Catto / Topic Map · Signals and Systems Level 2
Signals and Systems · 144 h · Topic 16 of 34

The complex exponential signal

If a complex exponential enters an LTI system, the same exponential comes out multiplied by a number. No other signal does that, and all of frequency analysis comes from it.

Eigenfunction Euler s-plane Phasors Negative frequency

01The most important function in the subject

There is a precise reason why all of signal analysis revolves around the complex exponential, and it is not an aesthetic one. It is this:

\[ x(t) = e^{st} \;\longrightarrow\; \boxed{\text{LTI system}} \;\longrightarrow\; y(t) = H(s)\,e^{st} \] The waveform comes out the same as it went in. The only thing that changes is a complex factor, \( H(s) \), which depends on the system and on the value of \( s \).

No other family of signals has that property. If a pulse goes in, something else comes out; if a triangle wave goes in, it comes out distorted. But if \( e^{st} \) goes in, \( e^{st} \) comes out, scaled and phase-shifted. Such signals are called eigenfunctions of the system, and they are the reason frequency analysis exists: if an arbitrary signal is written as a sum of complex exponentials, the output is built by adding up each one multiplied by its \( H(s) \). That is Fourier, and that is Laplace.

Where it comes from

Plugging \( e^{st} \) into the convolution integral: \( y(t) = \int h(\tau)\,e^{s(t-\tau)}d\tau = e^{st}\int h(\tau)e^{-s\tau}d\tau \). The exponential comes out because \( e^{s(t-\tau)} = e^{st}e^{-s\tau} \), and what is left inside no longer depends on \( t \): it is \( H(s) \), the transform of the impulse response.

02The purely imaginary case

With \( s = j\omega_0 \), the exponential stays on the unit circle and rotates forever:

\[ e^{j\omega_0 t} = \cos(\omega_0 t) + j\operatorname{sin}(\omega_0 t) \]

It is always periodic, with period \( T_0 = 2\pi/|\omega_0| \), and its magnitude is one at every instant: it is a phasor of constant amplitude. From this come the two Euler relations that are used constantly to go from sinusoids to exponentials and back:

\[ \cos(\omega_0 t) = \frac{e^{j\omega_0 t} + e^{-j\omega_0 t}}{2}, \qquad \operatorname{sin}(\omega_0 t) = \frac{e^{j\omega_0 t} - e^{-j\omega_0 t}}{2j} \] A real cosine is the sum of two phasors rotating in opposite directions. That is where “negative frequencies” come from: they are not an artifact, they are the phasor rotating the other way.

The harmonically related exponentials \( \phi_k(t) = e^{jk\omega_0 t} \) share the same period \( T_0 \) for every integer \( k \), and they form the basis from which a Fourier series is built. In continuous time there are infinitely many distinct ones; in discrete time, as we saw, there are only \( N \).

03The general case: the map of the s-plane

With the full \( s = \sigma + j\omega \), the exponential does two things at once:

\[ e^{st} = e^{\sigma t}\,e^{j\omega t} = \underbrace{e^{\sigma t}}_{\text{envelope}}\big(\cos\omega t + j\operatorname{sin}\omega t\big) \]

The real part of \( s \) sets the envelope and the imaginary part, the rotation speed. Each point of the \( s \)-plane corresponds to a different waveform, and that correspondence is the map you have to keep in your head for the rest of your degree.

Lab · each point of the plane, a waveform

Click the \( s \)-plane on the left and see which signal appears on the right. The marked points are the characteristic cases.

04In discrete time

The discrete version is \( x[n] = z^{n} \), and with \( z = e^{j\Omega_0} \) it becomes \( e^{j\Omega_0 n} \). The differences from the continuous case were already anticipated and are worth repeating because they are the source of half the mistakes:

Continuous timeDiscrete time
Distinct frequenciesInfinitely many: each \( \omega_0 \) gives a different signal Only an interval of \( 2\pi \)
PeriodicityAlways, with \( T_0 = 2\pi/\omega_0 \) Only if \( \Omega_0/2\pi \) is rational
Distinct harmonicsInfinitely manyExactly \( N \)
The “fastest” oneDoes not exist\( \Omega = \pi \): alternates +1 and −1

The third row is what gives rise to the DFT: since there are only \( N \) distinct harmonic exponentials, the representation of a periodic sequence of period \( N \) needs exactly \( N \) coefficients. Not an infinite sum: a finite sum, which a computer can calculate.

The z-plane, which is the s-plane in disguise

The mapping \( z = e^{sT} \) takes the imaginary axis of the \( s \)-plane to the unit circle of the \( z \)-plane, and the left half-plane to the interior of the circle. That is why the stability criterion changes in form but not in substance: where before it was “poles on the left”, now it is “poles inside the circle”.

05In the lab

Exercise 1 · Reading the s-plane on an oscilloscope

Build a series RLC circuit and adjust \( R \) to obtain three cases: overdamped, critically damped and underdamped. For each one, measure the step response, extract \( \sigma \) and \( \omega \) from the envelope and the period, and locate the poles in the plane.

Exercise 2 · The two phasors of a cosine

With a two-channel quadrature generator, add two signals that rotate in opposite directions and verify that the result is a real cosine. It is the Euler relation, built with instruments.

Exercise 3 · The eigenfunction, measured

Inject a sinusoid into an RC filter and measure the amplitude and phase of the output. Check that the waveform is identical and that the input-output relationship matches \( H(j\omega) = 1/(1 + j\omega RC) \) in magnitude and phase, for five frequencies.

06Common mistakes

  • Believing negative frequencies do not exist. They are the phasor rotating the other way, and they are needed to build a real cosine.
  • Confusing \( |e^{st}| \) with \( e^{|s|t} \). The magnitude depends only on \( \sigma \): \( |e^{st}| = e^{\sigma t} \).
  • Assuming \( e^{st} \) with \( \sigma \ne 0 \) is periodic. If there is an envelope, it does not repeat.
  • Forgetting the \( 2j \) of the sine. In the Euler relation the cosine carries a 2 and the sine a \( 2j \).
  • Carrying the properties of the continuous case over to the discrete case. Periodicity and the number of harmonics work differently.

07Self-assessment

What does it mean for \( e^{st} \) to be an eigenfunction of an LTI system?

That the output is the same exponential multiplied by a complex number \( H(s) \). The shape does not change.

What is \( |e^{(-2+j5)t}| \) at \( t = 1 \)?

\( e^{-2} \approx 0.135 \). The imaginary part only rotates; it does not affect the magnitude.

Why does a real cosine need two exponentials?

Because a single one rotates and has an imaginary part. Two rotating in opposite directions cancel their imaginary parts and leave a cosine.

What signal corresponds to a point on the imaginary axis?

An oscillation of constant amplitude: it neither grows nor dies out. It is the borderline case between stable and unstable.

How many distinct harmonic exponentials are there in discrete time?

Exactly \( N \). That is why the DFT has \( N \) coefficients and can be computed on a computer.

Where does the left half-plane go under \( z = e^{sT} \)?

To the interior of the unit circle. Hence the stability criterion for digital filters.

08Further reading

  • Alan V. Oppenheim and Alan S. Willsky. Signals and Systems. 2nd ed. (Spanish edition, “Señales y sistemas”), Prentice Hall, 1998. Sections 1.3 and 3.2: the complex exponential and its role as an eigenfunction.
  • John G. Proakis and Dimitris G. Manolakis. Digital Signal Processing. 4th ed. (Spanish edition, “Tratamiento digital de señales”), Pearson, 2007. The discrete case, with the N harmonics and the road to the DFT.
  • Katsuhiko Ogata. Modern Control Engineering. 5th ed. (Spanish edition, “Ingeniería de control moderna”), Pearson, 2010. Reading the s-plane in terms of time response, with the damping maps.
Development of the topic “The complex exponential signal” of Signals and Systems (Level 2), based on the curriculum of the UTN Electronic Engineering program, 2023 curriculum — Ordinance No. 1849 of the UTN Higher Council. Back to the Topic Map · catto.ar