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. Ask Allem, let's first derive the complex formula for G(jw), where w is the angular frequency. In this case, the input voltage is applied across the series combination of an inductor (L) and a resistor (R), and the output voltage is measured across the resistor. The impedance of the inductor is given by ZL = jwL, and the impedance of the resistor is ZR = R. Therefore, the total impedance in the circuit is Ztotal = ZL + ZR = jwL + R. a) Complex formula for G (jw): G(jw) == ZR Ziotal = R jwL+R b) Now, let's plot |G (jw) | vs. w on a linear scale, |G (jw) | vs. w on a decibel scale, and arg(G(jw)) vs. won a linear scale with limits of -π and +π. To sketch these plots, we'll calculate |G(jw), dB(|G(jw)|), and 2arg(G(jw)) for the specified frequencies: w → 0, w, w = wc, w = 6, and w = = 10wc. Given: ⚫ R = 5102 • ⚫ L = 0.068 H Let's start by calculating the cutoff frequency we: WC = RC RC = 510-0.068 We 23.374 rad/s Now, we can calculate |G(jw), dB(|G(jw)|), and 2arg(G(jw)) for the specified frequencies. We'll choose w values that cover a range from very low frequencies to very high frequencies, ensuring we capture the behavior of the low-pass filter across different frequency ranges. Let's calculate and sketch these plots.