Unit - 7
Introduction and Classification of filters and its uses
Q1) A LPF circuit consisting of a resistor of 40KΩ in series with capacitor 47nF across a 10V sinusoidal supply. Calculate VO at frequency 1Kz?
A1)
R = 40KΩ
C = 47nF
Vi = 10V
F = 1000 Hz
For LPF,
Capacitive Reactance is XC
The output voltage is given as
VO = Vi
XC = =
XC = 3.386KΩ
VO =
=10 x
= 10 x
VO = 0.843 V
(b) RL LPF
FC = R / 2nt
VO = Vi
Q2) An RL LPF consists of 5.6mH coil a 3.3 KΩ resistor. The output voltage is taken across resistor. Calculate the critical frequency?
A2) Given:
L = 5.6 x 10-3 H
R = 3.3 x 103Ω
For RL LPF
F ==
F = 93.78 KHz
Q3) A sinusoidal voltage with peak to peak value of 10 V is applied to an RC LPF. If reactance at input is zero, find output voltage?
A3)
VO =
XC =
F = XC = 0, F ->∞
Hence, VO = 0V.
Q4) An RC – LPF consists of 120 Ω resistor and 0.02 µF capacitor. The output is taken through capacitor. Calculate the critical frequency?
A4)
For RC LPF FC is given as
FC==
FC = 66.31 KHz
Q5) An RC HPF with input capacitor 10nf and the output resistor of 10K Ω. Find the cut off frequency.
A5)
FC = =
FC = 15.915 Hz
Q6) Calculate the break point (FC) for a simple passive HPF consisting of 82pF capacitor connected in series with 240 KΩ resistor?
A6)
FC = =
FC = 8KHz
Q7) The value of capacitor C1 required to give a cut off frequency fL = 1KHz and value of resistor is 10K for the BPF. The value of capacitor C2 and higher cut off frequency (FH) is 30 KHz with resistor 10KΩ. Also calculate the central resonant frequency?
A7) Given values for HPF
FL = 1000 Hz, R1 = 10 x 103Ω
C1 = == 15.9nF
For LPF,
FH = 30 x 103 Hz, R2 = 10 x 103Ω
C2 === 530nF
Central resonat frequency,
Fr =
=
Fr = 5.477 KHz
Q8) Design a constant K LPF T and ∏ section filter having Fc = 1000Hz and nominal characteristic impedance R0 = 400Ω. Also determine the frequency for this filter with attenuation of 19.1 dB.
A8) FC = 1000 Hz, R0 = 400Ω
FC = 1/∏√Lc, R0 = √L
R0/Fc = √LC/1/∏√LC
R0/Fc = ∏L
L = 400/∏ x 1000 = 127.32 mH
C = 1/∏FCRo = 1/∏ x 1000 x 400 = 0.795 µF
Attenuation in dB = 8.686 x Attenuation in nepers
Attenuation in nepers = 2.2 nepers
α = 2Cosh-1(F/FC)
F/2 = Cosh(1.1)
F = 3.3 KHz
The design of required filter for T and ∏ section with values L = 127.32 mH. C = 0.795µF
Q9) In a certain LPF circuit Fc = 4.5KHz R=300 ohm Find a) Its pass band b) If the input voltage is a 6v sine wave dc level calculate output voltage
A9) The pass band of LPF is from 0 to fC.
The output voltage is given as
VO =
XC ===0.035mΩ
VO = =6x=0.012V
Q10) For RC HPF with 325Ω resistor. Calculate the value of capacitor such that Xc is ten times less than the value of resistor at input frequency 12KHz?
A10) For RC HPF
FC =
XC =
Bur Xc is 10 times less than R. Hence, Xc=R/10=32.5Ω
Finding fC
fC ==1/(2x32.5)=4.89mHz
Now finding C
FC =
4.89x=
C=0.1F
Q11) For given -section LPF calculate Ro & Fc
A11) Given C/2=0.4F. The prototype -section LPF is given below
C=0.8F
L=200mH
FC = ==795.8Hz
R0 = ==500Ω
Q12) Design a constant K BPF with FC = 4KHz and 7KHz and R0 = 500Ω.
A12) Given:
FC1 = 4000Hz, FC2 = 7000Hz
For BPF,
L1 = R0/ ∏ (F2 – F1) = 500/∏ (7000 – 4000) = 53.05 mH
C1 = F2 – F1/4∏R0F1F2 = 7000 – 4000 / 4∏ x 500 x 7000 x 4000 = 0.017µF
L2 = R0(F2-F1)/ 4∏F1F2 = 500(7000 - 4000) / 4∏ (7000 x 4000) = 42.63 mH
C2 = 1/R0∏ (F2 - F1) = 1/500 x ∏ (7000 - 4000) = 0.212 µF
Q13) Design a constant – k BSF with cut-off frequency of 3KHz and 8KHz and R0 = 600Ω
A13) F1 = 3000Hz, F2 = 8000Hz, R0 = 600 Ω
L1 = (F2 – F1) R0 / ∏F1F2 = (8000 - 3000) x 600 / ∏ x 8 x 3 x 106 = 5 x 6 x 105 / ∏ x 8 x 3 x 106
L1 = 39.79mH
C1 = 1/4∏R0(F2 – F1) = 1/4∏ x 600(5000) = 0.0265µF
L2 = R0/4∏ (F2 – F1) = 600/4∏ (5000) = 9.54 x 10-3 H
C2 = F2 – F1 / ∏R0F1F2 = 5000 / ∏ x 600 x 8000 x 3000 = 1.105 x 10-3 H
L1/2 = 19.89mH, 2C1 = 0.053 µF
Q14) Design a content K LPF having Fc = 5.5 KHz and Ro = 500Ω
A14) For constant k LPF
FC =
R0 =
Given Fc = 5.5 KHz and Ro = 500Ω
=
= ∏L
L===28.93mH
C==0.116F
Q15) For Constant K T Section LPF shown, calculate Ro and fe
A15) The prototype T-section LPF is given below
Hence, C=0.2F
L/2=20mH
L=40mH
FC = ==3.56kHz
R0 = ==447.2Ω
Q16) Find the component values of a content LPF having Ro = 600Ω, Fc = 600Hz, Find the frequency at which attenuation of 38.2dB?
A16) FC =
R0 =
L===0.32H
C==0.88F
Attenuation in dB = 8.686 x Attenuation in nepers
Attenuation in nepers = 17.36 nepers
α = 2cosh-1()
F/2 = cosh(8.68)
F = 5.9 KHz
Q17) Calculate the components value of Section constant K HPF with Fc = 4 KHz and Ro = 600Ω Also calculate the characteristic impedance and phase constant at F = 10 KHz
A17) For constant k HPF
FC =
R0 =
L===47.75mH
C==0.133F
The characteristic impedance is given as
ZOT = RO = 600 =550Ω
ZO∏ = ==654.6 Ω
Phase constant α = 2Cosh-1)=3.133
β = 2Sin-1=45.85
Q18) Find the characteristic impedance of a constant K HPF with Fc = 10KHz and Ro = 200Ω and phase constant at 25KHz?
A18) The characteristic impedance of a constant k HPF is give as
ZOT = RO = 200 =183.3Ω
ZO∏ = ==218.2 Ω
Q19) Design constant K HPF having Ro = 200Ω, Fc = 5KHz
A19) For constant k HPF
FC =
R0 =
=
= ∏L
L===12.73mH
C==0.32F