Consider the filter shown in Figure below. Derive an expression for the transfer

Question

Consider the filter shown in Figure below. Derive an expression for the transfer function H(f) = vout/vin. Use the computer program of your choice to obtain a Bode plot of the transfer-function magnitude for R1, = 9 k Ohm, R2 = 1kOhm, and C = 0.01 mu F. Allow frequency to range from 10 Hz to 1 MHz. At very low frequencies, the capacitance becomes an open circuit. In this case, determine an expression for the transfer function and evaluate for the circuit parameters of part (b). Docs the result agree with the value plotted in part (b)? At very high frequencies, the capacitance becomes short circuit. In this case, determine an expression for the transfer function and evaluate for the circuit parameters of part (b). Does the result agree with the value plotted in part (b)? At very high frequencies the capacitance becomes short circuit. In this case, determine an expression for the transfer function and evaluate for the circuit parameters of part(b).Does the result agree with the value potted in part(b)?

Explanation / Answer

Please ask if you have any doubt.I will help you.

a) Zout = R2 - j/2fC.

Ztotal = R1 + R2 -j/2fC

H(f) = Vout/Vin = Zout/Ztotal = (R2 - j/2fC)/(R1 + R2 -j/2fC)

c) At low frequencies ,f 0 therefore the factor j/2fC becomes infinite.

Thus H at low frequencies 1.

d) At high frequncies ,f the capacitance factor becomes 0.

Therefore,H = R2/(R1+R2)

Note : The plot is difficult to draw with the tools ,because i dont have any.

If you want i can draw manually .

PLease post another question and send me the link

i will draw the graph manually if you need.

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