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I'm not sure this question is appropriate for this site, but I feel that if anyone knows about a filter, a radio ham knows! And as a ham myself I feel I need to understand this important area better. To that end, I've been reading various filter texts (Zverev and Dimopoulos mainly).

I'm at the point where the low-pass approximation has been developed (I'm going to chose a Butterworth for this question) and I'm moving to the transformation of a low-pass to other types (high-pass, band-pass, etc).

Lets take a Butterworth low-pass given in the book by Dimopoulos: $$ \frac{17.5514}{s^3+3.2743\cdot s^2+5.3604 \cdot s+4.3878}$$

If you have Maxima installed, you can try it yourself:

load(draw);

maxima_tempdir:"C:/maxima-5.41.0a";

H_lpbut(s):=17.5514/(s^3+3.2743*s^2+5.3604*s+4.3878);

G_lp(omega):=abs(subst([s=%i*omega],H_lpbut(s)));

Omega_3dB(N):=(1/beta_m())^(1/N);

draw2d(
        xlabel="Omega" 
        ,ylabel="Lowpass G(Omega)"
        ,xaxis=true
        ,yaxis=true
        ,xrange=[-10,10]
        ,yrange=[0,5]
        ,explicit(G_lp(omega),
                    omega,-10,10
                    )
        ,color=red
        ,points([[Omega_3dB(3), G_lp(Omega_3dB(3))]])
        ,points_joined=true
        ,line_type=dashes
        ,points([
                [0,G_lp(Omega_3dB(3))]
                ,[Omega_3dB(3), G_lp(Omega_3dB(3))]
                ,[Omega_3dB(3), 0]
            ])
    );

As you can see, I get the magnitude of the transfer function and plot from the normalized frequencies -10 to 10. Notice how wonderfully symmetrical the plot is.

enter image description here

That sort of makes sense, we're plugging in negative frequencies (from -10 to zero) and we end up with a mirror image of the low-pass giving a band-pass shape.

In fact, that's pretty much how the lp to bp transform works, only this time they use a different function:

$$ \frac{\omega ^{2}-\omega_0^{2}}{\omega\cdot BW} $$

Again, using Maxima we can explore how this function maps band-pass frequencies to normalized low-pass frequencies. I've taken some liberties here to highlight it's properties (it uses a log scale and negative infinity is now positive infinity - but you get the idea). The function goes to infinity at one end (omega=0) and takes a more leisurely trip to infinity at the other (omega=inf). It converges at zero (or bp centre frequency), which is exactly what required for bp transformation - so I get that bit.

bpt(omega,omega_0,BW):=(omega^2-omega_0^2)/(omega*BW);

draw2d(
        xrange=[0,1000]
        ,yrange=[0.0001,1000]
        ,logy=true
        ,explicit(sqrt(bpt(2*%pi*f,2*%pi*500,2*%pi*100)^2),
                    f,0,1000
                    )
    );

enter image description here

and finally, here's the transformed band-pass using the standard lp-to-bp transform:

enter image description here

Notice it isn't as symmetrical as the previous plot (using -10 to 10).

But, here's what I don't get...why would they use a function like this, which is obviously non-symmetrical, to produce the band-pass? Why not use a function that works to produce the nice linear -10 to 10 input that we saw earlier?

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  • $\begingroup$ Is there actually something (of reasonable order) that produces a symmetrical-looking bandpass response? It's not fair to compare it to a reflection around zero, even your bandpass filter will be symmetrical (with two humps) if you plot from -800 to 800 rad/s. $\endgroup$ – tomnexus Jul 14 at 18:16
  • $\begingroup$ I understand that. The question is an alternative way of asking where the translation function came from. I offered the argument that you could (at least mathematically) produce a symmetrical BP filter using a simpler (I'm guessing) function. I acknowledge the translation has meaning and works (of course it does!), I'm simply trying to tease out what that meaning is. $\endgroup$ – Buck8pe Jul 14 at 18:47
  • $\begingroup$ I'm guessing the translation is designed to produce a response that can be implemented by a real circuit. But, it would be good to know why it does that. $\endgroup$ – Buck8pe Jul 14 at 19:10
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    $\begingroup$ This question was asked and answered on dsp.SE. $\endgroup$ – Brian K1LI Jul 15 at 11:58
  • $\begingroup$ @Brian K1LI, that is a slightly different question. I don't have a problem understanding how the function creates a bandpass from a lowpass and I think I covered that in my question. What I'm asking is why this particular "simple, mathematically continuous mapping". What is it that makes this very geometric looking mapping work better than any others? $\endgroup$ – Buck8pe Jul 15 at 12:17

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