{"id":838,"date":"2019-06-01T16:38:44","date_gmt":"2019-06-02T00:38:44","guid":{"rendered":"https:\/\/linearcontrol.info\/fundamentals\/?p=838"},"modified":"2019-06-09T13:11:57","modified_gmt":"2019-06-09T21:11:57","slug":"step-by-step-bode-plot-example-part-iii","status":"publish","type":"post","link":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/2019\/06\/01\/step-by-step-bode-plot-example-part-iii\/","title":{"rendered":"Step-by-step Bode plot example. Part III"},"content":{"rendered":"\n

We have gone step-by-step over how to sketch straight-line approximations for the magnitude<\/a> and phase<\/a> diagrams of the Bode plot for a simple rational transfer-function. This transfer-function had only left-hand side poles and zeros, that is it was minimum-phase<\/em> (see Section 7.2). In this post we consider a non-minimum phase<\/em> transfer-function with a right-hand side zero. See Section 7.1 for details on the approximations.<\/p>\n\n\n\n\n\n\n\n

We consider a variation on the second-order transfer-function we addressed before:<\/p>\n\n\n\n

$$G(s) = \\frac{s+2}{s^2-9s -10}$$<\/p>\n\n\n\n

The only difference it that the pole at $s=-10$ has been moved to the right-hand side to $s=10$. This is more evident after factoring and normalizing<\/p>\n\n\n\n

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

Note that the above change also affects the sign of the gain, which now has become negative.<\/p>\n\n\n\n

Because the sign of the gain and the poles or zeros have no effect in the magnitude of the frequency response, the magnitude straight-line approximation is the same as the one obtained before in this post<\/a>:<\/p>\n\n\n\n

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

As for the phase, we can proceed as in this post<\/a> to construct the phase diagram up until $\\omega =1$, at which point the effects of the pole at $s=10$ begin:<\/p>\n\n\n\n

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

Because $s=10$ is a pole on the right-hand side, a slope of $+45^\\circ$\/decade is added that lasts until $\\omega=100$. At $\\omega=10$ the value of the phase will be<\/p>\n\n\n\n

$$-13 +45 \\log_{10}(10\/1) = 32^\\circ$$<\/p>\n\n\n\n

and, because of the residual slope of $+45$ in $[10, 20]$, the combined slope becomes $+90^\\circ$ at that range, which leads to a phase value of<\/p>\n\n\n\n

$$32+90\\log_{10}(20\/10) \\approx 59$$<\/p>\n\n\n\n

at $\\omega=20$. At that point, the slope is back at $+45^\\circ$\/decade and then goes flat at $\\omega=100$ as shown below:<\/p>\n\n\n\n

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

Because the gain $-2\/10$ is negative, an additional phase of $-180^\\circ$ is added to obtain the red line sketch as in the next figure:<\/p>\n\n\n\n

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

Note that we expect that the phase reaches $-90$ at $\\omega=100$ because we have $n_- = 1$ poles one the left-hand side, $n_+=1$ poles on the right-hand side, and $m_-=1$ zeros on the left-hand side so that<\/p>\n\n\n\n

$$-180^\\circ + 90^\\circ \\times (m_- + n_+) \\, – \\, 90^\\circ \\times (n_-) = -180^\\circ + 180^\\circ – 90^\\circ=-90^\\circ$$<\/p>\n\n\n\n

See this previous post<\/a> for a discussion on what phase to expect for large values of $\\omega$. If you are wondering why we subtracted $180^\\circ$ and not added $180^\\circ$ to account for the negative gain you are not alone. We will discuss that in a future post.<\/p>\n","protected":false},"excerpt":{"rendered":"

We have gone step-by-step over how to sketch straight-line approximations for the magnitude and phase diagrams of the Bode plot for a simple rational transfer-function. This transfer-function had only left-hand side poles and zeros, that is it was minimum-phase (see Section 7.2). In this post we consider a non-minimum phase transfer-function with a right-hand side … Continue reading “Step-by-step Bode plot example. Part III”<\/span><\/a><\/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":[40,47,54,41,44,45,43,42],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/838"}],"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=838"}],"version-history":[{"count":14,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/838\/revisions"}],"predecessor-version":[{"id":950,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/838\/revisions\/950"}],"wp:attachment":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/media?parent=838"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/categories?post=838"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/tags?post=838"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}