In preparation for an exam in linear signals and system my instructed handed us a couple of conceptual questions that should prepare us. I ran into the following:

Which statements regarding the Laplace transformations are true?

  1. Multiplication of two signals in the time-domain corresponds to convoluting the signals in the Laplace-domain.
  2. A differential equation can not be Laplace transformed if the roots in its characteristic equation lie in the RHP.
  3. Opposite the Fourier transformation, the Laplace transformation of the system response includes both zero input response and zero state response.
  4. The Laplace transformation of a system's step response gives the systems transfer function \$H(s) \$.

My attempt at reasoning

1 is wrong. It's true that \$x_1(t) *x_2(t) = X_1(s)X_2(s) \$, but \$x_1(t)x_2(t) = \frac{1}{2\pi j}X_1(s) * X_2(s) \$, so it is not equivalent.

2 is wrong. If the poles lie in the RHP doesn't mean that a Laplace transformation \$H(s) \$ doesn't exist, but it means that the system is unstable, the frequency response doesn't exist and \$H(j\omega) \$ is meaningless.

4 is wrong. It is the Laplace transformation of the impulse response \$h(t) \$ that gives the system transfer function \$H(s) \$.

3 I believe is correct. I just think of transforming a capacitor with an initial voltage to the Laplace domain, which will result in an impedance \$Z_c=\frac{1}{sC} \$ in series with a voltage source \$\frac{V_c(0)}{s} \$. So both the zero input and zero state response is taken care of.

However, the only answers available for this question state that either all of the statements are wrong, or that at least two of the statements are correct, which doesn't agree with my reasoning. Can someone spot a flaw in my reasoning?

  • 5
    \$\begingroup\$ Hint: "corresponds to" doesn't necessarily mean "is exactly equivalent to with no need for a scaling adjustment". \$\endgroup\$
    – The Photon
    Commented May 14, 2021 at 18:26
  • \$\begingroup\$ @ThePhoton hmm I guess you are right. If that is the case I suppose 1. and 3 are the correct options. \$\endgroup\$
    – Carl
    Commented May 14, 2021 at 18:28
  • 1
    \$\begingroup\$ Actually, \$H(j\omega)\$ has meaning for an unstable system, and can even be measured on a physical system if you first wrap that system with feedback that renders the overall system stable. \$\endgroup\$
    – TimWescott
    Commented May 14, 2021 at 18:53
  • 2
    \$\begingroup\$ 1: It works. Control engineers stabilize unstable systems all the time, and they often do it using Laplace-domain analysis. Examples abound. 2: If the math can't keep up, that's the math's fault. 3: if you can't believe that -- ask a separate question; said question is certainly answerable. \$\endgroup\$
    – TimWescott
    Commented May 14, 2021 at 19:53
  • 2
    \$\begingroup\$ For the purposes of your question -- if it's a signal processing course, just give the signal processing answer. Google "The Return of the Archons" for how things can go bad if you don't ostensibly follow authority, even if it's misguided. \$\endgroup\$
    – TimWescott
    Commented May 14, 2021 at 19:55

1 Answer 1


Both 1 and 3 are correct. Both 2 and 4 are wrong.

1 is correct. By duality principle, multiplication in time domain corresponds to convolution in the frequency domain and vice versa.

3 is correct as LT provides us the complete response zero input (transient) and zero state response (Steady State). By contrast, FT gives us only the steady state response.

  • \$\begingroup\$ "duality principle" is for Fourier transforms. Laplace original and transform domain characterisation and convergence conditions are way too different to leave room for "duality". Due to the relationship with Fourier transforms, a few pieces of "duality" intuition may work out given specific convergence constraints, but any rigorous proof will require elements of convergence and analyticality that have no dual counterpart. \$\endgroup\$
    – user107063
    Commented Jan 8, 2023 at 4:57
  • \$\begingroup\$ Yeah, it's a poor multiple choice question. If I were answering on a test, I would choose 4 as the more clearly wrong answer. \$\endgroup\$ Commented Sep 13, 2023 at 11:21

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