Electrical – Why Deliyannis bandpass filter with high Q does not have the expected transfer function bode plot

band passbode plotdesignfiltermultisim

What i need is to design a delyiannis band-pass filter with Q enhancement technique.

The theoretical analysis of this technique is referred here in pages 107-112.

For those who don't know how to design delyiannis band pass-filer here is an example:

And here is the expected bode plot

In my case, i need to design a filter with f0 = 825 Hz and Q = 36.

Although i do the same process like the above i don't get the expected results for my case.

Here is my circuit:

And here is the bode plot (where 825Hz is not on the top as expected)

Best Answer

When plotting high-\$Q\$ transfer functions with any graphical viewer whether data are coming from a SPICE engine or a mathematical solver, it is important to maintain an adequate number of points per decade. 100 is usually enough for 1st-order responses but as soon as \$Q\$ is way above 1, you can potentially miss the peak if granularity is not good enough. For this purpose, you can push the number of points per decade to 1000 or even more if necessary.

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Electronic – Choosing between active opamp filtering or an LC filter

The first thing to do when looking to design a filter for a signal is to obtain the spectrum of the signal. This can be done in different ways. For example, you can measure or capture the signal with a scope. Most modern scopes can calculate an FFT of captured data and give a spectrum.

For a design though, it is nice to make an idealized waveform and then write the Fourier series of it. This way you get only the signal and no unwanted stuff, and so get a clean spectrum which will just be the coefficients of the harmonics. Then you can truncate or alter harmonic content (coefficient amplitudes) according to some filter type. Reconstruct a filtered signal from those altered coefficients and the filter effects will be evident.

From the comments of having a 5MHz carrier with ~1MHz BPSK, it is likely that many harmonics will be needed if a good representation of the original signal is desired. It wouldn't be surprising if 20 to 40 harmonics were needed for good representation. That would be a filter that started to roll off at > 20MHz, or maybe 40MHz. That's only 2 or 3 octaves from the 2 meter band (~150MHz). A filter with 1st order roll off isn't going to do that. You're going to need a filter with at least 3rd order performance for something like that. It's possible to get 2nd order performance out of a single stage Sallen-Key low pass filter.

As to the design shown in the schematic, it looks like the amplifiers are OK for what's being asked of them. LC filters are sensitive to termination. Best performance is when they are terminated into their characteristic impedance. The LC shown has \$Z_o\$ of 45 Ohms, and when used alone shows a resonant lobe of about 6dB. You can see the effects of impedance matching in case 4 when the LC is terminated into U2 with 100 Ohms. For a better match, R8 and R7 values could be halved, or L1 could be doubled and C5 could be halved.

Usually when active filters are used, inductors are not. It's because inductors are not needed to get complex poles and high Q roll offs, amplifier gain (and sometimes a little positive feedback, as in Sallen-Key) give that. Passive filters are good if the filter will be someplace where there is no bias voltage, or where frequencies and or filter order is high.

Electronic – How to know whether the bandpass filter is stable

This is a second order biquadratic band pass filter (see HERE). The transfer function is $$ \frac{-sC_2R_2R_3}{s^2C_1C_2R_1R_2R_3 + s\left ( C_1+C_2 \right )R_1R_2 + \left ( R_1+R_2 \right )} $$

If the poles (those values of s for which the denominator equals zero) all lie in the left half of the s-plane (x-axis, reals; y-axis, imaginary), then the system is stable

>> denom=[C1*C2*R1*R2*R3 (C1+C2)*R1*R2 R1+R2]

denom =

   1.0e+04 *

    0.0000    0.0000    1.2474

>> roots(denom)

ans =

   1.0e+04 *

  -0.0535 + 1.2129i
  -0.0535 - 1.2129i

The real part of your roots are negative, thus all the poles lie in the left half of the s-plane, and your system is stable.

The Bode plot I get is: enter image description here

Your routine plotted the 270 degree phase lead as a 90 degree lag, but otherwise the bodes are close to matching. Your system is stable.

Best Answer

Related Solutions

Electronic – Choosing between active opamp filtering or an LC filter

Electronic – How to know whether the bandpass filter is stable

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