7/30/2019 cng chng 1

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Chng1:I. Bit: 15 cu

1. b chuyn mch ton thng l b chuyn mch :B. Mi u vo c th kt ni n nhiu u bt k.

2. theo nh ngha th b chuyn mch vung lC. b chuyn mch c s u vo = u ra v l ton thng

3. Trng chuyn mch kgian 1 khuA. ni // ng tn cc u vo ca trng chuyn mch ny vi trng chuyn

mch khc4. trong KTCM, chuyn mch PAM 4 dy s dng trung k A, m tn5.thnh phn k thuc thnh phn ca mng vin thng l

D. Trung tm bo hiu6. mng vin thng truyn thng c phn cp v bao gm

C. 5 nt7. thut ng POTS ch mng C. mng in thoi cng cng8. Thut ng ISDN dng ch mng D. mng s tch hp dch v12. theo s khi ca tng i s SPC, GiGS l phn h

B, bo hiu13. trong s khi ca tng i s SPC, APS l phn h

C. ng dng14. Trong s khi ca tng i s SPC, CPS l phn h

D. x l trung tm15. trong s khi ca tng i s SPC, SwNS l phn h

A. chuyn mchII. Hiu: 10 cu

1. Yu cu chung i vi cc p.t chuyn mch ca b chuyn mch k.gian tng t l:A. khng lm suy hao tn hiu

2. c im ca trng chuyn mch k.gian 1 khu l:A. Mi yc c.mch gia 1 u vo vi 1 u ra ca trng c.mch c th qua nhiup.t ca trng c.mch v mi tng quan 1:1 gia u vo vi u ra.

3. Quy tc chung xy dng trng chuyn mch nhiu tng l:

B. S u ra ca mi b c.mch tng trc phi = s b c.mch tng u, s uvo ca b c.mch tng sau phi = s b c.mch tng trc, v tun theo nguyntc 1:1, dng tn

4. KTCM c c trng bi: D. kiu c.mch v c tuyn truyn (tng t hay s)5. Chc nng bo hiu khng bao gm: C: Chc nng qun l6. Cc nhm chc nng chnh ca 1 trung tm c.mch bao gm:

B: Chc nng c.mch, chc nng b.hiu.7. D.v QoS c hiu l d.v g? C. D.v cht lng h thng.8. xt kha cnh chc nng ca cc HT thit b trn mng th mng VT khng bao gm

C. Mng iu khin9. Mng hi t c hiu l:

D. Tch hp c s h tng ca cc mng truyn thng hin c10. Cng ngh DSL c s dng trong mng NGN thc cht l cng ngh tn dngmin tn s cao truyn tn hiu tc cao trn: A. i dy cp ng thng thng.

III Vn dng: 5 cu3. trong cu trc ca tng i SPC, phn h thc hin chc nng kt cui ng dy thubao l

A: phn h ng dng

5. Cc b c.mch kim tra, o th thng dng c.mch no:B. C.mch tng t