Unveiling the Mysteries of Fourier Transform: A Practical Series — Part 2

Welcome back to Part 2 of our journey into the world of Fourier transforms. If you have just joined us, consider catching up with Part 1, where we explored the Fourier transform of a complex signal. Today, we move into signal filtering, with a specific focus on a filter characterized by a real and even frequency response, $H(s)$.

The Puzzle: Filters and Frequencies

The question for today is: given a filter with a real and even transfer function $H(s)$, what happens when the input to this filter is

\[\cos(2\pi a t)?\]

The key is to combine the symmetry property of $H(s)$ with the Fourier transform of the cosine function.

Throughout this post, we use the Fourier transform convention

\[\mathcal{F}\{x(t)\}(s) = \int_{-\infty}^{\infty} x(t)e^{-2\pi i s t}\,dt.\]

With this convention, multiplication by a frequency response in the frequency domain corresponds to filtering in the time domain.

The Fourier Transform of Cosine: A Brief Detour

Before solving the filtering question, let us revisit the Fourier transform of $\cos(2\pi a t)$. Using Euler’s formula,

\[\cos(2\pi a t) = \frac{1}{2}\left(e^{2\pi i a t} + e^{-2\pi i a t}\right),\]

so its Fourier transform is

\[\mathcal{F}\{\cos(2\pi a t)\}(s) = \frac{1}{2}\left(\delta(s-a) + \delta(s+a)\right). \tag{1}\]

Equivalently, this may be written as

\[\frac{1}{2}\left(\delta(a-s) + \delta(a+s)\right),\]

since the Dirac delta is even in its argument.

Filtering the Signal

Let the input spectrum be $F(s)$. For a linear time-invariant filter with frequency response $H(s)$, the output spectrum is obtained by multiplying the input spectrum by the filter response:

\[Y(s) = H(s)F(s). \tag{2}\]

For the cosine input, substitute equation (1):

\[Y(s) = H(s)\cdot \frac{1}{2}\left(\delta(s-a) + \delta(s+a)\right).\]

Using the sampling property of the Dirac delta,

\[H(s)\delta(s-a) = H(a)\delta(s-a),\]

and

\[H(s)\delta(s+a) = H(-a)\delta(s+a).\]

Therefore,

\[Y(s) = \frac{1}{2}\left(H(a)\delta(s-a) + H(-a)\delta(s+a)\right).\]

Because $H(s)$ is even, $H(a)=H(-a)$. Thus the filtered output spectrum becomes

\[Y(s) = \frac{H(a)}{2}\left(\delta(s-a) + \delta(s+a)\right). \tag{3}\]

The Final Revelation

To recover the time-domain signal, take the inverse Fourier transform of $Y(s)$:

\[\mathcal{F}^{-1}\{Y(s)\} = H(a)\cos(2\pi a t). \tag{4}\]

So the output of the filter, when the input is $\cos(2\pi a t)$, is

\[H(a)\cos(2\pi a t).\]

In other words, a cosine at frequency $a$ remains a cosine at the same frequency. The filter simply scales its amplitude by the value of the frequency response at that frequency.

Concluding Thoughts: Filters and Fourier Unite

This result is simple but powerful: sinusoids are eigenfunctions of linear time-invariant systems. When a cosine passes through such a filter, the frequency does not change; only the amplitude, and in the more general complex case the phase, is modified by the filter response.

For a real and even response $H(s)$, the positive and negative frequency components are scaled equally, which is why the output remains the real cosine $H(a)\cos(2\pi a t)$.

This idea is one of the foundations of signal processing, telecommunications, and many practical filtering problems: understand the signal in the frequency domain, understand the filter response in the frequency domain, and the output follows directly.

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