# C Fourier Transforms

## Continuous-Time Fourier Transforms

The Fourier transform is an alternative representation of a signal. It describes the frequency content of the signal. It is defined as \[\begin{equation} G(f) \overset{\text{def}}{=} \int_{-\infty}^\infty g(t)e^{-j2\pi f t}\,dt. \tag{C.1} \end{equation}\] Notice that it is a function of frequency \(f\), rather than time \(t\). Notice also that it is complex-valued, since its definition involves the imaginary number \(j \overset{\text{def}}{=} \sqrt{-1}\).

The following video explains the visual intuition behind the Fourier transform. Note that this video uses \(i\) to denote the imaginary number \(\sqrt{-1}\), whereas we use \(j\) (which is common in electrical engineering, to avoid confusion with current).

To calculate the Fourier transform of a signal, we will rarely use (C.1). Instead, we will look up the Fourier transform of the signal in a table like Appendix D.1 and use properties of the Fourier transform (as shown in Appendix D.3).

**Example C.1**Calculate the Fourier transform of \(g(t) = 80e^{-20t} u(t)\).

*Solution. * This signal is most similar to \(e^{-t} u(t)\) in Appendix D.1, whose Fourier transform is
\(\frac{1}{1 + j2\pi f}\). The only differences are:

- Our signal has an extra factor of 80 in front. This factor comes along for the ride by linearity of the Fourier transform.
- Our signal is time-scaled by a factor of 20, so we have to apply the scaling property.
Since we
*multiplied*by 20 in the time domain, we have to*divide*by 20 in the frequency domain, on both the*inside*and the*outside*.

Putting everything together, we see that the Fourier transform is: \[ G(f) = 80 \frac{1}{20} \frac{1}{1 + j 2\pi \frac{f}{20}}. \]

This is a complex-valued function. That is, at each frequency \(f\), \(G(f)\) is a complex number, with a real and an imaginary component. To make it easier to visualize, we calculate its magnitude using Theorem B.1. \[\begin{align*} |G(f)| = \sqrt{|G(f)|^2} &= \sqrt{G(f) \cdot G^*(f)} \\ &= \sqrt{\frac{4}{1 + j2\pi \frac{f}{20}} \cdot \frac{4}{1 - j2\pi \frac{f}{20}}} \\ &= \sqrt{\frac{4}{1 + (2\pi \frac{f}{20})^2}} \end{align*}\]

Now, the*magnitude*of the Fourier transform is a function we can easily visualize. From the graph, we see that the signal has more low-frequency content than high-frequency content.

## Discrete-Time Fourier Transforms

Discrete-time signals are obtained by sampling a continuous-time signal at regular time intervals. The sampling rate \(f_s\) (in Hz) specifies the number of samples per second. The signal below is sampled at a rate of \(f_s = 16\) Hz.

We write \(x[n] = x(t_n)\) for the \(n\)th time sample, where \(t_n = \frac{n}{f_s}\).

It’s possible to choose a sampling rate \(f_s\) that is too low. In the graph below, the underlying continuous-time signal (in black) is a sinusoid with a frequency of 14 Hz. However, because the discrete-time signal (in red) was sampled at a rate of 16 Hz, the sinusoid appears to have a frequency of 2 Hz.

We say that the higher frequency is **aliased** by the lower one.

At a sampling rate of \(f_s\), any frequencies outside the range
\[ (-f_s / 2, f_s / 2) \]
will be aliased by a frequency inside this range. This “maximum frequency”
of \(f_s / 2\) is known as the **Nyquist limit**.

Therefore, the Fourier transform of a discrete-time signal is effectively only defined for frequencies from \(-f_s/2\) to \(f_s/2\). Contrast this with the Fourier transform of a continuous-time signal, which is defined on all frequencies from \(-\infty\) to \(\infty\).

Aliasing is not just a theoretical problem. For example, if helicopter blades are spinning too fast for the frame rate of a video, then they can look like they are not spinning at all!

When calculating the Fourier transform of a discrete-time signal, we typically work with
*normalized* frequencies. That is, the frequencies are in units of “cycles per *sample*” instead of
“cycles per *second*”. (Another way to think about this is that we assume all discrete-time signals are
sampled at 1 Hz.) As a result, the Nyquist limit is \(1/2\), so the **Discrete-Time Fourier Transform**
is defined only for \(|f| < 0.5\).

\[\begin{align} G(f) &\overset{\text{def}}{=} \sum_{n=-\infty}^\infty g[n] e^{-j2\pi f t}\,dt & -0.5 < f < 0.5 \tag{C.2} \end{align}\]

To calculate the Fourier transform of a signal, we will rarely use (C.2). Instead, we will look up the Fourier transform of the signal in a table like Appendix D.2 and use properties of the Fourier transform (as shown in Appendix D.3).

## Essential Practice

Calculate the Fourier transform \(G(f)\) of the continuous-time signal \[ g(t) = \begin{cases} 1/3 & 0 < t < 3 \\ 0 & \text{otherwise} \end{cases}. \] Graph its magnitude \(|G(f)|\) as a function of frequency.

Calculate the Fourier transform \(G(f)\) of the discrete-time signal \[ g[n] = \delta[n] + 0.4 \delta[n-1] + 0.4 \delta[n+1]. \] Graph \(G(f)\). (The Fourier transform should be a real-valued function, so you do not need to take its magnitude.)

Which of the following terms fill in the blank? The Fourier transform \(G(f)\) of a real-valued time signal \(g(t)\) is necessarily ….

- real-valued: \(G(f) \in \mathbb{R}\)
- positive-valued: \(G(f) > 0\)
- symmetric: \(G(-f) = G(f)\)
- conjugate symmetric: \(G(-f) = G^*(f)\)

(

*Hint:*Write out \(G(f)\) and \(G(-f)\) according to the definition (C.1), and use Euler’s identity (Theorem B.2). Simplify as much as possible, using the facts that \(\cos(-\theta) = \cos(\theta)\) and \(\sin(-\theta) = -\sin(\theta)\).)Which of the following terms fill in the blank? The Fourier transform \(G(f)\) of a real-valued and

*symmetric*time signal \(g(t)\) is necessarily ….- real-valued: \(G(f) \in \mathbb{R}\)
- positive-valued: \(G(f) > 0\)
- symmetric: \(G(-f) = G(f)\)
- conjugate symmetric: \(G(-f) = G^*(f)\)