Problem 1. Low-pass RL circuit In lecture we saw that a resistor and capacitor together act as a low-pass filter when the input voltage is applied across the two components in series and the output voltage is measured across the capacitor. Our derivation used the impedance divider formula, G(jw) = VOUT/VIN = Zc/(ZC+ZR), where V=V(jw) = {v(t)}. We then determine the magnitude and phase of the gain G(jw) at w=0, w, and the half-power frequency. The half-power frequency is also called the cutoff frequency fc or wc. For an R-C circuit, wc = 1/RC and fc = wc/2π. Using the technique described above, show that a resistor and inductor together act as a low-pass filter when the input voltage is applied across the two components in series and the output voltage is measured across the resistor. The triangle is our 0 V reference. 0000 + Vin Vout R₁ a) Generate the complex formula for G(jw). b) Make the following three plots, in which w is on a log scale. 1 |G(jw)| vs. w, where |G(jw)| is on a linear scale from 0 to 1, 2 |G(jw)l vs. w, where |G(jw)] is on a decibel scale from 0 downward, and ③ZG(jw) vs. w, where Label the increments on the frequency axis with the actual value (0.1, 1, 10, 100, etc.), not the log of the value (-1, 0, 1, 2, etc.). > For each of the plots, calculate |G(jw)], dB(|G(jw)]) or G(jw), as appropriate, for the following frequencies: w→0, w→oo, w = wc, w= wc/10, and w = 10-wc. > You cannot get for w to reach zero on a logarithmic scale, so just use some small value of frequency as your minimum. In a high-pass filter, the frequency that makes the gain -40 dB is reasonable; in a low-pass filter, go low enough to show a flat pass band.
Problem 1. Low-pass RL circuit In lecture we saw that a resistor and capacitor together act as a low-pass filter when the input voltage is applied across the two components in series and the output voltage is measured across the capacitor. Our derivation used the impedance divider formula, G(jw) = VOUT/VIN = Zc/(ZC+ZR), where V=V(jw) = {v(t)}. We then determine the magnitude and phase of the gain G(jw) at w=0, w, and the half-power frequency. The half-power frequency is also called the cutoff frequency fc or wc. For an R-C circuit, wc = 1/RC and fc = wc/2π. Using the technique described above, show that a resistor and inductor together act as a low-pass filter when the input voltage is applied across the two components in series and the output voltage is measured across the resistor. The triangle is our 0 V reference. 0000 + Vin Vout R₁ a) Generate the complex formula for G(jw). b) Make the following three plots, in which w is on a log scale. 1 |G(jw)| vs. w, where |G(jw)| is on a linear scale from 0 to 1, 2 |G(jw)l vs. w, where |G(jw)] is on a decibel scale from 0 downward, and ③ZG(jw) vs. w, where Label the increments on the frequency axis with the actual value (0.1, 1, 10, 100, etc.), not the log of the value (-1, 0, 1, 2, etc.). > For each of the plots, calculate |G(jw)], dB(|G(jw)]) or G(jw), as appropriate, for the following frequencies: w→0, w→oo, w = wc, w= wc/10, and w = 10-wc. > You cannot get for w to reach zero on a logarithmic scale, so just use some small value of frequency as your minimum. In a high-pass filter, the frequency that makes the gain -40 dB is reasonable; in a low-pass filter, go low enough to show a flat pass band.
Introductory Circuit Analysis (13th Edition)
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Can you please help me with this problem from my Biomedical Signals and sensors class?
![Problem 1. Low-pass RL circuit
In lecture we saw that a resistor and capacitor together act as a low-pass filter when the
input voltage is applied across the two components in series and the output voltage is
measured across the capacitor. Our derivation used the impedance divider formula,
G(jw) = VOUT/VIN = Zc/(ZC+ZR), where V=V(jw) = {v(t)}. We then determine the
magnitude and phase of the gain G(jw) at w=0, w, and the half-power frequency.
The half-power frequency is also called the cutoff frequency fc or wc. For an R-C circuit,
wc = 1/RC and fc = wc/2π.
Using the technique described above, show that a resistor and inductor together act as a
low-pass filter when the input voltage is applied across the two components in series and
the output voltage is measured across the resistor. The triangle is our 0 V reference.
0000
+
Vin
Vout
R₁
a) Generate the complex formula for G(jw).
b) Make the following three plots, in which w is on a log scale.
1 |G(jw)| vs. w, where |G(jw)| is on a linear scale from 0 to 1,
2 |G(jw)l vs. w, where |G(jw)] is on a decibel scale from 0 downward, and
③ZG(jw) vs. w, where <G(jw) is on a linear scale with limits of -π and +π.
Plotting instructions:
Each of these plots should be sketched by hand, since you might need to do this on the
midterm & final exam. You may use MATLAB to check your work, but a MATLAB plot
should not be the only one you turn in.
You may use frequency in Hz or rad/sec, but remember the factor of 2π. The cutoff
frequency that you calculate from R and L is in rad/sec. For this problem you may use
R = 510 2 and L = 0.068 H... or other values as long as you show the calculations!
> Label the increments on the frequency axis with the actual value (0.1, 1, 10, 100, etc.),
not the log of the value (-1, 0, 1, 2, etc.).
> For each of the plots, calculate |G(jw)], dB(|G(jw)]) or G(jw), as appropriate, for the
following frequencies: w→0, w→oo, w = wc, w= wc/10, and w = 10-wc.
> You cannot get for w to reach zero on a logarithmic scale, so just use some small value
of frequency as your minimum. In a high-pass filter, the frequency that makes the gain
-40 dB is reasonable; in a low-pass filter, go low enough to show a flat pass band.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F80769c37-70a7-41b1-a8db-2e9d30552768%2Fe4847af5-06ae-4b00-bd2d-6455d8cc5f57%2Fk7n8xi9_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 1. Low-pass RL circuit
In lecture we saw that a resistor and capacitor together act as a low-pass filter when the
input voltage is applied across the two components in series and the output voltage is
measured across the capacitor. Our derivation used the impedance divider formula,
G(jw) = VOUT/VIN = Zc/(ZC+ZR), where V=V(jw) = {v(t)}. We then determine the
magnitude and phase of the gain G(jw) at w=0, w, and the half-power frequency.
The half-power frequency is also called the cutoff frequency fc or wc. For an R-C circuit,
wc = 1/RC and fc = wc/2π.
Using the technique described above, show that a resistor and inductor together act as a
low-pass filter when the input voltage is applied across the two components in series and
the output voltage is measured across the resistor. The triangle is our 0 V reference.
0000
+
Vin
Vout
R₁
a) Generate the complex formula for G(jw).
b) Make the following three plots, in which w is on a log scale.
1 |G(jw)| vs. w, where |G(jw)| is on a linear scale from 0 to 1,
2 |G(jw)l vs. w, where |G(jw)] is on a decibel scale from 0 downward, and
③ZG(jw) vs. w, where <G(jw) is on a linear scale with limits of -π and +π.
Plotting instructions:
Each of these plots should be sketched by hand, since you might need to do this on the
midterm & final exam. You may use MATLAB to check your work, but a MATLAB plot
should not be the only one you turn in.
You may use frequency in Hz or rad/sec, but remember the factor of 2π. The cutoff
frequency that you calculate from R and L is in rad/sec. For this problem you may use
R = 510 2 and L = 0.068 H... or other values as long as you show the calculations!
> Label the increments on the frequency axis with the actual value (0.1, 1, 10, 100, etc.),
not the log of the value (-1, 0, 1, 2, etc.).
> For each of the plots, calculate |G(jw)], dB(|G(jw)]) or G(jw), as appropriate, for the
following frequencies: w→0, w→oo, w = wc, w= wc/10, and w = 10-wc.
> You cannot get for w to reach zero on a logarithmic scale, so just use some small value
of frequency as your minimum. In a high-pass filter, the frequency that makes the gain
-40 dB is reasonable; in a low-pass filter, go low enough to show a flat pass band.
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