{"id":1714,"date":"2024-08-23T04:55:14","date_gmt":"2024-08-23T04:55:14","guid":{"rendered":"https:\/\/neilfoxman.com\/?page_id=1714"},"modified":"2025-01-08T05:53:02","modified_gmt":"2025-01-08T05:53:02","slug":"discrete-systems-described-by-linear-constant-coefficient-difference-equations","status":"publish","type":"page","link":"https:\/\/neilfoxman.com\/?page_id=1714","title":{"rendered":"Discrete Systems Described by Linear Constant Coefficient Difference Equations"},"content":{"rendered":"\n

Instead of differentiation used for continuous systems, discrete systems generally use difference equations to capture time-varying signal or system information.<\/p>\n\n\n\n

<\/span>Euler Backwards Differentiation Approximation<\/span><\/h2>\n\n\n\n

<\/span>First Order<\/span><\/h3>\n\n\n\n

The Euler Backward Differentiation Approximation allows us to approximate a derivative in continuous time using discrete time samples.<\/p>\n\n\n\n

$$
\\begin{align}
\\frac{d y(t)}{dt} &\\approx \\frac{y[n] – y[n-1]}{T} \\\\
\\end{align}
$$<\/p>\n\n\n\n

where $T$ is a scaling factor that represents the continuous period between samples.<\/p>\n\n\n\n

<\/span>Second Order<\/span><\/h3>\n\n\n\n

The Euler Backwards Differentiation Approximation may be further extended for second derivatives.<\/p>\n\n\n\n

$$
\\begin{align}
\\frac{d^2 y(t)}{dt^2} &\\approx \\frac{ \\frac{y[n] – y[n-1]}{T} – \\frac{y[n-1] – y[n-2]}{T} }{ T } \\\\

\\frac{d^2 y(t)}{dt^2} &\\approx \\frac{ \\frac{y[n] – 2 y[n-1] + y[n-2]}{T} }{ T } \\\\

\\frac{d^2 y(t)}{dt^2} &\\approx \\frac{ y[n] – 2 y[n-1] + y[n-2] }{ T^2 } \\\\
\\end{align}
$$<\/p>\n\n\n\n

<\/span>$\\tau \/ T$ and Integer Implmentation<\/span><\/h2>\n\n\n\n

Many of the first order systems below will express coefficients in terms of the quantity $\\tau \/ T$ where $\\tau$ is the time constant of the analagous continuous time system, and $T$ is the sampling period. This quantity may also be derived using other known quantities.<\/p>\n\n\n\n

$$
\\begin{align}
\\frac{\\tau}{T} &= \\frac{1}{\\omega_n T} \\\\
\\frac{\\tau}{T} &= \\frac{1}{2 \\pi f_n T} \\\\
\\frac{\\tau}{T} &= \\frac{f_s}{2 \\pi f_n} \\\\
\\end{align}
$$<\/p>\n\n\n\n

The quantity $\\tau \/ T$ is especially useful when $\\tau \/ T \\gg 1$, and so coefficients may be generally represented using integers.<\/p>\n\n\n\n

Some additional tips for using integers are:<\/p>\n\n\n\n

    \n
  1. Apply a gain to the input signal, so the numbers used in calculation are larger<\/li>\n\n\n\n
  2. Perform all addition and multiplication first, then apply division in one step at the end.<\/li>\n<\/ol>\n\n\n\n

    <\/span>FIR Averaging Filter<\/span><\/h2>\n\n\n\n

    Averaging causal FIR filter is expressed as<\/p>\n\n\n\n

    $$
    \\begin{align}
    y[n] &= \\frac{1}{N} \\left[ x[n] + x[n-1] + \\dots + x[n-(N-1)] \\right] \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    From Time Shifting Property<\/p>\n\n\n\n

    $$
    \\begin{align}
    Y(z) &= \\frac{1}{N} \\left[ X(z) + X(z) z^{-1} + \\dots + X(z) z^{-(N-1)} \\right] \\\\
    Y(z) &= \\frac{1}{N} \\left[ 1 + z^{-1} + \\dots + z^{-(N-1)} \\right] X(z) \\\\
    \\\\
    H(z) &= \\frac{1}{N} \\left[ 1 + z^{-1} + \\dots + z^{-(N-1)} \\right] \\\\
    H(z) &= \\frac{1}{N} \\sum_{m=0}^{N-1} (z^{-1})^{m} \\\\
    H(z) &= \\frac{1}{N} \\frac{1 – z^{-N}}{1 – z^{-1}} \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    When $|z| = 1$ we get the frequency response<\/p>\n\n\n\n

    $$
    \\begin{align}
    H(e^{j \\Omega}) &= \\frac{1}{N} \\frac{1 – e^{-j \\Omega N}}{1 – e^{-j \\Omega}} \\\\

    H(e^{j \\Omega}) &= \\frac{1}{N} \\frac{e^{j \\Omega \/ 2}}{e^{j \\Omega N \/ 2}} \\frac{e^{j \\Omega N \/ 2}}{e^{j \\Omega \/ 2}} \\frac{1 – e^{- j \\Omega N}}{1 – e^{-j \\Omega}} \\\\

    H(e^{j \\Omega}) &= \\frac{1}{N} \\frac{e^{j \\Omega \/ 2}}{e^{j \\Omega N \/ 2}} \\frac{e^{j \\Omega N \/ 2} – e^{-j \\Omega N \/ 2 }}{e^{j \\Omega \/ 2} – e^{-j \\Omega \/ 2}} \\\\

    H(e^{j \\Omega}) &= \\frac{1}{N} \\frac{e^{j \\Omega \/ 2}}{(e^{j \\Omega \/ 2})^N} \\frac{2 j \\sin( \\Omega N \/ 2)}{2 j \\sin(\\Omega \/ 2)} \\\\

    H(e^{j \\Omega}) &= \\frac{1}{N} (e^{j \\Omega \/ 2})^{1-N} \\frac{\\sin( \\Omega N \/ 2)}{\\sin(\\Omega \/ 2)} \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Magnitude is then<\/p>\n\n\n\n

    $$
    \\begin{align}
    |H(e^{j \\Omega})| &= \\frac{1}{N} \\left| \\frac{\\sin( \\Omega N \/ 2)}{\\sin(\\Omega \/ 2)} \\right| \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    <\/span>First Difference Lowpass Filter<\/span><\/h2>\n\n\n\n

    Reference filter_algorithm.dvi (techteach.no)<\/a><\/p>\n\n\n\n

    Recall<\/a> that in continuous time, a first order lowpass filter with some gain $G$ takes the form<\/p>\n\n\n\n

    $$
    \\begin{align}
    H(s) &= G \\frac{1}{\\tau s + 1} \\\\
    \\\\
    Y(s)[\\tau s + 1] &= G X(s) \\\\
    \\tau s Y(s) + Y(s) &= G X(s) \\\\
    \\\\
    \\tau \\frac{d y(t)}{dt} + y(t) &= G x(t) \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Using the Euler Backwards Differentiation Approximation<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\tau \\frac{y[n] – y[n-1]}{T} + y[n] &= G x[n] \\\\
    \\frac{\\tau}{T} y[n] – \\frac{\\tau}{T} y[n-1] + y[n] &= G x[n] \\\\
    \\left( \\frac{\\tau}{T} + 1 \\right) y[n] – \\frac{\\tau}{T} y[n-1] &= G x[n] \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Putting this in terms of a formula that may be used in an algorithm.<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\left( \\frac{\\tau}{T} + 1 \\right) y[n] &= \\frac{\\tau}{T} y[n-1] + G x[n] \\\\
    y[n] &= \\frac{ \\frac{\\tau}{T} y[n-1] + G x[n] }{ \\frac{\\tau}{T} + 1 } \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    <\/span>Frequency Response<\/span><\/h3>\n\n\n\n

    Frequency response of this system is derived from the difference equation using $G=1$<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\left( \\frac{\\tau}{T} + 1 \\right) y[n] – \\frac{\\tau}{T} y[n-1] &= G x[n] \\\\
    \\frac{\\tau + T}{T} y[n] – \\frac{\\tau}{T} y[n-1] &= x[n] \\\\
    \\\\
    \\frac{\\tau + T}{T} Y(e^{j \\Omega}) – \\frac{\\tau}{T} Y(e^{j \\Omega}) e^{-j \\Omega} &= X(e^{j \\Omega}) \\\\

    Y(e^{j \\Omega}) \\left[ \\frac{\\tau + T}{T} – \\frac{\\tau}{T} e^{-j \\Omega} \\right] &= X(e^{j \\Omega}) \\\\

    \\frac{Y(e^{j \\Omega})}{X(e^{j \\Omega})} &= \\frac{1}{\\frac{\\tau + T}{T} – \\frac{\\tau}{T} e^{-j \\Omega}} \\\\

    H(e^{j \\Omega}) &= \\frac{T}{T} \\frac{1}{\\frac{\\tau + T}{T} – \\frac{\\tau}{T} e^{-j \\Omega}} \\\\

    H(e^{j \\Omega}) &= \\frac{T}{\\tau + T – \\tau e^{-j \\Omega}} \\\\

    H(e^{j \\Omega}) &= \\frac{T}{\\tau + T – \\tau \\cos(-\\Omega) – j \\tau \\sin(-\\Omega)} \\\\

    H(e^{j \\Omega}) &= \\frac{T}{T + \\tau [1 – \\cos(-\\Omega)] – j \\tau \\sin(-\\Omega)} \\\\

    H(e^{j \\Omega}) &= \\frac{T}{T + \\tau [1 – \\cos(\\Omega)] + j \\tau \\sin(\\Omega)} \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Magnitude is then<\/p>\n\n\n\n

    $$
    \\begin{align}
    |H(e^{j \\Omega})| &= \\frac{T}{\\sqrt{\\{T + \\tau [1 – \\cos(\\Omega)] \\}^2 + \\{ \\tau \\sin(\\Omega) \\}^2 } } \\\\

    |H(e^{j \\Omega})| &= \\frac{T}{\\sqrt{T^2 + 2 \\tau T [1 – \\cos(\\Omega)] + \\tau^2 [1 – \\cos(\\Omega)]^2 + \\tau^2 \\sin^2(\\Omega) } } \\\\

    |H(e^{j \\Omega})| &= \\frac{T}{\\sqrt{T^2 + 2 \\tau T [1 – \\cos(\\Omega)] + \\tau^2 [1 – 2 \\cos(\\Omega) + \\cos^2(\\Omega)] + \\tau^2 \\sin^2(\\Omega) } } \\\\

    |H(e^{j \\Omega})| &= \\frac{T}{\\sqrt{T^2 + 2 \\tau T – 2 \\tau T \\cos(\\Omega) + \\tau^2 – 2 \\tau^2 \\cos(\\Omega) + \\tau^2 \\cos^2(\\Omega) + \\tau^2 \\sin^2(\\Omega) } } \\\\

    |H(e^{j \\Omega})| &= \\frac{T}{\\sqrt{T^2 + 2 \\tau T – 2 \\tau T \\cos(\\Omega) + \\tau^2 – 2 \\tau^2 \\cos(\\Omega) + \\tau^2 [\\cos^2(\\Omega) + \\sin^2(\\Omega)] } } \\\\

    |H(e^{j \\Omega})| &= \\frac{T}{\\sqrt{T^2 + 2 \\tau T – 2 \\tau T \\cos(\\Omega) + \\tau^2 – 2 \\tau^2 \\cos(\\Omega) + \\tau^2 } } \\\\

    |H(e^{j \\Omega})| &= \\frac{T}{\\sqrt{(\\tau + T)^2 – 2 \\tau \\cos(\\Omega)(\\tau + T) + \\tau^2 } } \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    There does not appear to be a simpler representation of this (when looking around on internet, trying a couple methods using sympy, and asking ChatGPT). However, you may note that if we assume the value in the radical is always positive, then the magnitude reaches a maximum when $\\cos(\\Omega) = 1 \\implies \\Omega = 0, 2 \\pi, \\dots$ and a minimum when $\\cos(\\Omega) = -1 \\implies \\Omega = \\pm \\pi, \\pm 3 \\pi, \\dots $ as we would expect for a discrete lowpass filter.<\/p>\n\n\n\n

    <\/span>First Difference Highpass Filter<\/span><\/h2>\n\n\n\n

    Recall<\/a> that a highpass filter in continuous time with gain $G$ is described using<\/p>\n\n\n\n

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

    Using the Euler Backwards Differentiation Approximation, we can approximate this as a difference equation in discrete time.<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\frac{d x(t)}{dt} &\\approx \\frac{x[n] – x[n-1]}{T} \\\\
    \\\\
    \\tau \\frac{y[n] – y[n-1]}{T} + y[n] &= G \\tau \\frac{x[n] – x[n-1]}{T} \\\\

    \\frac{\\tau}{T} y[n] – \\frac{\\tau}{T} y[n-1] + y[n] &= G \\frac{\\tau}{T} x[n] – G \\frac{\\tau}{T} x[n-1] \\\\

    \\left( \\frac{\\tau}{T} + 1 \\right) y[n] – \\frac{\\tau}{T} y[n-1] &= G \\frac{\\tau}{T} x[n] – G \\frac{\\tau}{T} x[n-1] \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    If using an algorithm, this can be rewritten<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\left( \\frac{\\tau}{T} + 1 \\right) y[n] &= \\frac{\\tau}{T} y[n-1] + G \\frac{\\tau}{T} x[n] – G \\frac{\\tau}{T} x[n-1] \\\\
    \\left( \\frac{\\tau}{T} + 1 \\right) y[n] &= \\frac{\\tau}{T} \\left( y[n-1] + G x[n] – G x[n-1] \\right) \\\\
    y[n] &= \\frac{ ( \\tau \/ T ) \\left( y[n-1] + G \\left\\{ x[n] – x[n-1] \\right\\} \\right) }{ (\\tau \/ T ) + 1} \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    For small changes in $x$ i.e. $x[n] – x[n-1]$ is small, one may choose the gain to be large such that<\/p>\n\n\n\n

    $$
    \\begin{align}
    G &= G’ \\left( \\frac{\\tau}{T} + 1 \\right) \\\\
    \\\\
    \\left( \\frac{\\tau}{T} + 1 \\right) y[n] &= \\frac{\\tau}{T} y[n-1] + G’ \\left( \\frac{\\tau}{T} + 1 \\right) \\frac{\\tau}{T} x[n] – G’ \\left( \\frac{\\tau}{T} + 1 \\right) \\frac{\\tau}{T} x[n-1] \\\\

    y[n] &= \\frac{ \\tau \/ T }{ ( \\tau \/ T ) + 1 } y[n-1] + G’ \\frac{\\tau}{T} ( x[n] – x[n-1] ) \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Although the solution above does not lend itself well to an implementation, one may note that the coefficient for $y[n-1] \\to 1$ whereas the $G’ (\\tau \/ T) \\gg 1$ coefficient dominates.<\/p>\n\n\n\n

    <\/span>First Difference Differentiator<\/span><\/h2>\n\n\n\n

    This is an easy to implement filter that resembles differentiation.<\/p>\n\n\n\n

    $$
    \\begin{align}
    y[n] &= x[n] – x[n-1] \\\\
    \\\\
    Y(z) &= X(z) – X(z) z^{-1} \\\\
    Y(z) &= X(z) [1 – z^{-1}] \\\\
    H(z) &= 1 – z^{-1} \\\\
    \\\\
    H(e^{j \\Omega}) &= 1 – e^{-j \\Omega} \\\\
    H(e^{j \\Omega}) &= e^{-j \\Omega\/2} (e^{j \\Omega\/2} – e^{-j \\Omega\/2}) \\\\
    H(e^{j \\Omega}) &= 2 j e^{-j \\Omega\/2} \\sin(\\Omega\/2) \\\\
    \\\\
    |H(e^{j \\Omega})| &= 2 |\\sin(\\Omega\/2)| \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Note around $\\Omega = 0$, the magnitude is zero and slope is 1, which is similar to an ideal differentiator<\/a>. Plus superscript indicates we are just looking at positive frequencies.<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\frac{d}{d\\Omega} |H(e^{j \\Omega})|^+ &= 2 \\cos(\\Omega\/2) \\cdot \\frac{1}{2} \\\\
    \\frac{d}{d\\Omega} |H(e^{j \\Omega})|^+ &= \\cos(\\Omega\/2) \\\\
    \\left[ \\frac{d}{d\\Omega} |H(e^{j \\Omega})| \\right]_{\\Omega=0^+} &= 1 \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    <\/span>Ideal Differentiator<\/span><\/h2>\n\n\n\n

    In continuous time, a differentiating system has the form<\/p>\n\n\n\n

    $$
    \\begin{align}
    y_c(t) &= \\frac{d}{dt} x_c(t) \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    And we know from the differentiation property<\/a> that the system response is<\/p>\n\n\n\n

    $$
    \\begin{align}
    Y_c(s) &= s X(s) \\\\
    H_c(s) &= s \\\\
    H_c(j \\omega) &= j \\omega \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Although we will revisit this approach in more detail later<\/a>, for now let’s assume that in order to create a comparable version of this filter for discrete time systems we must assume a band limited input signal and a system response described by<\/p>\n\n\n\n

    $$
    \\begin{align}
    H_d(e^{j \\omega}) &= j \\Omega &&\\left|\\Omega\\right| < \\pi \\\\
    \\end{align}
    $$<\/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

    Impulse Response of this signal is<\/p>\n\n\n\n

    $$
    \\begin{align}
    h_d[n] &= \\frac{1}{2 \\pi}\\int_{2 \\pi} H_d(e^{j \\Omega}) \\cdot e^{j \\Omega n} d \\Omega \\\\

    h_d[n] &= \\frac{1}{2 \\pi} \\int_{-\\pi}^{\\pi} j \\Omega e^{j \\Omega n} d \\Omega \\\\

    h_d[n] &= j\\frac{1}{2 \\pi} \\int_{-\\pi}^{\\pi} \\Omega e^{j \\Omega n} d \\Omega \\\\

    \\end{align}
    $$<\/p>\n\n\n\n

    Using IBP<\/p>\n\n\n\n

    $$
    \\begin{align}
    (fg)’ &= f’g + fg’ \\\\
    f’g &= (fg)’- fg’ \\\\
    \\int_a^b f’g d\\Omega &= \\left[ fg \\right]_{\\Omega=a}^b – \\int_a^b fg’ d\\Omega \\\\
    \\\\
    f’ &= e^{j \\Omega n} \\\\
    f &= \\frac{1}{j n} e^{j \\Omega n} \\\\
    g &= \\Omega \\\\
    g’ &= 1 \\\\

    \\end{align}
    $$<\/p>\n\n\n\n

    Returning to the impulse response<\/p>\n\n\n\n

    $$
    \\begin{align}
    h_d[n] &= j\\frac{1}{2 \\pi} \\left[ \\left(\\frac{\\Omega}{j n} e^{j \\Omega n} \\right)_{\\Omega=-\\pi}^{\\pi} – \\int_{-\\pi}^{\\pi} \\frac{1}{j n} e^{j \\Omega n} d \\Omega \\right] \\\\

    h_d[n] &= j\\frac{1}{2 \\pi} \\left[ \\left(\\frac{\\Omega}{j n} e^{j \\Omega n} \\right)_{\\Omega=-\\pi}^{\\pi} – \\left( \\frac{1}{(j n)^2} e^{j \\Omega n} \\right)_{\\Omega=-\\pi}^{\\pi} \\right] \\\\

    h_d[n] &= j\\frac{1}{2 \\pi} \\left[ \\left(\\frac{\\pi}{j n} e^{j \\pi n} – \\frac{-\\pi}{j n} e^{-j \\pi n} \\right) – \\left( \\frac{1}{(j n)^2} e^{j \\pi n} – \\frac{1}{(j n)^2} e^{-j \\pi n} \\right) \\right] \\\\

    h_d[n] &= j\\frac{1}{2 \\pi} \\left[ \\frac{\\pi}{j n} \\left( e^{j \\pi n} + e^{-j \\pi n} \\right) – \\frac{1}{(j n)^2} \\left( e^{j \\pi n} – e^{-j \\pi n} \\right) \\right] \\\\

    h_d[n] &= j\\frac{1}{2 \\pi} \\left[ \\frac{\\pi}{j n} 2 \\cos(\\pi n) – \\frac{1}{(j n)^2} 2 j \\sin(\\pi n) \\right] \\\\

    h_d[n] &= \\frac{1}{n} \\cos(\\pi n) + \\frac{1}{\\pi (j n)^2} \\sin(\\pi n) \\\\

    h_d[n] &= \\frac{1}{n} \\cos(\\pi n) – \\frac{1}{\\pi n^2} \\sin(\\pi n) \\\\

    h_d[n] &= \\frac{\\pi n \\cos(\\pi n) – \\sin(\\pi n)}{ \\pi n^2} \\\\

    \\end{align}
    $$<\/p>\n\n\n\n

    For $n \\neq 0$ we may note that because $n \\in \\mathbb{Z}$ we can simplify the above to<\/p>\n\n\n\n

    $$
    \\begin{align}
    h_d[n] &= \\frac{\\pi n \\cos(\\pi n) }{ \\pi n^2} – \\frac{\\sin(\\pi n)}{\\pi n^2} &&n \\neq 0 \\\\

    h_d[n] &= \\frac{\\cos(\\pi n) }{n} – \\frac{\\sin(\\pi n)}{\\pi n^2} &&n \\neq 0 \\\\

    n \\in \\mathbb{Z} \\implies \\sin(\\pi n) = 0 \\implies
    h_d[n] &= \\frac{\\cos(\\pi n) }{n} &&n \\neq 0 \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Because both the numerator and the denominator approach 0 at $n=0$, we need to use more advanced analysis to determine the value at $n=0$. For the time being, ignore that we are solving for integer values of $n$ and instead let’s solve for the generic case where the independent variable $x \\in \\mathbb{R}$.<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\frac{\\pi x \\cos(\\pi x) – \\sin(\\pi x)}{ \\pi x^2} \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Now applying L\u2019H\u00f4pital\u2019s Rule just like for the $\\text{sinc}$ function<\/a> we get<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\lim_{x \\to c} \\frac{f(x)}{g(x)} &= \\lim_{x \\to c} \\frac{f'(x)}{g'(x)} \\\\
    \\\\
    f(x) &= \\pi x \\cos(\\pi x) – \\sin(\\pi x) \\\\
    \\\\
    f'(x) &= \\pi \\{1 \\cdot \\cos(\\pi x) + x \\cdot [-\\sin(\\pi x)] \\cdot \\pi \\} – \\cos(\\pi x) \\cdot \\pi \\\\
    f'(x) &= \\pi \\{\\cos(\\pi x) – \\pi x \\sin(\\pi x) \\} – \\pi \\cos(\\pi x) \\\\
    f'(x) &= \\pi \\cos(\\pi x) – \\pi^2 x \\sin(\\pi x) – \\pi \\cos(\\pi x) \\\\
    f'(x) &= – \\pi^2 x \\sin(\\pi x) \\\\
    \\\\
    g(x) &= \\pi x^2 \\\\
    \\\\
    g'(x) &= 2 \\pi x \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    This result again cannot be solved as both the numerator and denominator approach 0 as $x \\to 0$. However, we may re-use L\u2019H\u00f4pital\u2019s Rule to infer that<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\lim_{x \\to c} \\frac{f(x)}{g(x)} &= \\lim_{x \\to c} \\frac{f'(x)}{g'(x)} = \\lim_{x \\to c} \\frac{f”(x)}{g”(x)} \\\\
    \\\\
    f”(x) &= – \\pi^2 \\{ 1 \\cdot \\sin(\\pi x) + x \\cdot \\cos(\\pi x) \\cdot \\pi \\} \\\\
    f”(x) &= – \\pi^2 \\{ \\sin(\\pi x) + \\pi x \\cos(\\pi x) \\} \\\\
    f”(x) &= – \\pi^2 \\sin(\\pi x) – \\pi^3 x \\cos(\\pi x) \\\\
    \\\\
    g”(x) &= 2 \\pi \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    The denominator no longer approaches 0, and we can now solve<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\lim_{x \\to c} \\frac{f(x)}{g(x)} &= \\lim_{x \\to c} \\frac{- \\pi^2 \\sin(\\pi x) – \\pi^3 x \\cos(\\pi x)}{2 \\pi} \\\\
    \\lim_{x \\to 0} \\frac{f(x)}{g(x)} &= \\lim_{x \\to 0} \\frac{- \\pi^2 \\sin(\\pi x) – \\pi^3 x \\cos(\\pi x)}{2 \\pi} \\\\
    \\lim_{x \\to 0} \\frac{f(x)}{g(x)} &= \\frac{0}{2 \\pi} \\\\
    \\lim_{x \\to 0} \\frac{f(x)}{g(x)} &= 0 \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    and so for the ideal differentiator in discrete time, the impulse response is<\/p>\n\n\n\n

    $$
    h[n] = \\cases{
    \\begin{align}
    0 &&&n=0 \\\\
    \\frac{\\cos(\\pi n)}{n} &&&n \\neq 0
    \\end{align}
    }
    $$<\/p>\n\n\n\n

    Verifying using online calculator, the continuous version of the impulse response is plotted below<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\frac{\\pi x \\cos(\\pi x) – \\sin(\\pi x)}{ \\pi x^2} \\\\
    \\end{align}
    $$<\/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

    Below is that same plot with the discrete function superimposed over it i,e. in blue we have<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\frac{\\cos(\\pi x)}{x}
    \\end{align}
    $$<\/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 that the two functions overlap when $x$ is an integer except at $x=0$.<\/p>\n\n\n\n

    Additionally, note that $\\forall n \\in \\mathbb{Z}$, the numerator $\\cos(\\pi n)$ just alternates between 1 and -1, so depending on computation method it may help to conceptualize the impulse response as<\/p>\n\n\n\n

    $$
    h[n] = \\cases{
    \\begin{align}
    0 &&&n=0 \\\\
    \\frac{(-1)^{n}}{n} &&&n \\neq 0
    \\end{align}
    }
    $$<\/p>\n\n\n\n

    Using the sifting property<\/a>, the impulse response may also be written as<\/p>\n\n\n\n

    $$
    \\begin{align}
    h[n] &= \\dots +
    \\frac{-1}{-3} \\delta[n+3] +
    \\frac{1}{-2} \\delta[n+2] +
    \\frac{-1}{-1} \\delta[n+1] +
    0 \\cdot \\delta[n] +
    \\frac{-1}{1} \\delta[n-1] +
    \\frac{1}{2} \\delta[n-2] +
    \\frac{-1}{3} \\delta[n-3] +
    \\dots \\\\

    h[n] &= \\dots +\\frac{1}{3} \\delta[n+3] – \\frac{1}{2} \\delta[n+2] +\\delta[n+1] – \\delta[n-1] + \\frac{1}{2} \\delta[n-2] – \\frac{1}{3} \\delta[n-3] + \\dots \\\\

    h[n] &=
    (\\delta[n+1] – \\delta[n-1]) +
    \\frac{\\delta[n-2] – \\delta[n+2]}{2} +
    \\frac{\\delta[n+3] – \\delta[n-3]}{3} +
    \\dots \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    <\/span>Central Difference Differentiator<\/span><\/h2>\n\n\n\n

    Note that if we truncate\/ignore the higher order terms and can accept a latency of one sample, we have<\/p>\n\n\n\n

    $$
    \\begin{align}
    h_{diffapprox}[n] &= \\delta[n+1] – \\delta[n-1] \\\\
    \\\\
    y_{diffapprox}[n] &= x[n+1] – x[n-1] \\\\
    \\\\
    Y_{diffapprox}(e^{j \\Omega}) &= X(e^{j \\Omega}) (e^{j \\Omega} – e^{-j \\Omega} \\\\
    Y_{diffapprox}(e^{j \\Omega}) &= X(e^{j \\Omega}) 2 j \\sin(\\Omega) \\\\
    H_{diffapprox}(e^{j \\Omega}) &= 2 j \\sin(\\Omega) \\\\
    |H_{diffapprox}(e^{j \\Omega})| &= 2 \\sin(\\Omega) \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Ignoring the scaling factor of 2, this filter also resembles a differentiator at low frequencies. However, note that in this case there is some attenuation at high frequencies near $\\pi$ because $\\sin(\\pi) = 0$. This effect is sometimes desirable however, and functions to attenuate high frequency noise and harmonics.<\/p>\n\n\n\n

    Reference: Design of a Discrete-Time Differentiator | Wireless Pi<\/a><\/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>Second Order Bandpass Filter<\/span><\/h2>\n\n\n\n

    Recall<\/a> that a bandpass filter in continuous time may be represented as<\/p>\n\n\n\n

    $$
    \\begin{align}
    H(s) &= \\frac{ s }{ s^2 + 2 \\zeta \\omega_n s + \\omega_n^2 } \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Putting this in a form with derivatives using the inverse Laplace Transform, we have<\/p>\n\n\n\n

    $$
    \\begin{align}
    [s^2 + 2 \\zeta \\omega_n s + \\omega_n^2] Y(s) &= s X(s) \\\\
    \\frac{d^2 y(t)}{dt} + 2 \\zeta \\omega_n \\frac{d y(t)}{dt} + \\omega_n^2 y(t) &= \\frac{d x(t)}{dt} \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    We may then use Euler Backwards Differentiation Approximation to make a difference equation<\/p>\n\n\n\n

    $$
    \\begin{align}

    \\frac{ y[n] – 2 y[n-1] + y[n-2] }{ T^2 } + 2 \\zeta \\omega_n \\left( \\frac{y[n] – y[n-1]}{T} \\right) + \\omega_n^2 y[n] &= \\frac{x[n] – x[n-1]}{T} \\\\

    \\left( \\frac{1}{T^2} + \\frac{ 2 \\zeta \\omega_n}{T} + \\omega_n^2 \\right) y[n] +
    \\left( \\frac{-2}{T^2} + \\frac{ -2 \\zeta \\omega_n}{T} \\right) y[n-1] +
    \\frac{1}{T^2} y[n-2] &=
    \\frac{1}{T} x[n] +
    \\frac{-1}{T} x[n-1] \\\\

    \\end{align}
    $$<\/p>\n\n\n\n

    <\/p>\n","protected":false},"excerpt":{"rendered":"

    Instead of differentiation used for continuous systems, discrete systems generally use difference equations to capture time-varying signal or system information. Euler Backwards Differentiation Approximation First Order The Euler Backward Differentiation Approximation allows us to approximate a derivative in continuous time using discrete time samples. $$\\begin{align}\\frac{d y(t)}{dt} &\\approx \\frac{y[n] – y[n-1]}{T} \\\\\\end{align}$$ where $T$ is a […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":251,"menu_order":3,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-1714","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/1714","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1714"}],"version-history":[{"count":64,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/1714\/revisions"}],"predecessor-version":[{"id":2115,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/1714\/revisions\/2115"}],"up":[{"embeddable":true,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/251"}],"wp:attachment":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1714"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}