{"id":1284,"date":"2024-08-02T04:36:24","date_gmt":"2024-08-02T04:36:24","guid":{"rendered":"https:\/\/neilfoxman.com\/?page_id=1284"},"modified":"2024-08-19T03:41:41","modified_gmt":"2024-08-19T03:41:41","slug":"common-continuous-transforms-and-examples","status":"publish","type":"page","link":"https:\/\/neilfoxman.com\/?page_id=1284","title":{"rendered":"Common Continuous Transforms and Examples"},"content":{"rendered":"\n

<\/span>Exponential Decay<\/span><\/h2>\n\n\n\n

$$
\\begin{align}
x(t) &= e^{-at} u(t) \\\\
\\\\
X(s) &= \\int_{-\\infty}^{\\infty} e^{-at} u(t) e^{-s t} dt \\\\
X(s) &= \\int_{0}^{\\infty} e^{-at} e^{-s t} dt \\\\
X(s) &= \\int_{0}^{\\infty} e^{-(a + s) t} dt \\\\
X(s) &= \\left[ \\frac{1}{-(a + s)} e^{-(a + s) t} \\right]_{t=0}^{\\infty} \\\\
X(s) &= \\frac{1}{a + s} &Re\\{a + s\\} > 0\\\\
X(s) &= \\frac{1}{a + s} &Re\\{s\\} > -Re\\{a\\}\\\\
\\end{align}
$$<\/p>\n\n\n\n

Note that in the Fourier Transform where $s = j\\omega$<\/p>\n\n\n\n

$$
\\begin{align}
|X(j \\omega)| = \\frac{1}{\\sqrt{a^2 + \\omega^2}} \\\\
\\\\
\\angle X(j \\omega) = \\angle 1 – \\angle a + j \\omega \\\\
\\angle X(j \\omega) = – \\text{atan}(\\omega\/a) \\\\
\\end{align}
$$<\/p>\n\n\n\n

\"\"\n\t\t\t\n\t\t\t\t\n\t\t\t<\/svg>\n\t\t<\/button><\/figure>\n\n\n\n

<\/span>Linearity<\/span><\/h2>\n\n\n\n

$$
\\begin{align}
x_1(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} X_1(s) &&ROC: s \\in R_1 \\\\
x_2(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} X_2(s) &&ROC: s \\in R_2 \\\\
\\\\
a x_1(t) + b x_2(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} a X_1(s) + b X_2(s) &&ROC: (s \\in R) \\land (R_1 \\cap R_2 \\in R) \\\\
\\end{align}
$$<\/p>\n\n\n\n

Note that ROC contains $R_1 \\cap R_2$ but may also include additional values outside of $R_1 \\cap R_2$. Example of this shown below. Define $x_1$ as<\/p>\n\n\n\n

$$
\\begin{align}
x_1(t) &= e^{-t} u(t) \\\\
X_1(s) &= \\frac{1}{s + 1} \\\\
ROC&: Re\\{s\\} > -1 \\\\
\\end{align}
$$<\/p>\n\n\n\n

Then define $x_2$ as<\/p>\n\n\n\n

$$
\\begin{align}
x_2(t) &= [e^{-t} – e^{-2t}] u(t) \\\\
X_2(s) &= \\int_{-\\infty}^{\\infty} x_2(t) e^{-st} dt \\\\
X_2(s) &= \\int_{-\\infty}^{\\infty} [e^{-t} – e^{-2t}] u(t) e^{-st} dt \\\\
X_2(s) &= \\int_{0}^{\\infty} [e^{-t} – e^{-2t}] e^{-st} dt \\\\
X_2(s) &= \\int_{0}^{\\infty} e^{-t} e^{-st} dt – \\int_{0}^{\\infty} e^{-2t} e^{-st} dt \\\\
X_2(s) &= \\mathscr{L}\\left\\{ e^{-t} u(t) \\right\\} – \\mathscr{L}\\left\\{ e^{-2t} u(t) \\right\\} &&ROC: (Re\\{s\\} > -1) \\land (Re\\{s\\} > -2) \\\\
X_2(s) &= \\mathscr{L}\\left\\{ e^{-t} u(t) \\right\\} – \\mathscr{L}\\left\\{ e^{-2t} u(t) \\right\\} &&ROC: Re\\{s\\} > -1 \\\\
X_2(s) &= \\frac{1}{s+1} – \\frac{1}{s+2} &&ROC: Re\\{s\\} > -1 \\\\
X_2(s) &= \\frac{(s+2)-(s+1)}{(s+1)(s+2)} &&ROC: Re\\{s\\} > -1 \\\\
X_2(s) &= \\frac{1}{(s+1)(s+2)} &&ROC: Re\\{s\\} > -1 \\\\
\\end{align}
$$<\/p>\n\n\n\n

Note the ROC when we define $x_3$ as<\/p>\n\n\n\n

$$
\\begin{align}
x_3(t) &= x_1(t) – x_2(t) \\\\
x_3(t) &= e^{-t} u(t) – [e^{-t} – e^{-2t}] u(t) \\\\
x_3(t) &= e^{-2t} u(t) \\\\
X_3(t) &= \\frac{1}{s+2} &&ROC: Re\\{s\\} > -2 \\\\
\\end{align}
$$<\/p>\n\n\n\n

ROC for both constituent signals was restrictive, however ROC was modified as we were able to cancel a pole and expand the ROC.
<\/p>\n\n\n\n

<\/span>Time Shift<\/span><\/h2>\n\n\n\n

From Synthesis Equation<\/p>\n\n\n\n

$$
\\begin{align}
x(t) &= \\frac{1}{2 \\pi} \\int_{-\\infty}^{\\infty} X(s) e^{s t} d\\omega \\\\

x(t-t_0) &= \\frac{1}{2 \\pi} \\int_{-\\infty}^{\\infty} X(s) e^{s (t-t_0)} d\\omega \\\\

x(t-t_0) &= \\frac{1}{2 \\pi} \\int_{-\\infty}^{\\infty} X(s) e^{s t} e^{-s t_0} d\\omega \\\\

x(t-t_0) &= \\frac{1}{2 \\pi} \\int_{-\\infty}^{\\infty} \\left[e^{-s t_0} X(s)\\right] e^{s t} d\\omega \\\\
\\end{align}
$$<\/p>\n\n\n\n

In other words, a time delay of $t_0$ is equivalent to multiplying the non-delayed transform by $e^{-st_0}$.<\/p>\n\n\n\n

$$
\\begin{align}
x(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} X(s) \\\\
x(t-t_0) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} e^{-st_0} X(s) \\\\
\\end{align}
$$<\/p>\n\n\n\n

Note that for any finite $t_0$ and $s$, then $e^{-st_0}$ cannot be a pole and the ROC remains unchanged.<\/p>\n\n\n\n

<\/span>s-Domain Shift<\/span><\/h2>\n\n\n\n

This property can be conceptualized as shifting the surface plot of $X(s)$ defined over the entire $s$ plane by $s_0$. On the shifted surface, $X(s-s_0)$, when $s=s_0$ we get the same result that would have previously been at the origin, $X(0)$.<\/p>\n\n\n\n

$$
\\begin{align}
X(s) &= \\int_{-\\infty}^{\\infty} x(t) e^{-st} dt &&ROC: Re\\{s\\} \\in R \\\\
\\\\
X(s-s_0) &= \\int_{-\\infty}^{\\infty} x(t) e^{-(s-s_0)t} dt &&ROC: Re\\{s-s_0\\} \\in R \\\\
X(s-s_0) &= \\int_{-\\infty}^{\\infty} \\left[ x(t) e^{s_0 t} \\right] e^{-st} dt &&ROC: Re\\{s\\} \\in R + Re\\{s_0\\} \\\\
\\\\
e^{s_0 t} x(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} X(s-s_0) &&ROC: Re\\{s\\} \\in R + Re\\{s_0\\} \\\\
\\end{align}
$$<\/p>\n\n\n\n

In the special case where $s_0 = j \\omega_0$ and $x(t)$ is modulated by a sinusoidal signal of frequency $\\omega_0$, then we have<\/p>\n\n\n\n

$$
\\begin{align}
e^{j \\omega_0 t} x(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} X(s-j \\omega_0) &&ROC: Re\\{s\\} \\in R \\\\
\\end{align}
$$<\/p>\n\n\n\n

<\/span>Convolution Property<\/span><\/h2>\n\n\n\n

Recall<\/a> that the intent of using transforms is to more simply conceptualize the result of convolution in the time domain. The convolution property fully realizes this result.<\/p>\n\n\n\n

<\/span>Laplace Transform<\/span><\/h3>\n\n\n\n

$$
\\begin{align}
y(t) &= h(t) * x(t) \\\\
y(t) &= \\int_{-\\infty}^{\\infty} x(\\tau) h(t – \\tau) d\\tau \\\\
\\\\
Y(s) &= \\int_{-\\infty}^{\\infty} \\left[ \\int_{-\\infty}^{\\infty} x(\\tau) h(t – \\tau) d\\tau \\right] e^{-st} dt \\\\
Y(s) &= \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} x(\\tau) h(t – \\tau) e^{-st} d\\tau dt \\\\
Y(s) &= \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} x(\\tau) h(t – \\tau) e^{-st} dt d\\tau \\\\
Y(s) &= \\int_{-\\infty}^{\\infty} x(\\tau) \\left[ \\int_{-\\infty}^{\\infty} h(t – \\tau) e^{-st} dt \\right] d\\tau \\\\
\\end{align}
$$<\/p>\n\n\n\n

The inner integral may be conceptualized as the transform of a signal $h(t)$ that is time shifted by some value $\\tau$. Using the time shift property<\/a> we then have<\/p>\n\n\n\n

$$
\\begin{align}
Y(s) &= \\int_{-\\infty}^{\\infty} x(\\tau) \\left[ H(s) e^{-s\\tau} \\right] d\\tau \\\\
Y(s) &= H(s) \\int_{-\\infty}^{\\infty} x(\\tau) e^{-s\\tau} d\\tau \\\\
Y(s) &= H(s) X(s) \\\\
\\end{align}
$$<\/p>\n\n\n\n

The two responses $H$ and $X$ just scale each other. The ROC is inclusive of the previous ROC as we saw in the linearity example<\/a>.<\/p>\n\n\n\n

$$
ROC: R_H \\cap R_X \\in s
$$<\/p>\n\n\n\n

<\/span>Differentiation in Time Domain<\/span><\/h2>\n\n\n\n

<\/span>Synthesis Equation Approach<\/span><\/h3>\n\n\n\n

$$
\\begin{align}
x(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} X(s) \\\\
\\\\
x(t) &= \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} X(\\sigma+j\\omega) e^{(\\sigma+j\\omega)t} d\\omega \\\\
\\\\
\\frac{d}{dt} x(t) &= \\frac{d}{dt} \\left[ \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} X(\\sigma+j\\omega) e^{(\\sigma+j\\omega)t} d\\omega \\right] \\\\
\\frac{d}{dt} x(t) &= \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} X(\\sigma+j\\omega) \\frac{d}{dt} \\left[ e^{(\\sigma+j\\omega)t} \\right] d\\omega \\\\
\\frac{d}{dt} x(t) &= \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} X(\\sigma+j\\omega) \\left[ (\\sigma+j\\omega) e^{(\\sigma+j\\omega)t} \\right] d\\omega \\\\
\\frac{d}{dt} x(t) &= \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} \\left[ (\\sigma+j\\omega) X(\\sigma+j\\omega) \\right] e^{(\\sigma+j\\omega)t} d\\omega \\\\
\\\\
\\frac{d}{dt} x(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} (\\sigma+j\\omega) X(\\sigma+j\\omega) \\\\
\\frac{d}{dt} x(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} s X(s) \\\\
\\\\
ROC\\left[ X(s) \\right]: &s \\in R \\\\
ROC\\left[ s X(s) \\right]: &s \\in R’, R \\in R’ \\\\
\\end{align}
$$<\/p>\n\n\n\n

Similar to examples above, the ROC for $sX(s)$ may get larger if the added $s$ cancels a pole in $X(s)$ at $s=0$<\/p>\n\n\n\n

<\/span>Bounded Signal Differentiation, Analysis Equation Approach<\/span><\/h3>\n\n\n\n

Reference: https:\/\/www.tutorialspoint.com\/time-differentiation-property-of-laplace-transform<\/a><\/p>\n\n\n\n

We sometimes want to use a relation where the Laplace operation is a definite integral (when $x(t)=0$ when outside some bounds $[a, b]$).<\/p>\n\n\n\n

$$
\\begin{align}
x(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} X_0(s) \\\\
X_0(s) &= \\int_a^b x(t) e^{-st} dt \\\\
\\\\
\\frac{dx(t)}{dt} &\\stackrel{\\mathscr{L}}{\\leftrightarrow} X_1(s) \\\\
X_1(s) &= \\int_a^b \\left[ \\frac{dx(t)}{dt} \\right] e^{-st} dt \\\\
\\end{align}
$$<\/p>\n\n\n\n

From IBP<\/p>\n\n\n\n

$$
\\begin{align}
\\int_a^b f(t) g'(t) dt &= [f(t) g(t)]_{t=a}^b – \\int_a^b f'(t) g(t) dt \\\\
\\\\
f(t) &= e^{-st} \\\\
f'(t) dt &= – s e^{-st} dt \\\\
\\\\
g'(t) dt &= \\frac{dx(t)}{dt} dt \\\\
g(t) &= x(t) \\\\
\\\\
X_1(s) &= [e^{-st} x(t)]_{t=a}^b – \\int_a^b – s x(t) e^{-st} dt \\\\
X_1(s) &= [e^{-st} x(t)]_{t=a}^b + s \\int_a^b x(t) e^{-st} dt \\\\
X_1(s) &= [e^{-sb} x(b) – e^{-sa} x(a)] + s \\int_a^b x(t) e^{-st} dt \\\\
X_1(s) &= [e^{-sb} x(b) – e^{-sa} x(a)] + s X_0(s) \\\\
\\end{align}
$$<\/p>\n\n\n\n

<\/span>Initial and Final Value Theorems, Special Cases of Bounded Signal Differentiation in Time Domain<\/span><\/h3>\n\n\n\n

References:<\/p>\n\n\n\n