{"id":767,"date":"2019-05-31T15:42:43","date_gmt":"2019-05-31T23:42:43","guid":{"rendered":"https:\/\/linearcontrol.info\/fundamentals\/?p=767"},"modified":"2019-06-09T13:12:18","modified_gmt":"2019-06-09T21:12:18","slug":"step-by-step-bode-plot-example-part-i","status":"publish","type":"post","link":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/2019\/05\/31\/step-by-step-bode-plot-example-part-i\/","title":{"rendered":"Step-by-step Bode plot example. Part I"},"content":{"rendered":"\n

In this post we will go over the process of sketching the straight-line Bode plot approximations for a simple rational transfer-function in a step-by-step fashion. See Section 7.1 for details on the approximations. We will start with the magnitude plot and cover the phase plot in a future post.<\/p>\n\n\n\n\n\n\n\n

Consider the following second-order transfer-function:<\/p>\n\n\n\n

$$G(s) = \\frac{s+2}{s^2+11 s + 10}.$$<\/p>\n\n\n\n

The first step to produce a Bode plot sketch is to factor the numerator and denominator in terms of its poles and zeros:<\/p>\n\n\n\n

$$G(s) = \\frac{s+2}{(s+1)(s+10)}.$$<\/p>\n\n\n\n

You might need the help of a numeric calculator here for large order transfer-functions. In this example all poles and zeros are real. We will consider the case of complex-poles and zeros in another post.<\/p>\n\n\n\n

Once the poles and zeros have been factored, normalize the numerator and denominator:<\/p>\n\n\n\n

$$G(s) = \\frac{2}{10} \\frac{1 + s\/2}{(1 + s)(1 + s\/10)}.$$<\/p>\n\n\n\n

This step is very important and not performing it before proceeding with the sketch is a common mistake that leads to an incorrect gain.<\/p>\n\n\n\n

We proceed to identifying the pole or zero with smallest magnitude, which is this case is the pole at $s=-1$. Such a pole will show up on the magnitude Bode plot as straight-line approximation that is $0$ before $\\omega=1$ and have a $-20$dB\/decade slope after the pole, as shown in the figure:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

After noticing that the next pole or zero happens at $s=-2$ we can proceed to draw the sketch as the red line in the next diagram, stopping right before $\\omega=2$.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The value of $-6$ marked above can be calculated by using a standard $\\Delta_y = \\text{slope} \\times \\Delta_x$ argument:<\/p>\n\n\n\n

$$\\Delta_y = -20 \\times (\\log_{10}(2) -\\log_{10}(1))= -20 \\log_{10}(2\/1) \\approx -6.$$<\/p>\n\n\n\n

Past $\\omega=2$, it is necessary to take into account the contribution of the zero at $s=-2$, which adds to the current slope of $-20$dB\/decade its own slope of $+20$dB\/decade. The resulting slope of $-20+20=0$dB\/decade is the straight line in the next figure:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

We then proceed with the red line until right before $\\omega=10$, at which point the second pole at $s=-10$ starts contributing $-20$dB\/decade, as shown next:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Since there is no other pole or zero, we conclude the diagram by following the $-20$dB\/decade slope line initiated at $\\omega=10$:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Last but not least, we need to offset the diagram to take into account the gain<\/em> of $2\/10$, which corresponds to<\/p>\n\n\n\n

$$20 \\log_{10}(2\/10)\\approx-13\\text{dB}$$<\/p>\n\n\n\n

to obtain the final sketch:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Finally, here are some sanity checks that you should perform on your sketch:<\/p>\n\n\n\n

  1. If there are no poles nor zeros at the origin then your magnitude diagram will start flat at $20 \\log_{10} |G(0)|$. In this example $20 \\log_{10}|2\/10|\\approx -13$.<\/li>
  2. It will always end with a slope of $-20(n-m)$dB\/decade, where $n$ is the number of poles and $m$ the number of zeros. In this example, $n=2$, $m=1$, and the plot ends with a $-20(2-1)=-20$dB\/decade slope.<\/li><\/ol>\n\n\n\n

    We will consider the sketch of the phase on a followup post.<\/p>\n","protected":false},"excerpt":{"rendered":"

    In this post we will go over the process of sketching the straight-line Bode plot approximations for a simple rational transfer-function in a step-by-step fashion. See Section 7.1 for details on the approximations. We will start with the magnitude plot and cover the phase plot in a future post.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[55],"tags":[40,47,54,41,42],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/767"}],"collection":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/comments?post=767"}],"version-history":[{"count":21,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/767\/revisions"}],"predecessor-version":[{"id":948,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/767\/revisions\/948"}],"wp:attachment":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/media?parent=767"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/categories?post=767"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/tags?post=767"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}