{"id":42549,"date":"2025-06-29T06:10:06","date_gmt":"2025-06-29T06:10:06","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=42549"},"modified":"2025-06-29T06:10:08","modified_gmt":"2025-06-29T06:10:08","slug":"find-the-percent-overshoot-settling-time-rise-time-and-peak-time-for-a","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-percent-overshoot-settling-time-rise-time-and-peak-time-for-a\/","title":{"rendered":"Find the percent overshoot, settling time, rise time, and peak time for a."},"content":{"rendered":"\n
Find the percent overshoot, settling time, rise time, and peak time for a.<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Here is the solution for the given control systems problem.<\/p>\n\n\n\n
This problem requires calculating key performance metrics for two systems based on their transfer functions. These metrics, including percent overshoot, settling time, rise time, and peak time, are standard measures for the transient response of a system, typically analyzed using a second-order system model.<\/p>\n\n\n\n
a. For T(s) = 14.145 \/ ((s\u00b2 + 1.204s + 2.829)(s + 5))<\/strong><\/h3>\n\n\n\n
This is a third-order system. To find the performance metrics, we can use a second-order approximation because one pole is much farther from the imaginary axis than the other two dominant complex conjugate poles. The poles are at s = -5 and s \u2248 -0.602 \u00b1 j1.57. Since the real pole at -5 is more than five times farther than the real part of the complex poles (-0.602), we can approximate the system by considering only the dominant poles from the quadratic term.<\/p>\n\n\n\n
The approximating second-order transfer function is:
T_approx(s) = 2.829 \/ (s\u00b2 + 1.204s + 2.829)
We match this to the standard form T(s) = \u03c9n\u00b2 \/ (s\u00b2 + 2\u03b6\u03c9n s + \u03c9n\u00b2).<\/p>\n\n\n\n
    \n
  1. Natural Frequency (\u03c9n):<\/strong>
    \u03c9n\u00b2 = 2.829 => \u03c9n \u2248 1.682 rad\/s<\/li>\n\n\n\n
  2. Damping Ratio (\u03b6):<\/strong>
    2\u03b6\u03c9n = 1.204 => \u03b6 = 1.204 \/ (2 * 1.682) \u2248 0.358<\/li>\n<\/ol>\n\n\n\n
    Using these values, we calculate the performance metrics:<\/p>\n\n\n\n
      \n
    • Percent Overshoot (%OS):<\/strong>\u00a0%OS = 100 * e^(-\u03b6\u03c0 \/ \u221a(1-\u03b6\u00b2)) \u2248 30.0%<\/li>\n\n\n\n
    • Settling Time (Ts, 2% criterion):<\/strong>\u00a0Ts \u2248 4 \/ (\u03b6\u03c9n) = 4 \/ (0.602) \u2248 6.64 s<\/li>\n\n\n\n
    • Peak Time (Tp):<\/strong>\u00a0Tp = \u03c0 \/ (\u03c9n\u221a(1-\u03b6\u00b2)) = \u03c0 \/ 1.57 \u2248 2.00 s<\/li>\n\n\n\n
    • Rise Time (Tr, 10-90%):<\/strong>\u00a0Tr \u2248 (\u03c0 – arccos(\u03b6)) \/ (\u03c9n\u221a(1-\u03b6\u00b2)) \u2248 1.24 s<\/li>\n<\/ul>\n\n\n\n
      b. For T(s) = (1.05 \u00d7 10\u2077) \/ (s\u00b2 + 1.6 \u00d7 10\u00b3 s + 1.05 \u00d7 10\u2077)<\/strong><\/h3>\n\n\n\n
      This is a standard second-order system, so we can directly extract the parameters by comparing it to the standard form.<\/p>\n\n\n\n
        \n
      1. Natural Frequency (\u03c9n):<\/strong>
        \u03c9n\u00b2 = 1.05 \u00d7 10\u2077 => \u03c9n \u2248 3240 rad\/s<\/li>\n\n\n\n
      2. Damping Ratio (\u03b6):<\/strong>
        2\u03b6\u03c9n = 1.6 \u00d7 10\u00b3 => \u03b6 = (1.6 \u00d7 10\u00b3) \/ (2 * 3240) \u2248 0.247<\/li>\n<\/ol>\n\n\n\n
        Now, we calculate the performance metrics:<\/p>\n\n\n\n
          \n
        • Percent Overshoot (%OS):<\/strong>\u00a0%OS = 100 * e^(-\u03b6\u03c0 \/ \u221a(1-\u03b6\u00b2)) \u2248 44.9%<\/li>\n\n\n\n
        • Settling Time (Ts, 2% criterion):<\/strong>\u00a0Ts \u2248 4 \/ (\u03b6\u03c9n) = 4 \/ 800 = 0.005 s or 5.0 ms<\/li>\n\n\n\n
        • Peak Time (Tp):<\/strong>\u00a0Tp = \u03c0 \/ (\u03c9n\u221a(1-\u03b6\u00b2)) = \u03c0 \/ 3140 \u2248 0.001 s or 1.0 ms<\/li>\n\n\n\n
        • Rise Time (Tr, 10-90%):<\/strong>\u00a0Tr \u2248 (\u03c0 – arccos(\u03b6)) \/ (\u03c9n\u221a(1-\u03b6\u00b2)) \u2248 0.00058 s or 0.58 ms<\/li>\n<\/ul>\n\n\n\n
          thumb_upthumb_down<\/p>\n\n\n\n
          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
          Find the percent overshoot, settling time, rise time, and peak time for a. The Correct Answer and Explanation is: Here is the solution for the given control systems problem. This problem requires calculating key performance metrics for two systems based on their transfer functions. These metrics, including percent overshoot, settling time, rise time, and peak […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-42549","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42549","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=42549"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42549\/revisions"}],"predecessor-version":[{"id":42552,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42549\/revisions\/42552"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42549"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42549"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42549"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}