Please help with this RLC bandpass filter design. 2. Design a series RLC bandpass filter as shown in Figure 1 with center frequency 1000HZ andbandwidth 300OHZ using a 10µF capacitor. For this filter:a. Determine L, and Rs.b. Compute the quality factorV„(s)c. Plot the amplitude and phase of the transfer function H,(s) =V¼(s)OS1,(s)and compute its polesV¼(s)d. Derive the transfer function Hcur, (8) = L,Si,(t)+Vis(t)R,Vos(t)Figure 1: Series RLC bandpass filter for problem 2.+

RLC Bandpass Filter A bandpass filter allows a range of frequencies and reject the other. A low pass filter is used to isolate the signals which have frequencies higher than the cutoff frequency. Similarly, a high pass filter is used to isolate the signals which have frequencies lower than the cutoff frequency. Now by cascading these two a bandpass filter is created. It has two cutoffs a high frequency one which will decide the higher frequency limit and low a low frequency one for deciding the lower frequency limit. Now in a RLC band pass filter these functions are carried out by a resistor, capacitor and an inductor. Inductor blocks high level frequencies and capacitor blocks low level frequencies. now, voltage across the resistor VR(s) driven by the source Vs(s) is given by, VR(s) = (RsL+1sC+R)Vs(s) using this we can find transfer function T(s)= VR(s)VS(s)= (RL)×(ss2+(RL)s+1LC)putting s =jω we get,T(jω)=(RL)×(jω(1LC-ω2)+(RL)jω) now T(jω) becomes maximum when denominator is minimum so 1LC-ω2=0ω0=1LC The frequency ω0 is called the center frequency. Now cutoff frequencies ωc1 and ωc2 is given by ωc1=-R2L+(R2L)2+1LCωc2=R2L+(R2L)2+1LC So bandwidth BW is BW = ωc2-ωc1 = RL There is one more way of how narrow or wide the filter is, given by its quality factor Q. The quality factor is defined as the ratio of the center frequency to the bandwidth Q = ω0BW=1LCRL= 1RLCTo determine Ls and Rs we know that , C = 10μF ω0 = 1000Hz C = 10μF ω0 = 1000Hzω0=1LsC Ls = 1ω02×C= 1(1000Hz)2×(10×10-6F)= 0.1H Now for Rs we know that, BW = RsLRs=BW×L=(300Hz)×(0.1H)=30Ω Answer: Rs = 30Ω , Ls = 0.1H Quality Factor Q is given by, Q = 1RLC= 130Ω×0.1H10×10-6F= 3.33 Answer: Q = 0.33Now to find out amplitude and phase of transfer function lets find out three different values of ω in transfer function T(jω)=(RL)×(jω(1LC-ω2)+(RL)jω) Case 1 : ω = 0 T(jω)=(RL)×(j×0(1LC-02)+(RL)j0) =0 Case 2 : ω=1LC T(jω)=(RL)×(j×1LC(1LC-1LC2)+(RL)j1LC) =1 Case 3: ω = ∞ T(jω)=(RL)×(j×∞(1LC-∞2)+(RL)j∞) ≅∞∞4+∞2≅1∞=0 Follow the same steps in introduction part

Answered: 2. Design a series RLC bandpass filter…