ELEC 6218 · Lecture 2 companion

Laplace and Fourier transforms compared

Definition and scope

Laplace transformFourier transform
Forward\(X(s)=\displaystyle\int_{-\infty}^{\infty}x(t)\,e^{-st}\,dt\)
Unilateral: lower limit \(0^-\)
\(X(j\Omega)=\displaystyle\int_{-\infty}^{\infty}x(t)\,e^{-j\Omega t}\,dt\)
Inverse\(x(t)=\dfrac{1}{2\pi j}\displaystyle\int_{\sigma-j\infty}^{\sigma+j\infty}X(s)\,e^{st}\,ds\), with \(\sigma\) inside the ROC\(x(t)=\dfrac{1}{2\pi}\displaystyle\int_{-\infty}^{\infty}X(j\Omega)\,e^{j\Omega t}\,d\Omega\)
Basis\(e^{st}=e^{\sigma t}e^{j\Omega t}\): growing, decaying or constant-envelope sinusoids\(e^{j\Omega t}\): constant-envelope sinusoids only
Where it livesA region of the \(s\)-plane (the ROC), a vertical strip \(\sigma_1<\operatorname{Re}\{s\}<\sigma_2\)The \(j\Omega\)-axis
Exists when\(x(t)e^{-\sigma t}\) is absolutely integrable for some \(\sigma\)\(x(t)\) is absolutely integrable (or finite energy); periodic and step-like signals only with impulses in \(\Omega\)
Must be stated\(X(s)\) and its ROC; the same expression with different ROCs gives different signals\(X(j\Omega)\) alone; no ROC
LinkIf the ROC contains the \(j\Omega\)-axis, \(X(j\Omega)=X(s)\big|_{s=j\Omega}\). If it does not, the Laplace transform exists but the Fourier transform does not (e.g. unstable signals), or exists only with impulses (e.g. \(u(t)\)).
System readingPoles and zeros of \(H(s)\); ROC gives causality (right of the rightmost pole) and stability (ROC contains the \(j\Omega\)-axis)Frequency response \(H(j\Omega)\): magnitude and phase seen by a sinusoid in steady state
Typical useSolving LCCDEs with initial conditions (unilateral), transients, stability, filter design in the \(s\)-planeSpectra, bandwidth, filtering, modulation, sampling
σ jΩ s = −a ROC: σ > −a right-sided, causal f(t) = e −at u(t) σ jΩ s = −a ROC: σ < −a left-sided, anti-causal f(t) = −e −at u(−t)
ROC must be stated. \(1/(s+a)\): ROC to the right gives \(e^{-at}u(t)\), to the left \(-e^{-at}u(-t)\). (L2, Fig. 2.2)
STABLE (LHP) UNSTABLE (RHP) Re{s} Im{s} MARGINAL
Stability from the poles. Causal: poles in the left-half plane. (L2, Fig. 2.4.a)
ROC of X(s): σ > −2 σ = Re{s} jΩ-axis (σ = 0) p₁ = −2+j5 p₂ = −2−j5 0
The link. \(H(s)=\frac{5s+25}{s^2+4s+29}\), poles \(-2\pm j5\): ROC \(\sigma>-2\) contains the \(j\Omega\)-axis, so \(H(j\Omega)\) exists. (L2, Fig. 2.6.a)

Properties

Laplace: \(x(t)\leftrightarrow X(s)\), ROC \(R\)Fourier: \(x(t)\leftrightarrow X(j\Omega)\)
Linearity\(aX_1(s)+bX_2(s)\), ROC contains \(R_1\cap R_2\)\(aX_1(j\Omega)+bX_2(j\Omega)\)
Time shift \(x(t-t_0)\)\(e^{-st_0}X(s)\), ROC \(R\)\(e^{-j\Omega t_0}X(j\Omega)\)
Frequency shift\(e^{s_0t}x(t)\leftrightarrow X(s-s_0)\), ROC shifted by \(\operatorname{Re}\{s_0\}\)\(e^{j\Omega_0t}x(t)\leftrightarrow X(j(\Omega-\Omega_0))\)
Time scaling \(x(at)\)\(\dfrac{1}{|a|}X\!\left(\dfrac{s}{a}\right)\), ROC scaled by \(a\)\(\dfrac{1}{|a|}X\!\left(\dfrac{j\Omega}{a}\right)\)
Differentiation \(\dfrac{dx}{dt}\)\(sX(s)\); unilateral: \(sX(s)-x(0^-)\)\(j\Omega\,X(j\Omega)\)
Integration \(\displaystyle\int_{-\infty}^{t}x(\tau)d\tau\)\(\dfrac{X(s)}{s}\), ROC contains \(R\cap\{\operatorname{Re}\{s\}>0\}\)\(\dfrac{X(j\Omega)}{j\Omega}+\pi X(0)\,\delta(\Omega)\)
Convolution \(x*h\)\(X(s)H(s)\), ROC contains \(R_x\cap R_h\)\(X(j\Omega)H(j\Omega)\)
Multiplication \(x(t)y(t)\)Complex convolution in \(s\); rarely used\(\dfrac{1}{2\pi}\,X(j\Omega)*Y(j\Omega)\)
Initial / final valueUnilateral: \(x(0^+)=\lim_{s\to\infty}sX(s)\); \(x(\infty)=\lim_{s\to0}sX(s)\) if the poles of \(sX(s)\) are in the left-half planeNo direct equivalent
EnergyNo direct equivalentParseval: \(\displaystyle\int|x(t)|^2dt=\frac{1}{2\pi}\int|X(j\Omega)|^2d\Omega\)
DualityNo direct equivalent\(X(jt)\leftrightarrow 2\pi\,x(-\Omega)\)

Common pairs side by side

\(x(t)\)Laplace \(X(s)\), ROCFourier \(X(j\Omega)\)
\(\delta(t)\)\(1\), all \(s\)\(1\)
\(u(t)\)\(\dfrac{1}{s}\), \(\operatorname{Re}\{s\}>0\)\(\pi\delta(\Omega)+\dfrac{1}{j\Omega}\): ROC edge on the axis, so an impulse appears
\(e^{-at}u(t)\), \(a>0\)\(\dfrac{1}{s+a}\), \(\operatorname{Re}\{s\}>-a\)\(\dfrac{1}{a+j\Omega}\): ROC contains the axis, so substitute \(s=j\Omega\)
\(e^{at}u(t)\), \(a>0\)\(\dfrac{1}{s-a}\), \(\operatorname{Re}\{s\}>a\)Does not exist: ROC excludes the axis
\(e^{-a|t|}\), \(a>0\)\(\dfrac{-2a}{(s-a)(s+a)}\), \(-a<\operatorname{Re}\{s\}<a\)\(\dfrac{2a}{a^2+\Omega^2}\)
\(\operatorname{rect}(t/T)\)\(\dfrac{e^{sT/2}-e^{-sT/2}}{s}\), all \(s\)\(T\,\dfrac{\sin(\Omega T/2)}{\Omega T/2}\)
\(1\) (all \(t\))Does not exist: no common ROC\(2\pi\,\delta(\Omega)\)
\(\cos(\Omega_0t)\) (all \(t\))Does not exist: no common ROC\(\pi\bigl[\delta(\Omega-\Omega_0)+\delta(\Omega+\Omega_0)\bigr]\)
\(e^{-at}\cos(\Omega_0t)\,u(t)\), \(a>0\)\(\dfrac{s+a}{(s+a)^2+\Omega_0^2}\), \(\operatorname{Re}\{s\}>-a\)\(\dfrac{a+j\Omega}{(a+j\Omega)^2+\Omega_0^2}\)

Reading the pairs. The Laplace transform handles signals that grow, because \(e^{-\sigma t}\) can tame them; the Fourier transform handles two-sided periodic signals, which have no ROC, by allowing impulses in \(\Omega\). Where both exist as ordinary functions, the Fourier transform is the Laplace transform evaluated on the \(j\Omega\)-axis.

ELEC 6218 · University of Southampton