8/13/2019 Bai Tap DTTT_final

1/7

Bi 1 : Thit k b lc thng thp bc 1 Butterworth, fc=300Hz, C=0,1F, /Ao/= 10trong 2 trng hp:

a. Tn a vo ng + ca opamp (dng K khng o) b. Tn a vo ng-ca opamp (dng K o)

Bi 2: Thit k b lc thng cao bc 1, Butterworth, fc=5kHz, C=10nF, /Ao/= 10trong 2 trng hp: a. Tn a vo ng + ca opamp (dng K khng o) b. Tn a vo ng-ca opamp (dng K o)

Bi 3: Thit k b lc thng thp bc 2 Butterworth, c tn s ct fc=1khz, A=1,Trong 2 trng hp:

a. Mch LPF Sallen Key b. Mch LPF a hi tip (MBF)

Bi 4: Thit k b lc thng cao bc 2 c tn s ct fc=1khz,/A /=1 , Trong 2 trng

hp: a. Mch HPF Sallen Key b. Mch HPF a hi tipMBF

Bi 5: Thit k b lc thng di (BPF) c tn s cng hng fm=10khz, BW=200hz,Am= 10. Trong 2 trng hp:

a. Mch BPF Sallen Key b. Mch BPF MBF

Bi 6: Thit k b lc thng di (BPF) c tn s cng hng fm=5khz, Q=50, Am=20. Trong 2 trng hp:

c.

Mch BPF Sallen Key d. Mch BPF MBF

Bi 7: Thit k b lc thng thp (LPF) bc 3, c tn s ct fc=1khz, Ao= 3.

Bi 8: Thit k b lc thng thp (LPF) bc 4, c tn s ct fc=1khz, Ao= 1.

Bi 9: Thit k b lc thng thp (LPF) bc 4, c tn s ct fc=2khz, Ao= 1, tt cin tr u bng 2k.

Bi 10: Thit k b lc thng thp (LPF) bc 5, c tn s ct fc=2khz, Ao= 3.

Bi 11: Cho mch khuch i cng sut y ko lp B, dng bin p, Vcc = 20V, R L = 10 . Sng vo vi c gi tr 16V peak -to- peak. Bit rng BJT VBE v V CE (sat) rtnh.

(a) V dng sng vi, ic1, v i c2 .(b) Tnh P in , P out (ti tn s c bn) v hiu sut.(c) NuVCE (sat) = 0.2 V th hiu sut cc i s nh th no?

8/13/2019 Bai Tap DTTT_final

2/7

Bi 12: Cho dng sng b khuch i cng sut lp C nh hnh sau:

.

Bi 13: Cho b khuch i cng sutlp C c Vcc = 48 V, 1 = 60 o vPout = 25W.(a) Nu s dng tn hiu 100kHz,Tnh P in , I CQ , and I dc . V dng sng iC v v o.(b) Tnh in p ra hi bc 2. Sdng gi tr RL cu a.

Bi 14: Cho ngun tn hiu Vi, c in tr ni Ri= 150, cn truyn cng sut nti RL= 50. Hy tnh ton b phi hp tr khng cng sut truyn n ti tcc i. Bit fi=1Mhz.

Bi 15: Cho ngun tn hiu Vi, c in tr ni Ri= 50, cn truyn cng sut n tiRL= 100. Hy tnh ton b phi hp trkhng cng sut truyn n ti t cci. Bit fi=1Mhz.

Bi 16: Cho mch KCS RF nh hnh bn,

fo=50Mhz, Vcc=13,5V, Vce(sat)= 1V,P1=26W. Tnh L, C. V dng in p tcthi cc C theo Vi.

Bit Vcc = 24V.(a) Tnh gi tr R L v cng sut hi c bn

phn phi trn ti R L .(b) Tnh cng sut ngun cung cp v hiu

sut ca mch khuch i

VCC

Vi

Q1Cb

RFC

RFC

Ranten

L

C

8/13/2019 Bai Tap DTTT_final

3/7

Bi 17: Cho mch KCS RF nh hnh bn, vmch nguyn l y vi phn cc =900 .Tnh mch phi hp tr khng vo ra. Vdng in p tc thi Vb(t), Vc(t), dng ic(t),ib(t) theo Vi(t),

Bi 18: Hy thit k mch lc thng thp bc 4 li n v c tn s ct 20kHz (v mch, vit hm truyn, xc nh tn s ct, v ptuyn tim cn ca mch).

Bi 19:

a. Hy v s khi ca my thu AM i tn mt ln, gii thch chc nng tngkhi. Hy cho bit tn s trung tn v di tn s ng vo ca my thu trn. b. Hy gii thch nguyn l hot ng mch tch sng AM dng diode, iu kin

chn thi hng ca mch tch sng.

Bi 20: Cho tn hiu AM, vi tn hiu cao tn vo(t) = 1,5cos(12 .10 5t) (V) v tn hium tn v (t) = sin(8 .10 3t) (V).

a. Vit biu thc v v tn hiu AM trn. b. V ph cng sut, xc nh bng thng ca tn hiu AM trn v tnh t scng

sut mt bin trn cng sut AM, bit rng cng sut ca sng mang l P0.

Bi 21: a. V s khi v ph tn hiu ca mch m ho FM stereo, gii thch chc

nng tng khi. b. di tn l g ? Vit biu thc tnh di tn. Trnh by quan h gia bng

thng tn hiu FM v di tn.

Bi 22: Cho tn hiu FM c biu thc nh sau :VFM(t)= 10cos[2 .10 6t +3sin(2 .20.10 3t)] (V), thng s t l K f = 8kHz/V. Tn hiu FMny c tiu th bi antenc R A=50.

a. Xc nh biu thc tn hiu tin tc (m tn), tnh di tn cc i. b. V ph bin in p, xc nh bng thng ca tn hiu FM trn.

c. Tnh cng sut pht ca tn hiu FM trn. Bi 23:

a. Hy thit k b lc thng thp tch cc bc 1 c tn s ct-3dB l 460kHz vc li Avo =20dB khi tn s tin v zr.

b. Hy thit k b lc thng cao tch cc bc 1 c tn s ct-3dB l 450kHz v c li Avo = 20dB khi tn s tin v v cng.

L2

C2 RantenL1

Vi(t)

Q1

C1

8/13/2019 Bai Tap DTTT_final

4/7

Bi 24: Cho tn hiu sng mang vo(t) = 100sin(7.10 6t) (mV), hai tn hiu m tnv1 (t) = 50cos(2 .10 3t) (mV) v v2 (t) = 80sin(5 .10 3 t) (mV). S dng ph ng php iu

bin AM- DSBFC cho hai tn hiu tin tc trn. a.Vit biu thc iu ch AM-DSBFC cho hai tn hiu m tn trn.

b.V ph bin tn hiu AM v tnh t s cng sut bin trn cng sut AM. Nhn xt. c.Xc nh bng thng ca tn hiu AM trn.

Bi 25: .a. Hy xy dng biu thc iu ch FM cho tn hiu tin tc m tn iu ha. Nu cccch xc nh bng thng ca tn hiu FM.

b . V s khi ca my thu FM i tn mt ln, gii thch chc nng tng khi. Xc nh tn ca b dao ng ni v tn s nh ca my thu nu tn s trung tm ca tnhiu FM cn thu l 99,9MHz

Bi 26: Cho s mch ca my thu thanh (trang sau), hy vs khi ca my thuthanh ny. Trnh by chc nng cc transistor t T1 n T8. Phn tch hot ng canhm mch khuch i trung tn v tch sng.

8/13/2019 Bai Tap DTTT_final

5/7

C 1 4

1 0 0 u F

R 1

6 2

K

3 3

R 1 8

C 9

0 . 0 2 2 T 3

3 D G 2 0 1

C 2 0

1 0 0 u F

C 4

1 0

R 8

L 3

L 1 0

C 1 6

4 .7 u F

D 1

2 A 1 9

T 6

3 D G 2 0 1

2 0 k

R 1 3

3 K

R 1 5

R 1 0

4 .7 k

C X 2

T

1 3DG201

D 2

1 N 4 1 4 8

1 k

R 6

C 1

L 1

6 2 K

R 7

4 7 0 R

1 6

T 2

3 D G 2 0 1

C 7 L

7

C x 1

C 1 3

0 . 0 2 2

L 6

1 K

R 2

3 3 0

R 1 2

T 2

1 5 k R 1

1

E 1

A N T E N N A

C 1 7

0 . 0 2 2

T 5

3 D G 2 0 1

L 8

C 1 5

4 .7 u F

C 1 0 L 9

6 0 0

R 9

C 1 9

1 0 0 u F

T 7

9 0 1 2

C 1 8

0 . 0 2 2

2 0 K

R 4

1 5

R 1 4

C 6

3 0 u F

L 2

R 3

T 4

3 D G 2 0 1

C 5 L

5

C 3

0 . 0 2 2

T 8

9 0 1 2

C 8

0 . 0 2 2

L 1 1

L 1 2

C 1 2

0 . 0 2 2

L 1 4

L 1 3

1 5 0

R 5

L O A

C 1 1

0 . 0 2 2

3 V

B A

T T E R Y

C 2

0 . 0 2 2

2 2 0

R 1 7

Bi 27: Hy thit k b lc LPF bc 1 c tn s ct 3,4KHz c li Avo =15dB vHPF tch cc bc 1 c tn s ct 340Hz v c li Av = 15dB.

8/13/2019 Bai Tap DTTT_final

6/7

Bi 28: a. Hy v mch gii iu ch AM - DSBFC, trnh by chc nng cc linh kin v giithch nguyn l hot ng. b. Gi s c tn hiu cao tn Vo(t) = sin(14 .10 5 t) (V) v tn hiu m tn c biu thc

nh sau: V (t) = 300sin( 103

t) + 200sin(2.103

t) + 400sin(4.103

t) (mV). Hy thc hincc yu cu sau : - Vit biu thc iu ch AMDSBFC cho tn hiu m tn trn. - Tnh cng sut ca sng mang cc thnh phn bin, gi s tn hiu AMDSBFC

trn c tiu th in tr ti 1 ohm ?- V ph cng sut tn hiu AMDSBFC trn, tnh t s cng sut bin trn cng

sut AMDSBFC v nhn xt. Bi 29:a.V s khi ca my thu FM i tn hai ln, gii thch chc nng tng khi.

b.Xc nh phm vi thay i tn s ca b dao ng ni my thu my thu FM c ththu

c cc tn hiu c tn s t 90Mhz n 100Mhz.

Bi 30: Hy v s khi ca my thu AM i tn mt ln, gii thch chc nng tngkh i. Hy cho bit tn s trung tn ca my thu.

Bi 31: Cho tn hiu AM, vi tn hiu cao tn vo(t) = 1,5sin(12 .10 5t) (V) v tn hium tnv 1 (t) = sin(8 .10 3t) (V).

a. Vit biu thc v v tn hiu AM trn. b. V ph cng sut, xc nh bng thng ca tn hiu AM trn v tnh t s cng

sut mt bin trn cng sut AM, bit rng cng sut ca sng mang l P0.

Bi 32: Cho mch nh hnh sau. Nu chc nng ca mch, v s khi ca mch, nuchc nng tng khi, nu chc nng ca cc transistor trong mch.

8/13/2019 Bai Tap DTTT_final

7/7