#22 Signals & Systems #22 | Fourier Coefficients - Analysis Equations & Worked Example

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

1. 1. Fix the series convention

   Corrected solution notes. The audio is the original recording and may contain errors or incomplete phrases. Where they differ, use the corrected written solution.

   Write the real series as a0 plus positive-index cosine and sine terms.

   Then a0 is the mean itself. State convergence assumptions for ordinary signals; impulse trains are handled as periodic distributions.

   Narration transcript

   In our previous lesson, we introduced the Fourier Series synthesis equation: x of t equals a zero plus the sum of a n cosine n omega zero t plus b n sine n omega zero t. We saw that any periodic signal can be built from harmonically related sinusoids. But how do we actually find these coefficients? Today, we derive the analysis equations.

2. 2. Analysis coefficients

   Over any complete period T, a0 is the integral of x divided by T.

   For k≥1, ak and bk are 2/T times the integrals against cos(kω0t) and sin(kω0t), with ω0=2π/T.

   Narration transcript

   Here are the Fourier Series analysis equations. The DC component a zero equals one over T times the integral of x of t over one full period. For the cosine coefficients, a k equals two over T times the integral of x of t times cosine of k omega zero t. For the sine coefficients, b k equals two over T times the integral of x of t times sine of k omega zero t. These integrals exploit the orthogonality of sinusoids: when you multiply a signal by a specific harmonic and integrate, only that harmonic's contribution survives.

3. 3. Complex convention

   X[k] is 1/T times the integral of x(t)exp(−jkω0t) over one period. Synthesis uses exp(+jkω0t).

   For k>0, X[k]=(ak−jbk)/2 and X[−k]=(ak+jbk)/2; X[0]=a0 for real-valued signals.

   Narration transcript

   We can also write the Fourier Series using complex exponentials. The synthesis equation becomes x of t equals the sum from k equals minus infinity to infinity of X k times e to the j k omega zero t. The analysis equation is X k equals one over T times the integral of x of t times e to the minus j k omega zero t. The complex coefficient X k encodes both amplitude and phase. The relationship is: X k equals one-half times a k minus j b k for k not equal to zero, and X zero equals a zero. This complex form is more compact and widely used in engineering.

4. 4. Two impulses per period

   Repeat unit impulses at t=−1 and t=1 every eight seconds.

   The distribution is even and has period eight; use one complete period containing each impulse once.

   Narration transcript

   Let's apply these formulas to a concrete example. Consider a signal whose one period consists of two impulses: delta of t plus one, plus delta of t minus one, with fundamental period T equals eight. This periodic signal has impulses at t equals plus and minus one, repeating every eight seconds. Since the signal is symmetric about t equals zero, it is an even function. Let's compute its Fourier coefficients.

5. 5. Compute by sifting

   $\displaystyle a0=1/4.$

   For k≥1, ak=(1/2)cos(kπ/4), and bk=0.

   Complex coefficients are X[k]=(1/4)cos(kπ/4) for every integer k.

   Narration transcript

   Starting with a zero, we integrate the signal over one period and divide by T. Applying the sifting property of the delta function, a zero equals one-eighth times two, which gives one-quarter. For a k, we integrate x of t times cosine k omega zero t. The sifting property extracts the cosine values at t equals minus one and t equals plus one. Since cosine is even, both terms contribute equally, giving a k equals one-half times cosine of k times two pi over eight. For b k, the sine terms cancel because the signal is even, giving b k equals zero for all k. This is a general rule: even signals have zero sine coefficients.

6. 6. Correct the spectral behavior

   These coefficients oscillate periodically with index k; they do not decay toward zero.

   For instance X[8m]=1/4 for every integer m. Plot signed coefficients separately from their nonnegative magnitudes.

   This impulse distribution does not satisfy the usual coefficient-decay conclusion for integrable ordinary functions.

   Narration transcript

   Now let's visualize these coefficients. Plotting a k versus k, we see a cosine-shaped envelope. The DC component a zero equals one-quarter. The coefficients decrease and oscillate as k increases. This plot is called the amplitude spectrum. It tells us how much of each harmonic frequency is present in our signal. In the complex form, X k equals one-quarter times cosine of k pi over four, which gives identical results.

7. 7. Review

   

   Keep DC and positive/negative harmonic conventions distinct.

   Evenness eliminates sine coefficients, but it does not imply that the remaining coefficients decay. The complete answer here is a nondecaying cosine sequence.

   Narration transcript

   In this lesson, we derived the Fourier Series analysis equations for finding coefficients from a periodic signal. We learned the trigonometric form with a k and b k integrals, and the compact complex exponential form with X k. We applied these formulas to compute the coefficients of a periodic impulse signal. Key insight: even signals have zero b k coefficients, and the amplitude spectrum reveals the frequency content of any periodic signal. This completes our introduction to Fourier Series analysis.

Source video: Signals & Systems #22 | Fourier Coefficients - Analysis Equations & Worked Example (4:36)