 Hello and welcome to today's lecture 6 of module 2. Now today we shall learn about a very interesting topic and instead of telling you verbally for a change let me show you a schematic, okay. So assume an elevated mountainous terrain say where you are standing after a successful say hiking trip. Now again assume a jet is moving towards you and that you are able to hear the sound of the engine. The jet has flown past you and even now you are able to hear the sound of engine but in a slightly different manner. So what just happened was you happen to hear a different frequency or a different pitch when the jet was approaching you and another pitch when the jet was moving away from you, okay. So you heard a different frequency, a different pitch when the jet was approaching you and another pitch when the jet was moving away from you and this apparent change in frequency of pitch when a sound source is either moving towards or away from a listener that is you in the example is called as Doppler effect. Remember this can also be observed when a listener is moving either toward or away from a sound source either ways. So Doppler effect is named after the Austrian physicist and mathematician Christian Doppler who was the first to describe this effect in the 19th century. So we just learned that a shift in frequency which is observed when either an observer is moving towards a sound source or a sound source is moving towards an observer is known as Doppler shift, the change in frequency that is observed is known as Doppler shift. Now this is a course on microwave remote sensing, is not it? So why are we discussing about Doppler shift? So let me take you back to one of the earlier lectures where you may recollect that I showed you this diagram and then I mentioned that the elliptical shaded region, illuminated region is known as the footprint that is the area on the ground that is viewed by a radar system as it is moving in the orbital track. And then this particular footprint you can see I have divided into two parts, one is a shaded region and the other is a non-shaded region, is not it? And this is the center of the beam as indicated. So we mentioned during the discussions that all of these that is towards the right that is the shaded region of the footprint they are moving towards the satellite and all of these that is towards the bottom trailing end left side it is moving away from the satellite as the radar is looking sideways and the footprint is the area illuminated on the ground which the radar sees as it is moving. Now compare this with the earlier example when a car say with a siren travels past you. So there is a change in pitch that you can clearly hear when the siren is approaching you and when the siren is moving away from you. And the pitch can be higher as it approaches you and lower when it recedes, is not it? Now similarly in the case of synthetic aperture radar with the platform being satellite. So the platform here is moving a synthetic aperture radar has side looking geometry and if we assume the target to be stationary for one instant. If we assume the target to be stationary as the footprint is moving along the target the return echo from the leading part of the footprint that is the shaded region here shall have a different frequency than the return echoes from the trailing part of the footprint that is the part which is not shaded. So this is what we discussed that as the footprint is moving because the footprint is the area that is viewed by a radar system which is on a platform let it be aircraft or satellite. So the footprint itself shall contain two parts. One part that is moving along the target that is the shaded region shown here and the other part which is a non shaded region it is the trailing part. So what happens is as the footprint moves along the target the return echoes from the leading part of the footprint shall have a different frequency and the return echoes from the trailing part of the footprint shall have a different frequency. So what we do is we can quantify this effect absolutely we can quantify this effect in the following manner. So in the screen in front of you I have written distance travelled by a wave from its source in a time interval t equals relative velocity into t. And as a relative velocity it is the relative velocity in the direction between the source and the detector remember there is a source and there is a detector. So now does this change in distance impact the face of a wave that is measured by a detector because till now we understood that there is a different frequency of the return echo that is coming from the leading part of the footprint and that is coming from the trailing part of the footprint. And then we understood that the distance travelled by a wave from its source in time interval t is nothing but velocity into time we are using relative velocity in the direction between source and detector. Now my question is does this change in distance impact the face of a wave that is measured by a detector. So the answer would be absolutely yes the change in distance is certainly going to have an impact on the face which is measured at the detector. So if phi be the additional phase change I can represent phi as nothing but d by lambda into 2 pi. So what is this the additional phase change the additional phase change I am expressing it in the form of phi equal d by lambda into 2 pi. So let me rewrite it as relative velocity into time by lambda into 2 pi simple expression is not it what does it represent it represents the additional phase change. Now just to reiterate what we discussed we are talking about Doppler shift and then we are trying to understand why Doppler shift is important with respect to synthetic aperture radar images and that is when I mentioned that there will be a change in frequency of the return echoes observed from the leading and trailing part of the footprint as it moves with the platform. And then we discussed that the change in distance is going to have a impact on the face and then we saw the expression to quantify this. Now please note that we are more interested in understanding the rate of change of phase with time okay which means phi is not the quantity we are interested in we are more interested in d phi by dt rate of change of phase with respect to time. And if you recollect one of our very earlier lectures where I was mentioning about wave you may remember that we defined the angular frequency denoted by omega I will try to put all these together in a storyline for now we can have expressions that help us calculate the Doppler frequency shift okay Doppler frequency shift as V relative that is relative velocity by lambda wavelength okay. Now with this simple background let us try to understand more about something known as aperture synthesis in synthetic aperture radar using Doppler interpretation remember we tried to understand the geometric interpretation the same manner we are going to understand about the Doppler interpretation as to how aperture synthesis is happening in a synthetic aperture radar. So by now we have the understanding that echoes returning from objects which are located in the front of the footprint shall be Doppler shifted to higher frequencies while the echoes from the receding or trailing end of the beam shall be Doppler shifted to lower frequencies okay and we also know the expression for Doppler frequency shift. Now remember even though the instrument is moving for convenience till now we have considered that the ground is stationary and the instrument is moving. And even though the instrument is moving for convenience we can consider that the ground surface is moving through the footprint and for one instant if you assume as the ground surface is moving through the footprint there are a history of return echoes because you know it happens in a continuous sequence. So as the ground surface is moving through the footprint there is a history of return echoes that are generated from the target and this shall approximately cause a linear shift in frequency okay linear shift in frequency. So a radar system should be capable of not only sending coherent pulses but it should also be capable of accumulating the return echoes over successive pulses for synthesizing an antenna which is very much longer than the physical antenna. Let me reiterate let me repeat a radar system should be of course capable of sending coherent pulses at the same time it should also have the ability to accumulate the return echoes over successive pulses and why are we doing this for synthesizing an antenna which is much much longer than the physical antenna that is why we call it synthetic aperture radar. Now let us consider in detail as to what actually happens when a target moves through a radar beam. Before that please don't forget that there is something known as a zero Doppler line which we discussed in one of the earlier lectures that is the line which is directly perpendicular to the flight path okay alright. So assume the Doppler shifted frequency is f suffix d okay let me write the expression Doppler shifted frequency suffix d and the original transmitted frequency I am going to denote it as f suffix naught. So what is this original transmitted frequency I am denoting it by suffix naught. I have already seen that the Doppler frequency can be written as Doppler frequency can be written as 2 into relative velocity by lambda you may ask where did the factor of 2 come from. So a factor of 2 has been added here now because the signal is Doppler shifted twice okay transmission reception. So the signal is Doppler shifted twice and hence a factor of 2 has been added in the expression for Doppler frequency. There is a full bandwidth of Doppler frequencies possible isn't it think about it for one instant. I have given you the original transmitted frequency as f naught the Doppler shifted frequency as f d which means the full bandwidth full bandwidth of Doppler frequencies, range of Doppler frequencies is nothing but f naught plus f d minus f naught minus f d highest lowest and full range is highest minus lowest. This gives you the full bandwidth of Doppler frequencies. Just to reiterate what we discussed about Doppler frequency and the signal is Doppler shifted twice. So you are adding a constant value of 2. We also discussed about the full bandwidth of Doppler frequencies. I can alter this expression isn't it because f naught gets cancelled ultimately it becomes 2 into f d but then we already know that the Doppler frequency is nothing but 2 into relative velocity by lambda. But then how well can a radar instrument differentiate between signals in time okay the factor of time comes in. How well can a radar instrument differentiate between signals in time that is a question we are trying to answer now. So to understand the time resolution okay it means that to understand time resolution we need to represent how well an instrument can differentiate between signals in time. But then we also can translate this to azimuth distance isn't it along track distance. Remember we discussed about azimuth direction along track direction and for translating we need to include something known as azimuthal speed of the sensor. Azimuthal speed of sensor please note that this is not equal to the speed of light. Azimuthal speed of sensor is nothing but the speed of platform carrying the sensor. It can be an aircraft it can be a satellite. So our aim is to understand azimuthal resolution okay azimuthal resolution why are we trying to understand all this because we want an answer to the question how well can a radar instrument differentiate between signals in time okay. So to understand that let us try to express azimuthal resolution as velocity in the azimuthal direction by Doppler bandwidth. So let me write that down azimuthal resolution is equal to velocity in the azimuthal direction Doppler relative velocity I am going to use the expression because by now we are familiar. So the relative velocity is nothing but azimuthal speed of sensor multiplied by sign of half of full beam width of antenna. So I am going to use this expression theta b to signify angle half of full beam width of antenna. Once we start our modules on antennas and the different types of antennas and the properties of antennas so that time it will be clear. But for now the relative velocity can be expressed as a function of azimuthal speed of sensor and sign of half of full beam width of antenna. But then when theta is very much less than 1 radian sin theta can be written nearly equal to theta b that is because the theta the value of the angle is very much less than 1 radian. So now I can write theta b is nothing but wavelength lambda by 2 into length of antenna, length of antenna. I can rewrite the azimuthal spatial resolution as azimuthal speed by Doppler bandwidth which means azimuthal spatial resolution I can write it as azimuthal speed Doppler bandwidth. This in turn can be written as length of antenna by 2. I am going to highlight this expression because it tells us that azimuthal spatial resolution equal to length of antenna by 2. Just to reiterate what we discussed, why are we discussing all this? We need an answer to the expression how good can a radar instrument differentiate between signals in time and that is when we started to understand about the speed of the platform. We denoted it as azimuthal speed or V suffix azimuthal and then we started to understand how it is related, how azimuthal spatial resolution is related to the length of antenna. The inferences what we get from this expression is that azimuthal spatial resolution is half the length of antenna. It is independent of platform whether it is aircraft or satellite it is independent of platform and it is independent of wavelength no matter which wavelength you use azimuthal spatial resolution is independent of wavelength very important results inferences I am going to star it here. So all in all we were trying to understand a few basics about Doppler frequency shift and how it is applicable to synthetic aperture radar images as part of this class. So let me hope that you found this lecture useful and I will see you in the next class. Thank you.