2. Continuous Wave Radar / Doppler

3. Objectives Distinguish between Pulsed radar and CW radar. Explain what is required for CW radar to “see” a contact. Explain and calculate Doppler Shift Calculate the combined speed between two contacts. List and identify the components of the basic CW radar system Solve for range for a CW radar system (FMCW). Distinguish between Pulsed radar and CW radar. Explain what is required for CW radar to “see” a contact. Explain and calculate Doppler Shift Calculate the combined speed between two contacts. List and identify the components of the basic CW radar system Solve for range for a CW radar system (FMCW).

4. Characteristics Duty Cycle = 1 P peak = P avg High SNR R min = 0 Uses 2 antennas –one transmit and one receive Basic CW radar cannot measure range! –No time delay because there are No “pulses” Duty Cycle = 1 P peak = P avg High SNR R min = 0 Uses 2 antennas –one transmit and one receive Basic CW radar cannot measure range! –No time delay because there are No “pulses”

5. Applications of CW Radar CW Illuminator –SPG-62 CWI for SM-2 (Aegis) –MK-92 FCS (FFG-7) TT/CWI Navigation (Doppler Navigation) Speed Guns Radar Proximity Fuze Altimeter (FMCW) CW Illuminator –SPG-62 CWI for SM-2 (Aegis) –MK-92 FCS (FFG-7) TT/CWI Navigation (Doppler Navigation) Speed Guns Radar Proximity Fuze Altimeter (FMCW)

6. Operation Must have Relative Motion Measure Frequency Shift (Doppler) Must have Relative Motion Measure Frequency Shift (Doppler)  Doppler Shift (  f) defined as: f D = 2* V LOS /  Doppler Shift (  f) defined as: f D = 2* V LOS /  “ V LOS ” defined as Speed in the Line of Sight V LOS = V 1 Cos    V 2 Cos    “ V LOS ” defined as Speed in the Line of Sight V LOS = V 1 Cos    V 2 Cos  

7. Doppler Shift Change in the frequency of the electromagnetic wave caused by motion of the transmitter, target, or both. If transmitter is moving: –wavelength reduced by fraction proportional to transmitter speed in direction of propagation. Speed of propagation is constant –therefore frequency must increase –c =  f (in this case) Change in the frequency of the electromagnetic wave caused by motion of the transmitter, target, or both. If transmitter is moving: –wavelength reduced by fraction proportional to transmitter speed in direction of propagation. Speed of propagation is constant –therefore frequency must increase –c =  f (in this case) stst stst ’ = (1 – s/c)  t = 1/f = /c ’

8. Doppler Classification Closing –  doppler  freq CPA –0 doppler Opening –  doppler  freq Closing –  doppler  freq CPA –0 doppler Opening –  doppler  freq

9. Doppler Review

10. LOS Speed Stationary Transmitter

11. Doppler shift can be calculated with knowledge of: transmitter / receiver speed & target speed (v 1, v 2 ), and the angles between respective directions of motion and connecting LOS (  1,  2 ) Combined speed (closure rate) is defined as instantaneous rate of change in range (range rate) Is simple as long as problem remains in 2 dimensions (x,y) Doppler shift can be calculated with knowledge of: transmitter / receiver speed & target speed (v 1, v 2 ), and the angles between respective directions of motion and connecting LOS (  1,  2 ) Combined speed (closure rate) is defined as instantaneous rate of change in range (range rate) Is simple as long as problem remains in 2 dimensions (x,y) v LOS = v 1 cos  1 + v 2 cos  2 LOS Speed 11 11 22 22 v 2 cos θ 2 v 1 cos θ 1

12. Own Course is 270 o T Own Speed is 400 Knots RF = 7.5 x 10 9 Hz Own Course is 270 o T Own Speed is 400 Knots RF = 7.5 x 10 9 Hz Doppler Shift Problem Target Course is 060 o T Target Speed is 600 Knots Target Course is 060 o T Target Speed is 600 Knots Always Draw A Picture

13. 000 090 180 270 000 090 180 270 Own Course is 270 o T Own Speed is 400 Knots RF = 7.5 x 10 9 Hz Target bears 300 o Own Course is 270 o T Own Speed is 400 Knots RF = 7.5 x 10 9 Hz Target bears 300 o Target Course is 060 o T Target Speed is 600 Knots Target Course is 060 o T Target Speed is 600 Knots 1. Draw Coordinate System 2. Draw v 1 and v 2 vectors 3. Draw connecting LOS

14. 000 090 180 270 000 090 180 270 Own Course is 270 o T Own Speed is 400 Knots RF = 7.5 x 10 9 Hz Target bears 300 o Own Course is 270 o T Own Speed is 400 Knots RF = 7.5 x 10 9 Hz Target bears 300 o Target Course is 060 o T Target Speed is 600 Knots Target Course is 060 o T Target Speed is 600 Knots  1 = Relative Bearing to target = difference between True Bearing and True Heading  1 = Relative Bearing to target = difference between True Bearing and True Heading  2 = Relative bearing from target’s perspective = Reciprocal of LOS bearing to target 4. Determine  1 and  2 300 o T 120 o T 22 11

15. 000 090 180 270 000 090 180 Own Course is 270 o T Own Speed is 400 Knots RF = 7.5 x 10 9 Hz Target bears 300 o T Own Course is 270 o T Own Speed is 400 Knots RF = 7.5 x 10 9 Hz Target bears 300 o T Target Course is 060 o T Target Speed is 600 Knots Target Course is 060 o T Target Speed is 600 Knots 300 o T 120 o T v LOS is interpreted as the rate of change in range and is frequently called the Range Rate v LOS is interpreted as the rate of change in range and is frequently called the Range Rate v LOS = 400 cos    + 600 cos   v LOS = 346.4 + 300 v LOS = 646.4 kts v LOS = 400 cos    + 600 cos   v LOS = 346.4 + 300 v LOS = 646.4 kts v LOS = v 1 cos  1 + v 2 cos  2  o  o 270 v LOS = 646.4 kts x (0.5144 m/s per kt) v LOS = 332.51 m/s v LOS = 646.4 kts x (0.5144 m/s per kt) v LOS = 332.51 m/s 5. Calculate V LOS

16. f D = (2)(332.51 m/s) x (1/0.04 m) f D = 16,625 Hz or 16.6 kHz f D = (2)(332.51 m/s) x (1/0.04 m) f D = 16,625 Hz or 16.6 kHz v LOS = 332.51 m/s and = 0.04 m f D = 2 (v LOS ) 000 090 180 270 000 090 180 Target Course is 060 o T Target Speed is 600 Knots Target Course is 060 o T Target Speed is 600 Knots 300 o T 120 o T  o  o 270 6. Calculate f D

17. Tips Relative Bearing = True Bearing - Heading Reciprocal Bearing = True Bearing + or - 180 degrees 240 o 060 o

18. While operating in the Arabian Gulf, you are flying your F-18 Hornet on a course of 325 o T at 450 knots, heading towards the nearest Tanker Track. An audio warning from your APG-73 all weather search and track radar, operating at 3.5 GHz, notifies you that you have a target that bears 300 o R, with a Doppler shift of 4500 Hz. You estimate the contact to be departing Al Amarah airfield on a course of 110 o T Step #1: Draw the PICTURE: While operating in the Arabian Gulf, you are flying your F-18 Hornet on a course of 325 o T at 450 knots, heading towards the nearest Tanker Track. An audio warning from your APG-73 all weather search and track radar, operating at 3.5 GHz, notifies you that you have a target that bears 300 o R, with a Doppler shift of 4500 Hz. You estimate the contact to be departing Al Amarah airfield on a course of 110 o T Step #1: Draw the PICTURE: Example Problem #1

19. You acquire an EA-6B bearing 070 degrees True on course 120 degrees True cruising along at 360 Kts. Your course is 350 degrees True and speed is 400 Kts. Your radar Frequency is 500 MHz. What is the Doppler Shift? You acquire an EA-6B bearing 070 degrees True on course 120 degrees True cruising along at 360 Kts. Your course is 350 degrees True and speed is 400 Kts. Your radar Frequency is 500 MHz. What is the Doppler Shift? Example Problem #2

20. Continuous Wave Radar Components Transmitter –Continuous RF oscillator –Supplies sample signal to mixer Transmitter –Continuous RF oscillator –Supplies sample signal to mixer  f Indicator Discriminator Receiver Transmitter Mixer Oscillator f f'f'

21.  f Indicator Discriminator Receiver Transmitter Mixer f f'f' Antenna – Transmit. – Sends out the signal in a specific direction. Antenna – Transmit. – Sends out the signal in a specific direction. Continuous Wave Radar Components Oscillator

22.  f Indicator Discriminator Receiver Transmitter Mixer f f'f' Antenna –Receive return signal Antenna –Receive return signal Continuous Wave Radar Components Oscillator

23.  f Indicator Discriminator Receiver Transmitter Mixer f f'f' Continuous Wave Radar Components Receiver Oscillator

24.  f Indicator Discriminator Receiver Transmitter Mixer f f'f' Continuous Wave Radar Components Mixer – Determines Doppler Shift Mixer – Determines Doppler Shift Oscillator

25.  f Indicator Discriminator Receiver Transmitter Mixer f f'f' Continuous Wave Radar Components Discriminator – Amplifies signal – Eliminates return signal from stationary targets Discriminator – Amplifies signal – Eliminates return signal from stationary targets Oscillator

26.  f Indicator Discriminator Receiver Transmitter Mixer f f'f' Continuous Wave Radar Components Indicator Oscillator

27.  f Indicator Discriminator Receiver Transmitter Mixer Oscillator f f'f' Continuous Wave Radar Components Power Supply

28. Range Determination – CW Radar Frequency Modulation Continuous Wave (FMCW) –Linear increase in frequency (changed at a constant rate) –Frequency changes over specific period (T) “Time Stamp” on transmitted signal –Returned signal compared to transmitted signal Difference in measured frequency gives a TIME (Δt) Using Δt gives RANGE Frequency Modulation Continuous Wave (FMCW) –Linear increase in frequency (changed at a constant rate) –Frequency changes over specific period (T) “Time Stamp” on transmitted signal –Returned signal compared to transmitted signal Difference in measured frequency gives a TIME (Δt) Using Δt gives RANGE

29. Range Determination – CW Radar

30. Knowing frequency difference, can calculate time difference and can therefore determine range! Such that and Knowing frequency difference, can calculate time difference and can therefore determine range! Such that and Range Determination – CW Radar

32. FMCW Range Determination Example Example: The frequency sweep is 420-455 Mhz with a period of 12 microseconds. The first return occurs at 435 MHz what is the range? Example: The frequency sweep is 420-455 Mhz with a period of 12 microseconds. The first return occurs at 435 MHz what is the range?

33. = 771.42 m  f = 435 Mhz - 420 Mhz = 15 Mhz F = 455 Mhz - 420 Mhz = 35 Mhz T = 12  sec  t = 5.14  sec