Table of Transform Pairs

Table of Properties

Impulse response

Sifting property of the impulse

\[\begin{align*} u(t) &= \int_{-\infty}^\infty \! u(\tau) \, \delta(t - \tau) \, d\tau = \int_{0^-}^t \! u(\tau) \, \delta(t - \tau) \, d\tau, & u(t) &= 0, \quad t < 0 \end{align*}\]

Linearity

\[\begin{align*} y(t) &= \int_{0^-}^\infty u(\tau) \, g(t, \tau) \, d\tau = \int_{0^-}^\infty g(t, \tau) \, u(\tau) \, d\tau, & g(t, \tau) &= G(\delta(t - \tau), t) \end{align*}\]

Causality

\[\begin{align*} g(t, \tau) &= 0, \quad t < \tau & & \implies & y(t) &= \int_{0^-}^t g(t, \tau) \, u(\tau) \, d\tau \end{align*}\]

Impulse response

Causality + Time-invariance

\[\begin{align*} g(t, \tau) &= g(t - \tau) = 0, \quad t < \tau & & \implies & y(t) &= \int_{0^-}^t g(t - \tau) \, u(\tau) \, d\tau \end{align*}\]

Laplace transform

\[\begin{align*} Y(s) &= G(s) \, U(s), & U(s) &= \mathcal{L} \{ u(t) \}, & G(s) &= \mathcal{L} \{ g(t) \}, & Y(s) &= \mathcal{L} \{ y(t) \} \end{align*}\]

\(G\) is the transfer-function

Example

Application: Inverse Laplace transform

\[\begin{align*} t &> 0: & \lim_{\rho \rightarrow \infty} \int_{C^\rho_-} F(s) \, e^{s t} \, ds &= 0 & & \implies & \mathcal{L}^{-1}\{F(s)\} &= \frac{1}{2 \pi j} \lim_{\rho \rightarrow \infty} \int_{\Gamma_-^\alpha} F(s) \, e^{s t} \, ds \\ & & & & & & &= \sum_{k = 1}^{n} \operatorname{Res}_{s = p_k} \left (F(s) \, e^{s t} \right ) \\ t &< 0: & \lim_{\rho \rightarrow \infty} \int_{C^\rho_+} F(s) \, e^{s t} \, ds &= 0 & & \implies & \mathcal{L}^{-1}\{F(s)\} &= \frac{1}{2 \pi j} \lim_{\rho \rightarrow \infty} \int_{\Gamma_+^\alpha} F(s) \, e^{s t} \, ds = 0 \end{align*}\]

Residue

\[\begin{align*} \operatorname{Res}_{s = s_0} f(s) &= \int_{C \in D} f(s) \, ds, & D &= \{ s : 0 < |s - s_0| < R \} \end{align*}\]

Laurent series expansion

\[\begin{align*} f(s) &= \frac{ g(s)}{(s - s_0)^m}, \quad g(s_0) \neq 0, \quad g \text{ is analytic at } s_0 \end{align*}\]

Calculus of residues

\[\begin{align*} \operatorname{Res}_{s = s_0} f(s) &= \begin{cases} g(s_0), & m = 1 \\[1ex] \displaystyle \frac{g^{(m -1)}(s_0)}{(m-1)!} , & m \geq 2 \end{cases} \end{align*}\]

Simple poles

\[\begin{align*} F(s) \, e^{s t} &= \frac{ G(s)}{(s - p_k)}, & G(s) &= (s - p_k) \, F(s) \, e^{s t}, & G(p_k) &\neq 0 \end{align*}\]

Residues

\[\begin{align} \operatorname{Res}_{s=p_k} \left ( F(s) \, e^{s t} \right ) &= G(p_k) = k_k \, e^{p_k t}, & k_k &= \lim_{s \rightarrow p_k} (s - p_k) F(s) \end{align}\]

Inverse Laplace transform

\[\begin{align*} \mathcal{L}^{-1}\{F(s)\} &= \sum_{k = 1}^{n} k_k e^{p_k t}, \quad t \geq 0 \end{align*}\]

Expansion in partial fractions

\[\begin{align*} F(s) &= \mathcal{L} \left \{ \sum_{k = 1}^{n} k_k e^{p_k t} \right \} = \sum_{k = 1}^{n} k_k \mathcal{L} \left \{ e^{p_k t} \right \} = \sum_{k = 1}^{n} \frac{k_k}{s - p_k} \end{align*}\]

Example: car model step response

Step input

\[\begin{align*} u(t) &= \tilde{u} \, 1(t), & U(s) &= \frac{\tilde{u}}{s} \end{align*}\]

Laplace transform of the output

\[\begin{align*} Y(s) &= \frac{\tilde{u}}{s} \times \frac{\frac{p}{m}}{s + \frac{b}{m}} + \frac{1}{s + \frac{b}{m}} \, y_0 \end{align*}\]

Inverse Laplace transform

\[\begin{align*} k_1 &= \lim_{s \rightarrow 0} s \, Y(s) = \frac{p}{b} \tilde{u}, & k_2 &= \!\!\!\lim_{s \rightarrow -\frac{b}{m}} \left ( s + \frac{b}{m} \right ) Y(s) = y_0 - \frac{p}{b} \tilde{u} \end{align*}\] \[\begin{align*} y(t) &= \sum_{k = 1}^2 \operatorname{Res}_{s = p_k} \left (Y(s) e^{s t} \right ) = k_1 e^{p_1 t} + k_2 e^{p_2 t} = \frac{p}{b} \tilde{u} + \left ( y_0 - \frac{p}{b} \tilde{u} \right ) e^{-\frac{b}{m} t}, \quad t > 0 \end{align*}\]

Example: closed-loop car model ramp response

Ramp input

\[\begin{align*} \bar{y}(t) &= \mu \, t \, 1(t), & \bar{Y}(s) &= \frac{\mu}{s^2} \end{align*}\]

Laplace transform of the output

\[\begin{align*} Y(s) &= H(s) \bar{Y}(s) = \frac{b_1}{s + a_1} \times \frac{\mu}{s^2}, & a_1 &= \frac{b}{m} + \frac{p}{m} K, & b_1 &= \frac{p}{m} K \end{align*}\]

Inverse Laplace transform

\[\begin{align*} k_1 &= \lim_{s \rightarrow -a_1} ( s + a_1 ) Y(s) = \frac{\mu \, b_1}{a_1^2}, & k_{2,1} &= \lim_{s \rightarrow 0} s^2 Y(s) = \frac{\mu \, b_1}{a_1}, & k_{2,2} &= \lim_{s \rightarrow 0} \frac{d}{d s} s^2 Y(s) = - \frac{\mu \, b_1}{a_1^2} \end{align*}\] \[\begin{align*} y(t) &= \sum_{k = 1}^2 \operatorname{Res}_{s = p_k} \left (Y(s) e^{s t} \right ) = \frac{\mu \, b_1}{a_1^2} \left [ e^{-a_1 t} + a_1 \, t - 1 \right ], \quad t \geq 0 \end{align*}\]

Bounded-Input-Bounded-Output (BIBO) stability

\[\begin{align*} | u(t) | &\leq M_u < \infty & & \implies & | y(t) | &\leq M_y < \infty \end{align*}\]

Necessary and sufficient condition

\[\begin{align} \|g\|_1 := \int_{0^-}^\infty |g(\tau)| \, d \tau < M_g < \infty \end{align}\]

Proof of necessity

\[\begin{align*} u_T(t) &= \begin{cases} \operatorname{sign}(g(T - t)), & 0 \leq t \leq T \\ 0, & t < 0 \text{ or } t > T \end{cases} \end{align*}\] is such that \(|u_T(t)| \leq 1\) and \[\begin{align*} y_T(T) &= \int_{0^-}^T g(T - \tau) \, u_T(\tau) \, d \tau = \int_{0^-}^T |g(\tau)| \, d \tau \rightarrow \|g\|_1 \text{ for } T \text{ large } \end{align*}\]

Frequency-domain

Asymptotic stability

\(G(s)\) has no poles on the imaginary axis or on the right-hand side of the complex plane

Same as BIBO stability

One side is easy: because \(|e^{-s t}| \leq 1\) for all \(\operatorname{Re}(s) \geq 0\) \[\begin{align*} |G(s)| = | \mathcal{L} \{ g(t) \} | = \left | \int_{0^-}^\infty g(t) \, e^{-s t} \, dt \right | \leq \int_{0^-}^\infty |g(t)| |e^{-s t}| \, dt \leq \int_{0^-}^\infty |g(t)| \, dt = \|g\|_1 \end{align*}\] hence \(G\) has no poles such that \(\operatorname{Re}(s) \geq 0\) because \(G\) is bounded in \(\operatorname{Re}(s) \geq 0\)

Frequency-domain

Asymptotic stability

\(G(s)\) has no poles on the imaginary axis or on the right-hand side of the complex plane

Same as BIBO stability

The other side is more complicated, unless \(G\) is rational. In this case, if \(G\) is strictly proper and with simple poles \[\begin{align*} |g(t)| &\leq \sum_{k = 1}^{n} |k_k| |e^{p_k t}| = \sum_{k = 1}^{n} |k_k| e^{\operatorname{Re}(p_k) t}, & t &> 0 \end{align*}\] and because \(\operatorname{Re}(p_k) < 0\) \[\begin{align*} \| g \|_1 \leq \int_{0^-}^{\infty} |g(t)| \, dt \leq \sum_{k = 1}^{n} |k_k| \int_{0^-}^{\infty} e^{\operatorname{Re}(p_k) t} \, dt = \sum_{k = 1}^{n} \frac{|k_k|}{-\operatorname{Re}(p_k)} < \infty \end{align*}\] Conclusion is the same if \(G\) is proper and/or has repeated poles

Bode plots

Examples (from Section 7.8)

\[\begin{align*} L_0 &= \frac{67}{s (s^2 + 49)}, & L_\pi &= \frac{67}{s (s^2 - 49)} \end{align*}\]

See Chapter 7 for much more!

Some norms of systems

Plancherel and Parseval’s theorem

\[\begin{align*} \| y \|_2^2 &= \int_{0^-}^{\infty} y(t)^2 \, dt = \int_{-\infty}^{\infty} y^*(t) \, y(t) \, dt = \frac{1}{2 \pi} \int_{-\infty}^{\infty} Y(j \omega) Y(-j \omega) \, d\omega = \frac{1}{2 \pi} \int_{-\infty}^{\infty} |Y(j \omega)|^2 \, d\omega = \| Y \|^2_2 \end{align*}\]

\(H_2\) norm

\[\begin{align*} \| y \|_\infty &\leq \| G \|_2 \| u \|_2, & \| g \|_2 &= \| G \|_2 \end{align*}\]

Computation can be done using residues!

\(H_\infty\) norm

\[\begin{align*} \| y \|_2 &\leq \| G \|_\infty \|u \|_2, & \| G \|_\infty &= \sup_{\omega \in \mathbb{R}} | G(j \omega) | \end{align*}\]

Value can be obtained from a Bode plot!