Question

Given the second order filter in figure, with ? = 4 ?Ω, ? = 50 ??, ?₁ = 50 nF and ?₂ = 1 μF:

Vin C2 – RK Vo

a) Reason the value that the gain of this filter will have for very low frequencies, at the resonance frequency of the parallel LC circuit and for very high frequencies. In the latter case (high frequencies) keep in mind that, since there are two capacitors with a similar asymptotic behavior, we cannot assume that its impedance is null, although it is very low value.

Why do you think this filter is called Low Pass Notch (LPN)?

b) Calculate the normalized network function of the filter ? (?) = ?_o(?) ⁄ ?_i (?) and, from it, the frequency response of the filter ? (? = ??). Relate this result to those obtained in the previous section.

c) Calculate the bandwidth of the filter.

Answer 1

Answer :

a.)

The reason that that the value of the gain of this filter will have for very low frequencies, at the resonance frequency of the parallel LC Circuit is that the  the capacitor in parallel LC circuit is storing the maximum energy, and the inductor is storing no energy or vice versa .

Due to which the energy is being swapped back and forth between the capacitor and the inductor, and none is coming from the source .
As the impedance Z= 0 , the current into the parallel LC combination is also zero due to which the gain of this filter have low frequencies . ω<ωο .

This filter can be called as low paas notch Because here  the zero frequency is greater than the pole frequency . i.e. (ω​​​​​​0 > ω )

b. )
Page No.: Page ① γουν Date: b) to find: Normalised. Network function - ie Hest - Vo(s) 'Vils) Given; - R=4K2 y L=50 MH ; =50nF & G = 1 ME From the given fig :- Zte = LSX GS - is LS + LGS² + 1 Algo - RX TS RGS tl Thus the function is given by:- Vils) Network ZRC ZRCtZLE - R R Vo (s) (6 st) * (LES2+1) Vols) R (RGS tt RILGS+1) LIS(RGS+1) -CRGSED (LGS ) s canned with Eartscanner

Page No.: Page ② youva Date: R (LCS²+1) Vils) R(L452+1) + LS(RC25+1) put the given values in above function Vols) 4x10° (50x1õx 50x1095 + 1 Vits)_4x10(50x10x50x10x+1)+ 50x105f4x10* 1xco's +1) Voce) 10x1042 +42103 Vils) (10x10*s?+ 4x103)+6-2810**8 +L - put s= jw Vors) = loxioo Ejw) + 4x100 Vifs) 10x10(jw){+ 4x103 + 0-2x10'S TL ca Scanned witOR OP CamScanner

Page No.: VOUV page 6 Date: 5) Bandwith of the filter the resonant frequency is given by := We know ) . Ir Vsoxlox Soxom Box I 16400x400 – 400x106 27 v [fo = 12-328 kHz Thus the band with is given by : B.W B-W = t = - 12328X10 150X103 T4X103) Pi Bandwith (BW) - 61-64 Ane Scanned with CamScanner