## 电子工程代写|电子系统工程代写Digital Systems Engineering代考|Signal Distortion

As signals pass through non-ideal linear systems or channels, we can model them through their impulse responses or equivalently their frequency responses. The two of them are related by
$$h(t) \Leftrightarrow H(\omega)$$
where $h(t)$ is the impulse response and $H(\omega)$ is the frequency response of the system or channel.
If we observe the output of such a system, then
$$y(t)=h(t) * x(t)$$
or
$$Y(\omega)=H(\omega) X(\omega)$$
which implies that since $X(\omega)$ is multiplied by $H(\omega)$, the magnitude of and phase of $Y$ is changed from that of $X$ by multiplication of the factor $H(\omega)$. That is
$$|Y(\omega)|=|H(\omega)||X(\omega)|$$
We can see from this equation, (2.162), that the system introduces no distortion of the magnitude spectrum if only
$$|H(\omega)|=k, \quad \text { a constant }$$
Similarly, from Eq. (2.161), we can see that
$$\angle Y(\omega)=\angle H(\omega)+\angle X(\omega)$$
so the output phase too suffers a distortion. From our earlier discussion, it is clear that the phase of the output signal is not distorted if the phase
$$\angle H(\omega)=-k \omega, \quad k, \text { a constant }$$
In short, every system introduces some sort of distortion in the signal, and the question is whether the distortion is acceptable or unacceptable.

## 电子工程代写|电子系统工程代写Digital Systems Engineering代考|Distortionless LTI Systems

$$E_{T}=(A / z) \sin (\omega t-\beta z)$$

$$E_{R}=B \sin (\omega t-\theta)$$

$$\theta=\beta d \quad=(2 \pi / \lambda) d=(2 \pi f / c) d \quad(c \text { is the velocity of the wave; } c=f \lambda) \quad=(\omega / c) d=\omega(d / c)$$

1. 波的幅度不是频率的函数。在这种情况下，它是 $A / d$.
2. 波的相位是频率的线性函数。在这种情况下，它是 $\omega(d / c)$.
这些结果如图 $2.40$ 所示。该图显示了信号随频率变化的幅值和相位特性 $\omega$. 实际上，这就是理想无失真 里叶变换
$$B \delta\left(t-t_{0}\right)$$

## 电子工程代写|电子系统工程代写Digital Systems Engineering代考|Signal Distortion

$$h(t) \Leftrightarrow H(\omega)$$

$$y(t)=h(t) * x(t)$$

$$Y(\omega)=H(\omega) X(\omega)$$

$$|Y(\omega)|=|H(\omega)||X(\omega)|$$

$$|H(\omega)|=k, \quad \text { a constant }$$

$$\angle Y(\omega)=\angle H(\omega)+\angle X(\omega)$$

$$\angle H(\omega)=-k \omega, \quad k, \text { a constant }$$

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