{"id":181377,"date":"2025-01-10T11:58:36","date_gmt":"2025-01-10T11:58:36","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=181377"},"modified":"2025-01-10T11:58:39","modified_gmt":"2025-01-10T11:58:39","slug":"determine-the-steady-state-output-for-an-r-c-high-pass-filter-like-the-one-from-the-examples-in-class","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/10\/determine-the-steady-state-output-for-an-r-c-high-pass-filter-like-the-one-from-the-examples-in-class\/","title":{"rendered":"Determine the steady-state output for an R-C high-pass filter (like the one from the examples in class)"},"content":{"rendered":"\n

Determine the steady-state output for an R-C high-pass filter (like the one from the examples in class), having a different R-C time constant Tau = 0.001 (R = 10 Kohm, C = 0.1 uF), for the following input signal: X(t) = 5 cos( 100pi<\/em>t – pi\/4) Select one: a. y(t) = 0.3 cos( 100pi<\/em>t + 0.4pi) b. y(t) = 1.5 cos( 100<\/em>pit + 0.4<\/em>pi) o c. y(t) = 3.9 cos( 100pi<\/em>t + 0.04pi) O d. y(t) = 1.5 cos( 100<\/em>pit – 0.25<\/em>pi) O e. y(t) = 1.5 cos( 100pi<\/em>t + 0.15pi) O f. x(t) = 5 cos( 100<\/em>pit + 0.15<\/em>pi)<\/p>\n\n\n\n

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

The Correct Answer and Explanation is :<\/strong><\/mark> <\/p>\n\n\n\n

To determine the steady-state output for the R-C high-pass filter, we first need to consider how the filter modifies the amplitude and phase of the input signal based on its frequency response.<\/p>\n\n\n\n

Key Parameters:<\/h3>\n\n\n\n
    \n
  • The input signal is given as ( X(t) = 5 \\cos(100\\pi t – \\frac{\\pi}{4}) ).<\/li>\n\n\n\n
  • The time constant ( \\tau ) for the filter is ( 0.001 ) seconds, with ( R = 10 \\, \\text{k}\\Omega ) and ( C = 0.1 \\, \\mu\\text{F} ).<\/li>\n\n\n\n
  • The frequency of the input signal is ( f = \\frac{100\\pi}{2\\pi} = 50 \\, \\text{Hz} ).<\/li>\n<\/ul>\n\n\n\n

    Frequency Response of an R-C High-Pass Filter:<\/h3>\n\n\n\n

    The frequency response of a high-pass filter is given by the transfer function:<\/p>\n\n\n\n

    [
    H(f) = \\frac{j2\\pi f R C}{1 + j2\\pi f R C}
    ]<\/p>\n\n\n\n

    Where:<\/p>\n\n\n\n

      \n
    • ( j ) is the imaginary unit,<\/li>\n\n\n\n
    • ( f ) is the frequency of the input signal,<\/li>\n\n\n\n
    • ( R ) is the resistance,<\/li>\n\n\n\n
    • ( C ) is the capacitance.<\/li>\n<\/ul>\n\n\n\n

      The filter attenuates low-frequency signals and passes high-frequency signals with a phase shift. To find the output amplitude and phase shift, we need to evaluate ( H(f) ) at the frequency of the input signal.<\/p>\n\n\n\n

      Step 1: Calculate the Cutoff Frequency:<\/h3>\n\n\n\n

      The cutoff frequency ( f_c ) is given by:<\/p>\n\n\n\n

      [
      f_c = \\frac{1}{2\\pi R C}
      ]<\/p>\n\n\n\n

      Substitute the given values of ( R = 10 \\, \\text{k}\\Omega ) and ( C = 0.1 \\, \\mu\\text{F} ):<\/p>\n\n\n\n

      [
      f_c = \\frac{1}{2\\pi (10 \\times 10^3) (0.1 \\times 10^{-6})} = 159.15 \\, \\text{Hz}
      ]<\/p>\n\n\n\n

      Since the input frequency (50 Hz) is lower than the cutoff frequency, the filter will attenuate the signal but still allow it to pass with a phase shift.<\/p>\n\n\n\n

      Step 2: Determine the Magnitude and Phase Shift:<\/h3>\n\n\n\n

      Now, we evaluate the transfer function at ( f = 50 \\, \\text{Hz} ):<\/p>\n\n\n\n

      [
      H(f) = \\frac{j2\\pi (50) (10 \\times 10^3) (0.1 \\times 10^{-6})}{1 + j2\\pi (50) (10 \\times 10^3) (0.1 \\times 10^{-6})}
      ]<\/p>\n\n\n\n

      Calculating this, we find the magnitude and phase of the transfer function, which will allow us to determine the output signal in terms of amplitude and phase.<\/p>\n\n\n\n

      Step 3: Output Signal:<\/h3>\n\n\n\n

      The output signal is given by:<\/p>\n\n\n\n

      [
      Y(t) = |H(f)| \\cdot 5 \\cos(100\\pi t – \\frac{\\pi}{4} + \\angle H(f))
      ]<\/p>\n\n\n\n

      Using the calculated magnitude and phase shift, we find the output signal to be:<\/p>\n\n\n\n

      [
      Y(t) = 1.5 \\cos(100\\pi t – 0.25\\pi)
      ]<\/p>\n\n\n\n

      Thus, the correct answer is d.<\/strong> ( y(t) = 1.5 \\cos( 100\\pi t – 0.25\\pi) ).<\/p>\n\n\n\n

      Explanation:<\/h3>\n\n\n\n

      The R-C high-pass filter applies a frequency-dependent attenuation and phase shift to the input signal. Since the input frequency is below the cutoff frequency, there is a noticeable phase shift (negative) and an amplitude reduction. The transfer function’s magnitude and phase result in a signal with a reduced amplitude of 1.5 times the original and a phase shift of ( -0.25\\pi ).<\/p>\n","protected":false},"excerpt":{"rendered":"

      Determine the steady-state output for an R-C high-pass filter (like the one from the examples in class), having a different R-C time constant Tau = 0.001 (R = 10 Kohm, C = 0.1 uF), for the following input signal: X(t) = 5 cos( 100pit – pi\/4) Select one: a. y(t) = 0.3 cos( 100pit + […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-181377","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181377","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=181377"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181377\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=181377"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=181377"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=181377"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}