RADIATING ELEMENTS MAY BE: DIPOLES, SLOTS, POLYRODS, LOOPS, HORNS, HELIX, SPIRALS, LOG PERIODIC STRUCTURES AND EVEN DISHES Dipoles simple structures,

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1 ANTENNA ARRAYS Array - collection of radiating elements An array may be: 1D (linear), 2D (planar), 3D (frequency selective) structure of radiating elements Purpose More directivity, Steereable beams Radiation pattern of an array is controlled by: relative amplitudes and phases of signals applied to each element Scanning angle with maintaining reasonable gain depends on Pattern of individual radiating element Mutual coupling between elements causes input impedance of an individual element to be different from its own impedance in isolation Large mutual coupling results in: Poor radiation pattern, Raised side lobe level, Mismatched array An array can simultaneously generate: several search beams and /or tracking beams within the same aperture 2D Planar arrays are more popular and most commonly used A Planar arrays may be Flat (square or rectangular), Curved (cylindrical or spherical) in shapes. Rectangular planar arrays generate Fan Shaped beams Square & Circular planar arrays generate Pencil Shaped beams Planar arrays have lower SL than curved due to Superior illumination Planar arrays require less maintenance Planar arrays Scan 60 O 70 O angle - off broadside Several faces needed for full hemisphere coverage More coverage by single array by curved shaped arrays Elements in an array may be arranged in different forms in terms of grid structures and cells therein

2 RADIATING ELEMENTS MAY BE: DIPOLES, SLOTS, POLYRODS, LOOPS, HORNS, HELIX, SPIRALS, LOG PERIODIC STRUCTURES AND EVEN DISHES Dipoles simple structures, used in both mechanical and electrical scanning, broad band pattern, widely used Slots easier to construct at MW frequencies, broad band pattern, widely used, simple structures Both dipoles and slots are used where larger angular coverage with single array is desired Polyrods, Helix, Spirals, Log Periodic Structures more directive Horns used at UHF & MW frequencies, easier to construct ELEMENT SPACING Element spacing governs grating lobes Limits the angle to which a beam can be steered Elements May be: equispaced (normally used) or non- equispaced (rarely used) Advantages of un-equispaced for comparable beam-width lesser number of elements required broadband operation possible Pattern with lower SL possible NUMBER OF ELEMENTS Few to several thousands Limited only by practical constraints DUE TO MUTUAL COUPLING

3 Radiation pattern gets modified Radiation resistance changes Large coupling results in: poor pattern, raised side-lobes and mismatch with Tx or Rx circuit. Two element array Let an array contains two elements 0 and 1. These are separated by a distance d. Antenna o is fed I 0 and antenna 1 is fed I 1 current. These currents bear the relation: I 1 = K I 0. K is the ratio of magnitudes of the currents. K = I 1 / I 0 is the phase angle by which I 1 leads I 0 The phase difference between two radiators is given by: = d cos + Where, d = (2 / ) d E T = E 0 (1 + ke j ) If E 0 is the field strength due to antenna 0 alone the total field strength is given by: E T = E 0 (1 + ke j ) = E 0 (1 + K cos + jk sin ) = E 0 [(1 + k cos ) 2 + (K sin ) 2 ] If I 0 = I 1, K = 1 E T = 2 E 0 cos( /2) = 2 E 0 cos[{( d/ )cos } + /2) Case I Currents of same amplitudes (K = 1) and same phase ( = 0 o ) d = /2 or d = (2 / )d = = d cos + = cos + cos + 0 o E T = E 0 [(1 + k cos ) 2 + (K sin ) 2 ] E T = E 0 [{1 + cos( cos )} 2 + sin 2 ( cos )] Case II Currents of same amplitudes (K = 1) and opposite phases ( = 180 o ) d = /2 or d = (2 / )d = = d cos + = cos o = cos + E T = E 0 [{1 + cos( cos + )} 2 + sin 2 ( cos + )] Case III Currents of same amplitudes (K = 1) and in phase quadrature ( = - 90 o ) d = /4 or d = (2 / )d = /2 = d cos + = ( /2) cos + = ( /2) cos - 90 o E T = E 0 [{1 + cos[( /2 )cos - ( /2)]} 2 + sin 2 {( /2) cos +( /2)}] Case IV

4 Currents of same amplitudes (K = 1) and in phase quadrature ( = - 90 o ) d = /2 or d = (2 / )d = = d cos + = cos + = cos - 90 o E T = E 0 [{1 + cos[( )cos - ( /2)]} 2 + sin 2 {( ) cos -( /2)}] Case V Currents of same amplitudes (K = 1) and in phase ( = 0 o ) d = = d cos + = 2 cos + 0 o d = (2 / )d = 2 = 2 cos + E T = E 0 [{1 + cos(2 cos )} 2 + sin 2 (2 cos )] Three element array In a three antenna elements 0, 1 and 2 separated by distances d1 and d2. Antenna 0 is fed I 0 1 is fed I 1 and antenna 2 is fed I 2 current. These currents bear the relation: I 1 = K 1 I 0 1 and I 2 = K 2 I 0 2 K 1 and K 2 are the ratio of magnitudes of the currents (i.e. K 1 = I 1 / I 0 and K 2 = I 2 / I 0 ). 1 2 are the phase angles by which I 1 and I 2 lead I 0 REFERENCES 1. Jesic, H., Antenna Engineering Hand book, TMH, Jordon, E. C. Balmain, K. G., Electromagnetic Waves and Radiating Systems, PHI, ND, Kraus, J. D., Marhefka, R. J. and Khan A. S., Antennas For All Applications, III ed. TMH, ND, Linear (multi element) Arrays At HF for point to point communication the desired RP is a single narrow beam or lobe Since the radiation is concentrated into a single lobe more directivity can be obtained To obtain such a beam a multi-element linear array is usually employed An array is linear when elements are equispaced along a straight line If elements are fed with currents of equal magnitudes and having the uniform progressive phase shift along the line the array is called uniform linear array Let I 0, I 1, I 2,. be the currents fed to element 0, 1, 2,.. Assuming I 0 = I 1 = I 2 =. = I in terms of amplitudes.

5 And I 1 = I 0, I 2 = I 0 2, I 3 = I 0 3, in terms of phase variation. Since I 0 results in field E 0, I 1 results in field E 1 and so on, the total resulting field at P can be written in the form: E T = E 0 + E 1 e j + E 2 e j2 +. In view of equality of currents E 0 = E 1 = E 2 =. E T = E 0 (1+ e j + e j2 +.+ e j(n-1) ) = d cos +, = (2 / ) is the phase shift constant and d is the distance between equispaced elements. It is a geometric progression and can be written in following two ways: 1. Central source is taken as reference 2. End source is taken as reference Principal maxima of an array Nulls of Array BROADSIDE ARRAY (BSA) G 30 db [1000 s element for same gain/ complex feed] For this case = 0, cos = / d or = 90 o Principal max. is at = /2 If first null occurs at ( /2 + ) cos ( /2 + ) = sin = 1 / d = 1 /2 d = (2 /n). ( /2 d) = /nd For small, sin = /nd or 2 = 2 /nd Thus width of principal lobe for uniform BSA is reciprocal to the length of array i.e. nd END FIRE ARRAY (EFA) Gain (G) 15 db Narrow band For this case = - d, = 0 o, Principal max. is at = 0 o = d cos + = d cos - d = d (cos - 1)

6 Let the null occurs at 1 = + = d (cos - 1) = -2 /n cos - 1 = (-2 /n). (1/ d) = (-2 /n). ( /2 d) = - /nd since cos - 1 = 2 sin 2 ( /2) = - /nd for small, sin( /2) ( /2) Thus ( /2) 2 = /2nd or ( ) 2 = 2 /nd or 2 = 2 (2 /nd) Thus width of principal lobe for uniform EFA is greater than that of uniform BSA for an array of the same length, since [2 /nd] < [2 (2 /nd)] Using the above relations RP s for different n are obtained and plotted in three subsequent slides DIFFERENT TYPES OF ARRAYS AND THEIR PROPERTIES: BSA, EFA, CLA, PSA Broadside Array (BSA) All elements are along a line drawn to their respective axis All elements are identical All elements are equally spaced All element are fed with currents of equal amplitude and same phase BSA is bidirectional Gain G 30 db can be obtained Endfire Array (EFA All elements are along a line drawn to their respective axis All elements are identical All elements are equally spaced Element are fed with currents of equal amplitudes with progressively increasing phases EFA in general is unidirectional. It can be made bidirectional by feeding half of the elements with 180 o out of phase End fire is narrow band with gain (G) 15 db Parasitic Arrays (PSA) It contains one driven (feeding) element (DE) and some parasitic elements (PE) Yagi array falls in this category Amplitude and phase of the induced currents in PE depends on its tuning (length) and spacing between elements (PE) and DE to which it is coupled PE which is longer ( 5%) w.r.t. DE is called reflector PE which is shorter to DE is called director Reflector makes the radiation maximum in

7 the direction from PE to DE Director makes the radiation maximum in the direction from DE to PE. Collinear Array (CLA) Arrangement of elements is shown below Elements have collinear arrangement Antennas are mounted end-to-end in a straight line Elements are fed with equal and in-phase currents Its RP closely resembles RP of BSA Its RP to the principal axis has (everywhere) circular symmetry with the main lobe CLA is also called broadcast or omni-directional array Gain of CLA is max when spacing between elements is between 0.3 and 0.5 Its construction becomes at these lengths and thus elements are placed very close No. of elements is generally not more than 4. Further increase in number does not result in higher gain MULTIPLICATION OF PATTERNS Non isotropic but similar point sources Consider two sources 1 and 2, each source is a short dipole oriented parallel to x axis Each of these source has a field pattern given by E 0 = E 0 sin (1) (A) Let sources 1 & 2 have equal currents and any phase difference The total (combined) field due to two sources is: E T = 2 E 0 cos /2 (2) Where = d cos + = (2 d/ ) cos + Substitute (1) in (2) Set 2E 0 = 1 for normalized magnitude E T = 2E 0 sin cos /2 = sin cos /2 (3) Eqn. (3) is the multiplication of field of individual source by the field (pattern) of two isotropic sources (B) Let sources 1 & 2 have un-equal currents and any phase difference E T = E 0 [(1 + k cos ) 2 + k 2 sin 2 ] (4) Substitute (1) in (4) and normalize by putting 2E 0 = 1 we get E T = sin [(1 + k cos ) 2 + k 2 sin 2 ] (5) Eqn. (5) is again the multiplication of field of individual source by the field (pattern) of two isotropic sources Examples (A and (B) exhibit the principle of pattern multiplication which can be expressed as below The field pattern of an array of non-isotropic but similar point sources is the product of the pattern of an individual source and the pattern of an array of isotropic point sources having the same location, relative amplitudes and phase as the non-isotropic point sources If the field of non-isotropic source and the array of isotropic sources vary in phase with the space angle the above statement can be written as: The total field pattern of an array of non-isotropic sources is the product of the pattern of the individual source and the pattern of an array of isotropic point sources each located at the phase center of the individual source and having the same relative amplitude and phase, while the total phase pattern is the sum of the phase patterns of the individual source and the array of isotropic point sources In mathematical language the above statement can be written as:

8 E = f(, ) F(, ) {f p (, ) +F p (, )} Where, f(, ) is the field pattern of individual source f p (, ) is the phase pattern of individual source F(, ) is the field pattern of array of isotropic sources F p (, ) is the phase pattern of array of isotropic sources The number of elements are also referred as order of an array In linear arrays as order of an array increases, directivity increases but side lobes (SLs) also appear. SLs are always undesirable except in some special (military) applications If d > /2, SLs will appear in UP as well as in GP If either UP or GP contains SLs, RRP [also referred as total pattern (TP)] will also contain SLs Binomial arrays In Binomial arrays numbers of array elements has to be such that these can be represented in terms of 2 n. Thus for 4-elements (= 2 2 ) or n = 2 For 8 elements (= 2 3 ) or n = 3. For n elements with /2 spacing between elements the relative current in the r th element from any one end of the array is given as: n! / [r! (n-r)!], where r = 0, 1, 2, For 4-element array:

9 Current in 1 st element = 2! /[0! (2-0)!] = 1 Current in 2 nd element = 2! /[1! (2-1)!] = 2 Current in 3 rd element = 3! /[2! (2-2)!] = 1 Thus current ratio in elements is: 1:2:1 For 8-element array: Current in 1 st element = 3! /[0! (3-0)!] = 1 Current in 2 nd element = 3! /[1! (3-1)!] = 3 Current in 3 rd element = 3! /[2! (3-2)!] = 3 Current in 4 th element = 3! /[3! (3-3)!] = 1 Thus current ratio in elements is: 1:3:3:1 Binomial arrays have no SLs PATTERN SYNTHESIS BY PATTERN MULTIPLICATION Pattern synthesis is the process of finding the source or array of sources that produces a desirable pattern EXAMPLE A broadcasting station has to operate in KHz range. It needs a RP in horizontal plane. The RP has to meet three conditions shown Antenna producing this pattern contains an array of 4 vertical towers (radiators) Currents in these towers have to be of equal magnitudes, but the phases may be adjusted to any relationship. There is no restriction on spacing or geometrical arrangement of towers. Each tower may be taken as an isotropic point source. Problem thus requires to find space and phase relationship of 4 isotropic sources located in horizontal plane fulfilling the above requirements. Solution:- Look for a pair of isotropic sources whose RP has a broad lobe with maxima in N and a null in SW direction. Name this pattern as primary pattern. Such an arrangement of two isotropic sources which results in an endfire radiation pattern is shown From the consideration of pattern shape as a function of separation d and phase, spacing between /4 and 3 /8 is a suitable choice. Let d = 0.3 Then E = cos /2 (1) Where, = d cos + d = (2 / )d = (2 / )x(0.3 ) = 0.6 = 0.6 cos + (2)

10 For appearance of null at = 135 o cos135 o = sin 45 o = Thus d cos = 0.6 x = = (2k +1) k = 0, 1, 2,. (3) Equating (2) and (3), after substitution of relevant values, gives = (2k +1) , for k = 0, = 104 o or /2 = 52 o Next find an array of two isotropic point sources to produce a pattern having a null at = 270 o and a broad lobe in north. Pattern of such an array may be referred as secondary pattern. Fig. represents secondary isotropic source with ( =) 180 o phase difference between elements. The secondary pattern with assumed d = 0.6 is given by: E = cos /2 Where, = 1.2 cos + By principle of multiplication the total pattern is: E = cos (54 cos - 52 o ) cos (108 cos - 90 o ) These patterns satisfy the given requirements Complete array is obtained by replacing each of the isolated sources of the secondary pattern by two source array producing the primary pattern The mid point of each primary array is its phase center, so this point is placed at the location of a secondary source The complete antenna is then a linear array of 4 point sources. In above figure each source (dot) represent a single tower and all towers carry equal currents. Current of tower 2 leads current of 1 and current of tower 4 leads current of 3 by 104. Current in 1 & 3 and 2 & 4 are in phase opposition as shown by arrows. The phase variation around the primary, secondary and total array is shown with the phase center at the center point of each array and also at the southern most sources. The arrangement of the array with their phase centers for both the case is shown below

11 Non-isotropic and dissimilar point sources In this case: the principle of multiplication is not applicable the fields of sources must be added at each angle for which the total field is calculated Consider two dissimilar sources 1 and 2. Let source 1 is located at the origin and source 2 is located at x-axis (at x = d). The total field in general: E = E 1 + E 2 = E 0 [{f( ) + af( ) cos } 2 + {af( ) sin } 2 [f p ( ) + arc tan{[af( ) sin ]/ {f( ) + af( ) cos }] (1) Where Field of source 1 is E 1 = E 0 f( ) {f p ( )} Field of 2 is E 2 = ae 0 F( ) [F( ) + dr cos + ] E 0 is constant, a is the ratio of max amplitude of source 2 to source 1 (0 a 1) = dr cos + - f p ( ) + F( ) = relative phase of source 2 w.r.t 1 f( ) = relative field pattern of 1 f p ( ) = phase pattern of source 1 F( ) = relative field pattern of 2 F p ( ) = phase pattern of source 2 If the sources are in phase and equal in amplitudes such uniform distribution yields maximum directivity or gain. The pattern has HPBW of 23 o but SLL is relatively high. Amplitude of first SL is 24% of main lobe maximum. To reduce SLL of linear in-phase BSA of n sources, John Stone proposed that the source amplitudes should be proportional to coefficients of a binomial series of form: (a + b) n+1 = a n-1 + (n-1) a n-2 b +[(n-1) (n-2)/2!] a n-3 b 2 + ] For arrays of 3 to 6 sources the relative amplitudes are tabulated in Pascal s triangle form. Any inside number is the sum of the adjacent numbers in the row above Next slide shows normalized field patterns of BSA of 5 isotropic point sources /2 apart. All sources are in phase but relative amplitudes have four different distributions viz. edge, uniform, optimum and binomial. Only half of the pattern is shown. All patterns are adjusted to the same maximum amplitude. LINEAR ARRAY WITH NON UNIFORM AMPLITUDE DISTRIBUTIONS DOLPH-TECHEBYSCHEFF DISTRIBUTION Consider a linear array of n isotropic (in-phase) point sources of uniform spacing d.

12 Direction = 0 is taken to the array with origin at the center of array Amplitudes of individual sources are given as: A0, A1, A2, A3,, Ak Total field at a large distance in direction is the sum of the fields of symmetrical pairs of the source For n even (ne) the fields E ne can be written as: E ne = 2A 0 cos /2 + 2A 1 cos3 / A k cos[(ne -1) /2] (1) Where = d sin = [2 d/ ] sin Each term in (1) represents the field due to a symmetrically disposed pair of sources Let 2(k+1) = ne, where k = 0, 1, 2, K = 0 ne = 2, K = 1 ne = 4 so that (ne 1)/2 = (2k +1)/2 Ne = 2 corresponds to k = 0 ½ = ½ Thus (1) with N = ne/2 becomes: N = 2/2 =1 For n odd (no) the fields E no can be written as: E n0 = 2A 0 + 2A 1 cos +2A 2 cos A k cos[(no-1) /2] (3) Let 2k+1 = no, k = 0, 1, 2,. Eqn. (3) becomes where N = (no-1)/2 Equations 2 and 4 are termed as even number and odd number Fourier series respectively. Both of these can be recognized as finite Fourier series on N terms. EX: Let an array of 9 isotropic point sources spaced /2 apart. Sources have the same amplitudes and phase. Hence the coefficients are related as under. 2A 0 = A 1 = A 2 = A 3 = A 4 = 1/2 The expression for field pattern is given by (4) is: E 9 = 1/2 + cos + cos2 + cos3 + cos4 (5) In Eqn. (5) First term (k = 0) is a constant, field pattern is a circle of amplitude ½ Second term (k-1) may be regarded as fundamental term of Fourier series and gives the pattern of two sources A1 on either side of the center. This pattern has 4 lobes of max amplitude of unity Third term (k = 2) and fourth term (k = 3) can similarly be interpreted. The table illustrates detailed information about the same. In view of field pattern expressions the total pattern of an array of 9 isotropic sources can be resolved into 5 Fourier components The 9 sources include a center source and 4 pairs of symmetrically located sources

13 Development of Dolph-Tchebysceff distribution Consider Eqns. (2) and (4) to be as polynomials of degrees ne-1 and no-1. Consider the case of BSA where = 0. Thus = dr sin = d sin (5) By de Moivré s theorem e jm /2 = n cos(m /2) + j sin (m /2) = (cos /2 + j sin /2) m (6) Taking the real part of (6) cos (m /2) = Re(cos /2 + j sin /2) m (7) On expanding (7) in binomial series we get cos (m /2) = cos m /2 {m(m-1)/2!} cos m /2 sin 2 /2 + {m(m-1)(m-2)(m-3)/4!} cos m-4 /2 sin 4 / (8) Putting sin 2 /2 = 1 - cos 2 /2 and substituting particular values of m (8) reduces to the following: m = 0 cos (m /2) = 1 m = 1,, = cos /2 m = 2,, = 2cos 2 /2-1 m = 3,, = 4cos 3 /2-3 cos /2 m = 4,, = 8cos 4 /2-8cos 2 /2 +1 etc. (9) Let x = cos /2 Eqn. (9) reduces to: m = 0 cos (m /2) = 1 m = 1,, = x m = 2,, = 2x 2-1 m = 3,, = 4x 3 3x m = 4,, = 8X 4-8X 2 +1 etc. (10) Polynomials of Eqn. (10) are called Tchebyscheff polynomials which may be designated as: Tm(x) = cos (m /2) (11) Values of first 8 Tchebyscheff polynomials are given in next slide. T 0 (x) = 1 T 1 (x) = x T 2 (x) = 2x 2 1 T 3 (x) = 4x 3-3x T 4 (x) = 8x 4 8x T 5 (x) = 16x 5 20x 3 + 5x T 6 (x) = 32x 6 48x x 2-1 T 7 (x) = 64x 7 112x x 3 7x (12) The degree of Polynomial is the same as the value of m The roots of polynomials occur when cos m /2 = 0 or when m /2 = (2k-1) /2, k = 1, 2, 3, The roots of x, designated as x are: x = cos[(2k-1) /2m] Since cos m /2 can be expressed as a polynomial of degree m, Eqn. (2 and 4) can be expressed as polynomials of degree 2k+1 and 2k respectively. Eqns. 2 and 4 which express the field pattern of a symmetric in-phase equispaced linear array of n isotropic point sources are polynomials of degree equal to the number of sources minus 1.

14 Set array polynomial 2 & 4 equal to Tchebyscheyff polynomial of like degree [m = n-1] and equate array coefficients to the coefficient of Tchebyscheyff polynomial. From the Fig. of Tchebyscheyff polynomial for m = 0 to 5 following properties can be noted. 1. All polynomials pass through point (1,1) 2. For values of x in (-1 x +1) range all polynomials lie between ordinate values of +1 and All roots occur between -1 x +1 and all max. values in this range are 1. Dolph s method of applying Tchebyscheyff polynomial to obtain optimum pattern Consider an array of 6 sources. The Field pattern will have a degree of polynomial = 6 1 = 5 The ratio of main lobe max to the minor lobe level R is specified Equate polynomial of pattern to the Tchebyscheyff polynomial Point (x 0, R) on the T 5 (x) polynomial curve corresponds to main lobe max The minor lobes are confined to a max value of unity Roots of polynomial correspond to the nulls of the field pattern If ratio R (= main lobe max / side lobe level) is specified the beam width to the first null (x - x 1 ) is minimum or If beam width is specified the ratio R is maximized (i.e. SLL minimized) To summarize the procedure rewrite Eqns. 2 and 4 Both of these equations are functions of /2. To get the relative pattern, factor 2 may be dropped from these. Select no. of sources (say n), the order of array polynomial = n -1 Select Tchebyscheyff polynomial T n-1 (x) Let m = n 1, Choose R, solve T m (x 0 ) = R for x 0 For R > 1, x 0 is also >1 But since x = cos /2 is restricted to -1 x 1, a change of scale is to be introduced such that w = x/x 0 = cos /2, w is restricted to 1 The pattern polynomial (Eqns. 2 & 4) may now be expressed as a polynomial in w Equate Tchebyscheyff polynomial T n-1 (x) and the array polynomial by putting w = cos /2 in 2 and 4, T n-1 (x) =En Obtain coefficient of array polynomial yielding the Dolph-Tchebyscheyff (DT) amplitude distribution, which is optimum for specified SLL. Tchebyscheyff polynomial of degree 5 with relation to coordinate scales EX: D-T Distribution for 8 source array To find the amplitude distribution to produce minimum BWFN, HPBW and gain if n = 8, d = /2, SLL = 26 db. SLL below main lobe maxima is given by SLL = 26 db = 20 log 10 R, thus R = 20 Tchebyscheyff polynomial degree is: n -1 = 8 1 = 7, thus T 7 (x 0 ) = 20. T 7 (x) = 64 x x x 3-7x X 0 may be obtained by trial and error method or by using the relation: X 0 = (1/2)[{R+ (R 2-1)} 1/m + {R- (R 2-1)} 1/m ] For R = 20 and m = 7, x 0 = 1.15 which is > 1, thus scaling is required.

15 On using factor 2 for scaling T 0 (x) = 1 T 1 (x) = x T 2 (x) = 2x 2 1 T 3 (x) = 4x 3-3x T 4 (x) = 8x 4 8x T 5 (x) = 16x 5 20x 3 + 5x T 6 (x) = 32x 6 48x x 2-1 T 7 (x) = 64x 7 112x x 3 7x E 8 = A 0 w + A 1 (4w 3-3w) + A 2 (16w 5-20w 3 + 5w) + A 3 (64w 7 112w w 3-7w) (1) But w = x/x 0 E 8 = (64A 3 / x 7 0 ) x 7 + {(16A 2-112A 3 ) / x 5 0 } x 5 + {(4A 1-20A A 3 ) / x 3 0 } x 3 + {(A 0-3A 1 + 5A 2-7A 3 ) / x 0 } x (2) Equating (1) and (2), by taking w = x E 8 = T 7 (x) 64A 3 / x 7 0 = 64 or A3 = x 7 0 = (1.15) 7 = 2.66 Similarly A 2 = 4.56, A 1 = 6.82, A 0 = 8.25 The amplitudes of 8 sources are: The relative amplitudes of these sources are: