G(jw) = VOUT/VIN = ZC/(ZC+ZR), where V = V(jw) = F {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. Vin + 5000 R₁ + Vout 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)| vs. w, where |G(jw)| is on a decibel scale from 0 downward, and ③ZG(jw) vs. w, where ZG(jw) is on a linear scale with limits of - and +”. Plotting instructions: www 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. www>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 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 ZG(jw), as appropriate, for the following frequencies: w→0, w⇒∞, w = wc, w = wc/10, and w = 10·ως. > 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.

G(jw) = VOUT/VIN = ZC/(ZC+ZR), where V = V(jw) = F {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. Vin + 5000 R₁ + Vout 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)| vs. w, where |G(jw)| is on a decibel scale from 0 downward, and ③ZG(jw) vs. w, where ZG(jw) is on a linear scale with limits of - and +”. Plotting instructions: www 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. www>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 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 ZG(jw), as appropriate, for the following frequencies: w→0, w⇒∞, w = wc, w = wc/10, and w = 10·ως. > 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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G(jw) = VOUT/VIN = ZC/(ZC+ZR), where V = V(jw) = F {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.
Vin
+
5000
R₁
+
Vout
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)| vs. w, where |G(jw)| is on a decibel scale from 0 downward, and
③ZG(jw) vs. w, where ZG(jw) is on a linear scale with limits of - and +”.
Plotting instructions:
www
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.
www>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 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 ZG(jw), as appropriate, for the
following frequencies: w→0, w⇒∞, w = wc, w = wc/10, and w = 10·ως.
> 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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