The magnetron is a vacuum tube that is used to generate microwaves of high power.
Type of microwaves Tubes:-
Tubes |
Advantages |
Applications |
TWT |
Wide Bandwidth |
Radars; Communications; Jammers |
Magnetron Oscillator |
Low-cost |
Radars, Domestic Cooking, Industrial heating. |
Klystron amplifier |
High gain & High η |
Radar and Television Broadcast |
Maximum range of a radar system is given as
\({R_{max}} = {\left( {\frac{{{P_t}{G^2}\sigma {A_e}}}{{{{\left( {4\pi } \right)}^2}{S_{min}}}}} \right)^{1/4}}\)
Also,
\({A_e} = \frac{{\pi {D^2}}}{4}\)
Where,
Rmax can now be written as:
\({R_{max}} = {\left( {\frac{{{P_t}{G^2}\sigma \pi {D^2}}}{{4\left( {4{\pi ^2}} \right){S_{min}}}}} \right)^{1/4}}\)
We observe that Rmax ∝ P_{t}1/4
So, If the peak transmitted power is increased by a factor of 16, then radar range will be increased by a factor of 2.The range of the radar is given as:
\(R = {\left[ {\frac{{{P_t}{G^2}\sigma {A_e}}}{{{{\left( {4\pi } \right)}^2}{P_r}}}} \right]^{\frac{1}{4}}}\)
G= Antenna Gain
σ = Radar Cross-Section
Pr = Reflected Power
Ae = Effective Area of the Antenna
Observation:
We observe that, R ∝ (Pt)1/4
Now, to double the range R, Pt has to be increased by a factor of 16
Range resolution is the ability of a radar to distinguish between two or more targets which are very close to each other.
The degree of range resolution depends on:
There the theoretical range resolution of radar can be calculated as-
S_{r}(separation between targets) \(\ge \frac{{{C_o}.\tau }}{2}\)Where τ = Pulse width.
And Bandwidth of the pulse ∝ \(\frac{1}{\tau }\)
\(So,\;{S_r} \ge \frac{{{C_o}}}{{2.\left( {Bandwidth\;of\;the\;Pulse} \right)}}\;\)
This can be understood with the help of the following diagram:
The radar principle is used in the detection of aircraft.
Radar:
Radar systems have been used in military applications for
Doppler Effect:
The Doppler Effect is the change in the observed frequency of a wave when the source or the detector moves relative to the transmitting medium.
The following equation can be used to calculate the Doppler frequency shift (D):
\(D=2\times f_{carrier}.\frac{Relative~velocity}{Speed~of~light}\)
Calculation:
Given:
v = 108 km/hr = 30 m/sec
f_{carrier }= 5 GHz = 5 × 10^{9} Hz
Also, c = 3 × 10^{8} m/s
The Doppler frequency shift will be:
\(D=2\times 5\times 10^9.\frac{30}{3\times 10^8}\)
D = 1000 Hz
PRF Doppler has a range of ambiguities.
Important Points
1. Pulse DOPPLER RADAR:
2. MTI RADAR:
Concept:
In the RADAR system, the range of the target is:
\(R = \frac{{C\; \times \;PRT}}{2}m\)
And,
PRT = (PW + RT) μ sec
Where
C = 3 × 10^{8} m/s = speed of light at which EMW travel
PRT = Pulse repetition time
PW = Pulse width
RT = rest time i.e. time interval between sending a pulse and return from target to the receiver.
Calculation:
Given that,
PW = 2 μ sec
RT = 10 μ sec
∴ PRT = 2 + 10 = 12 μ sec
\(R = \frac{{3\; \times \;{{10}^8}\; \times \;12\; \times \;{{10}^{ - 6}}}}{2}\)
\(R = 1800\;m\)
If we take an approximate value of PRT, we can write:
PRT = 2 + 10 ≈ 10 μ sec
\(R = \frac{{3\; \times \;{{10}^8}\; \times \;10\; \times \;{{10}^{ - 6}}}}{2}\)
Concept:
Power density is defined as the amount of power flowing through a unit area. It has a unit of W/m^{2}.
Mathematically, the power density is defined as:
\(P_{density}=\frac{Power}{Area}\)
For a given power density, the total power through a given area will be:
Power = Power density × Area
Calculation:
Given: Power density = 200 μW/m2
Area of the receiving antenna = 0.2 m2
∴ The received power will be:
Power = 200 μW/m^{2} × 0.2 m^{2}
Power = 40 μW
A radar system uses magnetron as high-power RF source. Its transmitter emits 300kW RF power at 10% duty factor. If the efficiency of radar transmitter during pulse is 60% and the power required during pulse off period is 1kW. The average power dissipation in radar is,
Radar efficiency \( = \frac{{Power\;transmitted}}{{Power\;delivered}}\)
Power delivered \( = \frac{{Power\;transmitted}}{{0.6}} = \frac{{300}}{{.6}}\)
= 500 kW
Dissipated power = P_{delivered} - P_{transmitted}
= 500 – 300 = 200 kW
Since duty cycle is 10% →
Duty cycle \( = \frac{{ON\;time}}{{ON + OFF}}\)
Power dissipated in ON (time = 200 × 0.1 T)
= 20 T kW
Power dissipated in OFF time = 1 kW × 0.9 T
= 0.9 T kW
Average power dissipation \(\, = \frac{{20\;T\;kW + 0.9\;T\;kW}}{T}\)
= 20.9 kW
Pulse compression is a signal processing technique commonly used by radar, sonar and echography to increase the range resolution as well as the signal to noise ratio. This is achieved by modulating the transmitted pulse and then correlating the received signal with the transmitted pulse.
The Range Resolution of any Radar with sinusoidal pulse is defined as \(\left( R \right) = \frac{{CT}}{2}\) where C = velocity of wave, T = pulse duration.
Since Range resolution ∝ pulse duration
Frequency (f) or Band width \( = \frac{1}{T}\) so \(T = \frac{1}{f}\)
∴ Range Resolution \( \propto \frac{1}{f} \propto \frac{1}{{Pulse\;Band\;width}}\)
Pulse Bandwidth ↓ Range Resolution ↑
A bistatic radar system shown in figure has following parameters: f = 5 GHz, G_{dt} = 34 dB, G_{dr} = 22 dB. To obtain a return power of 8p W the minimum necessary radiated power should be
Concept:
For the Bistatic radar system, Power received at the radar is given by:
\({P_r} = \frac{{{G_{dt}}{G_{dr}}{\lambda ^2}\sigma {P_{rod}}}}{{{{\left( {4\pi } \right)}^3}R_{tx}^2R_{rx}^2}}\)
Where,
P_{r} = received power
G_{dt} = Gain of Transmitting Antenna
G_{dr} = Gain of Receiving Antenna
λ = operating wavelength
σ = Area of the cross-section of the Target
R_{tx} = distance of Target from transmitting Antenna
R_{rx} = distance of the target from Receiving Antenna
Calculation:
R_{tx} = 4 km
\({R_{rx}} = \sqrt {{4^2} + {3^2}} = \sqrt {25} = 5\;km\)
f = 5 GHz
∴ \(\lambda = \frac{C}{f} = \frac{{3 \times {{10}^8}}}{{5 \times {{10}^9}}} = 0.06\;m\)
G_{dt} = 34 dB
10 log_{10} (G_{dt}) = 34
∴ G_{dt} = 10^{3.4} = 2511.86 ≈ 2512
G_{dr} = 22 dB
10 log_{10} (G_{dr}) = 22
∴ G_{dr} = 158.5
P_{r} = 8 PW = 8 × 10^{-12} W
Therefore
\({P_r} = \frac{{{G_{dt}}{G_{dr}}{\lambda ^2}\sigma {P_{rad}}}}{{{{\left( {4\pi } \right)}^3}R_{tx}^2R_{rx}^2}}\)
\(8 \times {10^{ - 12}} = \frac{{2512 \times 158.5 \times {{\left( {0.06} \right)}^2} \times 2.4 \times {P_{rad}}}}{{{{\left( {4 \times 3.14} \right)}^3} \times {{\left( 4 \right)}^2} \times {{\left( 5 \right)}^2} \times {{10}^6} \times {{10}^6}}}\)
\(8 \times {10^{ - 12}} = \frac{{3440.03 \times {P_{rad}}}}{{792554.08 \times {{10}^{12}}}}\)
∴ \({P_{rad}} = \frac{{6340432.7}}{{3440.03}}\)
= 1843.13
P_{rad} = 1.843 kWatt
Since there is no option is matching.
Now if we are considering the R_{tx} as the distance between the transmitter and receiver Antenna.
\({P_r} = \frac{{{G_{dt}}{G_{dr}}{\lambda ^2}\sigma {P_{rad}}}}{{{{\left( {4\pi } \right)}^3}R_{tx}^2R_{rx}^2}}\)
\(8 \times {10^{ - 12}} = \frac{{2512 \times 158.5 \times {{\left( {0.06} \right)}^2} \times 2.4 \times {P_{rad}}}}{{{{\left( {4 \times 3.14} \right)}^3} \times {{\left( 5 \right)}^2} \times {{\left( 3 \right)}^2} \times {{10}^6} \times {{10}^6}}}\)
P_{rad} = 1.038 kWatt
Concept:
The maximum unambiguous range of pulse radar is given as:
\(maximum \ unambiguous \ range =\frac{c\times PRT}{2}=\frac{c}{2\times PRF}\)
Where,
c = speed of light
PRT = pulse repetition time
PRF = pulse repetition frequency
Calculation:
Given:
Range = 60 km
\(PRF=\frac{3\times10^8}{2\times 60\times10^3}\)
PRF = 2.5 × 103 pps
Concept:
The maximum range of a radar system is given as
\({R_{max}} = {\left( {\frac{{{P_t}{G^2}\sigma {A_e}}}{{{{\left( {4\pi } \right)}^2}{S_{min}}}}} \right)^{1/4}}\)
Where:
R_{max} = Maximum range of the radar
S_{min} = Minimum Detectable Signal Power
P_{t} = Peak transmitted power
A_{e} = Aperture Area
D = Antenna diameter
We observe that R_{max} ∝ D x P_{t}^{1/4 }
Calculation:
P(new) = 16 P(old)
D(new) = 2D(old)
Rmax ∝ D x Pt1/4
R(new) = 2 D(old) x (16P(old))^{1/4}
R(new) = 4 R(old)
Hence, If the peak transmitted power is increased by a factor of 16, and the antenna diameter is increased by a factor of two, the radar range will be increased by a factor of 4.
RADAR – RAdio Detection And Ranging
It is a detection system that uses radio waves to determine the speed and distance of a faraway object such as an aircraft.RADAR can be used to find location, surveillance of aircrafts/UAV, speed and direction of motion of far off objects using radio waves.
Radar systems have been used in military applications for
Concept:
The maximum range of a radar system is given as
\({R_{max}} = {\left( {\frac{{{P_t}{}\sigma {A_e}^2}}{{{{\left( {4\pi } \right)}^2}{S_{min}}}}} \right)^{1/4}}\)
Where:
Rmax = Maximum range of the radar
Smin = Minimum Detectable Signal Power
Pt = Peak transmitted power
Ae = Aperture Area
D = Antenna diameter
We observe that
Rmax ∝ D^{1/2} x Pt1/4
Calculation:
P(new) = 16 P(old)
D(new) = 2D(old)
Rmax ∝ D^{1/2} x Pt1/4
R(new) = √2D^{1/2} x 2P1/4
R(new) = 2√2R(old)
R(new) = √8R(old)
Hence, If the peak transmitted power is increased by a factor of 16, and the antenna diameter is increased by a factor of 2, the radar range will be increased by a factor of √8.
Doppler effect in physics is defined as the increase (or decrease) in the frequency of sound, light, or other waves as the source and the observer move towards (or away from) each other.
Waves emitted by a source traveling towards an observer gets compressed. In contrast, waves emitted by a source traveling away from an observer get stretched out.
Doppler effect formula:
\(λ_{o}=λ_{s}\left ( 1+\frac{v}{c} \right )\sqrt{\frac{1}{1-\frac{v^{2}}{c^{2}}}}\)
λ' = wavelength observed by the observer
λ = actual wavelength
v = relative velocity of the source and observer
c = velocity of the light(m/s)
The velocity v is positive for motion away from an observer and negative for motion toward an observer.
Redshift | Blueshift |
The observed wavelength λo of electromagnetic radiation is longer than that emitted by the source. |
The observed wavelength λo of electromagnetic radiation is shorter than that emitted by the source |
The source moves away from the observer. | The source moves towards the observer. |
A radar receiver has a detection SNR threshold of 10 dB for a 4 MHz bandwidth signal at 300 MHz frequency. If the transmit EIRP of the radar is 40 dBW and receive G/T is 10 dB/K, what is the minimum Radar Cross section (in dB-meter square) detectable at 10 km range?
(Given: 10 log (4π) = 11, log 10 (k) = -228.6, k is Boltzmann constant)Concept:
Radar range equation is given by:
\({P_r} = \frac{{{P_t}{G_t}{G_r}{\lambda ^2}\sigma }}{{{{\left( {4\pi } \right)}^3}{R^4}}}\)
Also, the detectable radar cross-section at a distance R is given by:
\(\sigma \le {P_r}\frac{{{{\left( {4\pi } \right)}^3}{R^4}}}{{{P_t}{G_t}{G_r}{\lambda ^2}}}\)
The signal to noise ratio (SNR) at the receiver is given by:
\(\frac{{{P_r}}}{{{N_0}B}} = \frac{{{P_r}}}{{kTB}} = \frac{{{P_t}{G_t}{G_r}{\lambda ^2}\sigma }}{{{{\left( {4\pi } \right)}^3}{R^4}kTB}}\)
SNR in dB will be:
\(SNR = 10\log ({P_t}{G_t}) + 10\log \left( {\frac{{{G_r}}}{T}} \right) + 10\log \left( {\frac{{{\lambda ^2}}}{{{{\left( {4\pi } \right)}^3}{R^4}}}} \right) + 10\log \left( \sigma \right) - 10\log \left( k \right) - 10\log \left( B \right)\;\;\;\;\;\)
\(SNR = EIRP + 10\log \left( {\frac{{{G_r}}}{T}} \right) + 10\log \left( {\frac{{{\lambda ^2}}}{{{{\left( {4\pi } \right)}^3}{R^4}}}} \right) + 10\log \left( \sigma \right) - 10\log \left( k \right) - 10\log \left( B \right)\) ---(1)
Calculation:
Given: SNR = 10 dB; Bandwidth B = 4 MHz; EIRP = 40 dB
\(10\log \left( {\frac{{{G_r}}}{T}} \right) = 10\;dB/K\)
R = 10 km
10 log(k) = - 228.6
For the given frequency f = 300 MHz, the wavelength will be:
\(\lambda = \frac{c}{f} = \frac{{3 \times {{10}^8}}}{{300 \times {{10}^6}}} = 1m\)
Using equation (1), the detectable radar cross-section in dB will be:
\(10\log \left( \sigma \right) = SNR - EIRP - 10\log \left( {\frac{{{G_r}}}{T}} \right) - 10\log \left( {\frac{{{\lambda ^2}}}{{{{\left( {4\pi } \right)}^3}{R^4}}}} \right) + 10\log \left( k \right) + 10\log \left( B \right)\)
\( = 10 - 40 - 10 - 10\log \left( {\frac{{{1^2}}}{{{{\left( {4\pi } \right)}^3}{{\left( {10 \times {{10}^3}} \right)}^4}}}} \right) - 228.6 + 10\log \left( {4 \times {{10}^6}} \right)\)
= 10 – 40 – 10 - (-3 × 11 - 160) - 228.6 + 66
= -9.6 dB-m^{2}