哈工大版信號(hào)與系統(tǒng)第1章習(xí)題

1、.11-1 繪出下列信號(hào)波形繪出下列信號(hào)波形1. tu(t)(ttut02. (t-1)u(t)3. tu(t-1)() 1(tutt01) 1( ttut0114. (t-1)u(t-1) 1() 1(tutt01.21-31-3、寫出如圖所示各波形的函數(shù)表達(dá)式、寫出如圖所示各波形的函數(shù)表達(dá)式(a)(tft0221)()() 2() 2(21)()() 2() 2(21)(tuttututtuttututtf.3(b)2() 1()()(tutututf)2()()2(21)()2()2(21)(tututtututtf)(tft02112.4(c)(tft0TE)()(2sin)(Ttut

2、utTEtf.51-41-4、試證明、試證明cost,cos2t,cosntcost,cos2t,cosnt是在區(qū)間是在區(qū)間(0,2(0,2 ) )的正交函數(shù)集的正交函數(shù)集解：根據(jù)兩函數(shù)正交條件：解：根據(jù)兩函數(shù)正交條件：2020coscos)()(jtdtitdttgtgji20)cos()cos(21dttjitji)(0)sin()sin(2120jijitjijitji.6202202cos)(itdtdttgi20 12cos21dtit2022sin21tiit因此，函數(shù)集在因此，函數(shù)集在(0,2 )區(qū)間上是正交函數(shù)集區(qū)間上是正交函數(shù)集.71-7、利用沖激信號(hào)的抽樣性質(zhì)，求下列表示式

3、的函數(shù)、利用沖激信號(hào)的抽樣性質(zhì)，求下列表示式的函數(shù)值值 dttttf)()(. 20)() 0(00tftfdttt)4)(3(. 21431-8、利用沖激信號(hào)的抽樣性質(zhì)，求下列表示式的函數(shù)、利用沖激信號(hào)的抽樣性質(zhì)，求下列表示式的函數(shù)值值 .81-9 已知已知f(t)的波形如圖，試畫出的波形如圖，試畫出)2 ()(1tftg和和) 32()(2tftg的波形的波形)(tft011231解：解：1、g1(t)的波形：的波形：)( tf t011123反褶：反褶：.9)2(tft011121移位：移位：f(2-t)=f-(t-2)2、g2(t)=-f(2t-3)的波形：的波形：1)2( tft01

4、123)(tft011231展縮：展縮：.10移位：移位：f(2t-3)=f2(t-1.5)1) 32(tft01123沿橫軸反褶：沿橫軸反褶：-f(2t-3)1) 32(tft01123.111-10 已知已知f(t)的波形如圖，試畫出的波形如圖，試畫出 下列函數(shù)的波形：下列函數(shù)的波形：1. f(3t)2. f(t/3)u(3-t)dttdf)(. 3tdf)(. 40tf(t)2311.120tf(t)12311. 解：解：0tf(3t)12312. 解：解：0tf(t/3)1231456789.130tf(t/3)u(3-t)12313. 解：將解：將f(t)表示為函數(shù)形式表示為函數(shù)形式

5、)3() 1()()(tutRtRtf所以，所以，)3() 1()()(ttututf0tf(t)1231(1)0tf(t)1231.144. 解：將解：將f(t)表示為函數(shù)形式表示為函數(shù)形式) 3() 1()()(tutRtRtf所以：所以：ttduRRdf)3() 1()()(tttddd3101) 1(當(dāng)當(dāng)t0時(shí)時(shí)0)(tdf當(dāng)當(dāng) 0t1時(shí)時(shí)221)(tdft0tf(t)2311.15當(dāng)當(dāng) 1=t3時(shí)時(shí)tttdddf10) 1()(21 t當(dāng)當(dāng) 3=t 時(shí)時(shí)ttttddddf3101) 1()(5 . 20tf(t)2311.16tdf)(10.532t2.5.171-14、計(jì)算卷積、計(jì)

6、算卷積)()(21tftf) 1()()()(. 221tututftfdtfftftf)()()()(2121 dtutuuu) 1()() 1()(dtuudtuu) 1()()()(dtuudtuu) 1() 1()() 1(解：解：.18)2(1) 1(1) 1(1)(1111100tudtudtudtudtttt)2()2() 1() 1() 1() 1()(tuttuttutttu)2()2() 1() 1(2)(tuttutttu.19) 1() 1()(,cos)(. 421tttfttf解：解：) 1(cos) 1(cos)()(21ttttftf) 1(cos) 1(cos

7、tt.201-17、用圖解法求卷積波形、用圖解法求卷積波形)(1tft0a1)(2tft0b21解：解：1、變量替換：將、變量替換：將t換為換為 ，并將，并將f2( )反褶：反褶：)(1f0a1)(2f0b21)(2f0b21.21)(1f0a1)(2f0b21)(1f0a1)(1f0a1)(1f0a1)()(21ff0)3 ()(21ff0)2 (2f0b21)(2tf0b112bt)()(21tff012btat)(2tf0b21t)()(21tff012/abt2) 1( tabt.22)()(21tff0t2abt1、當(dāng)、當(dāng)0t1時(shí)，時(shí)，)1()(221)(tututabttg)1()

8、(42tutuabt.232、當(dāng)、當(dāng)1t2時(shí)，時(shí)，)2() 1()2) 1(221)(tututababttg)2() 1(42tutuababt)()(21tff012abt2)1 (abt t.243、當(dāng)、當(dāng)2t3時(shí)，時(shí)，)3() 2()2(1 2) 1(21)(tututtababtg)3()2(42432tutuabtabtab)()(21tff01ab2)1 (abt 2t.251、當(dāng)、當(dāng)0t1時(shí)，時(shí)，)1()(4)(2tutuabttg2、當(dāng)、當(dāng)1t2時(shí)，時(shí)，)2() 1(42)(tutuababttg3、當(dāng)、當(dāng)2t3時(shí)，時(shí)，)3() 2(2443)(2tutuabtabtabtg

9、)(tgt0ab1234ab.26)(1tft0a1)(2tft0b111-17 b解：解：1、變量替換：將、變量替換：將t換為換為 ，并將，并將f2( )反褶：反褶：)(1f0a1)(2f0b11)(2f0b11.27)(2tf0b11 t1 tt)(1f0a1)(tgt0ab1234ab1.281-17 c)(2tft011)(1tft0111)(1f0111)(2f011)(2f011解：解：1、變量替換：將、變量替換：將t換為換為 ，并將，并將f2( )反褶：反褶：.29)(1f0111)()(21tff01t)(1f0111t1t)()(21tff01t)(1f0111t1t)()(21tff01t01t21) 1(21)(ttg10t5 . 0)(21tttg.30)(tgt0112214321t23)1 (21)(ttg)(1f0111t1t)()(21tff01t.311-17 d)(1