Файл: Ersoy O.K. Diffraction, Fourier optics, and imaging (Wiley, 2006)(ISBN 0471238163)(427s) PEo .pdf
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APPENDIX A: THE IMPULSE FUNCTION |
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Figure A.1. Finite pulse of unit area.
The impulse function can also be written as the derivative of the unit step function: dðtÞ ¼ ddt uðtÞ ðA:1-5Þ
The impulse function can be obtained by limiting operations on a number of functions whose integral has the value 1. Some examples are given below.
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8 a!1½1 |
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atu t |
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> alim |
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d t |
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A:1-6 |
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ð |
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> a |
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0þ p2pa |
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< a |
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sin at |
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> ! |
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> lim |
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It is observed that these functions in the limit approach 0 at t 6¼ 0 and 1 at t ¼ 0. In the last case, the limit at t 6¼ 0 does not approach 0, but the function oscillates so fast that Eq. (A.1-6) remains valid.
Another way to construct the impulse function is by using the triangular pulse shown in Figure A.2. Its area is 1. As a shrinks towards 0, the area of 1 remains
(a) |
(b) |
Figure A.2. (a) A triangular pulse, and (b) its derivative.
380 APPENDIX A: THE IMPULSE FUNCTION
Property 4. Sum of shifted impulse functions |
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k |
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jTj k |
1 |
d t |
¼ k |
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e j2pkt=T |
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1 |
cosð2pkt=TÞ |
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¼ 1 |
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¼ 1 |
¼ 1 þ 2 k |
¼ |
1 |
ðA:1-13Þ |
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EXAMPLE A.1 Evaluate the integral |
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1 |
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t |
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ð |
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cosð10tÞdt: |
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Solution: |
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d |
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¼ d |
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ðt |
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3Þ |
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¼ 3dðt 3Þ |
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Hence, |
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1 |
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ð d |
t |
1 cosð10tÞdt ¼ 3 |
ð |
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dðt 3Þ cosð10tÞdt |
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3 |
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¼ 3 cosð30Þ |
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¼ |
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p |
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EXAMPLE A.2 Evaluate |
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lim |
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e t2=32a2 |
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in terms of the impulse function.!1 |
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a |
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p2pa |
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Solution: |
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p1 |
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e t2 |
=32a2 ¼ p |
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e ðt=4Þ2=2a2 |
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2pa |
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2pa |
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Since |
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lim |
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e t2=2a2 |
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t |
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!1 |
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¼ dð Þ |
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a |
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p |
2pa |
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it follows that |
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1 |
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t2 |
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t |
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a!1 p2pa e |
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¼ d |
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¼ 4dðtÞ |
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4 |
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lim |
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APPENDIX A: THE IMPULSE FUNCTION |
381 |
EXAMPLE A.3 Evaluate
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a |
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lim |
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a!1 p ð0 |
cosðtxÞdx |
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in terms of the impulse function. |
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Solution: It is known that |
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a |
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1 |
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sin at |
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ð0 |
cosðtxÞdx ¼ |
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ð Þ |
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a > 0 |
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Since |
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lim |
sinðatÞ |
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a!1 |
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pt |
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¼ dð |
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it follows that |
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a |
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t |
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dð |
Þ ¼ a!1 p ð0 |
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cosðtxÞdx |
ðA:1-14Þ |
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This result is often written as |
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a |
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lim |
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e jtxdx |
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dðtÞ ¼ a!1 2p ð |
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a