{"id":112,"date":"2023-12-12T05:12:50","date_gmt":"2023-12-12T05:12:50","guid":{"rendered":"https:\/\/neilfoxman.com\/?page_id=112"},"modified":"2024-06-24T03:53:29","modified_gmt":"2024-06-24T03:53:29","slug":"discrete-fourier-series","status":"publish","type":"page","link":"https:\/\/neilfoxman.com\/?page_id=112","title":{"rendered":"Discrete Transforms"},"content":{"rendered":"\n

<\/span>Intent<\/span><\/h2>\n\n\n\n

Recall<\/a> we found that an input signal $x[n]$ fed into an LTI system with impulse response $h[n]$ generates an output $y[n]$ described by the convolution operation.<\/p>\n\n\n\n

$$
\\begin{align}
x[n] &\\rightarrow y[n] = x[n] * h[n] \\\\
x[n] &\\rightarrow y[n] = \\sum_{k=-\\infty}^{\\infty} x[k] h[n-k] \\\\
\\end{align}
$$<\/p>\n\n\n\n

As it can be fairly hard to conceptualize how a system will shape an input to an output using convolution in the $n$ domain, it is commonly beneficial to transform the representation into a new domain that is easier to conceptualize.<\/p>\n\n\n\n

First, let’s imagine the simple case in discrete time where the input signal is merely some complex number raised to the $n$, that is<\/p>\n\n\n\n

$$
\\begin{align}
x[n] &= z^n &&&z \\in \\mathbb{C} \\\\
\\\\
y[n] &= x[n] * h[n] \\\\
y[n] &= h[n] * x[n] \\\\
y[n] &= \\sum_{k=-\\infty}^{\\infty} h[k] x[n-k] \\\\
y[n] &= \\sum_{k=-\\infty}^{\\infty} h[k] z^{n-k} \\\\
y[n] &= z^n \\sum_{k=-\\infty}^{\\infty} h[k] z^{-k} \\\\
\\end{align}
$$<\/p>\n\n\n\n

The critical takeaway here is that the output is a scaled version of the input. In this sense, $z^n$ is called an “eigenfunction” and the scaling factor $\\sum_{k=-\\infty}^{\\infty} h[k] z^{-k}$ is the “eigenvalue”. This eigenvalue is dependent on both the original eigenfunction $z$ and the system impulse response, and is commonly expressed as<\/p>\n\n\n\n

$$
\\begin{align}
H(z) &= \\sum_{k=-\\infty}^{\\infty} h[k] z^{-k} \\\\
\\\\
y[n] &= H(z) z^n \\\\
y[n] &= H(z) x[n] \\\\
\\end{align}
$$<\/p>\n\n\n\n

Also note that the variable of summation for $H(z)$ can be changed to $n$ to better envision the calculation for the impulse response i.e. the eigenvalue may also be calculated using<\/p>\n\n\n\n

$$
\\begin{align}
H(z) &= \\sum_{n=-\\infty}^{\\infty} h[n] z^{-n} \\\\
\\end{align}
$$<\/p>\n\n\n\n

Assuming an LTI system, if a new input signal is instead represented as a sum (series or integral) of $z$s, the output can be calculated as the sum of scaled $z$s. That is<\/p>\n\n\n\n

$$
\\begin{align}
x[n] = z^n &\\rightarrow y[n] = H(z) z^n \\\\
x[n] = \\sum_q a_q z_q^n &\\rightarrow y[n] = \\sum_q a_q H(z_q) z_q^n \\\\
\\end{align}
$$<\/p>\n\n\n\n

Note that the eigenvalue is dependent on $z$, so if the input is expressed as a sum of $z$s, then $H(z)$ must be calculated for each $z$.<\/p>\n\n\n\n

The transformations here are all built on this principle and essentially just vary in what set of $z$s we can use to describe our input signal impulse response.<\/p>\n\n\n\n

<\/span>Discrete Time Fourier Series<\/span><\/h2>\n\n\n\n

The Discrete Time Fourier Series is a method to represent any periodic signal as a series of periodic exponentials.<\/p>\n\n\n\n

Assume that a discrete signal may be represented as a series of periodic harmonics such that<\/p>\n\n\n\n

$$
\\begin{align}
x[n] &= \\sum_k a_k e^{j k \\Omega_0 n} \\\\
\\end{align}
$$<\/p>\n\n\n\n

Recall<\/a> that there are only $N$ distinct harmonics due to discrete frequency periodicity. We can then represent this signal with<\/p>\n\n\n\n

$$
\\begin{align}
x[n] &= \\sum_{k= \\langle N \\rangle} a_k e^{j k \\Omega_0 n} \\\\
\\end{align}
$$<\/p>\n\n\n\n

We can solve for Fourier Series coefficients, $a_k$ by scaling both sides with another complex exponential and summing both sides over one period with respect to $n$.<\/p>\n\n\n\n

$$
\\begin{align}

x[n] e^{-j q \\Omega_0 n} &= \\sum_{k= \\langle N \\rangle } a_k e^{j k \\Omega_0 n} e^{-j q \\Omega_0 n} \\\\

\\sum_{n = \\langle N \\rangle} x[n] e^{-j q \\Omega_0 n} &= \\sum_{n = \\langle N \\rangle} \\sum_{k= \\langle N \\rangle } a_k e^{j k \\Omega_0 n} e^{-j q \\Omega_0 n} \\\\

\\sum_{n = \\langle N \\rangle} x[n] e^{-j q \\Omega_0 n} &= \\sum_{n = \\langle N \\rangle} \\sum_{k= \\langle N \\rangle } a_k e^{j (k-q) \\Omega_0 n} \\\\

\\sum_{n = \\langle N \\rangle} x[n] e^{-j q \\Omega_0 n} &= \\sum_{k= \\langle N \\rangle } a_k \\sum_{n = \\langle N \\rangle} e^{j (k-q) \\Omega_0 n} \\\\

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

From how we defined the harmonics of $x[n]$ we know that $k \\in \\mathbb{Z}$. Now if we further restrict our study to cases when $k – q \\in \\mathbb{Z} \\implies q \\in \\mathbb{Z}$, then we may note that the rightmost summation is 0 $ \\forall q, k: q \\neq k$, but the exponential is 1 when $q=k$.<\/p>\n\n\n\n

If we know the LHS must be nonzero, then<\/p>\n\n\n\n

$$
\\begin{align}

q &= k \\\\
\\\\
\\sum_{n = \\langle N \\rangle} x[n] e^{-j q \\Omega_0 n} &= \\sum_{k= q} a_k \\sum_{n = \\langle N \\rangle} 1 &&(k = q) \\land (k, q \\in \\mathbb{Z}) \\\\

\\sum_{n = \\langle N \\rangle} x[n] e^{-j q \\Omega_0 n} &= a_q N &&(k = q) \\land (k, q \\in \\mathbb{Z}) \\\\

a_q &= \\frac{1}{N} \\sum_{n = \\langle N \\rangle} x[n] e^{-j q \\Omega_0 n} &&(k = q) \\land (k, q \\in \\mathbb{Z})

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

and then replacing $q$ with $k$ for notation consistency, we have the synthesis and analysis equations:<\/p>\n\n\n\n

$$
\\boxed{
\\begin{align}
x[n] &= \\sum_{k = \\langle N \\rangle} a_k e^{j k \\Omega_0 n} \\\\
a_k &= \\frac{1}{N} \\sum_{n = \\langle N \\rangle} x[n] e^{-j k \\Omega_0 n}
\\end{align}
}
$$<\/p>\n\n\n\n

<\/span>Discrete Time Fourier Transform<\/span><\/h2>\n\n\n\n

The Fourier Transform is an extension of the Fourier series that may represent finite aperiodic sequences in addition to periodic sequences.<\/p>\n\n\n\n

Consider the finite duration signal $x[n]$ where $x[n] = 0$ when $-N_1 \\leq n \\leq N_2$. Then imagine periodic version of that signal $\\tilde{x}[n]$ that repeats every $N \\geq N_1 + N_2$.<\/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

Recall from the Fourier series that this periodic signal $\\tilde{x}[n]$ may be represented as<\/p>\n\n\n\n

$$
\\begin{align}
\\tilde{x}[n] &= \\sum_{k=\\langle N \\rangle} a_k e^{j k \\Omega_0 n} \\\\
\\end{align}
$$<\/p>\n\n\n\n

and so we then have Fourier Series coefficients that can be described by<\/p>\n\n\n\n

$$
\\begin{align}

a_k &= \\frac{1}{N} \\sum_{n=\\langle N \\rangle} \\tilde{x}[n] e^{- j k \\Omega_0 n} \\\\
a_k &= \\frac{1}{N} \\sum_{n=-N_2}^{N_1} \\tilde{x}[n] e^{- j k \\Omega_0 n} \\\\
a_k &= \\frac{1}{N} \\sum_{n=-N_2}^{N_1} x[n] e^{- j k \\Omega_0 n} \\\\
a_k &= \\frac{1}{N} \\sum_{n=-\\infty}^{\\infty} x[n] e^{- j k \\Omega_0 n} \\\\

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

Define the “envelope function” $X$ as<\/p>\n\n\n\n

$$
X(e^{j \\Omega}) = \\sum_{n=-\\infty}^{\\infty} x[n] e^{- j \\Omega n} \\\\
$$<\/p>\n\n\n\n

Using this definition of $X$ we may rewrite the Fourier Series coefficients as<\/p>\n\n\n\n

$$
a_k = \\frac{1}{N} X(e^{j k \\Omega_0})
$$<\/p>\n\n\n\n

and then the synthesis equation can be rewritten as<\/p>\n\n\n\n

$$
\\begin{align}

\\tilde{x}[n] &= \\sum_{k=\\langle N \\rangle} \\left[ \\frac{1}{N} X(e^{j k \\Omega_0}) \\right] e^{j k \\Omega_0 n} \\\\

\\tilde{x}[n] &= \\sum_{k=\\langle N \\rangle} \\left[ \\frac{\\Omega_0}{2 \\pi} X(e^{j k \\Omega_0}) \\right] e^{j k \\Omega_0 n} \\\\

\\tilde{x}[n] &= \\frac{1}{2 \\pi} \\sum_{k=\\langle N \\rangle} X(e^{j k \\Omega_0}) e^{j k \\Omega_0 n} \\Omega_0 \\\\

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

Now if we extend the “period” of our signal until $N \\rightarrow \\infty$, we essentially have a finite duration signal, and we may note the phenomena<\/p>\n\n\n\n

$$
\\begin{align}
\\Omega_0 \\rightarrow d \\Omega \\\\
k \\Omega_0 \\rightarrow \\Omega \\\\
\\tilde{x} \\rightarrow x \\\\
\\end{align}
$$<\/p>\n\n\n\n

So that the Analysis and Synthesis equation become, for any signal,<\/p>\n\n\n\n

$$
\\boxed{
\\begin{align}
x[n] &= \\frac{1}{2 \\pi}\\int_{2 \\pi} X(e^{j \\Omega}) e^{j \\Omega n} d \\Omega \\\\
X(e^{j \\Omega}) &= \\sum_{n=-\\infty}^{\\infty} x[n] e^{- j \\Omega n} \\\\
\\end{align}
}
$$<\/p>\n\n\n\n

Note that $X$ is essentially a function of $\\Omega$ and is periodic every $2 \\pi$ due to the frequency periodicity of discrete signals as shown below.<\/p>\n\n\n\n

$$
\\begin{align}
X(e^{j (\\Omega+2 \\pi)}) &= \\sum_{n=-\\infty}^{\\infty} x[n] e^{- j (\\Omega+2 \\pi) n} \\\\

X(e^{j (\\Omega+2 \\pi)}) &= \\sum_{n=-\\infty}^{\\infty} x[n] e^{- j \\Omega n} e^{- j \\cdot 2 \\pi \\cdot n} \\\\

X(e^{j (\\Omega+2 \\pi)}) &= \\sum_{n=-\\infty}^{\\infty} x[n] e^{- j \\Omega n} \\\\ \\end{align}
$$<\/p>\n\n\n\n

In summary, these equations suggest that<\/p>\n\n\n\n

    \n
  1. Any finite discrete signal (periodic or aperiodic) may be represented by a summation of complex exponentials in the frequency domain $\\Omega$<\/li>\n\n\n\n
  2. Unique complex exponentials used to describe a signal reside in a frequency range of width $2 \\pi$<\/li>\n\n\n\n
  3. The amplitude of a signal’s constituent complex exponentials are described by the function $\\frac{1}{2 \\pi} X(e^{j \\Omega})$ which may, itself, be complex.<\/li>\n<\/ol>\n\n\n\n

    <\/span>Z Transform<\/span><\/h2>\n\n\n\n

    Note that for the Discrete Fourier Transform, we have been using the analysis function defined by<\/p>\n\n\n\n

    $$
    X(e^{j \\Omega})
    $$<\/p>\n\n\n\n

    Which for any $\\Omega \\in \\mathbb{R}$ is defined on a unit circle in the imaginary plane and repeats every $2 \\pi$. If we further generalize the independent variable of this function to accept any complex input, this would be described by<\/p>\n\n\n\n

    $$
    \\begin{align}
    X(r e^{j \\Omega})
    \\end{align}
    $$<\/p>\n\n\n\n

    This quantity is often represented by $z = r e^{j \\Omega}$, and in the special case where $r=1$ we have the Fourier Transform. Re-doing the same analysis as above, but for a general complex variable $z$, we have (in “fast motion”)<\/p>\n\n\n\n

    $$
    \\begin{align}
    \\tilde{x}[n] &= \\sum_{k= \\langle N \\rangle} a_k (r e^{j k \\Omega_0})^n \\\\

    \\sum_{n = \\langle N \\rangle} \\tilde{x}[n] (r e^{j q \\Omega_0 })^{-n} &= \\sum_{n = \\langle N \\rangle} \\sum_{k= \\langle N \\rangle } a_k (r e^{j k \\Omega_0 })^n (r e^{j q \\Omega_0 })^{-n} \\\\

    \\sum_{n = \\langle N \\rangle} \\tilde{x}[n] (r e^{j q \\Omega_0 })^{-n} &= \\sum_{n = \\langle N \\rangle} \\sum_{k= \\langle N \\rangle } a_k r^n e^{j k \\Omega_0 n} r^{-n} e^{-j q \\Omega_0 n} \\\\

    \\sum_{n = \\langle N \\rangle} \\tilde{x}[n] (r e^{j q \\Omega_0 })^{-n} &= \\sum_{n = \\langle N \\rangle} \\sum_{k= \\langle N \\rangle } a_k r^{n-n} e^{j (k-q) \\Omega_0 n} \\\\

    \\sum_{n = \\langle N \\rangle} \\tilde{x}[n] (r e^{j q \\Omega_0 })^{-n} &= \\sum_{n = \\langle N \\rangle} \\sum_{k= \\langle N \\rangle } a_k e^{j (k-q) \\Omega_0 n} \\\\

    \\sum_{n = \\langle N \\rangle} \\tilde{x}[n](r e^{j q \\Omega_0 })^{-n} &= \\sum_{k= \\langle N \\rangle } a_k \\sum_{n = \\langle N \\rangle} e^{j (k-q) \\Omega_0 n} \\\\

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

    We effectively get the same implication as before: for the kth harmonic, $k, q \\in \\mathbb{Z} \\implies$<\/p>\n\n\n\n

    $$
    \\sum_{n = \\langle N \\rangle} \\tilde{x}[n] (r e^{j q \\Omega_0})^{-n} = \\cases{
    \\begin{align}
    \\sum_{k= \\langle N \\rangle } a_k \\sum_{n = \\langle N \\rangle} 1 && k = q \\\\
    0 && k \\neq q \\\\
    \\end{align}
    }
    $$<\/p>\n\n\n\n

    Assuming the LHS is nonzero, we can say<\/p>\n\n\n\n

    \\begin{align}
    \\sum_{n = \\langle N \\rangle} \\tilde{x}[n] (r e^{j q \\Omega_0})^{-n} &= \\sum_{k=q} a_k \\sum_{n = \\langle N \\rangle} 1 \\\\

    \\sum_{n = \\langle N \\rangle} \\tilde{x}[n] (r e^{j q \\Omega_0})^{-n} &= a_q N \\\\

    a_q &= \\frac{1}{N} \\sum_{n = \\langle N \\rangle} \\tilde{x}[n] (r e^{j q \\Omega_0})^{-n} \\\\

    a_k &= \\frac{1}{N} \\sum_{n = \\langle N \\rangle} \\tilde{x}[n] (r e^{j k \\Omega_0})^{-n} \\\\
    \\end{align}<\/p>\n\n\n\n

    Assuming $\\forall n \\in \\langle N \\rangle, \\tilde{x}[n] = x[n]$, we can expand the region of integration such that<\/p>\n\n\n\n

    $$
    \\begin{align}
    a_k &= \\frac{1}{N} \\sum_{n=\\langle N \\rangle} \\tilde{x}[n] (r e^{j k \\Omega_0})^{-n} \\\\
    a_k &= \\frac{1}{N} \\sum_{n=-\\infty}^{\\infty} x[n] (r e^{j k \\Omega_0})^{-n} \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Now we define<\/p>\n\n\n\n

    $$
    X(re^{j \\Omega}) = \\sum_{n=-\\infty}^{\\infty} x[n] (r e^{j \\Omega})^{-n} \\\\
    $$<\/p>\n\n\n\n

    And expanding upon our earlier findings<\/p>\n\n\n\n

    $$
    \\begin{align}
    a_k &= \\frac{1}{N} X(r e^{j k \\Omega_0}) \\\\
    \\\\
    \\tilde{x}[n] &= \\sum_{k= \\langle N \\rangle} a_k (r e^{j k \\Omega_0})^n \\\\

    \\tilde{x}[n] &= \\sum_{k=\\langle N \\rangle} \\left[ \\frac{1}{N} X(r e^{j k \\Omega_0}) \\right] (r e^{j k \\Omega_0})^n \\\\

    \\tilde{x}[n] &= \\sum_{k=\\langle N \\rangle} \\left[ \\frac{\\Omega_0}{2 \\pi} X(r e^{j k \\Omega_0}) \\right] (r e^{j k \\Omega_0})^n \\\\

    \\tilde{x}[n] &= \\frac{1}{2 \\pi} \\sum_{k=\\langle N \\rangle} X(r e^{j k \\Omega_0}) \\cdot (r e^{j k \\Omega_0})^n \\Omega_0 \\\\
    \\\\
    N &\\rightarrow \\infty \\\\
    \\Omega_0 &\\rightarrow d \\Omega \\\\
    k \\Omega_0 &\\rightarrow \\Omega \\\\
    \\tilde{x} &\\rightarrow x \\\\
    \\\\
    x[n] &= \\frac{1}{2 \\pi}\\int_{2 \\pi} X(r e^{j \\Omega}) \\cdot (r e^{j \\Omega})^n d \\Omega \\\\

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

    To put in terms of $z$, the above equation can be conceptualized as an integral along a contour with fixed $r$ and varying $\\omega$ along one full counterclockwise revolution.<\/p>\n\n\n\n

    $$
    \\begin{align}
    z &= r e^{j \\Omega} \\\\
    dz &= j r e^{j \\Omega} d \\Omega \\\\
    dz &= j z d \\Omega \\\\
    d \\Omega &= \\frac{1}{j z} dz \\\\
    \\\\
    x[n] &= \\frac{1}{2 \\pi} \\oint_{\\Omega = \\langle 2\\pi \\rangle} X(z) z^n \\frac{1}{j z} dz \\\\
    x[n] &= \\frac{1}{2 \\pi j} \\oint_{\\Omega = \\langle 2\\pi \\rangle} X(z) z^{n-1} dz \\\\
    \\end{align}
    $$<\/p>\n\n\n\n

    Integration along counterclockwise closed circular contour with fixed $r$ chosen from within ROC. As this approach is unusual, partial fraction expansion and transform tables are generally preferred. In closing, the Analysis and Synthesis equations for the z-Transform are then<\/p>\n\n\n\n

    $$
    \\boxed{
    \\begin{align}
    X(z) &= \\sum_{n=-\\infty}^{\\infty} x[n] z^{-n} \\\\
    x[n] &= \\frac{1}{2 \\pi j} \\oint_{\\Omega = \\langle 2\\pi \\rangle} X(z) z^{n-1} dz \\\\
    \\end{align}
    }
    $$<\/p>\n\n\n\n

    Note that the real part of $z$ in the analysis equation suggests that the summation does not always converge, therefore a region of Convergence or ROC must be defined for each transform.<\/p>\n\n\n\n

    <\/span>Region of Convergence<\/span><\/h3>\n\n\n\n

    \\begin{align}
    X(z) &= \\sum_{n=-\\infty}^{\\infty} x[n] z^{-n} \\\\
    X(r e^{j \\Omega}) &= \\sum_{n=-\\infty}^{\\infty} x[n] r^{-n} e^{-j \\Omega n} \\\\
    \\end{align}<\/p>\n\n\n\n

    Notes:<\/p>\n\n\n\n

      \n
    1. When $|z| = r = 1$, then we effectively get the Fourier Transform. Convergence in that case is determined solely by $x[n]$. Otherwise, convergence depends on both $x[n]$ and $r$.<\/li>\n\n\n\n
    2. ROC must never contain poles (doesn’t converge at those locations)<\/li>\n\n\n\n
    3. If $x[n]$ is finite duration and absolutely integrable, then the ROC is the entire $z$ plane except possibly $z=0$ and\/or $z \\rightarrow \\infty$<\/li>\n\n\n\n
    4. If $X(z)$ is rational, then ROC is bounded by poles, or extends to infinity.<\/li>\n\n\n\n
    5. If the signal is right sided and circle $|z| = r_0$ is in the ROC, then $|r_1| > |r_0|$ is also in the ROC.

      Note that $r_1^{-n}$ will converge faster than $r_0^{-n}$

      Given a right sided signal that starts at $n=N_1$, the ROC of $z$ will include infinity when $N_1 \\geq 0$ but will not include infinity when $N_1 < 0$ because then there will be some terms with positive exponent in the summation. The case when $N_1 < 0$ is further analyzed with

      $$
      \\begin{align}
      X(z) &= \\sum_{n=N_1}^{\\infty} x[n] z^{-n} &&N_1 < 0 \\\\
      X(z) &= \\sum_{n=N_1}^{-1} x[n] z^{-n} + \\sum_{n=0}^{\\infty} x[n] z^{-n} &&N_1 < 0 \\\\
      X(z) &= \\sum_{n=1}^{|N_1|} x[n] z^{n} + \\sum_{n=0}^{\\infty} x[n] z^{-n} &&N_1 < 0 \\\\
      \\end{align}
      $$

      Note that the leftmost term would then include values where $z=\\infty$ would be raised to a positive number which does not converge.
      <\/li>\n\n\n\n
    6. If the signal is left sided and circle $|z| = r_0$ is in the ROC, then all values $0 < |z| < r_0$ are also in the ROC. ROC of $z$ will include 0 when $N_2 \\leq 0$ but will not include 0 when $N_2 > 0$.

      Consider the left-sided signal where the rightmost nonzero value is at $n=N_2$. The z-Transform is

      $$
      \\begin{align}
      X(z) &= \\sum_{n=-\\infty}^{N_2} x[n] z^{-n} \\\\
      X(z) &= \\sum_{n=-\\infty}^{0} x[n] z^{-n} + \\sum_{n=1}^{N_2} x[n] z^{-n} \\\\
      \\end{align}
      $$

      ROC does not include $r=0$ because the rightmost summation would put 0 in the denominator.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

      Intent Recall we found that an input signal $x[n]$ fed into an LTI system with impulse response $h[n]$ generates an output $y[n]$ described by the convolution operation. $$\\begin{align}x[n] &\\rightarrow y[n] = x[n] * h[n] \\\\x[n] &\\rightarrow y[n] = \\sum_{k=-\\infty}^{\\infty} x[k] h[n-k] \\\\\\end{align}$$ As it can be fairly hard to conceptualize how a system will shape […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":251,"menu_order":1,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-112","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/112","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=112"}],"version-history":[{"count":170,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/112\/revisions"}],"predecessor-version":[{"id":1191,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/112\/revisions\/1191"}],"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=112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}