{"id":799,"date":"2019-05-31T21:43:39","date_gmt":"2019-06-01T05:43:39","guid":{"rendered":"https:\/\/linearcontrol.info\/fundamentals\/?p=799"},"modified":"2019-06-09T13:12:08","modified_gmt":"2019-06-09T21:12:08","slug":"step-by-step-bode-plot-example-part-ii","status":"publish","type":"post","link":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/2019\/05\/31\/step-by-step-bode-plot-example-part-ii\/","title":{"rendered":"Step-by-step Bode plot example. Part II"},"content":{"rendered":"\n

In a previous post<\/a> we have gone step-by-step over how to sketch straight-line approximations for the magnitude diagram of the Bode plot for a simple rational transfer-function. In this post we cover the sketch of the phase diagram. See Section 7.1 for details on the approximations.<\/p>\n\n\n\n\n\n\n\n

We consider the same second-order transfer-function:<\/p>\n\n\n\n

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

which we normalize to:<\/p>\n\n\n\n

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

As with the magnitude we start flat but, this time, the effect of the first pole at $s=-1$ will begin at $\\omega=1\/10=0.1$. Since this is a left-hand side pole, at that point there will be a $-45^\\circ$\/decade slope which will last until $\\omega=10 \\times 1= 10$ as shown in the next figure: <\/p>\n\n\n\n

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

We can proceed to draw the sketch as the red line and stop right before the effect of the next zero at $s=-2$ begins at $\\omega=2\/10=0.2$:<\/p>\n\n\n\n

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

The value at $\\omega = 0.2$ can be calculated as<\/p>\n\n\n\n

$$
\\begin{aligned}
\\Delta_y &= \\text{slope} \\times \\Delta_x \\\\
&= -45 \\times (\\log_{10} (0.2) – \\log_{10} (0.1)) \\\\
&= -45 \\times \\log_{10} (0.2\/0.1) \\approx -13.
\\end{aligned}
$$<\/p>\n\n\n\n

Because $s=-2$ is a zero on the left-hand side it will add a slope of $+45^\\circ$\/decade starting at $\\omega=0.2$ and ending at $\\omega=20$. Combining that with the existing slope of $-45^\\circ$deg\/decade we have a flat segment starting at $\\omega=0.2$ and, recalling that the effect of the phase due to the pole at $s=-1$ ends at $\\omega=10$, we obtain the following figure:<\/p>\n\n\n\n

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

We can proceed to draw the red lines until $\\omega=1$, when the effect of the last pole at $s=-10$ begins. At that point, since $s=-10$ is a pole on the left-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

$$-13 -45 \\log_{10}(10\/1) = -58^\\circ$$<\/p>\n\n\n\n

and, because of the residual slope of $+45$ in $[10, 20]$, the combined slope becomes flat again at that range before resuming at $-45^\\circ$\/decade at $\\omega=20$ and then going flat at $\\omega=100$ as shown below:<\/p>\n\n\n\n

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

Note that the value of the phase at $\\omega = 100$ should be<\/p>\n\n\n\n

$$-58-\\log_{10}(100\/20) \\approx -90^\\circ.$$<\/p>\n\n\n\n

However, it is not necessary to calculate this value in this way since, in this example where all poles and zeros are on the left-hand side, we expect that the value of the phase flattens out at<\/p>\n\n\n\n

$$(m-n)\\times 90^\\circ=(1-2) \\times 90^\\circ = -90^\\circ.$$<\/p>\n\n\n\n

The resulting sketch is the red line:<\/p>\n\n\n\n

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

Because the gain of $2\/10$ is a positive real number, therefore with zero phase, there is no need to add or subtract any phase from the above sketch.<\/p>\n\n\n\n

As with the magnitude, there are some easy sanity checks that you should perform on your phase sketch:<\/p>\n\n\n\n

  1. The phase diagram will always start flat.<\/li>
  2. It may have an offset of $-90^\\circ$ ($+90^\\circ$) per pole (or zero) at the origin and plus $180^\\circ$ in case of a negative gain.<\/li>
  3. It will always end flat at a phase equal to $$90^\\circ \\times (m_{-} + m_0 + n_{+} ) \\, – \\, 90^\\circ \\times (m_{+} + n_{-} + n_0 )$$ plus $180^\\circ$ in case of a negative gain. In this formula $n_-$ and $m_-$ are the number of poles and zeros on the left-hand side, $n_0$ and $m_0$ are the number of poles and zeros on the imaginary axis and $n_+$ and $m_+$ are the number of poles and zeros on the right-hand side.<\/li><\/ol>\n\n\n\n

    On a future post we will do consider more examples with poles and zeros on the right-hand side and complex poles and zeros.<\/p>\n","protected":false},"excerpt":{"rendered":"

    In a previous post we have gone step-by-step over how to sketch straight-line approximations for the magnitude diagram of the Bode plot for a simple rational transfer-function. In this post we cover the sketch of the phase diagram. See Section 7.1 for details on the approximations.<\/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":[40,47,54,41,43,42],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/799"}],"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=799"}],"version-history":[{"count":19,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/799\/revisions"}],"predecessor-version":[{"id":949,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/799\/revisions\/949"}],"wp:attachment":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/media?parent=799"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/categories?post=799"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/tags?post=799"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}