# Silly Interpretation of the Fourier Transform

$x[n] \longrightarrow\boxed{H}\longrightarrow y[n]$

Signals and systems courses typically explain the response of an LTI system $$H$$ to an input $$x[n]$$, producing output $$y[n]$$, by introducing the impulse response and using the system’s properties of linearity and time invariance.

The system’s response to a delta $$\delta[n]$$ is defined as the impulse response $$h[n]$$:

$\delta[n] \longrightarrow\boxed{H}\longrightarrow h[n]$

We can write down the by-definition properties of linearity and time-invariance (LTI) of an LTI system:

$\text{Linearity and Time Invariance}$ $ax_1[n] + bx_2[n] \longrightarrow\boxed{H}\longrightarrow ay_1[n] + by_2[n]$ $x[n-T] \longrightarrow\boxed{H}\longrightarrow y[n-T]$

Which implies:

$c \delta[n-T] \longrightarrow\boxed{H}\longrightarrow c h[n-T]$

where c is some constant.

If we decompose an arbitrary input $$x[n]$$ as a sum of scaled and shifted deltas:

$x[n] = \dots + x[-1]\delta[n+1] + x[0]\delta[n] + x[1]\delta[n-1] + x[2]\delta[n-2] + \dots$

We can use the above properties to express the response of the system, $$y[n]$$, as a sum of scaled and shifted impulse responses:

$x[n] \longrightarrow\boxed{H}\longrightarrow y[n]$ $\dots x[0]\delta[n] + x[1]\delta[n-1] \dots \longrightarrow\boxed{H}\longrightarrow \dots x[0]h[n] + x[1]h[n-1] \dots$

Which yields the familiar convolution expression:

$y[n] = \dots + x[-1]h[n+1] + x[0]h[n] + x[1]h[n-1] + x[2]h[n-2] + \dots$ $y[n] = (x * h)[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k]$

If you take things in the limit, the discrete-time delta $$\delta[n]$$ turns into a continuous-time delta $$\delta(t)$$, the sum $$\sum$$ into an integral $$\smallint$$, and you have the continuous-time version of the convolution operation: $$y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau$$.

Okay. The silly thing I realized, is that this same $$x[n]$$ to sum of deltas decomposition is a perfectly valid way to interpret the DTFT or CTFT!

If you start out with the Fourier transform pair of a delta and a constant, $$\mathcal{F} \{ \delta[n] \} = 1$$:

$\delta[n] \Leftrightarrow 1$

Invoke linearity and time-shifting properties of the Fourier transform:

$a x_1[n] + b x_2[n] \Leftrightarrow a \mathcal{F} \{ x_1[n] \} + b \mathcal{F} \{ x_2[n] \}$ $x[n-T] \Leftrightarrow e^{-jwT} \mathcal{F} \{ x[n] \}$

Which implies:

$c \delta[n-T] \Leftrightarrow c e^{-jwT}$

And use these properties on $$x[n]$$ decomposed as a sum of scaled and shifted deltas:

$\dots x[0]\delta[n] + x[1]\delta[n-1] \dots \Leftrightarrow \dots x[0] e^{-jw0} + x[1] e^{-jw1} \dots$

You get the Fourier transform itself!

$\mathcal{F} \{ x[n] \} = \dots + x[-1] e^{jw1} + x[0] e^{-jw0} + x[1] e^{-jw1} + x[2] e^{-jw2} + \dots$ $\mathcal{F} \{ x[n] \} = X(e^{jw}) = \sum_{n=-\infty}^{\infty} x[n] e^{-jwn}$

Which means that the Fourier transform can be interpreted as a sum of all-frequency ( $$-\infty$$ to $$\infty$$ ) constants with different linear phase shifts $$e^{-jwn}$$, which ultimately when summed, constructively or destructively, yield the possibly bounded-in-frequency Fourier transform of the sequence.

As before, if you take things in the limit, turning discrete-time deltas into continuous-time deltas, and sums into integrals, you’ll get the same thing in continuous-time case: $$X(e^{jw}) = \mathcal{F} { x(t) } = \int_{-\infty}^{\infty} x(t) e^{-jwt} dt$$.

I call this interpretation silly because it relies on already knowing the transform pair $$\delta[n] \Leftrightarrow 1$$, which is on its own very important and not obvious. But knowing that, it’s been hard for me the last two days not to think of the Fourier transform as just a sum of these linearly phase-shifted constants.