<aside> 🐳 We can’t use Fourier series on aperiodic signals

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Continuous-Time Fourier Transform (CTFT)

CTFT takes a continuous time signal $x(t)$ and transforms it into a complex-valued function $X(j \omega)$, which represents the signal’s frequency spectrum

$$ X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt. $$

• We multiply $x(t)$ by $e^{-j\omega t}$: how much overlap there is between $x(t)$ and that oscillation at that frequency
  • If there is large overlap, $X(t)$ is very large, big contribution of that frequency to the overall of x(t)
  • No overlap means $e^{-j\omega t}$ is not a good approximation and integrates to nearly zero
  • $e^{-j\omega t}$ can be though of as a test signal at frequency $\omega$, by multiplying $x(t)$ with this complex exponential, we are asking: how much of the frequency $\omega$ is present in our signal $x(t)$
• Magnitude and phase of coefficients allows us to plot it

Inverse CTFT takes the complex-valued function $X(j \omega)$ and reconstructs the original time-domain signal $x(t)$

$$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega)e^{j\omega t} d\omega. $$

Gaussian distribution

CTFT of a Gaussian of width $\sigma$ is a Gaussian of width $\frac{1}{\omega}$

• If $x$ is very concentrated in the time domain, the spectum $X$ will be very spread out in the frequency domain

Existence of the CTFT

If $x$ has finite energy, then:

1. $X$ exists and has finite energy

2. Signal and its spectrum satisfy Parseval’s relation

   $$ \int _{- \infty} ^{\infty} |x(t)|^2 \text{dt} = \frac{1}{2\pi} \int _{- \infty} ^{\infty} |X(j \omega)|^2 d \omega $$

Properties of CTFT

Discrete-Time Fourier Transform (DTFT)

For a DT signal $x$ defined pointwise by the following, we call $X$ the Fourier transform or spectrum of $x$

$$ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n]e^{-j\omega n} $$