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radar:exercises [2018/06/01 14:17] – [Exercise 2] iatcoradar:exercises [2026/04/28 15:13] (current) – external edit 127.0.0.1
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->it is ok  you can start--- //[[webmaster@localhost|DokuWiki Administrator]] 2018/04/19 11:23// +======EXERCISES====== 
- +
->  Where do the figures come from? Please cite the document as decribed in [[:start|Welcome!]] --- //[[webmaster@localhost|DokuWiki Administrator]] 2018/05/03 16:16// +
- +
-> please use caption for tables and figures  ---  //[[webmaster@localhost|DokuWiki Administrator]] 2018/05/03 16:13// +
- +
- +
-> please use numbered equation (if they are not in-line with the text ---  //[[webmaster@localhost|DokuWiki Administrator]] 2018/05/03 16:13// +
- +
-=====EXERCISES==== +
 =====Radar measurement===== =====Radar measurement=====
  
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  $\tau_A = 0.5\;\mu s$\\  $\tau_A = 0.5\;\mu s$\\
  $\tau_B = 3\;\mu s$\\   $\tau_B = 3\;\mu s$\\ 
- $D = 2\;m$  (Antenna diameter)\\  + $D = 2\;m$  (Antenna diameter)\\ \\  
-\\ +Compute the beam width $\theta_B$, the maximum gain $G_{max}$, the maximum unambiguous range $R_{max}$, the maximum unambiguous velocity $v_{max}$ and finally the duty cycle $d$ for both cases A and B. 
-Compute the beam width $\theta_B$, the maximum gain $G_{max}$, the maximum range non ambiguous $R_{max}$, the maximum velocity non ambiguous $v_{max}$ and finally the duty cycle $d$ for both cases A and B. +
 \\ \\  \\ \\ 
 **Solution** \\  **Solution** \\ 
-Considering a Gaussian beam and using the approximation for $D>>\lambda$.\\ \\ +Considering a Gaussian beam and using the approximation for $D>>\lambda$.
 \begin{equation} \begin{equation}
 \lambda=\frac{c}{f}=0.032\;m  \lambda=\frac{c}{f}=0.032\;m 
-\end{equation}\\+\end{equation}
 \begin{equation} \begin{equation}
 \theta_B=\varphi_B=65\frac{\lambda}{D}=1.04°\;(0.018\;rad) \theta_B=\varphi_B=65\frac{\lambda}{D}=1.04°\;(0.018\;rad)
-\end{equation}\\+\end{equation}
 \begin{equation} \begin{equation}
 G_{max}=10\log_{10}(\frac{26000}{\theta_B^2})=48.3\;dB G_{max}=10\log_{10}(\frac{26000}{\theta_B^2})=48.3\;dB
-\end{equation}\\ \\ +\end{equation}
 It is also possible to compute the gain using the effective area, $G_{max}=4\pi\frac{A_e}{\lambda^2}$\\  It is also possible to compute the gain using the effective area, $G_{max}=4\pi\frac{A_e}{\lambda^2}$\\ 
-Assuming an aperture efficiency equal to $\rho_A=0.6$ in order to compute $A_e=\rho_AA$, the value of the gain obtained is quite similar to that already calculated.\\  +Assuming an aperture efficiency equal to $\rho_A=0.6$ in order to compute $A_e=\rho_AA$, the value of the gain obtained is quite similar to that already calculated.
-\\ +
 \begin{equation} \begin{equation}
 R_{max,A}=\frac{c}{2PRF_A}=125\;Km R_{max,A}=\frac{c}{2PRF_A}=125\;Km
-\end{equation}\\+\end{equation}
 \begin{equation} \begin{equation}
 R_{max,B}=\frac{c}{2PRF_B}=625\;Km R_{max,B}=\frac{c}{2PRF_B}=625\;Km
-\end{equation}\\+\end{equation}
 \begin{equation} \begin{equation}
 |v_{max,A}|=\frac{\lambda PRF_A}{4}=9.6\;m/s |v_{max,A}|=\frac{\lambda PRF_A}{4}=9.6\;m/s
-\end{equation}\\+\end{equation}
 \begin{equation} \begin{equation}
 |v_{max,B}|=\frac{\lambda PRF_B}{4}=1.92\;m/s |v_{max,B}|=\frac{\lambda PRF_B}{4}=1.92\;m/s
-\end{equation}\\+\end{equation}
 \begin{equation} \begin{equation}
 d_A=\frac{\tau_A}{PRT_A}=6\times10^{-4} d_A=\frac{\tau_A}{PRT_A}=6\times10^{-4}
-\end{equation}\\+\end{equation}
 \begin{equation} \begin{equation}
 d_B=\frac{\tau_B}{PRT_B}=7.2\times10^{-4} d_B=\frac{\tau_B}{PRT_B}=7.2\times10^{-4}
-\end{equation}\\  +\end{equation} 
-\\  +How we can see in case B the pulse is large, the range is bigger and the radar is ambiguous in velocity. Instead in the case A the radar is ambiguous in range but the resolution in distance is higher, as we can see in the following calculations:
-How we can see in case B the pulse is large, the range is bigger and the radar is ambiguous in velocity. Instead in the case A the radar is ambiguous in range but the resolution in distance is higher, as we can see in the following calculations.\\ \\ +
 \begin{equation} \begin{equation}
 \Delta R_A=\frac{c\; \tau_A}{2}=74\;m \Delta R_A=\frac{c\; \tau_A}{2}=74\;m
-\end{equation}\\+\end{equation}
 \begin{equation} \begin{equation}
 \Delta R_B=\frac{c\; \tau_B}{2}=450\;m \Delta R_B=\frac{c\; \tau_B}{2}=450\;m
-\end{equation}\\ +\end{equation} 
-\\  +We can also compute the radar cell, in which the radar can not distinguish between 2 different targets.
-We can also compute the radar cell, in which the radar can not distinguish between 2 different targets.\\ +
 \begin{equation} \begin{equation}
 RC_A= R_{max,A}\; \theta_B \;R_{max,A} \; \varphi_B \; \Delta R_A = 1.2\times 10^{12}\;m^3 RC_A= R_{max,A}\; \theta_B \;R_{max,A} \; \varphi_B \; \Delta R_A = 1.2\times 10^{12}\;m^3
-\end{equation}\\+\end{equation}
 \begin{equation} \begin{equation}
 RC_B= R_{max,B} \; \theta_B \; R_{max,B} \; \varphi_B \; \Delta R_B = 1.9\times 10^{14}\;m^3 RC_B= R_{max,B} \; \theta_B \; R_{max,B} \; \varphi_B \; \Delta R_B = 1.9\times 10^{14}\;m^3
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 <figure> <figure>
 {{ :media:radar_cell.jpg?400 |}}  {{ :media:radar_cell.jpg?400 |}} 
-<caption>Resolution cell [(cite:TTR)]</caption> +<caption>Resolution cell. [(cite:TTR)]</caption> 
 </figure> </figure>
 ====Exercise 2==== ====Exercise 2====
 A Marine Radar with a fan beam used to detect ships, rotates at a given angular velocity $\omega$. The radar antenna is placed at a height $h_r$ from the ground, while a target placed on the horizon has a height equal to $h_v$. \\ A Marine Radar with a fan beam used to detect ships, rotates at a given angular velocity $\omega$. The radar antenna is placed at a height $h_r$ from the ground, while a target placed on the horizon has a height equal to $h_v$. \\
-Compute the dwell time $t_D$, the number of pulses transmitted to a target $N$, the maximum range non ambiguous $R_{max}$ and the gain for coherent integration $G_{INT}$.\\ +Compute the dwell time $t_D$, the number of pulses transmitted to a target $N$, the maximum unambiguous range $R_{max}$ and the gain for coherent integration $G_{INT}$.\\ 
 $\omega=40\;rpm$\\ $\omega=40\;rpm$\\
 $PRT=1\;ms$\\  $PRT=1\;ms$\\ 
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 \begin{equation} \begin{equation}
 t_D=\frac{\theta_B}{6 \omega}=0.01\;s t_D=\frac{\theta_B}{6 \omega}=0.01\;s
-\end{equation}\\ \\  +\end{equation}
 \begin{equation} \begin{equation}
 N=PRF\;t_D=\frac{t_D}{PRT}=10 N=PRF\;t_D=\frac{t_D}{PRT}=10
-\end{equation}\\ +\end{equation}
-\\+
 In the hypothesis that the noise present in each echo received is a thermal noise so it can be represented with a Gaussian p.d.f. with 0 mean and a known variance, the gain for a coherent integration is exactly equal to the number of pulses if the target is fixed (frequency doppler $f_D=0$).  In the hypothesis that the noise present in each echo received is a thermal noise so it can be represented with a Gaussian p.d.f. with 0 mean and a known variance, the gain for a coherent integration is exactly equal to the number of pulses if the target is fixed (frequency doppler $f_D=0$). 
 \begin{equation} \begin{equation}
 G_{INT}=N=10=10\;dB G_{INT}=N=10=10\;dB
-\end{equation}\\  +\end{equation} 
-We can compute the radar horizon assuming an equivalent radius of the earth $r_e = 8500\;Km$\\ +We can compute the radar horizon assuming an equivalent radius of the earth $r_e = 8500\;Km$:
 \begin{equation} \begin{equation}
 R_{max,H} \simeq \sqrt{2 r_e}(\sqrt{h_r}+\sqrt{h_v}) \simeq 71.29\;Km  R_{max,H} \simeq \sqrt{2 r_e}(\sqrt{h_r}+\sqrt{h_v}) \simeq 71.29\;Km 
-\end{equation}\\  +\end{equation} 
-But the maximum range non ambiguous has a bigger value\\  +But the maximum unambiguous range has a bigger value: 
 \begin{equation} \begin{equation}
 R_{max}=\frac{c\;PRT}{2}=150\;Km R_{max}=\frac{c\;PRT}{2}=150\;Km
-\end{equation}\\ +\end{equation} 
-Given that $R_{max} >> R_{max,H}$ so the radar sees at a distance longer than necessary, we can reduce the $PRT$ in order to reduce the maximum range non ambiguous with the advantage of reducing the radar cell and therefore increasing the radar resolution. +Given that $R_{max} >> R_{max,H}$ so the radar sees at a distance longer than necessary, we can reduce the $PRT$ in order to reduce the maximum unambiguous range with the advantage of reducing the radar cell and therefore increasing the radar resolution. 
  
 +====Exercise 3====
 +A radar works in X band, $f_0=9\;GHz$, and it has to see up to a distance of $R_{lim}=100\;Km$. The radar also must be able to measure speed up to $v_{lim}=2000\;km/h$.\\ 
 +  * Compute the range velocity product $RVP$ and check if the radar is not ambiguous in distance and in velocity.\\ 
 +  * Compute the $PRF$ in order to avoid the ambiguity in doppler.
 +  * Assuming that the dimension of the antenna is $D=0.8\;m$ and the radar changes the working frequency form from the X to the Ku band, $f_1=13.5\;GHz$. Check if the radar range changes and how much.
 +  * Taking into account that the earth is curved and the radar antenna is placed at a height of $h_r=900\;m$ instead the targets have a height of $h_t=0\;m$, compute the radar horizon. \\ \\ 
 +
 +**Solution**\\
 +The $RVP$ can be computed in this way:
 +\begin{equation}
 +RVP=\frac{c\lambda_0 }{8}=1.25\times 10^{6}\;m^2/s
 +\end{equation}
 +Instead of computing the range velocity product with the desired requirements:
 +\begin{equation}
 +RVP=v_{lim}\;R_{lim} = 5.55\times 10^{-7}\; m^2/s
 +\end{equation}  
 +The value obtained is small than the previous, this means that with the wavelength we have is not possible to reach the desired limits, so the radar is ambiguous both in velocity and in range.\\
 +In order to avoid the ambiguity in doppler the $PRF$ have to be:
 +\begin{equation}
 +PRF\geqslant \frac{4v_{lim}}{\lambda_0 } \simeq 67\;KHz
 +\end{equation}  
 +This is an high $PRF$ for a radar. \\ 
 +Using this $PRF$ the maximum unambiguous distance of the radar is:
 +\begin{equation}
 +R_{max}=\frac{c}{2PRF} = \frac{c\;PRT}{2}\simeq 2.24\;Km
 +\end{equation}
 +In order to understand what happens at the maximum range of the radar if we increase the working frequency up to $f_1=13.5\;GHz$, we have to find the relationship between the maximum range and the wavelength. \\ 
 +Using the radar equation:
 +\begin{equation}
 +R_{max}=\frac{P_t\;G^2\;\lambda ^2 \;RCS}{(4\pi)^3\;S_{min}}
 +\end{equation}
 +Where the gain is equal to $G=4\pi A/\lambda ^2$, so now neglecting the other terms we can compute the proportionality between $R_{max}$ and $\lambda$ for the two considered frequency.  
 +\begin{equation}
 +R_{max,0}\propto \frac{1}{\sqrt{\lambda_0}}=5.5
 +\end{equation}
 +\begin{equation}
 +R_{max,1}\propto \frac{1}{\sqrt{\lambda_1}}=6.7
 +\end{equation}
 +\begin{equation}
 +\frac{R_{max,1}}{R_{max,0}}=1.22
 +\end{equation}
 +Increasing the frequency the range increase by a factor of 1.22 due to the fact that the dimension of the antenna is fixed.\\ 
 +Placing the radar antenna at a height of 900 m, the radar horizon $d$ can be computed using an equivalent earth radius of $r_e=8500\;Km$.
 +\begin{equation}
 +d\simeq \sqrt{2r_e}(\sqrt{h_r}+\sqrt{h_t})=123.7\;Km
 +\end{equation}
 =====Radar Equation===== =====Radar Equation=====
  
 ====Exercise 1==== ====Exercise 1====
-Compute the power received $P_r$ for given target which is located at a distance $D$ from the radar ad have a given Radar Cross Section $RCS$. Knowing the power transmitted $P_t$, the gain $G$, the system temperature $T_s$ and the working frequency $f_0$ of the radar. Assuming that the radar works in a monostatic configuration so the gain for the receiving and transmitting antenna is the same. Finally compute the the maximum range $R_{max}$ of the radar knowing the equivalent noise band $B$ and the minimum signal to noise ratio that is associated to the minimum detectable signal $SNR_{min}$. \\ +Compute the power received $P_r$ for given target which is located at a distance $D$ from the radar ad have a given Radar Cross Section $RCS$. Knowing the power transmitted $P_t$, the gain $G$, the system temperature $T_s$ and the working frequency $f_0$ of the radar. Assuming that the radar works in a monostatic configuration so the gain for the receiving and transmitting antenna is the same. Finally, compute the maximum range $R_{max}$ of the radar knowing the equivalent noise band $B$ and the minimum signal to noise ratio that is associated to the minimum detectable signal $SNR_{min}$. \\ 
 $P_t = 10\;KW$\\ $P_t = 10\;KW$\\
 $RCS=5\;m^2$\\  $RCS=5\;m^2$\\ 
Line 120: Line 152:
 $SNR_{min}=13\;dB$\\  \\  $SNR_{min}=13\;dB$\\  \\ 
 **Solution**\\  **Solution**\\ 
-We can compute the power received using the deterministic radar equation\\ +We can compute the power received using the deterministic radar equation: 
 \begin{equation} \begin{equation}
 P_r=\frac{P_t\;G^2\;\lambda^2\;RCS}{(4\pi)^3\;D^4} P_r=\frac{P_t\;G^2\;\lambda^2\;RCS}{(4\pi)^3\;D^4}
-\end{equation}\\  +\end{equation}  
 where $\lambda=c/f_0=0.23\;m$\\  where $\lambda=c/f_0=0.23\;m$\\ 
-For the calculation of $P_r$ is better to use the dB method:\\+To compute $P_r$ is better to use the dB method:\\
 <table> <table>
 ^Par.         ^N       ^D       ^ ^Par.         ^N       ^D       ^
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 |             ^98.23   ^237.53  ^ |             ^98.23   ^237.53  ^
 ^TOT          ^  -139.3         ^^ ^TOT          ^  -139.3         ^^
-<caption>Calculation of the power received in dB</caption>+<caption>Calculation of the power received in dB.</caption>
 </table> </table>
 \begin{equation} \begin{equation}
 P_r\simeq-139.3\;dBW=-109.3\;dBm=1.17\times10^{-11}\;mW P_r\simeq-139.3\;dBW=-109.3\;dBm=1.17\times10^{-11}\;mW
-\end{equation}\\  +\end{equation} 
-In order to compute the maximum range we assume that the system temperature has been taken at the antenna exit and also we assume a total losses equal to 1 (no losses).\\ +In order to compute the maximum range we assume that the system temperature has been taken at the antenna exit and also we assume a total losses equal to 1 (no losses). 
 \begin{equation} \begin{equation}
 R_{max}^4=\frac{P_t\;G^2\;\lambda^2\;RCS}{(4\pi)^3\;S_{min}^4} R_{max}^4=\frac{P_t\;G^2\;\lambda^2\;RCS}{(4\pi)^3\;S_{min}^4}
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 |             ^98.23   ^-99.64  ^ |             ^98.23   ^-99.64  ^
 ^TOT          ^  197.87         ^^ ^TOT          ^  197.87         ^^
-<caption>Calculation of the maximum range $4 R_{max}$ in dB</caption>+<caption>Calculation of the maximum range $4 R_{max}$ in dB.</caption>
 </table> </table>
 \begin{equation} \begin{equation}
 R_{max}=\frac{197.87}{4}\;dBmeters=88.51\;Km R_{max}=\frac{197.87}{4}\;dBmeters=88.51\;Km
-\end{equation}\\   +\end{equation} 
-   +  
-=====RCS and Pulse Integration=====   +====Exercise 2==== 
-====Exercise 1==== +A surveillance radar with a beam width $\theta_B$ and that rotates at the angular seed $\omega$, has the characteristic to produce 180 false alarms every minute. Knowing the parameters showed below, compute the probability of false alarm $P_{fa}$, the ratio between the threshold and the standard deviation of the noise process $V_T/\sigma$, probability of detection $P_d$ in case of Swerling 1, the number of radar cells $N$ and finally the number of false alarme in one rotation $n_{fa}$\\ 
-A surveillance radar with a beam width $\theta_B$ and that rotates at the angular seed $\omega$, has the characteristic to produce 180 false alarms every minute. Knowing the parameters showed below compute the probability of false alarm $P_{fa}$, the ratio between the threshold and the standard deviation of the noise process $V_T/\sigma$, probability of detection $P_d$ in case of Swerling 1, the number of radar cells $N$ and finally the number of false allarme in one rotation $n_{fa}$\\ +
 $B=2\;MHz$\\  $B=2\;MHz$\\ 
 $SNR=20\;dB$\\  $SNR=20\;dB$\\ 
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 $\omega=30\;rpm$\\ \\  $\omega=30\;rpm$\\ \\ 
 **Solution**\\  **Solution**\\ 
-Remembering that the probability of false allarme is given by the ratio between the average duration of a false alarm ($\simeq1/B$) and the average time between two false alarms+Remembering that the probability of false allarme is given by the ratio between the average duration of a false alarm ($\simeq1/B$) and the average time between two false alarms:
 <figure> <figure>
 {{ :media:prob_false_alarm.jpg?400 |}}  {{ :media:prob_false_alarm.jpg?400 |}} 
-<caption>Voltage trend and false alarms [(cite:TTR>Gaspare Galati (2009), Teoria e Tecnica Radar.)] </caption> +<caption>Voltage trend and false alarms. [(cite:TTR>Gaspare Galati (2009), Teoria e Tecnica Radar.)] </caption> 
-</figure>\\ +</figure>
 \begin{equation} \begin{equation}
 P_{fa}=\frac{t_{fa}}{T_{fa}}\simeq \frac{1}{T_{fa}\;B}=\frac{1}{\frac{60}{180}\;2\times 10^6}=1.5\times 10^{-6} P_{fa}=\frac{t_{fa}}{T_{fa}}\simeq \frac{1}{T_{fa}\;B}=\frac{1}{\frac{60}{180}\;2\times 10^6}=1.5\times 10^{-6}
-\end{equation}\\  +\end{equation}  
 \begin{equation} \begin{equation}
 \frac{V_T}{\sigma}=\sqrt{-2\;ln(P_{fa})}=5.17 \frac{V_T}{\sigma}=\sqrt{-2\;ln(P_{fa})}=5.17
-\end{equation}\\  +\end{equation}
 This means that the threshold has to be 5.17 times bigger than the noise in order to have the probability of false alarm already calculated.  This means that the threshold has to be 5.17 times bigger than the noise in order to have the probability of false alarm already calculated. 
-Considering the case of SW1 so assuming that the echo has Gaussian statistics and the envelope of the signals it's Rayleigh;\\ +Considering the case of SW1 so assuming that the echo has Gaussian statistics and the envelope of the signals it's Rayleigh: 
 \begin{equation} \begin{equation}
 1+SNR=\frac{ln(P_{fa})}{ln(P_d)} 1+SNR=\frac{ln(P_{fa})}{ln(P_d)}
-\end{equation}\\ +\end{equation}
 from which  from which 
 \begin{equation} \begin{equation}
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 \end{equation} \end{equation}
 (87% can be considered a good value for surveillance radar)\\   (87% can be considered a good value for surveillance radar)\\  
-In order to compute the number of radar cell in one rotation we have to know the number of radar cell in a given direction $N_P$ and the number of position occupied by the radar in azimuth in one rotation $N_{\theta}$.\\+In order to compute the number of radar cells in one rotation we have to know the number of radar cells in the beam $N_P$ and the number of position occupied by the radar in azimuth in one rotation $N_{\theta}$.
 \begin{equation} \begin{equation}
-N_P=\frac{PRT}{\tau}=\frac{1}{\tau\;PRT}= 4000 +N_P=\frac{PRT}{\tau}=\frac{1}{\tau\;PRF}= 4000 
-\end{equation}\\ +\end{equation}
 \begin{equation} \begin{equation}
 N_{\theta}=360/1=360 N_{\theta}=360/1=360
-\end{equation}\\  +\end{equation}  
-Now we can compute the number of cells: \\+Now we can compute the number of cells: 
 \begin{equation} \begin{equation}
 N=N_P\;N_{\theta}=144\times10^4 N=N_P\;N_{\theta}=144\times10^4
-\end{equation}\\  +\end{equation} 
-The number of false alarms in one scan are:\\ +The number of false alarms in one scan are:
 \begin{equation} \begin{equation}
 n_{fa}=N\;P_{fa}=2.16 n_{fa}=N\;P_{fa}=2.16
-\end{equation}\\  +\end{equation} 
-This means that in one scan there are at least 2 cells were there is a false alarm, this value can be good for surveillance radar bat can be a problem for defense radar. +This means that in one scan there are at least 2 cells where there is a false alarm, this value can be good for surveillance radar but can be a problem for defence radar. 
  
 =====System and Antenna Temperature===== =====System and Antenna Temperature=====
 ====Exercise 1==== ====Exercise 1====
-Compute the system temperature of the illustrated component chain in point A, knowing the antenna temperature $T_A=132\;K$, the losses of the RF connection $L_{RF}=3\;dB$, the gain $G_{LNA}=300\;dB$ and the noise figure $F_{LNA}=1.5\;dB$ of the low noise amplifier.\\ \\  +Compute the system temperature in the point A of the illustrated component chain (Figure {{ref>chaincomp}}) , knowing the antenna temperature $T_A=132\;K$, the losses of the RF connection $L_{RF}=3\;dB$, the gain $G_{LNA}=300\;dB$ and the noise figure $F_{LNA}=1.5\;dB$ of the low noise amplifier.\\  
-<figure>+<figure chaincomp>
 {{ :media:sys_temp_ex1.jpg?300 |}}\\ {{ :media:sys_temp_ex1.jpg?300 |}}\\
-<caption>Radar components chain</caption>+<caption>Radar components chain.</caption>
 </figure> </figure>
 **Solution**\\  **Solution**\\ 
Line 233: Line 264:
 T_{RFe}=(L_{RF}-1)\;T_0 T_{RFe}=(L_{RF}-1)\;T_0
 \end{equation} \end{equation}
-while that of the amplifier is given by:\\ +while that of the amplifier is given by:
 \begin{equation} \begin{equation}
 T_{LNAe}=(F-1)\;T_0 T_{LNAe}=(F-1)\;T_0
 \end{equation} \end{equation}
-The total system noise temperature is equal to:\\ +The total system noise temperature is equal to: 
 \begin{equation} \begin{equation}
-T_{sys}=T_A+T_{RFe}+T_{LNAe}\;L_{RF}=661\;K\simeq 28.2\;dB+T_{sys}=T_A+T_{RFe}+T_{LNAe}\;L_{RF}=661\;K\simeq 28.2\;dBK
 \end{equation} \end{equation}
  
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 Compute the real noise temperature of an antenna taking into account the antenna losses $L_A=1.5\;dB$, the sky brightness noise temperature $T_{sky}=20\;K$, the ground noise temperature $T_{GND}=290\;K$ and the power fraction that illuminates the ground $\alpha=0.2$.  \\ \\  Compute the real noise temperature of an antenna taking into account the antenna losses $L_A=1.5\;dB$, the sky brightness noise temperature $T_{sky}=20\;K$, the ground noise temperature $T_{GND}=290\;K$ and the power fraction that illuminates the ground $\alpha=0.2$.  \\ \\ 
 **Solution**\\  **Solution**\\ 
-Considering the Figure {{ref>realant}} that represents a real antenna decomposed in a ideal antenna without losses and a block called L which symbolize the antenna losses, we want to compute the antenna noise temperature $T'_A$ after the block L.+Considering the Figure {{ref>realant}} that represents a real antenna decomposed in an ideal antenna without losses and a block called L which symbolize the antenna losses, we want to compute the antenna noise temperature $T'_A$ after the block L.
 <figure realant> <figure realant>
 {{ :media:ant_temp_ex2.jpg?200 |}} {{ :media:ant_temp_ex2.jpg?200 |}}
-<caption>Schema of real antenna</caption>+<caption>Schema of real antenna.</caption>
 </figure> </figure>
 \begin{equation} \begin{equation}
 T'_A=\frac{T_A + T_L}{L_A} T'_A=\frac{T_A + T_L}{L_A}
 \end{equation} \end{equation}
-where $T_L=(L_A - 1)T_0$ is the equivalent temperature of the block L, while the ideal antenna temperature considering the noise coming both from sky and from ground is equal to:+where $T_L=(L_A - 1)T_0$ is the equivalent temperature of the block L, while the ideal antenna temperature considering the noise coming both from the sky and from the ground is equal to:
 \begin{equation} \begin{equation}
 T_A=(1 - \alpha )T_{sky} + \alpha \;T_{GND} T_A=(1 - \alpha )T_{sky} + \alpha \;T_{GND}
Line 264: Line 295:
 =====System Temperature and Radar Detection===== =====System Temperature and Radar Detection=====
 ====Exercise 1==== ====Exercise 1====
-Considering the represented radar reception subsystem with: $T_A=104\;K$, $L_{RF}=1.4\;dB$, $G_{LNA}=30\;dB$, $F_{LNA}=2\;dB$ and the losses introduced by the mixer $L_{MIX}=0.8\;dB$. \\ \\  +Consider the Figure {{ref>recrad}} that represents radar reception subsystem with: $T_A=104\;K$, $L_{RF}=1.4\;dB$, $G_{LNA}=30\;dB$, $F_{LNA}=2\;dB$ and the losses introduced by the mixer $L_{MIX}=0.8\;dB$. \\  
-<figure> +<figure recrad> 
-{{ :media:rad_detect_ex1.jpg?350 }} \\  +{{ :media:rad_detect_ex1.jpg?350 }} 
-<caption>Radar reception subsystem</caption>+<caption>Radar reception subsystem.</caption>
 </figure> </figure>
-Assuming that the radar has the following characteristics:\\ +Assuming than that the radar has the following characteristics:\\ 
 $P_T=600\;KW$\\ $P_T=600\;KW$\\
 $f_0=1.5\;GHz$\\  $f_0=1.5\;GHz$\\ 
Line 277: Line 308:
 $L_P^2=1.5\;dB$ (two-way loss per antenna beam)\\ $L_P^2=1.5\;dB$ (two-way loss per antenna beam)\\
 $RCS=5\;m^2$\\  $RCS=5\;m^2$\\ 
-$SNR_{min}=13.2\;dB$\\ +$SNR_{min}=13.2\;dB$\\ \\ 
 Compute the maximum range $R_{max}$ for the following cases: fixed target, fluctuating target for both SW1 and SW2, and finally in case of coherent integration with $N=10$ pulses and a losses factor $L_{INT}=0.5\;dB$.\\ \\  Compute the maximum range $R_{max}$ for the following cases: fixed target, fluctuating target for both SW1 and SW2, and finally in case of coherent integration with $N=10$ pulses and a losses factor $L_{INT}=0.5\;dB$.\\ \\ 
 **Solution**\\ **Solution**\\
Line 288: Line 319:
 \frac{T_{sys}}{T_0}=1.546=1.89\;dB \frac{T_{sys}}{T_0}=1.546=1.89\;dB
 \end{equation} \end{equation}
-Using the Blake's formula and assuming the two-way loss for propagation negligible:+Using the Blake formula and assuming the two-way loss for propagation negligible:
 \begin{equation} \begin{equation}
 R_{max}^4=\frac{P_T\;G^2\;\lambda^2\;RCS}{(4\pi)^3\;SNR_{min}\;K\;T_{sys}\;B_{eq}\;L_T\;L_P^2} R_{max}^4=\frac{P_T\;G^2\;\lambda^2\;RCS}{(4\pi)^3\;SNR_{min}\;K\;T_{sys}\;B_{eq}\;L_T\;L_P^2}
Line 307: Line 338:
 |             ^116.79   ^-89.44  ^ |             ^116.79   ^-89.44  ^
 ^TOT          ^  206.23         ^^ ^TOT          ^  206.23         ^^
-<caption>Calculation of the maximum range $4 R_{max}$ in dB</caption>+<caption>Calculation of the maximum range $4 R_{max}$ in dB.</caption>
 </table> </table>
-The maximum range in case of fixed target is equal to:+The maximum range in case of a fixed target is equal to:
 \begin{equation} \begin{equation}
 R_{max}=\frac{206.23}{4}\;dBmeters=143.2\;Km R_{max}=\frac{206.23}{4}\;dBmeters=143.2\;Km
 \end{equation} \end{equation}
-For SW1 and SW2 case it's necessary to derive how much we have to increase the minimum signal to noise ration for a single pulse in order to have the same probability of detection $P_D$ as in the non fluctuating case. The probability of detection can be derived from the Marcum function assuming a probability of false alarm $P_{fa}=10^{-6}$ and using the $SNR_{min}$ in case of fixed target that we already have. \\ +For SW1 and SW2 case it's necessary to derive how much we have to increase the minimum signal to noise ration for a single pulse in order to have the same probability of detection $P_D$ as in the non fluctuating case. The probability of detection can be derived from the Marcum function assuming a probability of false alarm $P_{fa}=10^{-6}$ and using the $SNR_{min}$ in case of a fixed target that we already have. \\ 
 <figure> <figure>
 {{ :media:marcum_function.jpg?500 |}} {{ :media:marcum_function.jpg?500 |}}
-<caption>Marcum's function [(cite:TTR)]</caption>+<caption>Marcum's function. [(cite:TTR)]</caption>
 </figure> </figure>
 <figure sw1_2> <figure sw1_2>
 {{ :media:snr_sw1_2.jpg?350 |}} {{ :media:snr_sw1_2.jpg?350 |}}
-<caption>Increase of the SNR in order to detect a fluctuating target respect to a fixed targe [(cite:TTR)]</caption>+<caption>Increase of the SNR in order to detect a fluctuating target respect to a fixed targe. [(cite:TTR)]</caption>
 </figure> </figure>
-As we can see in the Figure {{ref>sw1_2}} for SW1 and SW2 and a $P_D=0.9$ it's necessary to increase the minimum signal to noise ratio of 7.5 dB, so now $SNR_{min,SW1/2}=20.7\;dB$. Changing now the $SNR_{min}$ in Table {{ref>tab3}} the maximum range is:\\ +As we can see in the Figure {{ref>sw1_2}} for SW1 and SW2 and a $P_D=0.9$ it's necessary to increase the minimum signal to noise ratio of 7.5 dB, so now $SNR_{min,SW1/2}=20.7\;dB$. Changing now the $SNR_{min}$ in Table {{ref>tab3}} the maximum range is: 
 \begin{equation} \begin{equation}
 R_{max}=49.68\;dBmeters=92.9\;Km R_{max}=49.68\;dBmeters=92.9\;Km
Line 346: Line 377:
 \end{equation} \end{equation}
 How we expected the relatively ranges are increased. How we expected the relatively ranges are increased.
 +
 ====Exercise 2==== ====Exercise 2====
-Considering the EX. 1 of //System Temperature and Radar Detection// compute the maximum range in case of fixed target with coherent integration assuming that the two-way loss for propagation $L_{prop}=0.006\;dB/Km$ due to the rain.+Considering the EX. 1 of //System Temperature and Radar Detection// compute the maximum range in case of a fixed target with coherent integration assuming that the two-way loss for propagation $L_{prop}=0.006\;dB/Km$ due to the rain.
 \\ \\  \\ \\ 
 **Solution**\\  **Solution**\\ 
Line 370: Line 402:
 R_2=\frac{R_0}{\sqrt[4]{L_1}}= 213.8\;Km R_2=\frac{R_0}{\sqrt[4]{L_1}}= 213.8\;Km
 \end{equation}  \end{equation} 
-Now the losses was underestimated, we have to iterate this process until the maximum range converges.\\+Now the loss was underestimated, we have to iterate this process until the maximum range converges.\\
 In general at the 3/4 attempt the value can be acceptable, the value obtained at the fourth attempt is: In general at the 3/4 attempt the value can be acceptable, the value obtained at the fourth attempt is:
 \begin{equation} \begin{equation}
Line 378: Line 410:
 =====Optimum Filter===== =====Optimum Filter=====
 ====Exercise 1==== ====Exercise 1====
-A radar with $PRF=400\;Hz$ looks to a deterministic target with frequency doppler $f_D=800\;Hz$. Assuming that there is a Gaussian clatter that have $f_c=0\;Hz$ (this is a mean value) and standard deviation $\delta_c=40\;Hz$, design the optimum processor computing the coefficient of the filter $K$ and the improving factor using 2 pulses. +A radar with $PRF=400\;Hz$ looks to a deterministic target with frequency doppler $f_D=800\;Hz$. Assuming that there is a Gaussian clutter that has $f_c=0\;Hz$ (this is a mean value) and standard deviation $\delta_c=40\;Hz$. Design the optimum processor computing the coefficient of the filter $K$ and the improving factor using 2 pulses. 
-Remember that the correlation coefficient for a Gaussian clatter is: $\rho=e^{-2\pi^2\;\delta_c^2\;T^2}$.+Remember that the correlation coefficient for a Gaussian clutter is: $\rho=e^{-2\pi^2\;\delta_c^2\;T^2}$.
  
 **Solution**\\  **Solution**\\ 
 <figure> <figure>
 {{ :media:fir_ex1.jpg?300 |}}\\  {{ :media:fir_ex1.jpg?300 |}}\\ 
-<caption>FIR filter</caption>+<caption>FIR filter.</caption>
 </figure> </figure>
 Assuming $\underline{Z}$ as the input of the FIR filter the output will be: Assuming $\underline{Z}$ as the input of the FIR filter the output will be:
Line 420: Line 452:
 \end{bmatrix} = \frac{1}{\delta^2\;(1+\rho)} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} \end{bmatrix} = \frac{1}{\delta^2\;(1+\rho)} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix}
 \end{equation} \end{equation}
-Before computing the improving factor we have to compute the output of the filter:  +The output of the filter will be:  
 \begin{equation} \begin{equation}
 O = \frac{1}{\delta^2\;(1+\rho)}\; \begin{bmatrix} 1 & 1 \end{bmatrix} \; \begin{bmatrix} Z_1 \\ Z_2 \\ \end{bmatrix} O = \frac{1}{\delta^2\;(1+\rho)}\; \begin{bmatrix} 1 & 1 \end{bmatrix} \; \begin{bmatrix} Z_1 \\ Z_2 \\ \end{bmatrix}
 \end{equation} \end{equation}
-Assuming $\underline{Z} = \underline{S}$ the output is:+It's necessary to put $\underline{Z} = \underline{S}$ in order to compute the signal to noise ratio in output:
 \begin{equation} \begin{equation}
-O=\frac{2}{\delta^2\;(1+\rho)}+\Bigg(\frac{S}{N}\Bigg)_O=\frac{2}{\delta^2\;(1+\rho)}=\frac{2}{(1+\rho)} \Bigg(\frac{S}{N}\Bigg)_I
 \end{equation} \end{equation}
 The improving factor is equal to the ratio between the signal to noise ration in output and te signal to noise ration in input. The improving factor is equal to the ratio between the signal to noise ration in output and te signal to noise ration in input.
Line 432: Line 464:
 \eta = \frac{(S/N)_O}{(S/N)_I} = \frac{O}{(1/\delta ^2)}= \frac{2}{1+\rho } = 1.09 \eta = \frac{(S/N)_O}{(S/N)_I} = \frac{O}{(1/\delta ^2)}= \frac{2}{1+\rho } = 1.09
 \end{equation} \end{equation}
-The improving factor is very low because the doppler frequency , double respect to the $PRF$, is folded in the origin and overlaps the clatter, in this way the filter is not able to distinguish between the clatter and the doppler frequency. \\  +The improving factor is very low because the doppler frequency , double respect to the $PRF$, is folded in the origin and overlaps the clutter, in this way the filter is not able to distinguish between the clutter and the doppler frequency.  
 +  
 ====Exercise 2==== ====Exercise 2====
 A radar with $PRF=500\;Hz$ looks at three targets that have different doppler frequencies; $f_{D1}=1000\;Hz$, $f_{D2}=750\;Hz$ and $f_{D3}=625\;Hz$.\\  A radar with $PRF=500\;Hz$ looks at three targets that have different doppler frequencies; $f_{D1}=1000\;Hz$, $f_{D2}=750\;Hz$ and $f_{D3}=625\;Hz$.\\ 
-Design the optimum processor computing the coefficient $K$ of the filter and the improving factor, assuming to use 2 pulses and a Gaussian clatter centered in zero with standard deviation $\delta _c=30\;Hz$.\\ +Design the optimum processor computing the coefficient $K$ of the filter and the improving factor, assuming to use 2 pulses and a Gaussian clutter centered in zero with standard deviation $\delta _c=30\;Hz$.\\ 
 The correlation coefficient is: $\rho=e^{-2\pi^2\;\delta_c^2\;T^2}$.\\ \\  The correlation coefficient is: $\rho=e^{-2\pi^2\;\delta_c^2\;T^2}$.\\ \\ 
 **Solution**\\ **Solution**\\
Line 472: Line 505:
 O_3 = \underline{Z}\;\underline{K_3} = \begin{bmatrix} z_1 & z_2 \end{bmatrix} \; \underline{K_3}= \frac{z_1(1+j\rho )-z_2(\rho + j)}{\delta ^2 (1-\rho ^2)} O_3 = \underline{Z}\;\underline{K_3} = \begin{bmatrix} z_1 & z_2 \end{bmatrix} \; \underline{K_3}= \frac{z_1(1+j\rho )-z_2(\rho + j)}{\delta ^2 (1-\rho ^2)}
 \end{equation}        \end{equation}       
-Now we can compute the improving factors putting $\underline{Z}$ equal to the respective expected signal $S_{1,2,3}$ and calculating the the correlation coefficient that is egual to $\rho=0.9314$.\\ +Now we can compute the improving factors putting $\underline{Z}$ equal to the respective expected signal $S_{1,2,3}$ and calculating the correlation coefficient that is egual to $\rho=0.9314$.\\ 
 \begin{equation} \begin{equation}
 \eta_1 = \frac{\frac{2}{\delta ^2 (1+\rho )}}{\frac{1}{\delta ^2}}= 1.04 \eta_1 = \frac{\frac{2}{\delta ^2 (1+\rho )}}{\frac{1}{\delta ^2}}= 1.04
Line 482: Line 515:
 \eta_3 = \frac{\frac{2}{\delta ^2 (1-\rho ^2)}}{\frac{1}{\delta ^2}}= 15.1 \eta_3 = \frac{\frac{2}{\delta ^2 (1-\rho ^2)}}{\frac{1}{\delta ^2}}= 15.1
 \end{equation} \end{equation}
-As we expect the improvement in the first case is very low because $f_{D1}$ is folded exactly over the clatter, the case two instead is the best case that we can have because $f_{D2}$ is folded in the middle between 0 and $PRF$ (at 250 Hz) so putting a filter just a small part of the clatter is collected (small part of the Gaussian clatter tails). While the third is an intermediate case between the two.+As we expect the improvement in the first case is very low because $f_{D1}$ is folded exactly over the clutter, the case two instead is the best case that we can have because $f_{D2}$ is folded in the middle between 0 and $PRF$ (at 250 Hz) so putting a filter just a small part of the clutter is collected (small part of the Gaussian clutter tails). While the third is an intermediate case between the two.
 <figure> <figure>
 {{ :media:opt_process_ex.jpg?600 |}} {{ :media:opt_process_ex.jpg?600 |}}
-<caption>Clatter, simple MTI and folded doppler frequencies</caption>+<caption>Clutter, simple MTI and folded doppler frequencies.</caption>
 </figure> </figure>
 +
 =====Radar Doppler===== =====Radar Doppler=====
 ====Exercise 1==== ====Exercise 1====
-A radar with MTI doble canceler operate in X band at frequency $f_0=9\;GHz$ and has $PRF=200\;Hz$.\\  +A radar with a MTI double canceler operate in X band at the frequency $f_0=9\;GHz$ and has $PRF=2000\;Hz$.\\  
-Sketch the FIR filter for the MTI double canceler and compute the improving factor $\eta_1$ considering a Gaussian clatter with zero mean and standard deviation $\delta_v=0.1\;m/s$. Than compute again the improving factor $\eta_2$ considering a clatter due to the wind with velocity $v=6\;m/s$ and the same dispersion.\\ +Sketch the FIR filter for the MTI double canceler and compute the improving factor $\eta_1$ considering a Gaussian clutter with zero mean and standard deviation $\delta_v=0.1\;m/s$. Than compute again the improving factor $\eta_2$ considering a clutter due to the wind with velocity $v=6\;m/s$ and the same dispersion.\\ 
 Additional question: Compute the improving factor $\eta_3$ considering a real COHO oscillator with a phase noise equal to 2° RMS (Root Mean Square). Additional question: Compute the improving factor $\eta_3$ considering a real COHO oscillator with a phase noise equal to 2° RMS (Root Mean Square).
  
Line 501: Line 535:
 <figure MTIcanc> <figure MTIcanc>
 {{ :media:mti_double_canc.jpg?500 |}} {{ :media:mti_double_canc.jpg?500 |}}
-<caption>Double MTI caneceler (FIR filter)</caption>+<caption>Double MTI caneceler (FIR filter).</caption>
 </figure> </figure>
 Before computing the correlation coefficients we have to know the dispersion in frequency:  Before computing the correlation coefficients we have to know the dispersion in frequency: 
Line 507: Line 541:
 \delta_f= \frac{2\;\delta_v}{\lambda} = 6\;Hz \delta_f= \frac{2\;\delta_v}{\lambda} = 6\;Hz
 \end{equation} \end{equation}
-Having a Gaussian clatter the correlation coefficients are given by:+Having a Gaussian clutter the correlation coefficients are given by:
 \begin{equation} \begin{equation}
 \rho_1(KT)=e^{-2\pi ^2 \delta_f ^2 (KT)^2} \rho_1(KT)=e^{-2\pi ^2 \delta_f ^2 (KT)^2}
Line 521: Line 555:
 \eta_1 = \frac{1}{1-\frac{4}{3} Re[\rho_1(T)] + \frac{1}{3} Re[\rho_1(2T)]}=30000=44.77\;dB \eta_1 = \frac{1}{1-\frac{4}{3} Re[\rho_1(T)] + \frac{1}{3} Re[\rho_1(2T)]}=30000=44.77\;dB
 \end{equation}  \end{equation} 
-In the second case we have the same shape of clatter but with mean equal to $v=6\;m/s$ (the clatter is shifted forward). +In the second case we have the same shape of clutter but with mean equal to $v=6\;m/s$ (the clutter is shifted forward). 
-Is not necessary to do the Fourier antitransformed of the Gaussian clatter, we can use the property of the frequency translation. Knowing that the correlation coefficients for the clatter with velocity $v=6\;m/s$ will be given by:+Is not necessary to do the Fourier antitransformed of the Gaussian clutter, we can use the property of the frequency translation. Knowing that the correlation coefficients for the clutter with velocity $v=6\;m/s$ will be given by:
 \begin{equation} \begin{equation}
 \rho_2(KT)=\rho_1(KT)e^{-j2\pi f_D t} \rho_2(KT)=\rho_1(KT)e^{-j2\pi f_D t}
 \end{equation} \end{equation}
-Where $f_D$ is the frequency doppler of the clatter:+Where $f_D$ is the frequency doppler of the clutter:
 \begin{equation} \begin{equation}
 f_{D}=\frac{2\;v}{\lambda}=360\;Hz f_{D}=\frac{2\;v}{\lambda}=360\;Hz
Line 541: Line 575:
 \eta_2 = \frac{1}{1-\frac{4}{3} Re[\rho_2(T)] + \frac{1}{3} Re[\rho_2(2T)]}=4.54=6.57\;dB  \eta_2 = \frac{1}{1-\frac{4}{3} Re[\rho_2(T)] + \frac{1}{3} Re[\rho_2(2T)]}=4.54=6.57\;dB 
 \end{equation} \end{equation}
-In the second case $\eta_2$ is much lower than $\eta_1$ because the clatter is not more centered in zero and most of its part is inside the filter, just a little part is deleted.\\ \\ +In the second case $\eta_2$ is much lower than $\eta_1$ because the clutter is not more centered in zero and most of its part is inside the filter, just a little part is deleted.\\ \\ 
 To answer to the additional question we consider the vector representation of the echoes coming from a fixed target. The vectors have different direction due to the phase error of the oscillator (the amplitude is the same). This phase difference $\Delta \varphi$ between the two vectors is a random variable whose standard deviation is equal to 2° RMS. To answer to the additional question we consider the vector representation of the echoes coming from a fixed target. The vectors have different direction due to the phase error of the oscillator (the amplitude is the same). This phase difference $\Delta \varphi$ between the two vectors is a random variable whose standard deviation is equal to 2° RMS.
 <figure> <figure>
 {{ :media:vector_rap_echos.jpg?500 |}} {{ :media:vector_rap_echos.jpg?500 |}}
-<caption>Vector representation of two received echoes [(cite:TTR)]</caption>+<caption>Vector representation of two received echoes. [(cite:TTR)]</caption>
 </figure> </figure>
 The difference between the two signals can be approximated in this way: The difference between the two signals can be approximated in this way:
Line 563: Line 597:
 =====Pulse Compression===== =====Pulse Compression=====
 ====Exercise 1==== ====Exercise 1====
-Consider a radar that use pulse compression and transmits linear chirp signal starting from $f_1=9259\;MHz$ up to $f_2=9450\;MHz$, and considering a pulse duration $T=10\;\mu s$. Compute the radar resolution without compression $R_T$, the pulse width of the compressed pulse $\tau$, the radar resolution with pulse compression $R_{\tau}$ and the compression ration $T/\tau$.\\  +Consider a radar that usees pulse compression and transmits linear chirp signal starting from $f_1=9259\;MHz$ up to $f_2=9450\;MHz$, and considering a pulse duration $T=10\;\mu s$. Compute the radar resolution without compression $R_T$, the pulse width of the compressed pulse $\tau$, the radar resolution with pulse compression $R_{\tau}$ and the compression ratio $T/\tau$.\\  
-Finally say if the radar is able to detect a target far 1Km and justify the answer.+Finally, say if the radar is able to detect a target far 1Km and justify the answer.
  
 **Solution**\\ **Solution**\\
-In a linear chirp the frequency is linear modulated, the pulse start with frequency $f_1$ which linearly grows until it reaches $f_2$ at time $T$ when the transmitting of the pulse ends. +In a linear chirp the frequency is linear modulated, the pulse starts with frequency $f_1$ which linearly grows until it reaches $f_2$ at the time $T$ when the transmitting of the pulse ends. 
 <figure> <figure>
 {{ :media:pulse_comp_linear.jpg?300 |}} {{ :media:pulse_comp_linear.jpg?300 |}}
-<caption>Trend of the instantaneous frequency Vs. pulse duration</caption>+<caption>Trend of the instantaneous frequency Vs. pulse duration.</caption>
 </figure> </figure>
 At the receiver we have a matched filter that has in output a sinc function, the pulse width of the compressed pulse is the width of the main beam of the sinc, that is related to the used band. At the receiver we have a matched filter that has in output a sinc function, the pulse width of the compressed pulse is the width of the main beam of the sinc, that is related to the used band.
Line 584: Line 618:
 R_{\tau} = \frac{c\tau }{2}= 0.75\;m R_{\tau} = \frac{c\tau }{2}= 0.75\;m
 \end{equation} \end{equation}
-Using a compressed pulse with the same power used in a non compressed radar pulse we can obtain a high radar resolution without losing coverage.\\  +Using a compressed pulse with the same power used in a uncompressed radar pulse we can obtain a high radar resolution without losing coverage.\\  
 The compression ratio used is an high compression. The compression ratio used is an high compression.
 \begin{equation} \begin{equation}
 \frac{T}{\tau}=\frac{10\;\mu s}{5\;ns} = 2000 \frac{T}{\tau}=\frac{10\;\mu s}{5\;ns} = 2000
 \end{equation}  \end{equation} 
-The radar can not detect a target far 1 Km because it is located in the blinde zone of the radar, the echoes coming from targets in blind zone will arrive while the radar is still transmitting and therefore will not be able to receive it. The blinde zone have a radius of:+The radar can not detect a target far 1 Km because it is located in the blind zone of the radar, the echoes coming from targets in the blind zone will arrive while the radar is still transmitting and therefore will not be able to receive it. The blind zone has a radius of:
 \begin{equation} \begin{equation}
 R_{min}=\frac{cT}{2}=1.5\;Km R_{min}=\frac{cT}{2}=1.5\;Km
-\end{equation}\\ +\end{equation}
  
 ====Exercise 2==== ====Exercise 2====
-Considering a radar that use pulse compression with a pulse length $T=12\;\mu s$ and a chirp signal that has the following expression:+Considering a radar that uses pulse compression with a pulse length $T=12\;\mu s$ and a chirp signal that has the following expression:
 \begin{equation} \begin{equation}
 S(t)=cos(2\pi f_0 t+\frac{at^2}{2})\;\;\;\;\;\;0<t<T S(t)=cos(2\pi f_0 t+\frac{at^2}{2})\;\;\;\;\;\;0<t<T
 \end{equation} \end{equation}
-Knowing that $a=6\times 10^{12}\;[1/s^2]$ compute the bandwidth B, the width of the compressed pulse $\tau $ and the compression ratio $T/\tau$. Finally sketch the output of the matched filter and the absolute value of the sent signal spectrum. \\ \\ +Knowing that $a=6\times 10^{12}\;[1/s^2]$ compute the bandwidth B, the width of the compressed pulse $\tau $ and the compression ratio $T/\tau$. Finally, sketch the output of the matched filter and the absolute value of the sent signal spectrum. \\ \\ 
  
 **Solution**\\ **Solution**\\
Line 623: Line 657:
 \frac{T}{\tau } = BT = 137.5 \frac{T}{\tau } = BT = 137.5
 \end{equation}\\  \end{equation}\\ 
-The spectrum of the transmitted signal is a rect with unitary amplitude, it's rapresented in Figure {{ref>specPulse}}. +The spectrum of the transmitted signal is a rect with unitary amplitude, it's represented in Figure {{ref>specPulse}}. 
 <figure specPulse> <figure specPulse>
 {{ :media:spectrum_sent_pulse.jpg?350 |}} {{ :media:spectrum_sent_pulse.jpg?350 |}}
-<caption> Absolute value of the sent signal spectrum </caption> +<caption> Absolute value of the sent signal spectrum. </caption> 
-</figure>\\  +</figure> 
-Neglecting the processing time the output of the matched filter will have a behavior like in Figure {{ref>outFilt}} that is related to the Fourier transform of the sent signal spectrum.+Neglecting the processing time the output of the matched filter will have a behaviour like in Figure {{ref>outFilt}} that is related to the Fourier transform of the sent signal spectrum.
 <figure outFilt> <figure outFilt>
 {{ :media:output_match_filter.jpg?500 |}} {{ :media:output_match_filter.jpg?500 |}}
-<caption>Output of the matched filter[(cite:TTR)]</caption>+<caption>Output of the matched filter.[(cite:TTR)]</caption>
 </figure> </figure>
-The amplitude of the main lobe is equal to 1 (in linear domain) and the beam width is $2\tau$ (at -3 dB is $\tau$ ). The Peak Side Lobe ratio is 13.26 dB.\\ \\ +The amplitude of the main lobe is equal to 1 (in the linear domain) and the beam width is $2\tau$ (at -3 dB is $\tau$ ). The Peak Side Lobe ratio is 13.26 dB.
  
 ====Exercise 3==== ====Exercise 3====
Line 639: Line 673:
 Knowing that the pulse duration is $T=35\;\mu s$, compute the compression ratio $BT$ than sketch the impulse response of the matched filter $h(t)$ and also the output of the filter $o(t)$\\ \\  Knowing that the pulse duration is $T=35\;\mu s$, compute the compression ratio $BT$ than sketch the impulse response of the matched filter $h(t)$ and also the output of the filter $o(t)$\\ \\ 
 **Solution**\\   **Solution**\\  
-In the Barker encoding with 7 elements the pulse is divided in 7 sub-pulses of which the 4th,5th and 7th have a phase difference of $\pi $ respect to the others; this is a PSK type of encoding. In Figure {{ref>bark7}} is represented the complex envelope of the transmitted Barker encoding signal, the coefficient of this type of filter can be 1 or -1.   +In the Barker encoding with 7 elements the pulse is divided in 7 sub-pulses of which the 4th, 5th and 7th have a phase difference of $\pi $ respect to the others; this is a PSK type of encoding. In Figure {{ref>bark7}} is represented the complex envelope of the transmitted Barker encoding signal, the coefficient of this type of filter can be 1 or -1.   
 <figure bark7> <figure bark7>
 {{ :media:barker7.jpg?400 |}}    {{ :media:barker7.jpg?400 |}}   
-<caption>Complex envelope of the transmitted Barker encoding signal</caption> +<caption>Complex envelope of the transmitted Barker encoding signal.</caption> 
-</figure>\\ +</figure>
 As we know in this case the encoding is done over a period of 7 times $\tau$, this means $T=7\tau$. As we know in this case the encoding is done over a period of 7 times $\tau$, this means $T=7\tau$.
 \begin{equation} \begin{equation}
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 <figure> <figure>
 {{ :media:barker_discrite.jpg?400 |}}  {{ :media:barker_discrite.jpg?400 |}} 
-<caption>Encoding signal in discrete domain</caption>+<caption>Encoding signal in the discrete domain.</caption>
 </figure> </figure>
 <figure> <figure>
 {{ :media:disc_impulse_resp.jpg?400 |}} {{ :media:disc_impulse_resp.jpg?400 |}}
-<caption>Impulse response of the matched filter in discrete domain</caption> +<caption>Impulse response of the matched filter in the discrete domain.</caption> 
-</figure>\\ +</figure>
 If we do the convolution between $h(n)$ and $s(n)$ we will obtain the output of the matched filter. If we do the convolution between $h(n)$ and $s(n)$ we will obtain the output of the matched filter.
 <figure> <figure>
 {{ :media:out_filter_1_.jpg?600 |}}  {{ :media:out_filter_1_.jpg?600 |}} 
-<caption>Discrete matched filter output</caption>+<caption>Discrete matched filter output.</caption>
 </figure> </figure>
-Doing the convolution of the $o(n)$ with triangle that have base $2\tau $ and unitary height we can obtain the output in the continues time domain.+Doing the convolution of the $o(n)$ with triangle that has base $2\tau $ and unitary height we can obtain the output in the continues time domain.
 <figure> <figure>
 {{ :media:out_flter_cont.jpg?600 |}} {{ :media:out_flter_cont.jpg?600 |}}
-<caption>Continues time matched filter output</caption> +<caption>Continues time matched filter output.</caption> 
-</figure>\\  +</figure> 
-It was been added one zero at the beginning of the discrete time output (this means a delay) in order to have a CAUSAL filter. \\ \\ +It was been added one zero at the beginning of the discrete time output (this means a delay) in order to have a CAUSAL filter. 
 =====RCS===== =====RCS=====
 ====Exercise 1==== ====Exercise 1====
-A radar see in the middle of a set of objects, composed of 4 spheres that form a square in the way shown in Figure {{ref>setobj}}. All the spheres are equal and have a radius of $r=5\lambda $ and the distance between two of them placed in the same side of the square is $d=20\lambda $. Assuming that we are between the MIE Region and Optical Region (regions of the $RCS$ behavior Vs. electrical dimension of the object) the $RCS$ of a sphere can be assumed equal to $RCS_{sph}=\pi r^2$.\\ +A radar see in the middle of a set of objects, composed of 4 spheres that form a square in the way shown in Figure {{ref>setobj}}. All the spheres are equal and have a radius of $r=5\lambda $ and the distance between two of them placed in the same side of the square is $d=20\lambda $. Assuming that we are between the MIE Region and Optical Region (regions of the $RCS$ behaviour Vs. electrical dimension of the object) the $RCS$ of a sphere can be assumed equal to $RCS_{sph}=\pi r^2$.\\ 
 Compute the total radar cross section $RCS_{TOT}$ coming from this set of objects. Compute the total radar cross section $RCS_{TOT}$ coming from this set of objects.
 <figure setobj> <figure setobj>
  {{ :media:set_objects.jpg?250 |}}  {{ :media:set_objects.jpg?250 |}}
-<caption>Four spheres at a distance $20\lambda $</caption>+<caption>Four spheres at a distance $20\lambda $.</caption>
 </figure> </figure>
  
-Assume than that all the set of spheres rotates along an axis placed vertically in the middle as shown in Figure {{ref>objrot}}. \\ +Assume then that all the set of spheres rotates along an axis placed vertically in the middle as shown in Figure {{ref>objrot}}. \\ 
 Derive the total radar cross section formula as a function of $\theta $.   Derive the total radar cross section formula as a function of $\theta $.  
  
 <figure objrot> <figure objrot>
 {{ :media:set_objects_rot.jpg?300 |}} {{ :media:set_objects_rot.jpg?300 |}}
-<caption>Four sphere rotating along an axis</caption>+<caption>Four sphere rotating along an axis.</caption>
 </figure> </figure>
  
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 <figure veccomp> <figure veccomp>
 {{ :media:vec_com.jpg?200 |}} {{ :media:vec_com.jpg?200 |}}
-<caption>Vector composition of the contributions coming from the 4 spheres</caption>+<caption>Vector composition of the contributions coming from the 4 spheres.</caption>
 </figure> </figure>
 We are interested in calculating $A^2$ for power reasons. We are interested in calculating $A^2$ for power reasons.
radar/exercises.1527862640.txt.gz · Last modified: (external edit)