{"id":1586,"date":"2024-08-15T03:42:56","date_gmt":"2024-08-15T03:42:56","guid":{"rendered":"https:\/\/neilfoxman.com\/?page_id=1586"},"modified":"2025-01-02T04:19:28","modified_gmt":"2025-01-02T04:19:28","slug":"continuous-systems-described-by-linear-constant-coefficient-differential-equations","status":"publish","type":"page","link":"https:\/\/neilfoxman.com\/?page_id=1586","title":{"rendered":"Continuous Systems Described by Linear Constant Coefficient Differential Equations"},"content":{"rendered":"\n

Consider the second order system where the input and output are described by<\/p>\n\n\n\n

$$
\\begin{align}
a_2 \\frac{d^2 y(t)}{dt^2} + a_1 \\frac{d y(t)}{dt} + a_0 y(t) &= b_1 \\frac{d x(t)}{dt} + b_0 x(t) \\\\
\\end{align}
$$<\/p>\n\n\n\n

Using the differentiation property<\/a>, we can take the Laplace transform of both sides to get<\/p>\n\n\n\n

$$
\\begin{align}
a_2 s^2 Y(s) + a_1 s Y(s) + a_0 Y(s) &= b_1 s X(s) + b_0 X(s) \\\\
(a_2 s^2 + a_1 s + a_0) Y(s) &= (b_1 s + b_0) X(s) \\\\
\\end{align}
$$<\/p>\n\n\n\n

Recall from the Convolution property<\/a> that the benefit of using Transforms is to describe the output using the impulse response<\/p>\n\n\n\n

$$
\\begin{align}
Y(s) &= H(s) X(s) \\\\
H(s) &= \\frac{Y(s)}{X(s)} \\\\
\\end{align}
$$<\/p>\n\n\n\n

We can rearrange the Laplace Transform above to determine $H(s)$.<\/p>\n\n\n\n

$$
\\begin{align}
\\frac{Y(s)}{X(s)} &= \\frac{b_1 s + b_0}{a_2 s^2 + a_1 s + a_0} \\\\
H(s) &= \\frac{b_1 s + b_0}{a_2 s^2 + a_1 s + a_0} \\\\
\\end{align}
$$<\/p>\n\n\n\n

Generalizing this pattern to higher order systems<\/p>\n\n\n\n

$$
\\begin{align}
\\sum_{k=0}^N a_k \\frac{d^k y(t)}{dt^k} &= \\sum_{k=0}^M b_k \\frac{d^k x(t)}{dt^k} \\\\
\\\\
H(s) &= \\frac{\\sum_{k=0}^M b_k s^k}{\\sum_{k=0}^N a_k s^k} \\\\
\\end{align}
$$<\/p>\n\n\n\n

Note that this result does not include specification for ROC. ROC is determined by stability or causality information that is provided with the problem.<\/p>\n\n\n\n

<\/span>First Order Systems<\/span><\/h2>\n\n\n\n

The generic first order lowpass system is represented by<\/p>\n\n\n\n

$$
\\begin{align}
\\tau \\frac{dy(t)}{dt} + y(t) &= b_0 x(t) \\\\
\\end{align}
$$<\/p>\n\n\n\n

where $\\tau$ is the “time constant” of the system as we will see shortly. Taking the Laplace Transform, we have<\/p>\n\n\n\n

$$
\\begin{align}
\\tau s Y(s) + Y(s) &= b_0 X(s) \\\\
Y(s) [\\tau s + 1] &= b_0 X(s) \\\\
\\frac{Y(s)}{X(s)} &= \\frac{b_0}{\\tau s + 1} \\\\
H(s) &= b_0 \\frac{1}{\\tau s + 1} \\\\
\\end{align}
$$<\/p>\n\n\n\n

<\/span>Impulse and Step Response<\/span><\/h3>\n\n\n\n

Noticing that this result resembles the exponential decay transform<\/a>, we can find the impulse response of the system.<\/p>\n\n\n\n

$$
\\begin{align}
e^{-at} u(t) &\\stackrel{\\mathscr{L}}{\\leftrightarrow} \\frac{1}{s + a} &&Re\\{s\\} > -Re\\{a\\} \\\\
\\\\
H(s) &= \\frac{b_0}{\\tau} \\frac{1}{s + 1\/\\tau} &&Re\\{s\\} > -Re\\{1\/\\tau\\} \\\\
h(t) &= \\frac{b_0}{\\tau} e^{-t\/\\tau} u(t) &&Re\\{s\\} > -Re\\{1\/\\tau\\} \\\\
\\end{align}
$$<\/p>\n\n\n\n

The step response may be found directly by integrating the impulse response.<\/p>\n\n\n\n

$$
\\begin{align}
s(t) &= \\int_{-\\infty}^{t} h(x) dx \\\\

s(t) &= \\frac{b_0}{\\tau} \\int_{-\\infty}^{t} e^{-x\/\\tau} u(x) dx &&Re\\{s\\} > -Re\\{1\/\\tau\\} \\\\

s(t) &= \\frac{b_0}{\\tau} u(t) \\int_{0}^{t} e^{-x\/\\tau} dx &&Re\\{s\\} > -Re\\{1\/\\tau\\} \\\\

s(t) &= \\frac{b_0}{\\tau} u(t) \\left[ -\\tau e^{-x\/\\tau} \\right]_{x=0}^{t} &&Re\\{s\\} > -Re\\{1\/\\tau\\} \\\\

s(t) &= -b_0 u(t) \\left[ e^{-x\/\\tau} \\right]_{x=0}^{t} &&Re\\{s\\} > -Re\\{1\/\\tau\\} \\\\

s(t) &= -b_0 u(t) \\left[ e^{-t\/\\tau} – 1\\right] &&Re\\{s\\} > -Re\\{1\/\\tau\\} \\\\

s(t) &= b_0 \\left[ 1 – e^{-t\/\\tau} \\right] u(t) &&Re\\{s\\} > -Re\\{1\/\\tau\\} \\\\
\\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

Note<\/p>\n\n\n\n