How would you go about this parallel resonant bandpass filter - Page 1
Forum section: electronics / beginners

author: John B
author_rank: Frequent Contributor
timestamp: January 16, 2020, 08:02:01 am
content: I've been working through a textbook example of a parallel resonant bandpass filter, and I'm fairly sure I've encountered some wrong or contradictory information. The question is asking to solve for which values of L and C will provide a resonant frequency of 1MHz, and a bandwidth of 20kHz. Firstly Ill go through how I think the book intends to solve the problem.The book defines the Q of the circuit as $$Q = \frac{f_{r}}{f_{2} - f_{1}} = \frac{f_{r}}{\Delta f} $$ where fr is the resonant frequency, f1 and f2 are the lower and upper cutoff frequencies, and delta f is the bandwidth. Subbing 1MHz and 20kHz gives Q = 50The book also defines Q as $$Q = \frac{Z_{max}}{X_{Lr}}$$ where Z max is the input impedance at resonance, and Xl is at the resonant frequency as well.I think the book is hinting at combining R1 and R2 in parallel to give a Zmax of 60k. Along with Q=50 this means Xl at resonance is 1200 ohms. Then solve for L and C by calculating with Xl=Xc=1200 at 1Mhz. The answers in the book are approximate, but this method effectively gives the right answer. However it's not clear to me how they expect to solve it, so I did it the long way using the complex plane.I solved as follows:1. Combine Xl and Xc into a single net impedance Zx.2. Derive an equation that expresses the Z of the entire circuit.3. Express the total current as a function of Zx.4. The output voltage at resonance is 60mV (resistor divider only, Zx is infinite) and at the cut off frequencies the voltage is (sqr(2)/2)*60mv with a phase of +/- 45 degrees.5. The lower and upper cutoff frequency, Ir2 is (0.03+j0.03)/150000 and (0.03-j0.03)/150000 respectively.6. Subtract Ir2 from total current to give Ix.7. Calculate Zx=Vx/Ix at each cutoff frequency.8. Discounting extraneous solutions, Zx is +j60000 at f1, and -j60000 at f2At this point, this is where I think I run into issues with the book's definition of things. It defines the cutoff frequencies as:$$ f_{cut} = f_{r} \pm \frac{\Delta f}{2} $$Giving f1 and f2 as 990kHz and 1010Khz.Since I know Zx at each frequency, I expressed Zx in terms of L,C and f, then solving for L:$$ Z_{X} = \frac{j \cdot 2\pi \cdot f \cdot L \cdot (\frac{-j}{2 \pi \cdot f \cdot C})}{j \cdot 2 \pi \cdot f \cdot L - j \cdot 2\pi \cdot f \cdot C} $$$$ L = \frac{j \cdot Z_{X}}{j \cdot Z_{X} \cdot 4 \pi^{2} \cdot f^{2} \cdot C - 2\pi \cdot f} $$Subbing each pair of values for Zx and frequency, then finding the intersection gives the values:$$ C = \frac{1}{24\pi\cdot 10^8} \; L = \frac{2}{3333\pi} $$Or approximately L = 191uH, C=132.63pFThese are the same answers that the book gives (keeping in mind the answers are still approximated), however, the resonant frequency is not 1Mhz exactly, and the various values of Xl and Xc differ slightly too. Digging around some more, I've read that the relationship between fr and the cutoff frequencies is in fact a geometric mean where $$f_{r} = \sqrt{f_{1}\cdot f_{2}}$$ rather than the previous definitions of fr, f1 and f2.To summarise, I think that for this circuit:1. You can have values of L and C that produce a resonant freq of 1Mhz, but the cutoff frequencies will not be 990kHz and 1010kHz2. You can have values of L and C that produce cutoff frequencies of 990kHz and 1010kHz, but the a resonant freq will not be 1Mhz.Sorry it's a bit of a long post, but it makes me question what the book is defining. Does anything look amiss, or how would you solve it?