Phasors, impedance and the Laplace transform

Intent

• Represent sinusoidal signals by phasors, and inductors and capacitors by impedances.
• Compute the impedance of an inductor and a capacitor at a given frequency.
• Find a first-order low-pass filter’s corner frequency and its attenuation above the corner, the building block of converter transfer functions.

Theory

Phasors and impedance

In sinusoidal steady state every voltage and current in a linear circuit is a sinusoid at the source frequency. Each can be represented by a complex number, its phasor, which carries the amplitude and the phase [Alexander & Sadiku 2017, Ch. 9]. Differentiation becomes multiplication by $j\omega$, and each element is described by its impedance, the ratio of its voltage phasor to its current phasor [Alexander & Sadiku 2017, Ch. 9]. For an inductor and a capacitor the magnitudes are

$$|Z_L| = 2 \pi f L$$

$|Z_L|$

inductor impedance

$L$

inductance

$f$

frequency

Impedance magnitude of an inductor✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• sinusoidal steady state (phasor analysis)

Notes

the impedance is $j 2\pi f L$: its magnitude grows in proportion to the frequency and the current lags the voltage by 90°

Source

Alexander & Sadiku 2017, Ch. 9 (Sinusoids and Phasors)

Derivation

step-by-step sympy derivation

Equation id

imp.ZL · equations.yaml entry

$$|Z_C| = \frac{1}{2 \pi f C}$$

$|Z_C|$

capacitor impedance

$C$

capacitance (output capacitor)

$f$

frequency

Impedance magnitude of a capacitor✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• sinusoidal steady state (phasor analysis)

Notes

the impedance is $1/(j 2\pi f C)$: its magnitude falls in inverse proportion to the frequency and the current leads the voltage by 90°

Source

Alexander & Sadiku 2017, Ch. 9 (Sinusoids and Phasors)

Derivation

step-by-step sympy derivation

Equation id

imp.ZC · equations.yaml entry

An inductor’s impedance rises with frequency and a capacitor’s falls. That is what lets a converter’s L-C filter pass the dc component of the switch-node voltage while it removes the switching harmonics [Erickson & Maksimović 2020, Ch. 2].

The Laplace transform and transfer functions

The Laplace transform extends the same idea from sinusoids to any signal: with the complex frequency $s$ in place of $j\omega$, a linear circuit is described by transfer functions, ratios of an output transform to an input transform [Alexander & Sadiku 2017, Ch. 15]. Evaluated at $j\omega$, a transfer function gives the frequency response, the gain and the phase at each frequency [Alexander & Sadiku 2017, Ch. 14].

The first-order low-pass filter

A resistor in series with a capacitor, with the output taken across the capacitor, is the simplest low-pass filter. Its gain is close to one at low frequency and falls above the corner frequency [Alexander & Sadiku 2017, Ch. 14]:

$$f_c = \frac{1}{2 \pi R_f C_f}$$

$f_c$

corner (-3 dB) frequency

$C_f$

filter capacitor

$R_f$

filter resistor

Corner frequency of a first-order R-C low-pass filter✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• first-order R-C low-pass filter with no load on its output

Notes

at $f_c$ the output magnitude is $1/\sqrt{2}$ of the input (the -3 dB point)

Source

Alexander & Sadiku 2017, Ch. 14 (Frequency Response)

Derivation

step-by-step sympy derivation

Equation id

rc.fc · equations.yaml entry

$$|G| = \frac{1}{\sqrt{1 + \left(\frac{f}{f_c}\right)^{2}}}$$

$|G|$

low-pass gain

$f$

frequency

$f_c$

corner (-3 dB) frequency

Gain magnitude of a first-order low-pass filter✓ verified by 8 tests

Assumptions, convention and source

Assumptions

• first-order R-C low-pass filter with no load on its output
• sinusoidal steady state (phasor analysis)

Notes

well above $f_c$ the gain falls in inverse proportion to the frequency: -20 dB per decade on a Bode plot

Source

Alexander & Sadiku 2017, Ch. 14 (Frequency Response)

Derivation

step-by-step sympy derivation

Equation id

rc.gain · equations.yaml entry

Well above the corner, every decade of frequency lowers the gain by a factor of ten: a slope of -20 dB per decade on a Bode plot [Alexander & Sadiku 2017, Ch. 14]. A converter’s L-C filter is of second order, and above its resonance it falls at -40 dB per decade [Erickson & Maksimović 2020, Ch. 8] (see the small-signal page).

Worked example

An example inductor, capacitor and R-C filter at one frequency:

Impedance and R-C filter example

Symbol	Value	How it is obtained
$f$ frequency	100 kHz	given
$L$ inductance	100 µH	given
$C$ capacitance (output capacitor)	1 µF	given
$R_f$ filter resistor	1 kΩ	given
$C_f$ filter capacitor	10 nF	given
$|Z_L|$ inductor impedance	62.8 Ω	Impedance magnitude of an inductor
$|Z_C|$ capacitor impedance	1.59 Ω	Impedance magnitude of a capacitor
$f_c$ corner (-3 dB) frequency	15.9 kHz	Corner frequency of a first-order R-C low-pass filter
$|G|$ low-pass gain	0.157	Gain magnitude of a first-order low-pass filter

At 100 kHz the inductor’s impedance is 62.8 Ω, while the capacitor’s is only 1.59 Ω. The R-C filter’s corner lies at 15.9 kHz, so at that frequency the filter passes only 0.157 of the input amplitude.

Try it

▶ Open in the equation explorer: Gain magnitude of a first-order low-pass filter
Preset from the example impedance-basic

Sweep $f$ on its logarithmic axis: the gain is flat below the corner, equals $1/\sqrt{2}$ at the corner and falls tenfold per decade above it. Then open the capacitor’s impedance and sweep $f$ again to see the same inverse proportion.

Bench exercise

Build an R-C low-pass filter with a corner in the audio range, drive it from a function generator with a sine wave, and measure the output amplitude and phase with an oscilloscope at several frequencies around the corner. Plot the gain in decibels against frequency on a logarithmic axis and compare it with the equation: at the corner the gain is down by 3 dB and the output lags the input by 45°, and a decade above it the gain is down by about 20 dB.

Gotchas

• Magnitudes hide the phase. An inductor and a capacitor with equal impedance magnitudes cancel in series instead of adding: at that frequency they are in series resonance [Alexander & Sadiku 2017, Ch. 14].
• Radians or hertz. $\omega$ is in radians per second and $f$ in hertz; mixing them moves a corner by a factor of $2\pi$.
• A loaded filter is a different filter. The corner-frequency equation assumes that nothing draws current from the output. A load in parallel with the capacitor lowers the gain and moves the corner.

Go deeper

• Lecture notes 6.003 Lecture Notes — D. Freeman, MIT OpenCourseWare · retrieved 2026-09-23
  Lecture slides: the Laplace transform (lecture 6), frequency response and Bode plots (lectures 9 and 11), Fourier series (lecture 15).
• Course 6.003 Signals and Systems (Fall 2011) — D. Freeman, MIT OpenCourseWare · retrieved 2026-09-23
  Signals and systems with video lectures: Laplace transform, frequency response, Bode plots and Fourier series.
• Lecture notes 6.002 Lecture Notes — A. Agarwal, MIT OpenCourseWare · retrieved 2026-09-23
  Lecture-by-lecture notes (from the Fall 2000 offering) on circuit laws, energy-storage elements and sinusoidal steady state.

Quiz

1. What is the impedance magnitude of a $10\,\mu\mathrm{H}$ inductor at $1\,\mathrm{MHz}$?

   1. $6.28\,\Omega$
   2. $62.8\,\Omega$
   3. $628\,\Omega$
   4. $0.0159\,\Omega$
   Answer and explanation

   Correct answer: B The magnitude is $2\pi f L$: $2\pi \times 10^6 \times 10^{-5}$.

2. The frequency is doubled. What happens to the impedance magnitude of a capacitor?

   1. It doubles
   2. It halves
   3. It is unchanged
   4. It falls to a quarter
   Answer and explanation

   Correct answer: B A capacitor's impedance magnitude is inversely proportional to the frequency.

3. An R-C low-pass filter has $10\,\mathrm{k}\Omega$ and $1\,\mathrm{nF}$. What is its corner frequency?

   1. $1.59\,\mathrm{kHz}$
   2. $15.9\,\mathrm{kHz}$
   3. $159\,\mathrm{kHz}$
   4. $100\,\mathrm{kHz}$
   Answer and explanation

   Correct answer: B The corner is $1/(2\pi R C)$, with $R C$ equal to $10\,\mu\mathrm{s}$.

4. What is the gain of a first-order low-pass filter at its corner frequency?

   1. 1
   2. $1/\sqrt{2}$, about -3 dB
   3. 1/2, about -6 dB
   4. 0
   Answer and explanation

   Correct answer: B At the corner the real and imaginary parts of the denominator are equal, so the magnitude is $1/\sqrt{2}$.

5. By roughly how much does a first-order low-pass filter attenuate a sinusoid one decade above its corner?

   1. By a factor of 2
   2. By a factor of about 10 (-20 dB)
   3. By a factor of 100 (-40 dB)
   4. Not at all
   Answer and explanation

   Correct answer: B The gain is $1/\sqrt{1 + 100}$, about $0.0995$: the slope is -20 dB per decade.

6. Why does a buck converter's L-C filter pass the dc component of the switch-node voltage but not the switching ripple?

   1. The inductor's impedance is zero at dc and large at the switching frequency, while the capacitor's is the opposite
   2. The capacitor blocks dc
   3. The inductor is a short circuit at every frequency
   4. The switch-node voltage has no harmonics
   Answer and explanation

   Correct answer: A At dc the inductor is a short and the capacitor an open circuit; at the switching frequency the inductor's impedance is large and the capacitor's small, so the ripple is divided down.