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radar:exercises [2018/06/05 19:39] – iatcoradar:exercises [2026/04/28 15:13] (current) – external edit 127.0.0.1
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-=====EXERCISES==== +======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 unambiguous range $R_{max}$, the maximum unambiguous velocity $v_{max}$ 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====
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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 unambiguous range 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 unambiguous range 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==== ====Exercise 3====
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 $SNR_{min}=13\;dB$\\  \\  $SNR_{min}=13\;dB$\\  \\ 
 **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} 
-   +  
-=====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 alarme in one rotation $n_{fa}$\\ 
 $B=2\;MHz$\\  $B=2\;MHz$\\ 
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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 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 $N_{\theta}$.\\+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\;PRF}= 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 where there is a false alarm, this value can be good for surveillance radar but can be a problem for defence 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**\\ 
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 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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 <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}
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 =====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:\\ 
 $P_T=600\;KW$\\ $P_T=600\;KW$\\
 $f_0=1.5\;GHz$\\  $f_0=1.5\;GHz$\\ 
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 $L_P^2=1.5\;dB$ (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**\\
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 |             ^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 a fixed target is equal to: The maximum range in case of a fixed target is equal to:
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 <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
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 \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 a 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.
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 =====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 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.+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 clutter is: $\rho=e^{-2\pi^2\;\delta_c^2\;T^2}$. Remember that the correlation coefficient for a Gaussian clutter is: $\rho=e^{-2\pi^2\;\delta_c^2\;T^2}$.
  
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 <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:
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 \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 clutter, in this way the filter is not able to distinguish between the clutter 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 $f_{D3}=625\;Hz$.\\ 
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 <figure> <figure>
 {{ :media:opt_process_ex.jpg?600 |}} {{ :media:opt_process_ex.jpg?600 |}}
-<caption>Clutter, 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 double canceler operate in X band at the 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 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.\\  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).
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 <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: 
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 <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:
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 <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.
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 \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====
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 <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 behaviour 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 the 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====
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 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 the 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 the 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 has 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====
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 <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>
  
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 <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.
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