Question

Please answer parts E, F and G given answers for previous steps:

2. The following circuit is to be used as a filter. C1 Vout 47nF V1 R1 10Vpk 1KHZ 1.5kQ 0° a) What is the name of this type of filter? b) Calculate the reactance of the capacitor at 1000 Hz c) What is the impedance of the circuit at 1000 Hz d) Calculate the break (cut-off) frequency for this filter. e) Calculate Vout at: Cut-off frequency (fc), 0.1fe, 0.5fc, and 10fc f) Calculate the voltage gain in decibels (AvidB)) at: fc, 0.1fo, 5fc, and 10fc g) Sketch the Bode Plot (decibels vs frequency) of this filter labelling all critical values.

a) High Pass Filter or HPF

b) Reactance X_c = 1/(2 x pi x f x C) = 1/(2 x 3.14 x 1000 x 47 x 10^-9) = 3386.28 Ohm

c) Impedance = $\sqrt{}$(1500² + 3386.28²) = 3703.63 Ohm

d) Cut off frequency = 1/ (2 x pi x R x C) = 1/(2 x 3.14 x 1500 x 47 x 10^-9) = 2258 Hz = 2.258 kHz

Answer 1

Phase = -tan^-1 (- 1/wcR/1) = tan^-1 (1/wcR) = tan^-1 (1/2 n times 2.258 times 10^3 times 47 times 10^-9 times 1500) = 45 degree to calculate 3 d beta freq Mag = 1/squareroot 2 so w^2 c^2 R^2 = 1 w c R = 1 therefore f_3 d beta = 1/2 pi c R = 2.258 times 10^3 Solution rightarrow this is high pass filter (H P F) rightarrow cut off frequency f_c = 2.258 k H z therefore v_out = v_in times R/R + 1/c s [applying voltage divider rule] = v_in R C S/R C S + 1 = 10 times 1500 times 47 times 10^-9 times s/1500 time 47 times 10^-9 s + 1 = 7.05 times 10^-4 (s)/7.05 times 10^-5 (s) + 1 in frequency domain (s = j w) therefore v_out = 7.05 times 10^-4 times j w/7.05 times 10^-5 times j w + 1 therefore v_out = 7.05 times 10^-4 times j 2 n f_c/7.05 times 10^-5 times j 2 pi f_c + 1 Now