Even and Odd Signal Decomposition
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Description
Any signal can be written as the sum of an even signal and an odd signal:
Three plots share one time axis from \(-5\) s to \(+5\) s:
- Top (blue): the signal \(x(t)\), with the purple dashed sum \(x_e(t) + x_o(t)\) lying exactly on top of it.
- Middle (green): the even part \(x_e(t)\), a mirror image about \(t = 0\).
- Bottom (red): the odd part \(x_o(t)\), which changes sign when reflected about \(t = 0\).
Dashed vertical lines mark the probe time \(t\) (filled dot) and its mirror \(-t\) (open dot). The panel below substitutes \(x(t)\) and \(x(-t)\) into the formulas and reports how the energy divides between the two parts, \(E = E_e + E_o\), over the displayed 10 s window.
Controls: Signal selects a preset (pulse, triangle, sawtooth, one-sided exponential) or a custom signal. Time shift \(t_0\) slides a preset. Probe time \(t\) picks the instant in the worked calculation. Show sum toggles the check curve. Drag in the top plot to draw \(x(t)\); Clear starts from zero.
Signals are sampled every 0.01 s, and a jump takes its midpoint value, so \(u(0) = 0.5\) in the exponential.
This simulation is used in Chapter 2: Introduction to Signals and Systems.
Lesson Plan
Learning Objectives
After using this MicroSim, students will be able to:
- Explain even and odd symmetry in terms of reflection about \(t = 0\).
- Apply the decomposition formulas to compute \(x_e(t)\) and \(x_o(t)\) at a given time.
- Verify that the even and odd parts sum to the original signal.
- Recognize the even and odd parts of pulses, triangles, sawtooth waves and one-sided exponentials.
Target Audience
College undergraduates in an introductory signals and systems course. Plan 15-20 minutes.
Prerequisites
- Even and odd functions from algebra or precalculus
- Time reversal \(x(-t)\) and time shifting \(x(t - t_0)\)
Activities
Activity 1: Symmetric signals (5 min). Select Pulse and set the time shift to 0. The odd part is zero everywhere and the panel reports 100% even. Now select Sawtooth and set the time shift to 0. The even part is zero and the panel reports 100% odd. Explain both results from the shape of the signal.
Activity 2: Hand calculation (5 min). Predict first, then read the panel.
- Pulse, time shift 1 s, probe time 1 s: \(x(1) = 1\) and \(x(-1) = 0\), so \(x_e(1) = 0.500\) and \(x_o(1) = 0.500\). Move the probe to \(-1\) s: \(x_e(-1) = 0.500\) and \(x_o(-1) = -0.500\).
- Exponential, time shift 0 s, probe time 1 s: \(x(1) = e^{-1} = 0.368\) and \(x(-1) = 0\), so \(x_e(1) = x_o(1) = 0.184\). At probe time \(-1\) s the even part is unchanged and the odd part flips to \(-0.184\).
Activity 3: Slide a pulse away from the origin (5 min). Select Pulse and watch the energy line while you move the time shift from 0 s upward. The split is 100% / 0% at 0 s, 75% / 25% at 0.5 s, and 50% / 50% at 1 s and beyond. Explain why it stops changing at 1 s.
Activity 4: Draw your own (5 min). Press Clear, then drag in the top plot to draw a signal that is zero for \(t < 0\) (for example a ramp or a bump on the right half only). The panel should report close to 50% even and 50% odd, and the even and odd parts should be equal for \(t > 0\) and opposite for \(t < 0\). Then clear again and draw a rising ramp that passes through the origin, from about \((-2, -1)\) to \((2, +1)\). The even part stays near zero and the signal is almost entirely odd.
Assessment
- Triangle, time shift 1 s, probe time 0.5 s: \(x(0.5) = 0.75\) and \(x(-0.5) = 0.25\). What are \(x_e(0.5)\) and \(x_o(0.5)\)? Answer: \(x_e = (0.75 + 0.25)/2 = 0.5\) and \(x_o = (0.75 - 0.25)/2 = 0.25\).
- Why is \(x_o(0) = 0\) for every signal, and what is \(x_e(0)\)? Answer: at \(t = 0\) the two samples \(x(t)\) and \(x(-t)\) are the same sample, so \(x_o(0) = [x(0) - x(0)]/2 = 0\) and \(x_e(0) = x(0)\).
- A signal is zero for \(t < 0\) and has total energy 8 J. How is the energy divided between \(x_e\) and \(x_o\)? Answer: 4 J each. For \(t > 0\) both parts equal \(x(t)/2\), and each part has the same energy on the negative side by symmetry, so \(E_e = E_o = E/2\).
- Write \(x_e(t)\) and \(x_o(t)\) for \(x(t) = e^{-t}u(t)\). Answer: \(x_e(t) = \tfrac{1}{2}e^{-|t|}\) and \(x_o(t) = \tfrac{1}{2}\operatorname{sgn}(t)\,e^{-|t|}\), that is \(\tfrac{1}{2}e^{-t}\) for \(t > 0\) and \(-\tfrac{1}{2}e^{t}\) for \(t < 0\).
- Can a nonzero signal be both even and odd? Answer: No. If \(x(t) = x(-t)\) and \(x(t) = -x(-t)\) then \(x(t) = -x(t)\), so \(x(t) = 0\) everywhere. This is also why the decomposition is unique.
Common Misconceptions
"Every signal is either even or odd." Most signals are neither. Select the pulse with a time shift of 1 s: both parts are nonzero and the panel reports 50% even and 50% odd.
"The even part is the part of the signal for \(t > 0\)." Both parts exist for all \(t\). With the pulse at a 1 s shift, \(x(-1) = 0\), yet \(x_e(-1) = 0.5\) and \(x_o(-1) = -0.5\). Each part borrows from the signal at \(-t\).
"Shifting a signal does not change its even and odd parts." Compare the pulse at shifts of 0 s and 1 s: the odd part goes from zero to a square wave of height 0.5 (negative for \(t < 0\), positive for \(t > 0\)), and the energy split moves from 100% / 0% to 50% / 50%.
References
- Even and odd functions - Wikipedia - definitions of even and odd symmetry and the decomposition of a general function into an even and an odd part
- Energy (signal processing) - Wikipedia - signal energy, used by the energy split in the read-out panel
- Fourier series - Wikipedia - the even part of a periodic signal needs only cosine terms and the odd part only sine terms
- The Scientist and Engineer's Guide to Digital Signal Processing - Steven W. Smith - free online textbook with a treatment of signal symmetry in the Fourier transform chapters
- Chapter 2: Introduction to Signals and Systems - this textbook - the chapter section that uses this MicroSim
- Oppenheim, A. V., Willsky, A. S., & Nawab, S. H. (1997). Signals and Systems (2nd ed.). Prentice Hall. - textbook - Chapter 1 covers even and odd signals and their decomposition (no link)