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    Fourier Transforms and Wave Analysis

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    The question is the example on page 2 of the attachment (entitled 'Uniform Transducer'). it states that the center of the finger is at z'=L/4. I assume this is an arbitrary position. For Eq (2.4.6), the contribution from the left-hand finger is added. I'm not entirely sure how this equation is arrived at. It does not look like a mere addition.

    The contributions from the other fingers are added to give Eq(2.4.7). Whilst I understand where the first line in (2.4.7) comes from (although unsure where lambda appeared from), I do not understand at all how the second line of (2.4.7) is arrived at from the integral given on the previous line. I understand that the Double Angle Formula is used to get line 3 of (2.4.7).

    Eq(2.4.8) perplexes me: if N is large then kl (zero response) ~ 2*pi.... which we know equals +/-2N. I'm missing something here.

    Finally, how is Eq (2.4.9) and Eq (2.4.10) arrived at?

    Please do not omit any lines (however simple).

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    Solution Preview

    About z'=l/4:

    You assume correctly that the choice z'=l/4 just before equation (2.4.5) is arbitrary, chosen for convenience of calculation (it is just a matter of convenient choice of the origin).

    Equation (2.4.6):

    In equation (2.4.6) there is a typo:
    The brackets as shown in red in the attached figure 246.jpg are mysteriously missing however they certainly should be there.
    If you correct this error, the continuation to the last line of equation (2.4.6) should not be a difficulty for you.

    λ in equation (2.4.7)

    To understand the λ in equation ...

    Solution Summary

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