21 consider an antenna array of n elements design an array...

21. Consider an antenna array of N elements. Design an array with one main beam at an angle f = 125\deg and a null at 75\deg . Compare the performance of the array using two designs, one with a three-element array and the other using a ten-element array. In your comparison, include half-power beamwidth, number of side lobes, spacing between the array elements, and the progressive phase difference between the array elements.

Transcript

00:01 In this question we have been given the number of elements n is equal to 3 and the spacing between the element that is d is lambda divided by 2 and then error factor which is given by af and this is equal to mode of sin n psi divided by 2 whole divided by sin psi divided by 2.

00:28 And we will consider this as equation first here this psi is equal to beta times b cos phi plus alpha where alpha is the phase between the elements so for for a broadside array of maximum radiation that occurs perpendicular to the line so therefore this phi will be of 90 degree and if phi is 90 degree then this alpha will be equal to 0 degree so this psi will become now beta times b cos phi and this can be also written as 2 pi divided divided by lambda and d is lambda divided by 2 and then this is cos phi so this lambda and lambda will get cancelled out and we will get this psi equal to phi times cos phi so now this error factor will be 0 when this n psi divided by 2 is equal to plus minus times of k times phi where k is equal to 1 2 3 and so on so these are the null pattern and this is n divided by 2 pi times cos phi equal to plus minus k times phi so this phi and phi will get cancelled out and we will get cos phi phi equal to plus minus 2k divided by n and then this direction of nulls we can write phi equal to cos inverse of plus minus 2k divided by n where k is varying from 1 to 3 and n we have been given 3.

02:38 So here we can write for n equal to 3 and k equal to 1 we will get phi equal to cos inverse of 2 divided by 3 and this is equal to plus minus 47 .93 degree then again if we put n equal to 3 and now k is equal to 2 so if we put all this value so this phi will will be equal to now cos inverse of 4 divided by 3.

03:14 And as we know, the maximum value of cos is 1...

Question

Please give Ace some feedback