#27 Periodic extension, transform samples and the inverse-transform limit

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. From periodic to aperiodic

   

   Recall Fourier series for periodic signals.

   Consider signals that do not repeat periodically.

   Examples include a single pulse and a decaying exponential.

   A Fourier series represents periodic repetition.

   Extend frequency analysis to aperiodic signals.

   Derive the transform by increasing the repetition period without bound.

   Introduce the FT and inverse FT pair; their convergence conditions still apply.

   Narration transcript

   In the previous videos, we explored Fourier Series and its eleven properties for periodic signals. But here's a natural question: what about signals that are NOT periodic? Aperiodic signals, like a single rectangular pulse or a decaying exponential, don't repeat forever. So we can't directly apply Fourier Series to them. In this video, we'll bridge that gap. We'll derive the Continuous-Time Fourier Transform by starting from Fourier Series and taking the period to infinity. By the end, you'll have the FT and Inverse FT formulas that work for any aperiodic signal.

2. 2. Repeat a finite-duration signal

   

   Begin with a finite-duration signal.

   Choose a full period containing its support.

   Repeat the signal every T seconds without overlap.

   Write the periodic extension as p:

   $\displaystyle p\left(t\right)=\sum _{k=-\infty }^{\infty }x\left(t-k\cdot T\right)$

   This tiles copies along the time axis.

   Inside the selected central period, the periodic extension equals the original.

   Apply Fourier series to the periodic extension.

   Write its coefficients as C:

   $\displaystyle C\left[k\right]=\left(\frac{1}{T}\right)\int _{-T/2}^{T/2}p\left(t\right)e^{-jk\omega _0t}\mathrm{d}t$

   Now use the finite support of the original signal.

   Since the original vanishes outside this interval, extend its integral limits.

   The original-signal integral is:

   $\displaystyle C\left[k\right]=\left(\frac{1}{T}\right)\int _{-\infty }^{\infty }x\left(t\right)e^{-jk\omega _0t}\mathrm{d}t$

   Narration transcript

   Here's the key idea. Suppose we have an aperiodic signal x of t, maybe a pulse that exists only for a finite duration. We can artificially make it periodic by repeating copies every T seconds. This creates a periodic signal x-tilde of t, defined as the sum from k equals negative infinity to positive infinity of x of t minus k T. Think of it as tiling x of t along the time axis. One period of x-tilde is exactly our original x of t. Since x-tilde is periodic with period T, we CAN apply Fourier Series to it. The FS coefficients are: x-tilde of k equals one over T times the integral over one period of x-tilde of t times e to the minus j k two pi over T t, dt. Now here's the crucial step. Since one period of x-tilde is just x of t, and x of t is zero outside that interval, we can extend the integration limits to negative infinity to positive infinity. So x-tilde of k equals one over T times the integral from negative infinity to positive infinity of x of t, e to the minus j k omega-zero t, dt.

3. 3. Define the Fourier transform

   

   Read a continuous-frequency function from the coefficient integral.

   Multiply both sides by the period.

   The scaled coefficients are:

   $\displaystyle T\cdot C\left[k\right]=\int _{-\infty }^{\infty }x\left(t\right)e^{-jk\omega _0t}\mathrm{d}t$

   Sample at harmonic angular frequencies:

   $\displaystyle \omega =k\omega _{0}$

   Define the standard forward transform:

   $\displaystyle X\left(\omega \right)=\int _{-\infty }^{\infty }x\left(t\right)e^{-j\omega t}\mathrm{d}t$

   This transforms the original aperiodic signal under the stated conditions.

   The transform samples and periodic coefficients satisfy:

   $\displaystyle X\left(k\omega _{0}\right)=T\cdot C\left[k\right]$

   Divide each transform sample by the period:

   $\displaystyle C\left[k\right]=\frac{X\left(k\omega _{0}\right)}{T}$

   Narration transcript

   Now let's extract something powerful from this expression. Multiply both sides by T. We get: T times x-tilde of k equals the integral from negative infinity to infinity of x of t, e to the minus j k omega-zero t, dt. The right side depends on the continuous variable omega, evaluated at discrete points omega equals k omega-zero. So we define a new function: X of omega equals the integral from negative infinity to infinity of x of t, e to the minus j omega t, dt. This is the Fourier Transform of the aperiodic signal x of t. Notice the key relationship: X of omega, evaluated at omega equals k omega-zero, gives us T times the FS coefficients of the periodic extension. In other words, the FS coefficients are samples of X of omega, divided by T: x-tilde of k equals X of k omega-zero, divided by T.

4. 4. Take the inverse-transform limit

   

   Derive the inverse transform.

   Start with periodic synthesis:

   $\displaystyle p\left(t\right)=\sum _{k=-\infty }^{\infty }C\left[k\right]e^{jk\omega _0t}$

   Substitute the sampled transform and frequency width:

   $\displaystyle p\left(t\right)=\left(\frac{1}{2\pi }\right)\sum _{k=-\infty }^{\infty }\omega _{0}X\left(k\omega _{0}\right)e^{jk\omega _0t}$

   Inspect the frequency-width factor in this sum.

   It has the form of a Riemann sum.

   As the period grows, the frequency spacing tends to zero:

   $\displaystyle \omega _{0}=\frac{2\pi }{T}$

   The grid becomes dense and the weighted sum tends to an integral when justified.

   The standard inverse formula is:

   $\displaystyle x\left(t\right)=\left(\frac{1}{2\pi }\right)\int _{-\infty }^{\infty }X\left(\omega \right)e^{j\omega t}\mathrm{d}\omega$

   This is the inverse Fourier transform, with the appropriate convergence interpretation.

   Narration transcript

   Now let's derive the inverse transform. The Fourier Series synthesis equation for x-tilde is: x-tilde of t equals the sum over k of x-tilde of k, times e to the j k omega-zero t. Substituting x-tilde of k equals X of k omega-zero over T, and using T equals two pi over omega-zero, we get: x-tilde of t equals the sum over k of omega-zero over two pi, times X of k omega-zero, times e to the j k omega-zero t. Look carefully. This is a Riemann sum! As T goes to infinity, omega-zero, which equals two pi over T, approaches zero. The discrete frequencies k omega-zero become the continuous variable omega, and the sum becomes an integral. In the limit, x-tilde of t approaches x of t, and we obtain: x of t equals one over two pi, times the integral from negative infinity to infinity of X of omega, e to the j omega t, d-omega. This is the Inverse Fourier Transform.

5. 5. Transform conventions

   

   Collect the transform pair.

   Forward transform under the stated conditions:

   $\displaystyle X\left(\omega \right)=\int _{-\infty }^{\infty }x\left(t\right)e^{-j\omega t}\mathrm{d}t$

   Inverse transform under the stated conditions:

   $\displaystyle x\left(t\right)=\left(\frac{1}{2\pi }\right)\int _{-\infty }^{\infty }X\left(\omega \right)e^{j\omega t}\mathrm{d}\omega$

   Observe the near symmetry of the two integrals.

   Their exponent signs are opposite; the inverse has one divided by two pi.

   For general angular-frequency conventions:

   $\displaystyle c_{1}\cdot c_{2}=\frac{1}{2\pi }$

   The standard convention is:

   $\displaystyle c_{1}=1, c_{2}=\frac{1}{2\pi }$

   The symmetric convention is:

   $\displaystyle c_{1}=c_{2}=1/\sqrt{2\pi }$

   The transform extends frequency analysis to aperiodic signals with suitable interpretations.

   Next: convergence conditions and the first transform examples.

   Narration transcript

   Let's put it all together. For an aperiodic signal x of t, the Fourier Transform pair is: Forward transform: X of omega equals the integral from minus infinity to infinity of x of t, e to the minus j omega t, dt. Inverse transform: x of t equals one over two pi, times the integral from minus infinity to infinity of X of omega, e to the plus j omega t, d-omega. Notice the beautiful near-symmetry. Both are integrals of the same form, but with opposite exponent signs and a factor of one over two pi in the inverse. In general, we can define the pair with constants c-one and c-two, where c-one times c-two must equal one over two pi. Our convention uses c-one equals 1 and c-two equals one over two pi. Another popular choice is the symmetric form, where both constants equal one over the square root of two pi. Regardless of convention, the Fourier Transform extends frequency analysis from periodic signals to aperiodic signals. In the next video, we'll discuss convergence conditions and compute our first Fourier Transform examples.

Source video: Signals & Systems #27 | CT Fourier Transform - From FS to FT (5:55)