Formula of Complex Analysis
Step Function (Heaviside Function)¶
\[
\theta(t)=\mathrm{i}\int_{-\infty}^{\infty} \, \frac{\mathrm{d}z}{2\pi} \frac{\mathrm{e}^{ -\mathrm{i}zt }}{z+i\varepsilon } \quad(\varepsilon \to 0 )
\]
Laplace Transform¶
Properties of the unilateral Laplace transform
Property | Time domain | \(s\) domain |
---|---|---|
Linearity | \(a f(t) + b g(t)\) | \({\displaystyle aF(s)+bG(s)\ }\) |
Frequency-domain derivative | \({\displaystyle tf(t)\ }\) | \({\displaystyle -F'(s)\ }\) |
Frequency-domain general derivatives | \(t^{n}f(t)\) | \((-1)^{n}F^{(n)}(s)\) |
General Derivative | \({\displaystyle f^{(n)}(t)\ }\) | \({\displaystyle s^{n}F(s)-\sum _{k=1}^{n}s^{n-k}f^{(k-1)}(0^{+})\ }\) |
Frequency-domain integration | \({\displaystyle {\frac {1}{t}}f(t)\ }\) | \({\displaystyle \int _{s}^{\infty }F(\sigma )\,d\sigma \ }\) |
Time-domain integration | \({\displaystyle \int _{0}^{t}f(\tau )\,d\tau =(u*f)(t)}\) | \({\displaystyle {1 \over s}F(s)}\) |
Frequency shifting | \({\displaystyle e^{at}f(t)\ }\) | \({\displaystyle F(s-a)\ }\) |
Time shifting | \({\displaystyle f(t-a)u(t-a)\ }\) | \({\displaystyle e^{-as}F(s)\ }\) |
Time scaling | \({\displaystyle f(at)}\) | \({\displaystyle {\frac {1}{a}}F\left({s \over a}\right)}\) |
Multiplication | \({\displaystyle f(t)g(t)}\) | \({\displaystyle {\frac {1}{2\pi i}}\lim _{T\to \infty }\int _{c-iT}^{c+iT}F(\sigma )G(s-\sigma )\,d\sigma \ }\) |
Convolution | \({\displaystyle (f*g)(t)=\int _{0}^{t}f(\tau )g(t-\tau )\,d\tau }\) | \({\displaystyle F(s)\cdot G(s)\ }\) |
Complex conjugation | \({\displaystyle f^{*}(t)}\) | \({\displaystyle F^{*}(s^{*})}\) |
Cross-correlation | \({\displaystyle (f\star g)(t)=\int _{0}^{\infty }f(\tau )^{*}\,g(t+\tau )\,d\tau }\) | \({\displaystyle F^{*}(-s^{*})\cdot G(s)}\) |
Periodic function | \({\displaystyle f(t)}\) | \({\displaystyle {1 \over 1-e^{-Ts}}\int _{0}^{T}e^{-st}f(t)\,dt}\) |
Periodic summation | \(\({\displaystyle \sum _{n=0}^{\infty }f(t-Tn)u(t-Tn)}\quad{\displaystyle \sum _{n=0}^{\infty }(-1)^{n}f(t-Tn)u(t-Tn)}\)\) | \({\displaystyle {\frac {1}{1-e^{-Ts}}}F(s)}\quad{\displaystyle {\frac {1}{1+e^{-Ts}}}F(s)}\) |