What is the transfer function for a series RL circuit with output taken across the resistor (high-pass or low-pass behavior)?

What is the transfer function for a series RL circuit with output taken across the resistor (high-pass or low-pass behavior)?

• A. H(jω) = 1 / (1 + j ω L / R)
• B. H(jω) = (R + j ω L) / R
• C. H(jω) = j ω L / (R + j ω L)
• D. H(jω) = R / (R + j ω L)

Answer: (D)

When output is taken across the resistor in a series RL circuit, the transfer function is the ratio of the resistor voltage to the input voltage. With Vin = I(R + jωL) and Vout = IR, the transfer function becomes H(jω) = Vout/Vin = R / (R + jωL). This form shows the circuit behaves as a low-pass filter: at low frequencies (ω → 0), the inductor looks like a short, so Vout ≈ Vin and the magnitude is near 1; at high frequencies (ω → ∞), the inductor dominates and most of the input drops across L, so Vout → 0 and the magnitude drops. A common equivalent way to write the same function is 1 / (1 + jωL/R); it's just the above expression divided by R. The form with jωL in the numerator would correspond to the output across the inductor, not the resistor, so that isn’t correct for this configuration.

When output is taken across the resistor in a series RL circuit, the transfer function is the ratio of the resistor voltage to the input voltage. With Vin = I(R + jωL) and Vout = IR, the transfer function becomes H(jω) = Vout/Vin = R / (R + jωL).

This form shows the circuit behaves as a low-pass filter: at low frequencies (ω → 0), the inductor looks like a short, so Vout ≈ Vin and the magnitude is near 1; at high frequencies (ω → ∞), the inductor dominates and most of the input drops across L, so Vout → 0 and the magnitude drops. A common equivalent way to write the same function is 1 / (1 + jωL/R); it's just the above expression divided by R.

The form with jωL in the numerator would correspond to the output across the inductor, not the resistor, so that isn’t correct for this configuration.