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Below I have the loop gain and the Bode plot for it, however I do not understand the phase plot. Why does it start at -90 degrees instead of 90 degrees since we have a zero? Please explain. Loop Gain & Bode Plot

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  • \$\begingroup\$ Why are you calculating -AB and not AB as the open loop TF? If you're including the minus at the summing junction then that's wrong. \$\endgroup\$ – Chu Oct 24 '19 at 19:05
  • \$\begingroup\$ @Chu The circuit is a Schmitt Trigger, therefore the feedback is positive. \$\endgroup\$ – Hector Oct 24 '19 at 19:11
  • \$\begingroup\$ For the numerator term, \$Arctan(\frac{-j\omega}{0})=-90^o\$ \$\endgroup\$ – Chu Oct 24 '19 at 19:14
  • \$\begingroup\$ @Chu You can not divide by zero! \$\endgroup\$ – Jan Oct 24 '19 at 19:23
  • \$\begingroup\$ @Jan That fraction is indeterminate. You can find solutions to expressions including it, as the denominator approaches zero from either side (with differing results, on occasion, depending on from which side you approach.) You may find it defined in some math systems, too. (Riemann, for ex.) So context matters. Also, Abraham Robinson's nonstandard analysis (mid-60's) is worth studying. But the ratio \$\frac00\$ is specifically undefined (necessary for group theoretic reasons -- given \$z=\frac ab\$, no \$z\$ satisfies \$0\cdot z=a\$ if \$a\ne0\$ but all \$z\$ satisfies \$0\cdot z=0\$.) \$\endgroup\$ – jonk Oct 24 '19 at 20:38
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Well, we have the following transfer function:

$$\mathcal{H}\left(\text{s}\right)=-\frac{\text{GLCs}}{\text{LCs}^2+\text{CRs}+1}\tag1$$

Now, in order to plot the the bode-diagram we need to use \$\text{s}=\text{j}\omega\$, where \$\text{j}^2=-1\$:

$$\underline{\mathcal{H}}\left(\text{j}\omega\right)=-\frac{\text{GLC}\text{j}\omega}{\text{LC}\left(\text{j}\omega\right)^2+\text{CR}\text{j}\omega+1}=\frac{\text{GLC}\omega\text{j}}{\text{LC}\omega^2-1-\text{CR}\omega\text{j}}\tag2$$

Now, the amplitude can be plot using:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\frac{\left|\text{GLC}\omega\text{j}\right|}{\left|\text{LC}\omega^2-1-\text{CR}\omega\text{j}\right|}=\frac{\text{GLC}\omega}{\sqrt{\left(\text{LC}\omega^2-1\right)^2+\left(\text{CR}\omega\right)^2}}\tag3$$

And for the phase we get (assuming that all the variables are positive and real):

$$\arg\left(\underline{\mathcal{H}}\left(\text{j}\omega\right)\right)=\arg\left(\frac{\text{GLC}\omega\text{j}}{\text{LC}\omega^2-1-\text{CR}\omega\text{j}}\right)=$$ $$\arg\left(\text{GLC}\omega\text{j}\right)-\arg\left(\text{LC}\omega^2-1-\text{CR}\omega\text{j}\right)=$$ $$\frac{\pi}{2}-\begin{cases} 0,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space\text{LC}\omega^2-1=0\\ \\ \frac{3\pi}{2}+\arctan\left(\frac{\text{LC}\omega^2-1}{\text{CR}\omega}\right),\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space\text{LC}\omega^2-1>0\\ \\ \pi+\arctan\left(\frac{\text{CR}\omega}{\left|\text{LC}\omega^2-1\right|}\right),\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space\text{LC}\omega^2-1<0 \end{cases}\tag4$$

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Because the plot is of -BA (minus sign included) I would have expected the phase plot to start at -180 as a result of the minus sign before almost immediately adding +90 from the +20dB/decade zero giving -90 during the rising slope. At the apex the rising slope levels off removing the 90 degrees of lead thereby reverting the phase back to -180 (as shown). Next the -20dB negative slope kicks in giving an extra 90 lag resulting in -270 in total. The double pole gives a change in slope of -40dB/decade changing the slope from +20dB/decade to -20dB/decade and the phase from -90 to -270. (With -180 of that coming from the minus in -BA). That is to say the double pole changes the slope by -40dB/decade and adds 180 degrees of phase lag.

When the minus is included in the loop gain (-BA) it means that the complete loop is taken into account including the inverting input on the difference amplifier which changes the Nyquist criteria to loop gain must be less than unity before the loop phase lag reaches -360 degrees. (instead of -180 degrees).

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