A meteor radar has the following parameters:
$f_0 = 9400\;MHz$ (X Band)
$P_T = 200\;KW$
$PRF_A = 1200\;Hz$
$PRF_B = 240\;Hz$
$\tau_A = 0.5\;\mu s$
$\tau_B = 3\;\mu s$
$D = 2\;m$ (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.

Solution
Considering a Gaussian beam and using the approximation for $D>>\lambda$. \begin{equation} \lambda=\frac{c}{f}=0.032\;m \end{equation} \begin{equation} \theta_B=\varphi_B=65\frac{\lambda}{D}=1.04°\;(0.018\;rad) \end{equation} \begin{equation} G_{max}=10\log_{10}(\frac{26000}{\theta_B^2})=48.3\;dB \end{equation} 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. \begin{equation} R_{max,A}=\frac{c}{2PRF_A}=125\;Km \end{equation} \begin{equation} R_{max,B}=\frac{c}{2PRF_B}=625\;Km \end{equation} \begin{equation} |v_{max,A}|=\frac{\lambda PRF_A}{4}=9.6\;m/s \end{equation} \begin{equation} |v_{max,B}|=\frac{\lambda PRF_B}{4}=1.92\;m/s \end{equation} \begin{equation} d_A=\frac{\tau_A}{PRT_A}=6\times10^{-4} \end{equation} \begin{equation} d_B=\frac{\tau_B}{PRT_B}=7.2\times10^{-4} \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: \begin{equation} \Delta R_A=\frac{c\; \tau_A}{2}=74\;m \end{equation} \begin{equation} \Delta R_B=\frac{c\; \tau_B}{2}=450\;m \end{equation} We can also compute the radar cell, in which the radar can not distinguish between 2 different targets. \begin{equation} RC_A= R_{max,A}\; \theta_B \;R_{max,A} \; \varphi_B \; \Delta R_A = 1.2\times 10^{12}\;m^3 \end{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 \end{equation}

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 unambiguous range $R_{max}$ and the gain for coherent integration $G_{INT}$.
$\omega=40\;rpm$
$PRT=1\;ms$
$\theta_B=2.5\;deg$
$h_r=200\;m$
$h_v=10\;m$

Solution
The width of the time interval in which the target is illuminated by the beam at -3 dB can be computed in the following way since we have the angular velocity in $rpm$ and the beam width in degrees. \begin{equation} t_D=\frac{\theta_B}{6 \omega}=0.01\;s \end{equation} \begin{equation} N=PRF\;t_D=\frac{t_D}{PRT}=10 \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 $f_D=0$). \begin{equation} G_{INT}=N=10=10\;dB \end{equation} We can compute the radar horizon assuming an equivalent radius of the earth $r_e = 8500\;Km$: \begin{equation} R_{max,H} \simeq \sqrt{2 r_e}(\sqrt{h_r}+\sqrt{h_v}) \simeq 71.29\;Km \end{equation} But the maximum unambiguous range has a bigger value: \begin{equation} R_{max}=\frac{c\;PRT}{2}=150\;Km \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.

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}

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 $SNR_{min}$.
$P_t = 10\;KW$
$RCS=5\;m^2$
$G=32\;dB$
$D=130\;Km$
$T_s=664\;K$
$f_0=1.3\;GHz$
$B=0.3\;MHz$
$SNR_{min}=13\;dB$

Solution
We can compute the power received using the deterministic radar equation: \begin{equation} P_r=\frac{P_t\;G^2\;\lambda^2\;RCS}{(4\pi)^3\;D^4} \end{equation} where $\lambda=c/f_0=0.23\;m$
To compute $P_r$ is better to use the dB method:

\begin{equation} P_r\simeq-139.3\;dBW=-109.3\;dBm=1.17\times10^{-11}\;mW \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). \begin{equation} R_{max}^4=\frac{P_t\;G^2\;\lambda^2\;RCS}{(4\pi)^3\;S_{min}^4} \end{equation} where \begin{equation} S_{min}=SNR_{min}\;P_n \end{equation} $P_n$ is the power noise so defined: \begin{equation} P_n=k\;B\;T_s \end{equation} The Boltzmann constant is equal to $k=1.38\times10^{-23}\;J/K$

\begin{equation} R_{max}=\frac{197.87}{4}\;dBmeters=88.51\;Km \end{equation}

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 $n_{fa}$
$B=2\;MHz$
$SNR=20\;dB$
$\tau=0.5\;\mu s$
$\theta_B=1\;deg$
$PRF=500\;Hz$
$\omega=30\;rpm$

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:

\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} \end{equation} \begin{equation} \frac{V_T}{\sigma}=\sqrt{-2\;ln(P_{fa})}=5.17 \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. Considering the case of SW1 so assuming that the echo has Gaussian statistics and the envelope of the signals it's Rayleigh: \begin{equation} 1+SNR=\frac{ln(P_{fa})}{ln(P_d)} \end{equation} from which \begin{equation} P_d=exp(\frac{ln(P_{fa})}{1+SNR})= 0.87 \end{equation} (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 $N_{\theta}$. \begin{equation} N_P=\frac{PRT}{\tau}=\frac{1}{\tau\;PRF}= 4000 \end{equation} \begin{equation} N_{\theta}=360/1=360 \end{equation} Now we can compute the number of cells: \begin{equation} N=N_P\;N_{\theta}=144\times10^4 \end{equation} The number of false alarms in one scan are: \begin{equation} n_{fa}=N\;P_{fa}=2.16 \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.

Compute the system temperature in the point A of the illustrated component chain (Figure 3) , 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.

Solution
The equivalent noise temperature of the RF component is: \begin{equation} T_{RFe}=(L_{RF}-1)\;T_0 \end{equation} while that of the amplifier is given by: \begin{equation} T_{LNAe}=(F-1)\;T_0 \end{equation} The total system noise temperature is equal to: \begin{equation} T_{sys}=T_A+T_{RFe}+T_{LNAe}\;L_{RF}=661\;K\simeq 28.2\;dBK \end{equation}

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
Considering the Figure 4 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.

\begin{equation} T'_A=\frac{T_A + T_L}{L_A} \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 the sky and from the ground is equal to: \begin{equation} T_A=(1 - \alpha )T_{sky} + \alpha \;T_{GND} \end{equation} putting everything together the real antenna noise temperature is: \begin{equation} T'_A =\frac{(1.412-1)\cdot290 + (1-0.2)\cdot 20 + 0.2\cdot 290}{1.412} = 137\;K \simeq 21.37\;dBK \end{equation}

Consider the Figure 5 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 $L_{MIX}=0.8\;dB$.

Assuming than that the radar has the following characteristics:
$P_T=600\;KW$
$f_0=1.5\;GHz$
$G_T=G_R=G=33\;dB$
$B_{eq}=2\;MHz$
$L_T=2\;dB$ (losses in transmission)
$L_P^2=1.5\;dB$ (two-way loss per antenna beam)
$RCS=5\;m^2$
$SNR_{min}=13.2\;dB$

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 $L_{INT}=0.5\;dB$.

Solution
In order to compute the maximum range it's necessary to know the system temperature, that can be computed as in the exercise 1 of System and Antenna Temperature but considering also the losses introduced by the mixer. \begin{equation} T_{sys}= T_A + (L_{RF}-1)T_0\;L_{RF} + \frac{(L_{MIX}-1)T_0\;L_{RF}}{G_{LNA}} = 448.4\;K \end{equation} It's better to compute the ration $T_{sys}/T_0$ in order to facilitate the maximum range calculation in dB \begin{equation} \frac{T_{sys}}{T_0}=1.546=1.89\;dB \end{equation} Using the Blake formula and assuming the two-way loss for propagation negligible: \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} \end{equation}

The maximum range in case of a fixed target is equal to: \begin{equation} R_{max}=\frac{206.23}{4}\;dBmeters=143.2\;Km \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 a fixed target that we already have.

As we can see in the Figure 7 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 3 the maximum range is: \begin{equation} R_{max}=49.68\;dBmeters=92.9\;Km \end{equation} It is observed that we lost 1/3 of the maximum range respect to the case of fixed target.
In case of coherent integration the gain is equal to the number of sent pulses: \begin{equation} G_{INT}=N=10=10\;dB \end{equation} We calculate the minimum signal to noise ratios for coherent integration, necessary to detect with the same $P_D$ and $P_{fa}$ for both fixed target and fluctuating target (SW1 and SW2). \begin{equation} SNR_{min,INT}=(L_{INT} + SNR_{min}) - G_{INT}= 3.7\;dB \end{equation} \begin{equation} SNR_{min,INT,SW1/2}=(L_{INT} + SNR_{min,SW1/2}) - G_{INT}= 11.2\;dB \end{equation} Again substituting the new signal to noise ratios obtained in Table 3 we can compute the maximum range in case of coherent integration for both fixed and fluctuating target. \begin{equation} R_{max,INT}=247\;Km \end{equation} \begin{equation} R_{max,INT,SW1/2}=160.3\;Km \end{equation} How we expected the relatively ranges are increased.

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
Taking the value obtained in the previous exercise and putting it equal to $R_0$: \begin{equation} R_0=R_{max,INT}=247\;Km \end{equation} we can compute the losses for $R_0$ in this way: \begin{equation} L_0 = 2\;L_{prop}\;R_0 = 2.96\;dB = 1.98\ (linear) \end{equation} The multiplication by 2 is due to the fact that are considered two way. Now it is computed again the maximum range $R_1$ considering the losses $L_0$. \begin{equation} R_1=\frac{R_0}{\sqrt[4]{L_0}}= 208.2\;Km \end{equation} The value obtained is too low because we have overestimated the losses computing it with $R_0$, we have to compute again the losses $L_1$ using now $R_1$ and then calculate the new range $R_2$. \begin{equation} L_1 = 2\;L_{prop}\;R_1 = 2.5\;dB = 1.78\ (linear) \end{equation} \begin{equation} R_2=\frac{R_0}{\sqrt[4]{L_1}}= 213.8\;Km \end{equation} 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: \begin{equation} R_4=213.18\;Km \end{equation}

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: $\rho=e^{-2\pi^2\;\delta_c^2\;T^2}$.

Solution

Assuming $\underline{Z}$ as the input of the FIR filter the output will be: \begin{equation} O = \underline{Z}^T\;\underline{K} \end{equation} Where $\underline{K}$ are the coefficiente of the FIR filter: \begin{equation} \underline{K} = \underline{M}^{-1}\;\underline{S}^* \end{equation} Substituting the covarianca matrix of noise plus clutter we obtain: \begin{equation} \underline{K}=\frac{1}{\delta^2\;(1-\rho^2)} \begin{bmatrix} 1 & -\rho \\ -\rho & 1 \\ \end{bmatrix} \underline{S}^* \end{equation} Assuming that $\underline{S}$ (expected signal) is a sinusoidal wave form with unitary amplitude and zero initial phase, this mean that the analytic vector is oriented along the I axis of the the I and Q chart. The second analytic vector will be given by: \begin{equation} S_2=e^{j2\pi f_DPRT} \end{equation} So now we know also $\underline{S}=\begin{bmatrix} 1 \\ 1 \\ \end{bmatrix}$ and we can compute the coefficients of the filter, not numerically because we don't have the power of the clutter $\delta ^2$. \begin{equation} \underline{K}=\frac{1}{\delta^2\;(1-\rho^2)} \begin{bmatrix} 1 & -\rho \\ -\rho & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} = \frac{1}{\delta^2\;(1+\rho)} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} \end{equation} The output of the filter will be: \begin{equation} O = \frac{1}{\delta^2\;(1+\rho)}\; \begin{bmatrix} 1 & 1 \end{bmatrix} \; \begin{bmatrix} Z_1 \\ Z_2 \\ \end{bmatrix} \end{equation} It's necessary to put $\underline{Z} = \underline{S}$ in order to compute the signal to noise ratio in output: \begin{equation} \Bigg(\frac{S}{N}\Bigg)_O=\frac{2}{\delta^2\;(1+\rho)}=\frac{2}{(1+\rho)} \Bigg(\frac{S}{N}\Bigg)_I \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. \begin{equation} \eta = \frac{(S/N)_O}{(S/N)_I} = \frac{O}{(1/\delta ^2)}= \frac{2}{1+\rho } = 1.09 \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.

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 clutter centered in zero with standard deviation $\delta _c=30\;Hz$.
The correlation coefficient is: $\rho=e^{-2\pi^2\;\delta_c^2\;T^2}$.

Solution
It's necessary to compute the expected signal for the 3 different doppler frequencies with the following general expression: \begin{equation} S = e^{j2\pi (k\;f_D/PRF)} \end{equation} Assuming $k=0$ for the first pulse and $k=1$ for the second one, and using the respective $f_{D 1,2,3}$ \begin{equation} \underline{S_1}= \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} \end{equation} \begin{equation} \underline{S_2}= \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} \end{equation} \begin{equation} \underline{S_3}= \begin{bmatrix} 1 \\ j \\ \end{bmatrix} \end{equation} As in the previous exercise the coefficients are: \begin{equation} \underline{K_1}=\underline{M}^{-1} \underline{S_1}^* = \frac{1}{\delta^2\;(1-\rho^2)} \begin{bmatrix} 1 & -\rho \\ -\rho & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} = \frac{1}{\delta^2\;(1+\rho)} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} \end{equation} \begin{equation} \underline{K_2}=\underline{M}^{-1} \underline{S_2}^* = \frac{1}{\delta^2\;(1-\rho^2)} \begin{bmatrix} 1 & -\rho \\ -\rho & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} = \frac{1}{\delta^2\;(1-\rho)} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} \end{equation} \begin{equation} \underline{K_3}=\underline{M}^{-1} \underline{S_3}^* = \frac{1}{\delta^2\;(1-\rho^2)} \begin{bmatrix} 1 & -\rho \\ -\rho & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ -j \\ \end{bmatrix} = \frac{1}{\delta^2\;(1-\rho ^2)} \begin{bmatrix} 1+j\rho \\ -\rho -j \\ \end{bmatrix} \end{equation} The outputs of the FIR filter for the 3 targets are: \begin{equation} O_1 = \underline{Z}\;\underline{K_1} = \begin{bmatrix} z_1 & z_2 \end{bmatrix} \; \underline{K_1}= \frac{z_1+z_2}{\delta ^2 (1+\rho)} \end{equation} \begin{equation} O_2 = \underline{Z}\;\underline{K_2} = \begin{bmatrix} z_1 & z_2 \end{bmatrix} \; \underline{K_2}= \frac{z_1-z_2}{\delta ^2 (1-\rho)} \end{equation} \begin{equation} 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} 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} \eta_1 = \frac{\frac{2}{\delta ^2 (1+\rho )}}{\frac{1}{\delta ^2}}= 1.04 \end{equation} \begin{equation} \eta_2 = \frac{\frac{2}{\delta ^2 (1-\rho )}}{\frac{1}{\delta ^2}}= 29.15 \end{equation} \begin{equation} \eta_3 = \frac{\frac{2}{\delta ^2 (1-\rho ^2)}}{\frac{1}{\delta ^2}}= 15.1 \end{equation} 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.

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.
Additional question: Compute the improving factor $\eta_3$ considering a real COHO oscillator with a phase noise equal to 2° RMS (Root Mean Square).

Solution
Assume: \begin{equation} T=PRT=\frac{1}{PRF} \end{equation} The schema of the double canceler with the relative weights is represented in the Figure 10.

Before computing the correlation coefficients we have to know the dispersion in frequency: \begin{equation} \delta_f= \frac{2\;\delta_v}{\lambda} = 6\;Hz \end{equation} Having a Gaussian clutter the correlation coefficients are given by: \begin{equation} \rho_1(KT)=e^{-2\pi ^2 \delta_f ^2 (KT)^2} \end{equation} \begin{equation} \rho_1(T)=0.9998 \end{equation} \begin{equation} \rho_1(2T)=0.9993 \end{equation} For a double canceler the improving factor is equal to: \begin{equation} \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} 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 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} \rho_2(KT)=\rho_1(KT)e^{-j2\pi f_D t} \end{equation} Where $f_D$ is the frequency doppler of the clutter: \begin{equation} f_{D}=\frac{2\;v}{\lambda}=360\;Hz \end{equation} In order to compute the improving factor we have to take the real part of the correlation coefficients: \begin{equation} Re[\rho_2(T)]=Re[e^{-2\pi ^2\delta_f^2 T^2 } e^{-j2\pi f_D T}]=\rho_1(T)\; cos(2\pi f_D T)= 0.4257 \end{equation} \begin{equation} Re[\rho_2(2T)]=Re[e^{-2\pi ^2\delta_f^2 (2T)^2 } e^{-j2\pi f_D 2T}]=\rho_1(2T)\; cos(2\pi f_D 2T)= -0.63697 \end{equation} As before improving factor worth: \begin{equation} \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} 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.

The difference between the two signals can be approximated in this way: \begin{equation} \Delta V \simeq 2V\;sen(\frac{\Delta \varphi}{2})\simeq V\;\Delta \varphi \end{equation} Also $\Delta V$ is a random variable, and we can compute: \begin{equation} \eta_{\phi}^{-1} =\frac{\overline{\Delta V ^2}}{\overline{V^2}}=\frac{V^2\Delta \phi^2}{V^2}=\Delta \phi^2=(\frac{2\pi}{360} 2)^2 =0.00122 \end{equation} \begin{equation} \eta_{\phi} = 819.67 = 29.14\;dB \end{equation} This is an upper bound for the improving factor, because the phase noise is like a virtual clutter internally to the radar that can not be removed. This means that for the first case in which we got $\eta_1=44.77\;dB$ in reality we can not get more than 29.14 dB due to the phase error introduced by the oscillator.

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.

Solution
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.

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. \begin{equation} \tau=\frac{1}{B}=5\;ns \end{equation} The radar resolution as in the first exercises is: \begin{equation} R_T=\frac{cT}{2}=1.5\;Km \end{equation} Instead the radar resolution with pulse compression, so putting the threshold after the matched filter it's much smaller: \begin{equation} R_{\tau} = \frac{c\tau }{2}= 0.75\;m \end{equation} 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. \begin{equation} \frac{T}{\tau}=\frac{10\;\mu s}{5\;ns} = 2000 \end{equation} 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} R_{min}=\frac{cT}{2}=1.5\;Km \end{equation}

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} S(t)=cos(2\pi f_0 t+\frac{at^2}{2})\;\;\;\;\;\;0<t<T \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.

Solution
We know that the instantaneous frequency is given by the derivate of the phase: \begin{equation} f_i(t)=f_0+\frac{1}{2\pi }\frac{d}{dt} \bigg[ \frac{at^2}{2} \bigg] =f_0 + \frac{at}{2\pi }\;\;\;\; 0<t<T \end{equation} This means the chirp is linear, the frequency increase linearly with the pulse duration. In order to find $B$ we have to make the subtraction between the frequency at the end and the frequency at the beginning of the pulse (respectively for $t=T$ and $t=0$). \begin{equation} f_2 = f_0 + \frac{aT}{2\pi }\;\;for\;\;t=T \end{equation} \begin{equation} f_1 = f_0 \;\;for\;\;t=0 \end{equation} \begin{equation} B=f_2-f_1=\frac{aT}{2\pi }=11.46\;MHz \end{equation} The width of the compressed pulse and the compression ratio are respectively: \begin{equation} \tau = \frac{1}{B}=87.2\;ns \end{equation} \begin{equation} \frac{T}{\tau } = BT = 137.5 \end{equation}
The spectrum of the transmitted signal is a rect with unitary amplitude, it's represented in Figure 13.

Neglecting the processing time the output of the matched filter will have a behaviour like in Figure 14 that is related to the Fourier transform of the sent signal spectrum.

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.

A radar uses pulse compression with $N=7$ elements Barker encoding (+ + + - - + -).
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
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 15 is represented the complex envelope of the transmitted Barker encoding signal, the coefficient of this type of filter can be 1 or -1.

As we know in this case the encoding is done over a period of 7 times $\tau$, this means $T=7\tau$. \begin{equation} BT=\frac{T}{\tau }= N = 7 \end{equation} We can also compute $\tau $: \begin{equation} \tau =\frac{T}{N} = 0.5\;\mu s \end{equation} Knowing that the impulse response of a matched filter is equal to: \begin{equation} h(t)=s^*(\tau -t) \end{equation} We can represent the transmitted signal and the impulse response of the filter in a discrete domain

If we do the convolution between $h(n)$ and $s(n)$ we will obtain the output of the matched filter.

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.

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.

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 20. 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.

Assume then that all the set of spheres rotates along an axis placed vertically in the middle as shown in Figure 21.
Derive the total radar cross section formula as a function of $\theta $.

Solution
Assuming that the scatterers are independent and using the definition of radar cross section we can write, that: \begin{equation} RCS_{TOT}= | \sum\limits_{i=1}^4 \alpha _i \;e^{j \frac{4\pi }{\lambda} \delta _i} |^2 \end{equation} Due to the symmetry of the 4 spheres the sum of $exp(j \frac{4\pi }{\lambda} \delta _i)$ will be 1, because the distance between each sphere and the radar is the same (no phase difference). While the field amplitude $\alpha_i$ is related to the $RCS$ by $RCS\simeq \alpha_i ^2\simeq \pi r^2$. Given that the total radar cross section is: \begin{equation} RCS_{TOT}= (4\alpha)^2 = 16 RCS_{sph}=16\pi r^2 \end{equation} In order to solve the second part of the exercise we can analyze the problem using the vector composition of the contributions coming from the 4 spheres, Figure 22.
Two spheres will have a phase difference respect to the other two of $\Delta \varphi$ due to the different path that they have respect to the radar. \begin{equation} \Delta \varphi = \frac{4\pi }{\lambda}\;20\lambda sin(\theta) \end{equation} The amplitudes are all the same $\alpha_1 = \alpha_2 = \alpha_3 =\alpha_4 =\alpha$.

We are interested in calculating $A^2$ for power reasons. \begin{equation} \frac{A}{2}^2 = (2\alpha )^2 cos^2(\Delta \varphi /2) = 4 \alpha ^2 \frac{1+cos(\Delta \varphi)}{2} \end{equation} The total radar cross section as a function of $\theta $ will be: \begin{equation} RCS_{TOT}=A^2=8 RCS_{sph}\bigg[1+cos\Big(\frac{4\pi}{\lambda}\; 20\lambda sin(\theta)\Big)\bigg] \end{equation}