武汉科技大学 - 信号与系统习题精解第1章(3)

2019-08-01 23:58

f1?t?11f2(t)-2.5-1.500.51.52.5t-3-2-101234tf3(t)1-3-2-101234-1-11f4?t?t01tf5?t?11f6?t?-101234t-3-2-10-11234t

题11解图

12.求下列积分。 (1) ????cos?4t[??(t)??(t)]dt; (2) ?t???[?(t?2)??(t?2)]dt

(3) ?(4?t2)??(t?4)dt; (4) ??36???(t?2)?(x?t)dt

解:

(a) ?(b) ????cos?4t[??(t)??(t)]dt??1;

t??6[?(t?2)??(t?2)]dt?u(t?2)?u(t?2)

(c) ?(4?t2)??(t?4)dt?8

?3?(d) ??(t?2)?(x?t)dt???????(x?2)0x?2x?2

13. 画出下列各信号的波形。

(1) f1(n)?(n?1)u(n)

(2) f2(n)?n[u(n)?u(n?5)] (3) f3?n????0.5?u?n?

?n(4) f4(n)?2u?n?

?n解:各波形如题13解图所示。

11

ff2?n?1?n?44332?2111234n1234nf3?n?42f4?n?32?11?1234n1234n-2题13解图

14. 对于题14图中的信号f(t),为以下各式作图。 f(t) (a) f1(t)?f(t?3); 2 (b) f2(t)?f(2t?2); (c) f3(t)?f(2?2t); (d) f?204t4(t)?f(?0.5t?1); 题14图

(e) fe(t)(偶分量);

(f) f0(t)(奇分量)。 解: 各波形如题14解图所示。

f1?t?f2?t?2211-5-31t13tf3?t?f4?t?2211-112t-10-22tfe?t?fo?t?2211-4-224t-4-224t-1

题14解图

12

15.求下列函数的卷积积分f1?t??f2?t? (1) f1?t??e?3tu?t?,f2?t??u?t?;

(2) f1?t??f2?t??e?3tu?t? (3)f1?t??tu?t?,f2?t??e?tu?t?

(4) f1?t??u?t?1?,f2?t??u?t?5? (5) f1?t??tu?t?,f2?t??u?t?1??u?t?2?

现求解如下: (1) f1?t??e?3tu?t?,f2?t??u?t?;

解:

f1?t??f2?t?????3???eu????u?t???d???t?3t0e?3?d???1t?3?3e?1?

03?1?e?u?t(2) f?3t1?t??f2?t??eu?t? 解:

ft1?t??f2?t???t3??t???t?3t0e??e?3d???0ed??e?3t??3t0d??teu?t?

(3)f?t1?t??tu?t?,f2?t??eu?t?

解:

f1?t??f2?t??f?1?1?t??f??1?2?t??u?t???t0e??d??u?t???1?e?t?u?t???t?1?e??0?d?????e???t??t?e?t0?1?u?t?(4) f1?t??u?t?1?,f2?t??u?t?5?

解:

f1?t??f2?t??u?t?1??u?t?5???t?6?u?t?6?

(5) f1?t??tu?t?,f2?t??u?t?1??u?t?2?

解: f??1?'1?t??122tu?t?,f2?t????t?1????t?2?f21?t??f2?t??12tu?t?????t?1????t?2???12?t?1?2u?t?1??12?t?2?2u?t?2? 16.已知

13

(1)f1?t??tu?t???t?e?t?1?u?t?

(2)f1?t???e?tu?t????1?e?t?u?t???1?e?(t?1)?u?t?1? 求f1?t? 现求解如下:

(1)f1?t??tu?t???t?e?t?1?u?t?,求f1?t? 解:

把f1?t??tu?t???t?e?t?1?u?t?求导2次 f1?t????t??1?e??t???eu?t?

?t

(2)f1?t???e?tu?t????1?e?t?u?t???1?e?(t?1)?u?t?1?,求f1?t? 解: 左式:

?f?t???eu?t?????f?t???e??t??eu?t???f?t?????t??eu?t???t?t?t?t111?f1?t??f1?t??eu?t??t

右式:

ddt??1?e?u?t???1?e?t?(t?1)?u?t?1?????t??eu?t??e??t????t?1??e?t?t?t?0??t?1?u?t?1??e??t?1???t?1???t?1????t??eu?t??e??t????t?1??e?eu?t??e?t??t?1???t?1?u?t?1??e??1?1?u?t?1?所以

f1?t??f1?t??eu?t??eu?t??e?t?t??t?1?u?t?1?

?(t?1)把f1?t???eu?t????1?e?t?t?u?t???1?e?(t?1)?u?t?1?代入上式,得

??t?1?f1?t??1?e??t?u?t???1?e?u?t?1??eu?t??e?tu?t?1?f1?t??u?t??u?t?1?

17.已知下列f1?n?,f2?n?的值,求f1?n??f2?n?。 (1)f1?n??f2?n??u?n? (2)f1?n??u?n?,现求解如下:

f2?n????n????n?1?

14

(1)f1?n??f2?n??u?n? 解:

f1?n??f2?n??u?n??u?n???n?1?u?n?

(2)f1?n??u?n?,f2?n????n????n?1?

解:

f1?n??f2?n??u?n?????n????n?1???u?n??u?n?1?

18.已知fn1?n??anu?n?,f2?n??bu?n?,求f1?n??f2?n?。

解:

nfn1?n??f2?n??au?n??bn?n???aibn?i

i?0当a?b时

nnif1?n??f2?n??anu?n??bn?n???aibn?i?bn???a??i?0i?0?b?i?11???a??

n?1?bn?b??b?an?11?ab?ab当a?b时

nnfnn1?n??f2?n??au?n??b?n???aibn?i??bn??n?1?bn

i?0i?0上二式在n?0成立,故得

?bn?1?an?1fn??anu?n??bn?n????b?au?n?,a?b1?n??f2?

???n?1?bnu?n?a?b当a?b?1时

u?n??u?n???n?1?u?n?

19.已知f1?n??sinn?,fn2?n??au?n?,f3?n????n??a??n?1?,求f1?n??f2?n??f3?n?。 解:

15

f1?n??f2?n??f3?n??sinn??au?n??[??n??a??n?1?]n这里用到性质: u?n????n??u?n?1?

an??n??a0??n????n?

?sinn???anu?n????n??anu?n??a??n?1???sinn???anu?n??an?1u?n????n?1???sinn???anu?n??an?1u?n????n?1?? ?sinn???anu?n??anu?n?1???sinn???an[??n??u?n?1?]?anu?n?1???sinn????n??sinn?16


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