Question

a) Design a low-pass filter using the given circuitry with a cut-off value of 1 kHz and plot the frequency response curve on the given axes 1.0 0.7 0.5 in out 0.0 101 102 103 104 10s Hz b) Design a band-pass filter using the given circuitry with a bandwidth of 500 Hz and a lower cut-off value of 100 Hz, and draw the frequency response curve. Keep all resistors at the same value (i.e. Ri-R-R3-R4). 1.0 0.7 0.5 0.0 101 102 103 104 105 Hz c) For the circuit below, insert the missing simple circuit element, E3 (which is a resistor, capacitor, or inductor), to create an inverting band pass filter with a high cut-off frequency of 20 kHz. Draw the frequency response curve on the given axes ЕЗ R2 100K 0.7 Inverting Amp Band Pass Filter 0.5 C1 R1 .8uF 10K Vout 0.0 101 102 103 104 0 Hz

Answer 1

H2 The cutoft nmula nequen Given R C = o. 0001591 The capacitos? value must be arge nesiston be cause it nesfs to mone rt atteruates the input sronal (Om t° the filte»,

Vin o.lbHF So, Theや ty Nermelized velfge equen nes ponse 's net in -31B d B -20dB/decade D-5 フ . 0-0 Phose 0

Givenband pass filtes ith a equency value is loot z4 band wid th is500H? SooH2 s te be destgnes ive n in the desrqn tigh pass Hiter Je (an sealize a band pass ustry hjy*h pass and lou pass fltes n Cas cade hioh pass ilten hequenc the uppen cut eft frequenty the iS above bandpasx ff tte and. utfu n and cut above bonipass flites,

Given band width Soot2 aswe Knobu andん[ H冫600 Hz 60 0 Rt C 0.000 26S But as pi eft ects fhe oain of the ket p be some 2 ks

similaaly (appmaximat 0.8시 F So the cincuit wlbe 132nt Vou hequeny nesponse i SooH 근 Banduidth Phase tTo neqt n

To the ab ove cinuit to be band pss 3 must be 4bove cct s notting but impnoved Capacttas. because the Given чрре cut. Hequent CE,X Pin - SnF

SnF CE, o suF lok 2-1 Veut Noomaliged Vout (vj o-5 o、o (2.$ 2gueny equ?n OLS 幻eguenl

Any doubts leave them in comments. please don't forget to like.