Question

Inductors Resistors 2700 3300 390024700 56002 Capacitors 1.OnF 1.5nF 2.2nF 1500 1800 22002 2.7mH 4.7nF 6.8nF 6800 3.3kΩ 82002 1.00 1.260 1.560 1.8k 2.20 2.750 3.3nF 3.90 4.7ko 5.660 6.8k0 8.2ko 10k 12k2 100F 18kΩ 2260 270 33ko 39ko 47k 56kΩ 33nF 15nF 22nF 15k02 47nF 68nF 68k 82k 100k 120k150k 2150k02180k 220ko 100nF 150nF 270k02 330k 390k22 470K22560526802 82021.OMO Table 1 7.0 kHz 3.0 kHz Low-Pass High-Pass Band- Pass Band- Stop 28 kHz 13 kHz [10 kHz, 59kHz] [48k Hz, 88kHz] 23 kHz 16 kHz 11 kHz 9.0 kHz 5.0 kHz 4.0 kHz [5.0 kHz, (3.0kHz, (2.0kHz, 29kHz] 17kHz] 12kHz) (38kHz, 68kHz] [28kHz, 70kHz] [20kHz, 47kHz] [16kHz, 27kHz] Table 2

Answer 1

selected L=2.7mH Given: from table 1 it is clearly Ang 2. Selected lower cut off frequency fa 5 KHz Upper Cut off frequency fcz = 29 kHz Aus 3 To calculate C we have to have to find centre frequeny fo = 1 fax fer as centre frequency is geometric of corner frequencies fo= √5 +29 With the help of fo and L, C Can be calculated mean = 12 kHz to share 22 J L XC 12 x103 = o 22 12.7 X1033 XC 3 С. (2X A X 12x109)? X109) x2.7810 C QX* X12X103)? x 2.7X103 C = 65.14 nf n 68 nf as 6&nt is given in Table 1.

2 .95 R L Ans4 To Calculate R we need to find Bandwidth BW= for fi = (295) KHz 24K Hz The value of R can be Calculated as Bandwidth = R R = BWXL - 24x103 x2.7 x 103 R = 64.82 which can approximately (+672) accomplished by parallel combination of resistors Ra isoll 18011680/1820 ito teot to R = 672 1-1 820