The RC circuit can serve as either a low-pass or a high-pass filter; that is, either low-frequency or high-frequency si
Posted: Fri Jul 01, 2022 6:00 am
The RC circuit can serve as either a low-pass or a high-passfilter; that is, either low-frequency or high-frequency signals are passed from the input to the output withlow attenuation (signal loss), depending on whether the output is the voltage across the capacitoror the resistor. The purpose of this experiment is to examine output voltage and phase asfunctions of the input frequency.
The RC circuit can serve as either a low-pass or a high-pass filter; that is, either low-frequency or high-frequency signals are passed from the input to the output with low attenuation (signal loss), depending on whether the output is the voltage across the capacitor or the resistor. The purpose of this experiment is to examine output voltage and phase as functions of the input frequency. Part I. Construct the following series RC circuit. Start with the function generator's output set to 100 Hz, high load impedance, and 10 Vp-p. V₁ 5.1 ΚΩ ww 10 nF + V₂ A. Measure the amplitude response for V₂/V₁ from 100 Hz to 100 kHz. Use the cursors on the oscilloscope to measure peak-to-peak voltage. It may be more efficient to do Part I.B as you are doing Part I.A. 1 11/1³ = √2/22 B. Also using an oscilloscope, measure the phase response, 1, from 100 Hz to 100 kHz. Take enough data to make accurate graphs of the voltage and phase responses for the circuit. This means that most of the data should be taken in regions in which quantities (V₂ and/or 1) are changing rapidly as a function of frequency. Be sure to take several data points near the half-power frequency, which, for this circuit, occurs when You will need to plot your data using either a spreadsheet, like Microsoft Excel, or specialized plotting software. Because the frequency changes over a much wider range than the voltage or phase, you need to create semi-log plots; that is, plots in which one axis, the frequency axis in this case, is on a logarithmic scale, while the other axis is linear. Three cycles (decades) are sufficient to go from 100 to 100,000. An example of an appropriate plot will be provided.
C. From your amplitude response curve, determine the experimental value of the half-power frequency and compare it with the expected theoretical value, f = 1 2πRC D. From your phase response data, determine the experimental value of the phase at the half- power frequency, as determined in (C), and compare it with the expected theoretical value. E. Is this circuit a low-pass or a high-pass filter? V₁ Part II. Repeat Part I for the following series RC circuit; 1 will now be 2, of course. 10 nF 5.1 ΚΩ Report Put both sets (Parts I and II) of amplitude response data on the same graph. Use your software to draw smooth curves through your data, while also showing the individual data points. V₂ For the phase response in Parts I and II, plot the theoretical curve without showing the individual data points, then show your experimental data on that plot. Use the sign convention for phase from Experiment 1 to show that the phases measured in Part I (1) and Part II (2) are related by $₂ = $₁ + TT The two sets of data can be compared with the same theoretical curve by shifting the vertical axis of 2. Distinguish the two sets of data by using one shape (circle, square, triangle, etc.) for the Part I data points, and a different shape for the Part II data points.
For Part I, following the Experiment 1 sign convention, show that For Part II, show that = 開 1 √1 + (RCw)² = RCw √1+ (RCw)² For each part, use the definition that, at half-power, M- do not use but, rather, 20 log10 Suppose that the maximum value of ; tan ₁ = -RCw to derive the expressions for the half-power frequency and the phase at that frequency. This relationship motivates the derivation in the next section, which shows that the half-power points on a Bode plot occur 3.01 dB below the maximum (note that 3.01 = 10-log10 2). ; tan ₂ Amplitude and Phase Response Plots Semi-log paper has one linear scale (the divisions are equally spaced) and one logarithmic scale (the divisions are not equally spaced), as shown in the supplied plots. In making amplitude and phase response plots, use the logarithmic scale for frequency. For phase response, label the vertical axis in degrees; however, for amplitude response, for example, H. Such a quantity is called a decibel (dB), in honor of Alexander Graham Bell. The resulting graph is known as a Bode (usually pronounced BÖ-dee, in honor of H.W. Bode) plot. 20 log10 max M 1 RCw H = M; then, on the amplitude response plot, the = corresponding value is 20 log₁0 | = 20 log10 M. The half-power frequencies are those frequencies that correspond to the value: 20(log10 M - log10 √2) = 20 log10 M - 20 log10 √2 20 log10 M - 20(0.1505) = 20 log10 M - 3.01 dB
- The Rc Circuit Can Serve As Either A Low Pass Or A High Pass Filter That Is Either Low Frequency Or High Frequency Si 1 (542.05 KiB) Viewed 50 times
C. From your amplitude response curve, determine the experimental value of the half-power frequency and compare it with the expected theoretical value, f = 1 2πRC D. From your phase response data, determine the experimental value of the phase at the half- power frequency, as determined in (C), and compare it with the expected theoretical value. E. Is this circuit a low-pass or a high-pass filter? V₁ Part II. Repeat Part I for the following series RC circuit; 1 will now be 2, of course. 10 nF 5.1 ΚΩ Report Put both sets (Parts I and II) of amplitude response data on the same graph. Use your software to draw smooth curves through your data, while also showing the individual data points. V₂ For the phase response in Parts I and II, plot the theoretical curve without showing the individual data points, then show your experimental data on that plot. Use the sign convention for phase from Experiment 1 to show that the phases measured in Part I (1) and Part II (2) are related by $₂ = $₁ + TT The two sets of data can be compared with the same theoretical curve by shifting the vertical axis of 2. Distinguish the two sets of data by using one shape (circle, square, triangle, etc.) for the Part I data points, and a different shape for the Part II data points.
For Part I, following the Experiment 1 sign convention, show that For Part II, show that = 開 1 √1 + (RCw)² = RCw √1+ (RCw)² For each part, use the definition that, at half-power, M- do not use but, rather, 20 log10 Suppose that the maximum value of ; tan ₁ = -RCw to derive the expressions for the half-power frequency and the phase at that frequency. This relationship motivates the derivation in the next section, which shows that the half-power points on a Bode plot occur 3.01 dB below the maximum (note that 3.01 = 10-log10 2). ; tan ₂ Amplitude and Phase Response Plots Semi-log paper has one linear scale (the divisions are equally spaced) and one logarithmic scale (the divisions are not equally spaced), as shown in the supplied plots. In making amplitude and phase response plots, use the logarithmic scale for frequency. For phase response, label the vertical axis in degrees; however, for amplitude response, for example, H. Such a quantity is called a decibel (dB), in honor of Alexander Graham Bell. The resulting graph is known as a Bode (usually pronounced BÖ-dee, in honor of H.W. Bode) plot. 20 log10 max M 1 RCw H = M; then, on the amplitude response plot, the = corresponding value is 20 log₁0 | = 20 log10 M. The half-power frequencies are those frequencies that correspond to the value: 20(log10 M - log10 √2) = 20 log10 M - 20 log10 √2 20 log10 M - 20(0.1505) = 20 log10 M - 3.01 dB