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## 电气工程代写|数字系统设计作业代写Digital System Design代考|Receiver Antenna Gain

The gain of the receiver antenna (in $\mathrm{dB}$ ) is calculated in the same way as the transmitter antenna gain:
$$G_r=10 \log \left[n(\pi(D) / \lambda)^2\right]$$
where
\begin{aligned} n &=\text { efficiency factor }<1 \ D &=\text { diameter of the parabolic dish } \ \lambda &=\text { wavelength } \end{aligned}
The antenna receiver gain is achieved by focusing the received antenna pattern in a specific direction toward the transmitter, which reduces the amount of noise that is received. There are basically two ways to increase the signal-to-noise ratio (SNR): increasing the signal level or reducing the noise. The transmitter increases the signal level by focusing the available power in the direction of the receiver. The receiver reduces the overall noise level by focusing the antenna beam toward the transmitter.

The receiver antenna is not required to have the same antenna as the transmitter. The receiver can use an omnidirectional antenna and receive transmissions from a transmitter that uses a parabolic dish antenna, or the transmitter can be an omnidirectional antenna and the receiver can use a parabolic dish. However, if this is a direct line-of-sight system, the antennas should have the same polarization. For example, if a transmitter is using a vertically polarized antenna, the receiver should have a vertically polarized antenna. An exception to this rule is that if the system is using the ionosphere to bounce the signal for maximum range using lower frequencies, then the polarization can be reversed by the reflection off the ionosphere. For example, if a transmitter is using a horizontally polarized antenna, the optimal receiver antenna may be vertically polarized if the reflection off the ionosphere causes reversal of the polarization. The gain of the antenna is a direct gain in the communications link: a $1 \mathrm{~dB}$ gain equals a $1 \mathrm{~dB}$ improvement in the link.

## 电气工程代写|数字系统设计作业代写Digital System Design代考|Receiver Component Losses

Any components between the antenna and LNA will reduce the SNR of the system. For example, often a limiter is placed in the line between the antenna and the LNA to protect the LNA from damage by high-power signals. This can be included in the NF of the receiver or viewed as a loss of the signal. Both methods are used in the industry, with the same end results, since the first method increases the noise and the second method decreases the signal level, producing the same SNR results for the receiver.

However, since this loss does not add noise to the system but only attenuates the signal, a more straightforward approach would be to treat it as a loss in signal level and calculate the NF separately. The noise before and after the lossy devices is the same, since the temperature before and after the device is the same. Only the signal level is attenuated. The noise on the front end of the receiver before the LNA is equal to $k T B$, where
\begin{aligned} &k=\text { Boltzmann constant }\left(1.38 \times 10^{-23} \mathrm{~J} /{ }^\rho \mathrm{K}\right) \ &T=\text { nominal temperature }\left(290^{\circ} \mathrm{K}\right) \ &B=\text { bandwidth } \end{aligned}
For a $1 \mathrm{~Hz}$ bandwidth,
$$k T B=1.38 \times 10^{-23} \mathrm{~J} /{ }^0 \mathrm{~K} * 290^{\circ} \mathrm{K} * 1 \mathrm{~Hz}=4.002 \times 10^{-21} \mathrm{~W}=4.002 \times 10^{-18} \mathrm{~mW}$$

Converting to $\mathrm{dBm}$,
$$10 \log (k T B)=10 \log \left(4.002 \times 10^{-18}\right)=-174 \mathrm{dBm}$$
A convenient way to calculate the $k T B$ noise is to use the aforementioned $1 \mathrm{~Hz}$ bandwidth number and simply take $10 \log (B)$ and add to it. For example, for a $1 \mathrm{MHz}$ bandwidth,
\begin{aligned} -174 \mathrm{dBm}+10 \log (1 \mathrm{MHz}) \mathrm{dB} &=-174 \mathrm{dBm}+10 \log \left(10^6 \mathrm{~Hz}\right) \mathrm{dB} \ &=-174 \mathrm{dBm}+60 \mathrm{~dB}=-114 \mathrm{dBm} \end{aligned}
The LNA amplifies the input signal and noise, and during this amplification process additional noise is present at the output, mainly due to the active transistors in the amplifier. This additional noise is referred to as the NF of the LNA and increases the overall noise floor, which reduces the SNR. Since this is a change in noise level, NF is in $\mathrm{dB}$. The resultant noise floor is equal to $k T B F$, where $F$ is the increase in the noise floor due to the LNA. The noise factor $\mathrm{F}$ is the increase in noise due to the amplifier, and the noise figure $\mathrm{NF}$ is the increase in noise converted to $\mathrm{dB}$. The rest of the components of the receiver increase the $k T B F$ noise floor, but this contribution is relatively small.

# 数字系统设计代考

## 电气工程代写|数字系统设计作业代写Digital System Design代考|Receiver Antenna Gain

$$G_r=10 \log \left[n(\pi(D) / \lambda)^2\right]$$

$$n=\text { efficiency factor }<1 D \quad=\text { diameter of the parabolic dish } \lambda=\text { wavelength }$$

## 电气工程代写|数字系统设计作业代写Digital System Design代考|Receiver Component Losses

$k=$ Boltzmann constant $\left(1.38 \times 10^{-23} \mathrm{~J} /{ }^\rho \mathrm{K}\right) \quad T=$ nominal temperature $\left(290^{\circ} \mathrm{K}\right) B=$ bandwidth

$$k T B=1.38 \times 10^{-23} \mathrm{~J} /{ }^0 \mathrm{~K} * 290^{\circ} \mathrm{K} * 1 \mathrm{~Hz}=4.002 \times 10^{-21} \mathrm{~W}=4.002 \times 10^{-18} \mathrm{~mW}$$

$$10 \log (k T B)=10 \log \left(4.002 \times 10^{-18}\right)=-174 \mathrm{dBm}$$

$$-174 \mathrm{dBm}+10 \log (1 \mathrm{MHz}) \mathrm{dB}=-174 \mathrm{dBm}+10 \log \left(10^6 \mathrm{~Hz}\right) \mathrm{dB} \quad=-174 \mathrm{dBm}+60 \mathrm{~dB}=-114 \mathrm{dBm}$$
LNA 放大输入信号和噪声，在此放大过程中，输出端会出现额外的噪声，这主要是由于放大器中的有源晶体管造成的。这种额外的噪声称为 LNA 的 $N F$ ，它会增加整体本底噪声，从而降低 SNR。由于这是噪声水平的变化，因此 NF 在dB. 所得的本底噪声等于 $k T B F$ ，在哪里 $F$ 是由于 LNA 导致的 本底噪声的增加。噪声系数 $\mathrm{F}$ 是由于放大器引起的噪声增加，噪声系数 $\mathrm{NF}$ 是噪声的增加转换为 $\mathrm{dB}$. 接收器的其余组件增加了 $k T B F$ 本底噪声，但这 种贡献相对较小。

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