DFT Explorer: Probing a Signal with Rotating Phasors
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Description
The discrete Fourier transform
is a sum of \(N\) complex numbers, and this MicroSim draws every one of them. The signal \(x[n]\) (blue stems) is the sum of up to three sinusoids at whole-number bins \(m\). Each drop-down list picks a cosine (\(m = 0 \ldots N/2\)) or a sine (\(m = 1 \ldots N/2-1\)), and each slider sets its amplitude \(A\) (0 to 2). The Bin k slider chooses the probing frequency, and N sets the length to 8, 16 or 32.
The unit circle shows the phasor \(e^{-j2\pi kn/N}\) for every \(n\). It turns clockwise by \(360^\circ\,k/N\) per sample, and samples that land on the same point share a label. On screens at least 600 px wide, a panel below the signal plots its real part \(\cos(2\pi kn/N)\) and imaginary part \(-\sin(2\pi kn/N)\).
The Running sum plane places the products \(x[n]\,e^{-j2\pi kn/N}\) head to tail (green). The maroon arrow from the origin to the end of the chain is \(X[k]\), and its length is the highlighted bar of the magnitude spectrum \(|X[k]|\). When \(k\) matches a frequency in the signal, the products keep pointing the same way and add up. When it does not, they wind around into a closed loop and \(X[k] = 0\). Animate adds the products one sample at a time; Step adds one per click and starts again at \(n = 0\) after a finished sum. Clicking a bar selects that bin.
Inverse DFT rebuilds the signal from the bins you keep, \(x_r[n] = \frac{1}{N}\sum_{k\ \text{kept}} X[k]\,e^{+j2\pi kn/N}\), and draws its real part as maroon rings. Click bars to remove or restore bins. The sim shows the error \(e[n] = x[n] - \mathrm{Re}\,x_r[n]\), the fraction of the energy kept, and a warning when \(x_r[n]\) is complex.
This simulation is used in Chapter 6: Fourier Analysis Fundamentals.
Lesson Plan
Learning Objectives
After using this MicroSim, students will be able to:
- Explain how \(X[k]\) is formed by multiplying each sample \(x[n]\) by a phasor that turns \(k/N\) of a revolution per sample and adding the products
- Predict the non-zero DFT values of a sum of whole-bin sinusoids: \(AN/2\) at bins \(m\) and \(N - m\), or \(AN\) when \(m = 0\) or \(m = N/2\)
- Distinguish a matching bin, where the products line up and add, from a non-matching bin, where they close into a loop, and relate this to the orthogonality of the DFT phasors
- Interpret the phase of \(X[k]\): a cosine gives a real \(X[m]\) and a sine gives \(X[m] = -jAN/2\)
- Analyze an inverse DFT built from a subset of bins by predicting its error and energy, and explain why bins \(k\) and \(N - k\) must be kept together for a real result
Target Audience
Undergraduate students in a first digital signal processing or signals and systems course (sophomore or junior engineering, physics or computer science). Plan 25 minutes for the activities below.
Prerequisites
- Complex numbers in polar form and Euler's formula \(e^{j\theta} = \cos\theta + j\sin\theta\)
- Discrete-time sinusoids such as \(\cos(2\pi mn/N)\)
- Summation notation
Activities
Activity 1: Products that add up and products that cancel (7 min). Start from the defaults: \(N = 16\), \(x[n] = \cos(2\pi\cdot 2n/16) + 0.5\cos(2\pi\cdot 5n/16)\) and \(k = 2\). The green products march to the right and the arrow ends at \(X[2] = 8.00 + j0.00\) ("the products add up"). Move \(k\) to 3: the chain closes into a loop and the read-out shows \(X[3] = 0.00 + j0.00\) ("the products cancel"). Sweep \(k\) from 0 to 15 and list the non-zero bins: \(X[2] = X[14] = 8.00\) and \(X[5] = X[11] = 4.00\). Check them against \(AN/2\): \(1 \times 16/2 = 8\) and \(0.5 \times 16/2 = 4\).
Activity 2: One sample at a time (6 min). With \(k = 2\), press Step and watch the gold point on the unit circle move \(-45.0^\circ\) per sample. After \(n = 0\) the partial sum is \(1.50 + j0.00\), which is just \(x[0]\) because the phasor starts at 1. After \(n = 0\ldots3\) it is \(2.04 + j0.16\), after \(n = 0\ldots7\) it is \(4.50 + j0.32\), and after all 16 samples it is \(8.00 + j0.00\). The highlighted spectrum bar grows toward its dashed final height. Once the sum is complete, set \(k = 3\) and press Animate: the bar rises and then falls back to zero as the chain closes.
Activity 3: Phase, negative frequencies and special bins (6 min). Set \(A_3 = 1.00\) (component 3 is \(\sin(2\pi\cdot 3n/16)\)) and \(k = 3\). The arrow points straight down: \(X[3] = 0.00 - j8.00\), phase \(-90.0^\circ\). Move to \(k = 13\): \(X[13] = 0.00 + j8.00\), phase \(90.0^\circ\), and the circle reports "\(-292.5^\circ = +67.5^\circ\) per sample", a counter-clockwise turn: bins above \(N/2\) are negative frequencies. Set \(A_3\) back to 0 and \(A_2\) to 0, choose "cos, m = 8" for component 1 and set \(k = 8\). Now \(x[n] = (-1)^n\) and \(X[8] = 16.00\), not 8: bins \(m\) and \(N - m\) are the same bin. Choosing "cos, m = 0" with \(k = 0\) also gives \(X[0] = 16.00\).
Activity 4: Rebuilding the signal (6 min). Reload the page to restore the defaults and check Inverse DFT. With all 16 bins kept, the rings sit on the stems: max \(|e[n]| = 0.000\) and energy kept 100.0%. Click bars 5 and 11 to remove them: the rings follow \(\cos(2\pi\cdot 2n/16)\) alone, max \(|e[n]| = 0.500\) and energy kept 80.0%. Click bar 5 to restore it while bar 11 stays removed: the warning reports that \(x_r[n]\) is complex (max \(|\mathrm{Im}| = 0.250\)), max \(|e[n]| = 0.250\) and energy kept 90.0%.
Assessment
- With \(N = 16\), what are the DFT values of \(x[n] = 2\cos(2\pi\cdot 3n/16)\)? Answer: \(X[3] = X[13] = AN/2 = 2 \times 16/2 = 16\), and every other bin is 0.
- For the default signal, \(X[3] = 0\) although \(x[n]\) is not zero. Why exactly zero? Answer: multiplied by the probing phasor, each cosine becomes two phasors that make \(m - k\) and \(m + k\) whole turns over the \(N\) samples, and \(\sum_n e^{j2\pi qn/N} = 0\) for any whole \(q\) that is not a multiple of \(N\). For \(k = 3\) the turn counts are \(-1\) and 5 (from \(m = 2\)) and 2 and 8 (from \(m = 5\)), so every part sums to zero and the chain returns to the origin.
- Why does \(x[n] = \cos(\pi n)\) with \(N = 16\) give \(X[8] = 16\) instead of \(N/2 = 8\)? Answer: here \(m = N/2 = 8\), so bins \(m\) and \(N - m\) are the same bin and both halves of the cosine land in it: \(X[8] = AN = 16\).
- What is \(X[3]\) for \(x[n] = \sin(2\pi\cdot 3n/16)\), and what does its phase mean? Answer: \(X[3] = -j8\), phase \(-90^\circ\): a sine is a cosine delayed by a quarter cycle. The mirror bin is \(X[13] = +j8\).
- In Inverse DFT mode with the default signal, you remove bins 2 and 14 and keep the rest. Predict the largest error and the energy kept. Answer: \(x_r[n] = 0.5\cos(2\pi\cdot 5n/16)\) and \(e[n] = \cos(2\pi\cdot 2n/16)\), so max \(|e[n]| = 1.000\); the energy kept is \((4^2 + 4^2)/(8^2 + 8^2 + 4^2 + 4^2) = 32/160 = 20.0\%\).
Common Misconceptions
"A cosine at bin \(m\) makes one spike at \(m\)." The spectrum always shows two bars, at \(m\) and \(N - m\), each \(AN/2\) tall. At \(k = N - m\) the unit circle turns counter-clockwise, which is the negative-frequency half of \(\cos\theta = (e^{j\theta} + e^{-j\theta})/2\). Removing only one of the pair in Inverse DFT mode makes \(x_r[n]\) complex, and the sim warns you.
"If \(k\) does not match, \(X[k]\) is small but not zero." For whole-bin sinusoids the chain closes exactly and the read-out shows \(0.00 + j0.00\). Non-zero values between peaks (spectral leakage) appear only when a frequency falls between bins, which this MicroSim does not generate.
References
- Discrete Fourier transform - Wikipedia - definition of \(X[k]\) and the inverse DFT, with the orthogonality of the basis phasors
- Root of unity - Wikipedia - the points \(e^{-j2\pi kn/N}\) on the unit circle are \(N\)-th roots of unity
- Euler's formula - Wikipedia - why a cosine is the sum of two counter-rotating phasors
- Negative frequency - Wikipedia - why bins above \(N/2\) mirror the positive frequencies of a real signal
- Parseval's theorem - Wikipedia - the energy-kept read-out in Inverse DFT mode
- The Scientist and Engineer's Guide to Digital Signal Processing - Steven W. Smith - free online textbook with chapters on the DFT and its properties