Why phasors work for same-frequency sinusoids
Derive the phasor mapping from Euler's relation rather than memorizing it. Separate the real waveform from the complex helper, define AoN's peak-amplitude underlined-lowercase phasor, prove addition and differentiation rules, show why linear time-invariant differential equations preserve one frequency in steady state, and state exactly when a single phasor description fails.
- Derive v(t)=Re{v̲ e^{jωt}} from a real cosine and Euler's relation, with v̲=Vm e^{jφ} under AoN's peak-amplitude convention.
- Explain why the imaginary component is mathematical bookkeeping rather than a second physical voltage or current.
- Prove that same-frequency sinusoid addition maps to complex-number addition of phasors.
- Prove that time differentiation maps to multiplication by jω and state the condition attached to integration.
- Explain the deeper linear-time-invariant reason: e^{jωt} is an eigenfunction of constant-coefficient differential operators.
- Separate sinusoidal steady state from transient response and explain which part the phasor represents.
- Use counterexamples to show why nonlinear multiplication and time-varying systems can generate new frequencies outside a one-phasor description.
- State AoN's phasor scope and notation limits before any capacitor/inductor impedance table is introduced.
Why is it legitimate to replace a real waveform such as v(t)=Vmcos(ωt+φ) by one complex number, do algebra on that number, and recover the correct real waveform afterward?
The answer is not “because engineers use phasors.” It follows from Euler’s relation, linearity, and the special way complex exponentials behave under linear time-invariant differential operations.
01.1Euler’s relation creates a useful complex helper
Electrical engineering uses j=√(−1) for the imaginary unit so that the symbol i remains available for current. Euler’s relation is
Taking the real part gives
Therefore the real sinusoid
can be written exactly as
Nothing physical has become complex. We have embedded one real waveform in a complex mathematical helper whose real projection is the quantity of interest.
01.2AoN’s phasor is the fixed complex amplitude
AoN now fixes the amplitude convention deferred by EF-01. For
the phasor is the underlined lowercase complex amplitude
and the mapping is
This course uses peak-amplitude phasors: the phasor magnitude equals the sinusoidal peak amplitude Vm. Some power-engineering texts use root-mean-square (RMS) phasors instead; that is a different convention and must never be mixed silently with the peak convention.
When one angular frequency is fixed throughout a derivation, AoN may omit the explicit (ω) after the convention has been stated.
Figure 1.1 — The phasor is the coefficient; the rotating vector is the complex time helper
The one question is: which part of the complex object gives the physical real voltage v(t)?
The underlined phasor is the fixed complex amplitude. Multiplication by the rotating factor exp(jωt) creates the complex time helper; its projection onto the real axis is v(t). The imaginary axis is not a second physical voltage.
01.3Same-frequency sinusoid addition becomes complex-number addition
Suppose two real sinusoids share the same angular frequency:
Because taking the real part is a linear operation,
The common time factor ejωt is untouched. Only the complex amplitudes need to be added.
Worked example.
Their phasors are 3 and j4, so
and the real sum is
This works because the two terms have the same frequency. If their frequencies differ, there is no single common rotating factor to remove.
01.4Differentiation becomes multiplication by jω
Start from the complex helper
where the phasor is constant in time. Differentiate:
Taking the real part afterward gives the derivative of the real waveform. Thus the phasor associated with the derivative is
Multiplication by j rotates a complex number by +90°, while multiplication by ω scales its magnitude. This matches the time-domain fact that differentiating a cosine produces a sinusoid shifted by +90° in cosine notation.
For integration of a zero-mean sinusoidal component at nonzero ω,
01.5The deeper reason: complex exponentials are eigenfunctions of linear time-invariant differential operators
Consider a constant-coefficient differential operator
Applying it to a complex exponential gives
so
The operator changes only the complex coefficient; the exponential keeps the same angular frequency. In systems language, ejωt is an eigenfunction of a linear time-invariant (LTI) system.
This is the mathematical core behind phasor analysis: for a linear time-invariant lumped circuit in sinusoidal steady state, the common time dependence can be factored out and the differential equation becomes algebra in jω.
01.6A generic differential equation becomes an algebraic equation
Suppose a linear circuit model produces
and we seek the sinusoidal steady-state response at one angular frequency. Write
Using the derivative rule yields
A differential equation in time has become an algebraic equation in complex amplitudes. The next chapters will identify the circuit-specific factors that we call impedance and admittance; this chapter deliberately derives the machinery before naming those shortcuts.
01.7Phasors describe sinusoidal steady state, not the transient
Take a generic stable first-order linear model
Under a sinusoidal input, the full solution can contain two qualitatively different pieces:
For this stable example the homogeneous transient has the form Ce−t/τ; it depends on initial conditions and decays. The steady-state particular solution is a sinusoid at the forcing frequency.
The phasor describes the sinusoidal steady-state term. It does not magically contain the initial-condition transient.
01.8Where one phasor fails
A compact nonlinear counterexample is
Squaring one sinusoid creates a DC term and a second harmonic. A single phasor at ω cannot represent that operation.
01.9Worked checks: mapping in both directions
Example A — waveform to phasor.
Example B — phasor to waveform. At f=5 kHz,
Example C — derivative. If
then
which matches the direct derivative −4ω sin(ωt−20°)=4ω cos(ωt+70°).
01.10Mastery exercises
- Starting from Euler’s relation, derive A cos(ωt+φ)=Re{Aejφejωt}.
- Under AoN’s peak convention, map v(t)=5 cos(1000t+40°) V to its phasor.
- At ω=2000 rad/s, convert i=3∠−60° mA back to a real waveform.
- Add 2 cosωt and 2 cos(ωt+120°) using phasors.
- If v=10∠25°, find the phasor of dv/dt.
- Why does the integration rule 1/(jω) not capture an arbitrary integration constant?
- Explain in one paragraph why a linear time-invariant differential equation driven at one frequency has a same-frequency steady-state particular solution.
- Use cos²ωt to explain why a nonlinear square-law element cannot in general be solved with only the original-frequency phasor.
Solutions
1. Substitute θ=ωt+φ into Euler’s relation and take the real part; factor Aejφ as the time-independent complex amplitude.
2. v=5∠40° V.
3. i(t)=3 cos(2000t−60°) mA.
4. The phasors are 2 and 2∠120° = −1+j√3. Their sum is 1+j√3 = 2∠60°, so the waveform is 2 cos(ωt+60°).
5. jωv=2000·10∠(25°+90°)=20,000∠115°, with units per second relative to the original quantity.
6. The harmonic phasor represents only the nonzero-frequency sinusoidal component. An integration constant is a separate DC component.
7. Every derivative of ejωt returns the same exponential times a power of jω, so a constant-coefficient linear differential operator changes only its complex coefficient, not its frequency.
8. cos²ωt=(1+cos2ωt)/2 contains 0 and 2ω; the nonlinear operation generates frequencies absent from the original one-phasor description.
Chapter 02 will formalize complex-amplitude arithmetic, polar/rectangular conversion and AoN notation discipline before capacitor and inductor impedance appear.