 this class. So in this class we shall focus on numericles because we need to know what is the practical implication or importance about the relationships that we have been studying so far. So we will learn how to practically interpret the relationships as I said which you have been learning throughout the course of last few classes. Starting with radar equation. So just a quick recap, we have tried to understand that radar equation gives you the fundamental relationship between the power received and the power transmitted and the received power is a function of antenna gain that is g, wavelength, lambda, radar cross section that is sigma and the range that is r. Now whenever we try to solve numericles pertaining to radar equation, there are things to be careful about mainly in terms of units. So here p has units energy per time, sigma will be distance squared because it is radar cross section Rcs. R and lambda will have units of distance, okay. Now let us slowly start to solve the first problem. Given in the screen in front of you is the question, it states that the operating center frequency of a radar is 10 gigahertz and it is transmitting 100 watts peak power. The system has an antenna gain of 9 decibels and the pulse repetition frequency that is Prf of 10 kilohertz pulse width of 1 mu second. The question is find the power that is received by the radar if it observes a target having Rcs, radar cross section of 10 meter square at a range of 0.7 kilometers. So firstly what we have to do is we should understand what is given as part of the question. So we have the frequency that is given, frequency is given as 10 gigahertz okay and then gain of antenna is given as 9 decibels, pulse width is given, pulse width is given as 1 mu second. Now 1 mu second means 1 microsecond which is 1 millionth of a second which is nothing but 10 power minus 6 seconds. So always after reading the question we should try to relate it to the notations we have been using as part of the lecture. So I am going to use the notation tau p to represent pulse width which is given as 10 power minus 6 seconds and then the radar cross section sigma it is given as 10 meter square okay. What else is given? The range R is given as 0.7 kilometers okay given details and the question is what is the power received? Now is the wavelength given somewhere? No right, but the frequency is given. So we know the relationship that c equal to nu lambda where c is the speed of light, lambda is the wavelength, nu is frequency which means I can write 3 into 10 power 8 equal to 10 into 10 power 9 into lambda. Why did I write 10 into 10 power 9? Because frequency is given as 10 gigahertz, gigahertz okay. So I will get the value of lambda as 0.03 meters which is also equal to 3 centimeters. So now we have all the details required for us to use the radar equation alright. But then I am a little confused about gain of the antenna you know it is given as 9 decibels. I want to know what is decibels because we have not discussed that in detail while covering the derivation of a radar equation. So decibel is 1 tenth of a bell and sound is measured in decibels and it is also measured in logarithmic units because it gives the ratio between 2 powers okay 9 decibels and 1 decibel can be written as 10 into log to the base 10 ratio of 2 powers P1 by P2 okay which means I can write 9 decibels equal to 10 into log to the base 10 ratio of 2 powers P1 by P2 and I need the ratio of 2 powers to be used in the radar equation. So if you rearrange a few terms you are going to get P1 by P2 equal to 10 raised to 9 by 10 which is nothing but 7.94. So what is this? This is the gain of the antenna that is 9 decibels represented in the form of P1 by P2 ratio of 2 powers because gain is measured in decibels. Decibels as I mentioned earlier is 1 tenth of a bell it is expressed as logarithmic units and you can use the expression 1 decibel equals 10 log 10 of P1 by P2. You can use this expression to find the value of gain, antenna gain alright. So now let us try to solve the radar equation. We already know that the power received is nothing but transmitted power into gain of the antenna square into wavelength square into radar cross section that is sigma the whole divided by 4 pi cube multiplied by range raised to 4 okay. So we can start plugging in the values for each of the terms now. So we will get Pt it is 100 into 7.94 okay gain expressed as P1 by P2 ratio of 2 powers multiplied by lambda meters 0.03 square into 10 meter square which was given as the radar cross section sigma whole divided by 4 pi cube multiplied by range 700 raised to 4 0.7 kilometers 700 meters 700 raised to 4. So you will get the answer as 11.94 into 10 to the power minus 14 easy. See once you take care of the units of each term that make up the radar equation, the rest becomes direct straightforward alright. Now let us move to the second question. Remember we are trying to solve problems related to radar equation which we have derived as part of the previous lectures okay. So the second question reads like this a 4 gigahertz radar system has an unambiguous range of 30 kilometers and uses a parabolic dish antenna of a Pulture efficiency 0.6 and diameter 5 meters. There are two parts to this question. So the first part reads find the gain of the radar's antenna if the system achieves a range resolution of 10 meters and the second part reads also find the range to target if the target gets detected at a time delay of 100 mu second okay. So as before I will give it some time so that you can note down the values and then we will start solving. So we have to first write the given details. So here range is given as 30 kilometers okay. It is given that we have a parabolic dish antenna of a Pulture efficiency 0.6. Let us write that down. A Pulture efficiency is given as 0.6 and it is also mentioned that a parabolic dish antenna is used of diameter 5 meter. So let us write that down. Antenna diameter 5 meters. The first question is find the gain if the system achieves a range resolution of 10 meters. Now to solve this let us try to recollect the definition of gain of an antenna, antenna gain. If you remember there was a part in the previous lecture where I compared the radar to ears okay. Like the ears are trying to point at different directions to get sound waves. In a similar manner we had the dish the antenna of a radar that is having directional sensitivity which is rotating at 360 degrees at the same time which is trying to collect the information, the received power okay. Now we also discussed that gain of an antenna can be written as 4 pi by lambda square lambda wavelength into physical area of antenna. Let us write that down. Physical area antenna multiplied by antenna efficiency, antenna efficiency. So we have 4 pi by lambda square into physical area of antenna into antenna efficiency. Are all the terms given in the question? Because we have seen range is given, frequency is given which means lambda needs to be computed. So as before let us try to compute the value of lambda. We know that C is nothing but new lambda. Lambda can be written as 3 into 10 power 8 speed of light by 4 into 10 power 9. So you get the value of lambda as 0.075 meters okay. Now let us try to come back to the relationship. Gain g equal to 4 pi by lambda square that is 0.075 square into physical area of the antenna okay. It is given that the antenna has a diameter of 5 meters and it is also mentioned that a parabolic dish antenna is used, is not it? Which means the physical area of the antenna is nothing but pi into 5 square by 4 physical area of the antenna. Antenna efficiency was given as 0.6 okay 60% efficiency 0.6. So we have a direct calculation here. Once you complete it you are going to get the answer something like 26292 gain of the antenna okay. So we have completed the first part of the question just to reiterate. The second part was mentioning that find the range to target if the target is getting detected at a time delay of 100 mu second okay 100 mu second. Which means the question is find the value of R if tau P equal to 100 mu second. So we can also write tau P as 100 into 10 power minus 6 seconds. We have seen that in an earlier question, isn't it? If you remember the range can be written as C tau P by 2. We have covered this as part of an earlier lecture C tau P by 2 range speed of light tau P is already given which means I can further write it as 3 into 10 power 8 into 100 into 10 power minus 6 by 2 which means range can be obtained as 15 kilometers, isn't it? 15 kilometers second part of the question. So throughout this particular course there are very many relationships equations that we have been covering as part of lectures time and again. It is enough if you understand the concepts you do not have to buy hard the equations every time. So just have an understanding of what is range and then what are the parameters that range is going to depend on velocity is distance by time. If so, we have range nothing as C into tau P by 2. Now let us go to one more question. This time I will try to complicate things a little bit. So the question 3 reads given below are the radar parameters used in an airport. As you hypothetical radar parameters are given which is used in an airport, the question is can it detect or get a reasonable signal to noise ratio S by N ratio for a small aircraft at 110 kilometers. That is a question. So the radar parameters as given range is 110 kilometers. The aircraft cross section is given as 1 meter square that is nothing but sigma. The peak power is given as 1.5 megawatts. Pulse width tau P is given as 0.6 microseconds. Bandwidth B is given as 1.67 megahertz remember and always using the terms which were used as part of this course. For example, we use the term B to refer to bandwidth or to refer to range aircraft cross section RCS sigma pulse width tau P. I am using the same notation so that it is easy for you to recollect. Again frequency is given as 2800 megahertz pulse repetition frequency or PRF is given as 1200 hertz. Antenna size is given as 4.9 meters wide by 2.7 meters high. Azimuth beam width is given as 1.35 degrees and if you see there is a new term that is given, is not it? System noise temperature 800 degree Kelvin. We have not come across system noise temperature so far. Antenna gain is given as 30 decibels and it is also mentioned that you can assume losses as 9 decibels. See before solving this question, first of all we need to understand what are the different information that is shared with us. So, let us try to summarize the given details and then I will try to mention a little bit more about signal to noise ratio. So, first of all to understand what is given we have PT that is 1.5 into 10 power 6 watts. PT gain is nothing but 30 decibels which is nothing but 10 log to the base 10 P1 by P2. We are more interested to get the ratio of two powers P1 by P2. So, if you rearrange you will get P1 by P2 equal to 10 raise to 30 by 10 which is nothing but 1000 and then frequency is given as 2800 megahertz as before. Now, we know how to compute wavelength lambda which is nothing but 3 into 10 to the power 8 by 2800 into 10 to the power 6. You will get the value as 0.107 meters and of course, sigma is given as 1 meter square range is given as 110 kilometers 110 into 10 power 3 meters. So, this is what is given. At this point let us have a little bit of discussion about a few terms that were not discussed as part of lecture. See as far as radar equation is concerned we discussed when we can rearrange the terms of the radar equation to represent signal to noise ratio PR by N. At this point it is worthwhile to mention that there are different sources that can contribute to noise in a radar. Assume that what you see here is a Doppler weather radar DWR. There are different sources of noise that can contribute to the radar such as a few sources are the solar noise, we can have the atmospheric noise, we can have man-made interference, man-made interference, we can have noise from the ground as well. The noise power at receiver is denoted as capital N which can be written as nothing but multiplication of K which is nothing but the Boltzmann constant given here multiplied by B which is nothing but bandwidth of receiver and multiplied by T s which is system noise temperature. So, what is this? This is the noise power at receiver because while discussing about radar equation to get signal to noise ratio we just divided PR by N and mentioned that it represents signal to noise ratio but then we never expanded N. So, N that is noise power at receiver can be written as K into B into T s, K is nothing but the Boltzmann constant, B is nothing but bandwidth of the receiver and T s is nothing but the system noise temperature. So, we are discussing about the signal to noise ratio of radar here and that is when I mentioned that there are few sources that contribute to noise in a radar which can be either from solar noise, it can be atmospheric noise, man-made interference, noise from ground these are just a few sources that contribute to noise and we need to account that in the relationship as well whenever we are trying to estimate the signal to noise ratio of a radar. So, let us move forward. So, we know the given details that is P T was given transmitted power 1.5 into 10 power 6 watts and then G was given as 30 decibels. We have found out the P 1 by P 2 ratio of 2 powers as 1000 and if you recollect the frequency was given as 2800 megahertz we got lambda value as 0.107 meters and we have already written sigma which is given as 1 meter square, range is given as 110 kilometers that is 110 into 10 power 3 meters and then now our aim is to use it in the radar equation. So, let us try to incorporate the new terms that we just discussed that is K which is nothing but 1.38 into 10 power minus 23 watts per hertz degree Kelvin. We know T s was given as 800 degree Kelvin, V bandwidth was given as 1.67 megahertz which is nothing but 1.67 into 10 power 6 hertz. The losses we were asked to assume it as 9 decibels. So, now we know how it can be converted to the ratio of power P 1 by P 2 which is going to be 7.94 alright. So, now we can try to plug in the relationship, use the relationship for signal to noise ratio. It can be written as s by n signal to noise ratio equal to P t transmitted power into g square to lambda square into sigma the whole by 4 pi cube into r to the power 4. Remember instead of just writing n, I am going to use K T s V L. So, we can substitute the values now that is 1.5 into 10 to the power 6 into 10 to the power 3 into 10 to the power 3 into 0.107 square into 1 that is sigma the whole by 4 pi cube into 110 kilometer in meters we have written into 1.38 into 10 to the power minus 23 that is K. And now we can write the value of T s that is 800 into 1.67 into 10 to the power 6 that is B into 7.94 losses 9 decibels. I can further do the calculations and write the expression as 1.7 into 10 to the power 10 that is the numerator and in the denominator we can get 4.24 into 10 to the power 12 into 10 to the power minus 23 into 10 to the power 6 into 10 to the power 15. Now, what we do is we try to calculate the ratio you will get a value as 0.4, but then 0.4 can be converted to decibel. Remember till now we were trying to convert decibel into ratio of power P 1 by P 2. Now, when I am calculating signal to noise ratio I am getting a value of 0.4. So, 0.4 can be converted into decibels which will come to minus 3.94 decibel per pulse. I am getting a negative value, negative signal to noise ratio it means that the signal power is lower than the noise power. So, we have the answer here signal power lower than the noise power because we have got a negative signal to noise ratio. So, in this section we have tried to cover few numericals pertaining to radar equation and then we also got to understand a little bit about the signal to noise ratio. Let me hope that you could follow these numericals and I shall meet you in the next class. Thank you.