# Passive bandpass center-frequency gain alteration

Discussion in 'Homework Help' started by Starhowl, Feb 13, 2014.

1. ### Starhowl Thread Starter New Member

Dec 3, 2013
25
0
Modify this bandpass circuit to change its center-frequency gain from 1 to 0.5 without changing $\omega_0$ or $Q$.

The general transfer-function of a bandpass is
$T(s) = \frac{a_1 s}{s^2 + s \frac{\omega_0}{Q} (\omega_0)^2}$ where the natural modes can be found by finding the poles of any response-function without altering its natural structure.

As the natural modes of the left-hand side are $s^2 + s\frac{1}{RC} + \frac{1}{LC}$ and the natural structure isn't changed on the right-hand side of the alteration, the natural modes should stay the same as follows: $s^2 +s\frac{1}{(2R||2R)C}+\frac{1}{LC}.

But by calculation
= \frac{s\frac{1}{2CR}}{s^2+s\frac{3}{2}\frac{1}{CR}+\frac{1}{CL}}" alt="T(s) = \frac{\frac{1}{\frac{1}{R} + \frac{1}{sL} + sC}}{2R + \frac{1}{\frac{1}{R} + \frac{1}{sL} + sC}} = \frac{1}{1 + 2R({\frac{1}{R} + \frac{1}{sL} + sC})} = \frac{1}{1 + 2 + \frac{2R}{sC} + sC2R} = \frac{sL}{sL3 + 2R + s^2C2RL} = \frac{\frac{L}{C2RL}}{s^2 + s \frac{3}{2}\frac{1}{CR} + \frac{2R}{C2RL}}
= \frac{s\frac{1}{2CR}}{s^2+s\frac{3}{2}\frac{1}{CR}+\frac{1}{CL}}" />

As one can see, the center-frequency gain indeed has been halfed - but despite theory about natural modes $\frac{\omega_0}{Q}$ has been altered.

(Problem 16.37 from Microelectronic Circuits, 6th edition.)" alt="s^2 +s\frac{1}{(2R||2R)C}+\frac{1}{LC}.

But by calculation
$T(s) = \frac{\frac{1}{\frac{1}{R} + \frac{1}{sL} + sC}}{2R + \frac{1}{\frac{1}{R} + \frac{1}{sL} + sC}} = \frac{1}{1 + 2R({\frac{1}{R} + \frac{1}{sL} + sC})} = \frac{1}{1 + 2 + \frac{2R}{sC} + sC2R} = \frac{sL}{sL3 + 2R + s^2C2RL} = \frac{\frac{L}{C2RL}}{s^2 + s \frac{3}{2}\frac{1}{CR} + \frac{2R}{C2RL}}
= \frac{s\frac{1}{2CR}}{s^2+s\frac{3}{2}\frac{1}{CR}+\frac{1}{CL}}$

As one can see, the center-frequency gain indeed has been halfed - but despite theory about natural modes $\frac{\omega_0}{Q}$ has been altered.

(Problem 16.37 from Microelectronic Circuits, 6th edition.)" />

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2. ### Starhowl Thread Starter New Member

Dec 3, 2013
25
0
Whoooooops, I'm so sorry, I didn't see the R instead of 2R in my calculation - everything is fine! Thread can be closed! Or deleted.. Whatever. Can't I do this?