Problem 1: Sinusoidal Steady State

The circuits shown below are driven by sinusoidal voltage sources and operate in the sinusoidal steady state with transfer function 𝐻 𝜔 = 𝑉%&'

𝑉()where 𝑉() and 𝑉%&' are complex voltage amplitudes. For each circuit, write an expression for the transfer function, then select from the list of plots provided, the plot that represents the transfer function magnitude and the plot that represents the transfer function phase.

Transfer Function

𝐻 𝑗𝜔 =𝑉()𝑉%&'

=𝑍,

𝑍, + 𝑍. + 𝑍/

𝐻 𝑗𝜔 =𝑅

𝑅 + 𝑗𝜔𝐿 + 1𝑗𝜔𝐶

=𝑗𝜔𝑅𝐿

−𝜔5 + 𝑗𝜔 𝑅𝐿 +1𝐿𝐶

This is a band-pass filter. Magnitude: A B C D E F

Phase: G H I J K L

Transfer Function

𝐻 𝑗𝜔 =𝑉()𝑉%&'

=𝑍/

𝑍, + 𝑍. + 𝑍/

𝐻 𝑗𝜔 =

1𝑗𝜔𝐶

𝑅 + 𝑗𝜔𝐿 + 1𝑗𝜔𝐶

=1𝐿𝐶

−𝜔5 + 𝑗𝜔 𝑅𝐿 +1𝐿𝐶

This is a low-pass filter. Magnitude: A B C D E F

Phase: G H I J K L

Transfer Function

𝐻 𝑗𝜔 =𝑉()𝑉%&'

=𝑍/

𝑍,6 + 𝑍. + 𝑍/

This is the same as the previous example with effective resistance 𝑅7 = 𝑅 ∥ 𝑅 = ,

5

𝐻 𝑗𝜔 =

1𝑗𝜔𝐶

𝑅2 + 𝑗𝜔𝐿 +

1𝑗𝜔𝐶

=1𝐿𝐶

−𝜔5 + 𝑗𝜔 𝑅2𝐿 +

1𝐿𝐶

This is a low-pass filter. Here, the effective R is less than the previous example causing a higher Q (𝑄 = 𝜔;𝐿/𝑅). Magnitude: A B C D E F

Phase: G H I J K L

Circuit1

𝐻 𝑗𝜔 =𝑅

𝑅 + 𝑗𝜔𝐿 + 1𝑗𝜔𝐶

=𝑗𝜔𝑅𝐿

−𝜔5 + 𝑗𝜔 𝑅𝐿 +1𝐿𝐶

𝐻 𝑗𝜔 =𝜔𝑅𝐿

( 1𝐿𝐶 − 𝜔5)5 + (𝜔𝑅𝐿 )

5

∠𝐻 𝑗𝜔 =𝜋2 − tan

DE𝜔𝑅𝐿

1𝐿𝐶 − 𝜔

5

𝜔 → 0 𝐻 ≈ 0 ∠𝐻 =

𝜋2

𝜔 → ∞ 𝐻 ≈ 0 ∠𝐻 = −𝜋2

𝜔 = 𝜔; =1𝐿𝐶

𝐻 = 1 ∠𝐻 = 0

Circuit2

𝐻 𝑗𝜔 =

1𝑗𝜔𝐶

𝑅 + 𝑗𝜔𝐿 + 1𝑗𝜔𝐶

=1𝐿𝐶

−𝜔5 + 𝑗𝜔 𝑅𝐿 +1𝐿𝐶

𝐻 𝑗𝜔 =1𝐿𝐶

( 1𝐿𝐶 − 𝜔5)5 + (𝜔𝑅𝐿 )

5

∠𝐻 𝑗𝜔 = − tanDE𝜔𝑅𝐿

1𝐿𝐶 − 𝜔

5

𝜔 → 0 𝐻 ≈ 1 ∠𝐻 = 0𝜔 → ∞ 𝐻 ≈ 0 ∠𝐻 = −𝜋

𝜔 = 𝜔; =1𝐿𝐶

𝐻 =𝐿

𝑅 𝐶 ∠𝐻 = −

𝜋2

Circuit3

𝐻 𝑗𝜔 =

1𝑗𝜔𝐶

𝑅2 + 𝑗𝜔𝐿 +

1𝑗𝜔𝐶

=1𝐿𝐶

−𝜔5 + 𝑗𝜔 𝑅2𝐿 +

1𝐿𝐶

𝐻 𝑗𝜔 =1𝐿𝐶

( 1𝐿𝐶 − 𝜔5)5 + (𝜔𝑅2𝐿 )

5

∠𝐻 𝑗𝜔 = − tanDE𝜔𝑅2𝐿

1𝐿𝐶 − 𝜔

5

𝜔 → 0 𝐻 ≈ 1 ∠𝐻 = 0𝜔 → ∞ 𝐻 ≈ 0 ∠𝐻 = −𝜋

𝜔 = 𝜔; =1𝐿𝐶

𝐻 =2 𝐿𝑅 𝐶

∠𝐻 = −𝜋2

Problem 2: By analyzing the plot, the voltage across the device in volts is

and the current is (current peaks 0.001 seconds before voltage.

With a period of 0.01 seconds, this is a phase lead in current (or a phase lag).

In Phasor form both of these could be written as and , for the voltage

and current, respectively. Current is leading voltage in this situation. It is incorrect to say the

component must be a capacitor (that would be implied if the lead is exactly ), however

Impedance takes the form:

Therefore the impedance of would be:

In rectangular form, which will prove useful on this problem this can be written as:

To insert an impedance such that the overall current lines up with voltage we need to insert a

value in series with which results in all imaginary impedance canceling out. There is no

restriction (other than extreme results) on the real value of the total impedance, but the phase

needs to be 0 (which comes about from having no total imaginary part). So the part we add

must have an offseting imaginary component.

The two parts are in series, so they add easily: .

The imaginary terms will offset if has a value at the current frequency (100Hz) of

. What value could provide a positive, purely real impedance? An inductor! (A

capacitor on its own can't without negative frequency).

What inductor value would accomplish this?

Therefore:

Fall 2019 6.002 Midterm 2 7

Problem 3.

Consider the circuit below:

A) Assume switch 𝑆1 has been closed for a long time. Determine the current in the inductor 𝑖𝐿 and

voltage across the inductor 𝑣𝐿.

𝑖𝐿 =

𝑣𝐿 =

Fall 2019 6.002 Midterm 2 8

B) After sitting with switch S1 closed for a long time, at 𝑡 = 0 the switch 𝑆1 opens, leading to the

circuit as shown below. Determine the current in the inductor and voltage across the inductor at

𝑡 = 0+.

𝑖𝐿(0+) =

𝑣𝐿(0+) =

Fall 2019 6.002 Midterm 2 9

C) Determine the time constant governing the change of the current through the inductor when

switch S1 is open. The circuit from part B is reproduced below for convenience.

𝜏 =