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At what frequency is the phase shift introduced by the circuit of Example 6.7 equal to -10 degrees.

At what frequency is the phase shift introduced by the circuit of Example 6.7 equal to -10 degrees.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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At what frequency is the phase shift introduced by the circuit of Example 6.7 equal to -10 degrees. 

Known Quantities: R 1 k; C = 0.47 µF; vi(t) = 5 cos(@t) V. be close to the ones computed above. Find: The output voltage v.(t) at each frequency. This low-pass filter would pass only the bass range of the audio spectrum. EXAMPLE 6.7 Frequency Response of RC Low-Pass Filter Compute the response of the RC filter of Figure 6.20 to sinusoidal inputs at the frequencies of Analysis: In this problem, the input signal voltage and the frequency response of the circuit cies. If the voltages are represented in phasor form, the frequency response can be used for (equation 6.43) are known, and the output voltage must be found at two different frequen- Part I Circuits 347 Problem LO Compute 60 and 10,000 Hz. the mesponse of the RC filter of Figure 6.20 to sinusoidal inputs at the frequencies of Solution %3D Assumptions: None. calculation: Vo ja) = Hv (ja) = Vi %3D 1+ jwCR V.(ja) = Hy(jw)V,(ja) = 1 V,(jw) %3D 1+ jwCR The cutoff frequency of the filter is wo = 1/RC = 2,128 rad/s such that the expression for the %3D frequency response in the form of equations 6.45 and 6.46 is: 1 1 |Hy (jw)| = W Hy (jw) = %3D %3D 1+ jw/wo V1+ (@/@0)? RA8.0 = - arctan Wo Next, recognize that at w = 60 Hz = 1207 rad/s, the ratio w/wo = 0.177. At w = 10,000 Hz = 20,000T, w/wo = 29.5. Thus, the output voltage at each frequency can be computed as %3D %3D %3D follows: V,(@ = 27 60) = 1 V; (@ = 27 60) = (0.985 x 5)Z-0.175 V %3D %D %3D 1+ j0.177 V.(@ = 2n 10,000) = 1 V; (@ = 27 10,000) = (0.0339 × 5)2-1.537 V %3D %3D %3D 1+ j29.5 And finally we write the time-domain response for each frequency: v,(t) = 4.923 cos(27 60t – 0.175) V at w = 2n 60 rad/s %3D o(t) = 0.169 cos(2n 10,000t - 1.537) V at w = 27 10,000 rad/s %3D This Lo plots that only the low-frequency components of the signal are of the audio spectrum. passed by the filter. be close to the ones computed above.
Transcribed Image Text:Known Quantities: R 1 k; C = 0.47 µF; vi(t) = 5 cos(@t) V. be close to the ones computed above. Find: The output voltage v.(t) at each frequency. This low-pass filter would pass only the bass range of the audio spectrum. EXAMPLE 6.7 Frequency Response of RC Low-Pass Filter Compute the response of the RC filter of Figure 6.20 to sinusoidal inputs at the frequencies of Analysis: In this problem, the input signal voltage and the frequency response of the circuit cies. If the voltages are represented in phasor form, the frequency response can be used for (equation 6.43) are known, and the output voltage must be found at two different frequen- Part I Circuits 347 Problem LO Compute 60 and 10,000 Hz. the mesponse of the RC filter of Figure 6.20 to sinusoidal inputs at the frequencies of Solution %3D Assumptions: None. calculation: Vo ja) = Hv (ja) = Vi %3D 1+ jwCR V.(ja) = Hy(jw)V,(ja) = 1 V,(jw) %3D 1+ jwCR The cutoff frequency of the filter is wo = 1/RC = 2,128 rad/s such that the expression for the %3D frequency response in the form of equations 6.45 and 6.46 is: 1 1 |Hy (jw)| = W Hy (jw) = %3D %3D 1+ jw/wo V1+ (@/@0)? RA8.0 = - arctan Wo Next, recognize that at w = 60 Hz = 1207 rad/s, the ratio w/wo = 0.177. At w = 10,000 Hz = 20,000T, w/wo = 29.5. Thus, the output voltage at each frequency can be computed as %3D %3D %3D follows: V,(@ = 27 60) = 1 V; (@ = 27 60) = (0.985 x 5)Z-0.175 V %3D %D %3D 1+ j0.177 V.(@ = 2n 10,000) = 1 V; (@ = 27 10,000) = (0.0339 × 5)2-1.537 V %3D %3D %3D 1+ j29.5 And finally we write the time-domain response for each frequency: v,(t) = 4.923 cos(27 60t – 0.175) V at w = 2n 60 rad/s %3D o(t) = 0.169 cos(2n 10,000t - 1.537) V at w = 27 10,000 rad/s %3D This Lo plots that only the low-frequency components of the signal are of the audio spectrum. passed by the filter. be close to the ones computed above.
348 Frequency Response and System Concepts Chapter 6 1.0 0.5 102 104 10 Radian frequency, rad/s (logarithmic scale) 100 103 105 106 Phase response of RC filter of Example 6.7 -50 -100 100 101 102 103 104 105 10 Radian frequency, rad/s (logarithmic scale) Figure 6.24 Response of RC filter of Example 6.7 proupal lo mot Degrees
Transcribed Image Text:348 Frequency Response and System Concepts Chapter 6 1.0 0.5 102 104 10 Radian frequency, rad/s (logarithmic scale) 100 103 105 106 Phase response of RC filter of Example 6.7 -50 -100 100 101 102 103 104 105 10 Radian frequency, rad/s (logarithmic scale) Figure 6.24 Response of RC filter of Example 6.7 proupal lo mot Degrees
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