本文格式為Word版，下載可任意編輯——積分變換習(xí)題解答11-2

1．求矩形脈沖函數(shù)f(t)??解：

?j?t?j?tA1?e????eF(?)?F?f(t)???f(t)e?j?tdt??Ae?j?tdt?A?????0j???j??0?A,0?t???的Fourier變換.

??0,其他????2.設(shè)F???是函數(shù)f?t?的Fourier變換，證明F???與f?t?有一致的奇偶性.

證明：F???與f?t?是一個(gè)Fourier變換對(duì)，即F???????1??f?t?e?j?tdt，f?t??F???ej?td????2π??假使F???為奇函數(shù)，即F??????F???，則

f??t??1??1??j???t?j????tF?ed???F??ed???????2π???2π???1??（令???u）??F?u?ejutdu

2π??（換積分變量u為?）??所以f?t?亦為奇函數(shù).

1??j?tF?ed???f?t????2π??假使f?t?為奇函數(shù)，即f??t???f?t?，則

F???????????j??t?j??tf?t?e??dt???f??t?e??dt??????（令?t?u）??f?u?e?j?udu

??（換積分變量u為t）?????f?t?e?j?tdt??F?????所以F???亦為奇函數(shù).

同理可證f?t?與F???同為偶函數(shù).

4．求函數(shù)f?t??e?t?t?0?的Fourier正弦變換，并推證

???0?sin??π??d??e???0?1??22解：由Fourier正弦變換公式，有

?t??ftsin?tdt?eFs(?)?Fs?ft?????0sin?tdt???0????e?t??sin?t??cos?t??????1??21??20由Fourier正弦逆變換公式，有

f?t??Fs?1??Fs(?)???2??2???sin?tF(?)sin?td??d?s2??00ππ1??由此，當(dāng)t???0時(shí)，可得

???0?sin??ππ??d??f????e???0?21??225．設(shè)F??f?t????F(?)，試證明：

1）f?t?為實(shí)值函數(shù)的充要條件是F(??)?F(?)；2）f?t?為虛值函數(shù)的充要條件是F(??)??F(?).

證明：在一般狀況下，記f?t??fr?t??jfi?t?其中fr?t?和fi?t?均為t的實(shí)值函數(shù)，且分別為f?t?的實(shí)部與虛部.因此

F??????????f?t?e?j?tdt??fr?t??jfi?t??cos?t?jsin?t?dt????????????f?t?cos?t?fi?t?sin?t????dt?j?????fr?t?sin?t?fi?t?cos?t??dt???r?Re??F??????jIm??F?????

????fr?t?cos?t?fi?t?sin?t?其中Re?F?????????dt，?a???Im??F???????????f?t?sin?t?fi?t?cos?t??dt?b????r1）若f?t?為t的實(shí)值函數(shù)，即f?t??fr?t?,fi?t??0.此時(shí)，?a?式和?b?式分別為

Re??F?????????f?t?cos?tdt??r????Im?F???????f?t?sin?tdt

??r所以

F?????Re??F???????jIm??F??????

?Re??F??????jIm??F??????F???

反之，若已知F?????F???，則有

Re??F???????jIm??F???????Re??F??????jIm??F?????

此即說(shuō)明F???的實(shí)部是關(guān)于?的偶函數(shù)；F???的虛部是關(guān)于?的奇函數(shù).因此，必定有

F?????????fr?t?cos?tdt?j?f?t?sin?tdt????r亦即說(shuō)明f?t??fr?t?為t的實(shí)值函數(shù).從而結(jié)論1）獲證.

2）若f?t?為t的虛值函數(shù)，即f?t??jfi?t?,fr?t??0.此時(shí)，?a?式和?b?式分別為

??Re??F?????????fi?t?sin?tdtIm??F?????????fi?t?cos?tdt??所以

F?????Re??F???????jIm??F??????

??Re??F??????jIm??F?????

??Re??F??????jIm??F?????

????F???

反之，若已知F??????F???，則有

Re??F???????jIm??F????????Re??F??????jIm??F?????

此即說(shuō)明F???的實(shí)部是關(guān)于?的奇函數(shù)；F???的虛部是關(guān)于?的偶函數(shù).因此，必定有

F??????????fi?t?sin?tdt?j?f?t?cos?tdt，????i亦即說(shuō)明f?t??jfi?t?為t的虛值函數(shù).從而結(jié)論2）獲證.

6.已知某函數(shù)的Fourier變換F(?)?解：F(?)?f?t???sin?sin??，求該函數(shù)f?t?.

?為連續(xù)的偶函數(shù)，由公式有

π??1??sin?j?tF?ed??cos?td?????2??π0?1??sin?1?t??1??sin?1?t??d??d???2π0?2π0?但由于當(dāng)a?0時(shí)

??sina???sina???sintπd??d(a?)?dt??0??0??0t2當(dāng)a?0時(shí)

??sina???sin(?a)?πd???d????0??0?2?1t?1?2，???sina??1d??0,所以得f?t???，當(dāng)a?0時(shí)，?t?1

0??4?0，t?1??7．已知某函數(shù)的Fourier變換為F????π??δ????0??δ????0???，求該函數(shù)f?t?.

解：由函數(shù)δ?t?t0?g?t?dt?g?t0?，易知

1??f?t??F???ej?td??2π??1??1??j?tj?t?πδ???ed??πδ???ed?????00??2π??2π??

11?ej?t?ej?t?cos?0t

????02???028．求符號(hào)函數(shù)（又稱正負(fù)號(hào)函數(shù)）sgn?t???變換.

??1,t?0?1,t?0的Fourier

解：簡(jiǎn)單看出sgn?t??u?t??u??t?，而F[u(t)]?F(?)?1??a1?πδ(?).j????9．求函數(shù)f?t???δ?t?a??δ?t?a??δ?的t??δt??????222?????a?Fourier變換.

解：

1????a??a???j?t?F????F?ft?δt?a?δt?a?δt??δt???ed????????????2????2?2??????1??j?t?e?e?j?t?e?j?t?e?j?ta2?t??at?at??t??2???a?2??

a?cosa??cos?.

210.求函數(shù)f?t??costsint的Fourier變換.解:已知

F??sin?0t???jπ??δ????0??δ????0???

由f?t??costsint?sin2t有F?f?t????δ???2??δ???2??????2211.求函數(shù)f?t??sin3t的Fourier變換.

j?t?解:已知F?e???2πδ????0?,由

01πj?ejt?e?jt?j3jt3f?t??sint???e?3ejt?3e-jt?e?3jt???8?2j?3即得

πj?F?ft?????4??δ???3??3δ???1??3δ???1??δ???3???

π??12.求函數(shù)f?t??sin?5t??的Fourier變換.

?3?解:由于

π?13?f?t??sin?5t???sin5t?cos5t

3?22??3π?πj????δ???5??δ???5??.故F?f?t????δ???5??δ???5????22??j??t???F???，其中?e14.證明：若F????t?為一實(shí)數(shù)，則

1??cos?t?F????F?????F???????2F??sin??t????1?F????F??????2j?其中F????為F???的共軛函數(shù).

證明：由于F???????ej??t??e?j?tdt

F????????????ej??t?j?tedt??????e?j??t??e?j?tdte?j?tdt??cos??t?e?j?tdt?F??cos??t?????????e1?F????F??????????2?j??t??e2?j??t?同理可證另一等式.

17．求作如圖的鋸齒形波的頻譜圖.（圖形見(jiàn)教科書）.

?12π?ht,0?t?T解：?0?,f?t???T

T?0,其他?1C0?T?T01f?t?dt?T1T?T01hhtdt?T21TC