Introduction

A coupled oscillator is a system of two or more oscillators that exchange energy through a shared element, such as a spring, a mutual inductance or a capacitor. In electronics, the simplest case is a pair of LC resonant circuits joined by a small coupling capacitor or by a mutual inductance. A coupled oscillator experiment measures how that coupling changes the resonant behaviour of the pair.

The central result is mode splitting. A single LC circuit has one natural frequency. Two identical circuits that are coupled have two, and the gap between them is a direct measure of the coupling strength. The same physics appears in transformer-coupled amplifiers, double-tuned filters, wireless power transfer and quadrature oscillators.

This article describes how to carry out the experiment with standard laboratory equipment: a function generator, a dual-channel oscilloscope and a handful of passive components. It covers the theory, the circuit, the measurement procedure, worked numerical examples, common sources of error and the cross coupled oscillator, a term that names a different circuit and is often confused with the coupled oscillator experiment.

Fundamentals: What a Coupled Oscillator Really Is

Start with a single LC tank. Energy sloshes between the capacitor (as electric field) and the inductor (as magnetic field), and the rate of that exchange is the natural frequency f₀ = 1/(2π√(LC)), where L is the inductance in henry and C is the capacitance in farad. Real components lose a little energy every cycle, so the oscillation slowly dies out unless something supplies energy.

Now place a second, identical tank next to the first and link them with a weak connection. Energy can now leak from tank one into tank two. A mechanical analogy helps here. Two identical pendulums hung from a slightly sagging string behave the same way. Set one swinging and, after a while, the other takes over while the first comes almost to rest. Then the energy flows back. The electrical version does exactly this, with voltage in place of displacement.

The formal description uses normal modes. A normal mode is a pattern of motion in which every part of the system oscillates at the same single frequency. Two identical coupled oscillators have exactly two: a symmetric mode where both move together (in phase) and an antisymmetric mode where they move in opposition (out of phase). Any other motion is a mixture of these two. When both modes are present at once, their slightly different frequencies produce beats, and the beats are what the eye sees as energy moving from one oscillator to the other.

The strength of the link is expressed by the coupling coefficient k, a dimensionless number between 0 and 1. For two coils it is k = M/√(L₁L₂), where M is the mutual inductance. For the capacitively coupled circuit used in this experiment, it is k = Cc/(C + Cc). The stronger the coupling, the wider the gap between the two normal-mode frequencies.

One more idea decides whether the experiment is visible at all. The quality factor Q of a tank measures how many cycles it rings before the energy fades, and it can be estimated from a single tank as Q = f₀/Δf₃dB, where Δf₃dB is the width of the resonance peak at the half-power points. As a rule of thumb, the two modes appear as separate peaks only when k is greater than about 1/Q, a condition called critical coupling. Below that, the two peaks merge into one broad hump.

The Circuit Used in the Coupled Oscillator Experiment

The recommended arrangement uses two parallel LC tanks, each with its inductor and capacitor connected from a node to ground. The nodes are called A and B. A coupling capacitor Cc links A to B, and the function generator feeds node A through a large series resistor.

This topology is chosen deliberately. Every tank is referenced to ground, so ordinary single-ended oscilloscope probes can measure both nodes without any differential tricks. The large Rs keeps the generator from loading the tank, so the circuit rings at its own natural frequencies rather than at frequencies shaped by the source.

Deriving the Two Normal Modes

Let vA and vB be the node voltages, with C1 = C2 = C and L1 = L2 = L. Applying Kirchhoff's current law at node A, the current through the first capacitor, the coupling capacitor and the inductor must sum to zero:

C·dvA/dt + Cc·d(vA − vB)/dt + (1/L)∫vA dt = 0

Differentiating once removes the integral and gives

(C + Cc)·d²vA/dt² − Cc·d²vB/dt² + vA/L = 0

Node B gives the same equation with A and B swapped. Adding the two equations eliminates the cross term, and the sum vA + vB obeys C·d²(vA + vB)/dt² + (vA + vB)/L = 0. That is a plain LC tank, so the in-phase mode has

f₁ = 1/(2π√(LC))

In this mode the voltages are equal, no current flows in Cc, and the coupling capacitor has no effect. Subtracting the two equations instead gives (C + 2Cc)·d²(vA − vB)/dt² + (vA − vB)/L = 0, so the out-of-phase mode has

f₂ = 1/(2π√(L(C + 2Cc)))

In this mode the coupling capacitor is charged from both ends and acts as if 2Cc were added to each tank. Because f₂ is lower than f₁, the in-phase mode is the higher-frequency one in this circuit. Taking the ratio of squares gives a neat measurement formula, valid exactly for this circuit:

k = (f₁² − f₂²)/(f₁² + f₂²) = Cc/(C + Cc)

The same formula works for an inductively coupled pair. Two series LC loops that share a mutual inductance M ring at 1/(2π√((L + M)C)) when the currents are in phase and at 1/(2π√((L − M)C)) when they are in opposition. Here the in-phase mode is the lower one, and the same expression returns k = M/L. Knowing which mode is higher in which circuit is a favourite viva question.

Equipment and Components

A value of 10 nF for C is a practical choice. Breadboard strips and probes add stray capacitance in the range of picofarads, which is only a tiny fraction of 10 nF, so the measured frequencies stay close to the calculated ones.

Step-by-Step Procedure

Step 1: Measure and match the components. Use the LCR meter to measure L1, L2, C1, C2 and CC. From a handful of capacitors, pick two whose values agree within about 1%, and do the same for the inductors if possible. This step matters more than any other, for a reason explained in the error section below.

Step 2: Characterise one tank on its own. Build tank 1 only, drive node A through Rs and sweep the generator across roughly 30 kHz to 70 kHz while watching CH1. The amplitude peaks at f₀. Compare it with the calculated 1/(2π√(LC)), and note the peak width to estimate Q. This quick check confirms that the components, probes and generator are behaving before the coupling is added.

Step 3: Add the second tank and the coupling capacitor. Connect tank 2 and CC, keep the generator amplitude modest (a couple of volts peak to peak is enough) and sweep slowly again. Node B now shows two peaks instead of one. Record the frequency of the lower peak as f₂ and the upper peak as f₁. Node A will show a dip between the two peaks, which is a useful sign that both modes are present.

Step 4: Confirm the mode shapes in XY mode. Set the oscilloscope to XY display, with node A on the X input and node B on the Y input. Tune the generator to the upper peak. The trace collapses to a line rising to the right, meaning A and B move in phase. Tune to the lower peak and the line tilts the other way, meaning A and B are in opposition. With well-matched tanks the lines are close to 45 degrees. Visible ellipses or unequal tilts point to mismatch or heavy damping.

Step 5: Watch the beats directly. Replace the sine wave with a square wave at about 200 Hz and trigger on CH1. Each edge kicks the tanks, which then ring. Both modes are excited at nearly equal strength, so the envelope of node A shrinks while node B grows, then the process reverses. The time for the envelope to complete one full cycle is 1/(f₁ − f₂), and the energy moves completely from one tank to the other in half that time.

Step 6: Repeat for different coupling. Change Cc to 0.47 nF, then 1 nF, then 2.2 nF. The gap between the peaks should widen steadily. Record each result in a table like the one below and compare the measured k with Cc/(C + Cc).

Worked Example 1: Capacitively Coupled Tanks

Problem. Two identical parallel LC tanks have L = 1 mH and C = 10 nF. They are coupled by Cc = 1 nF. Find the two normal-mode frequencies, the coupling coefficient, and the beat period. Decide whether the modes will be resolved if Q = 60.

Step 1, the in-phase mode. Coupling has no effect, so f₁ = 1/(2π√(LC)). The product LC = 10⁻³ × 10×10⁻⁹ = 1×10⁻¹¹ s², whose square root is 3.162×10⁻⁶ s. Then f₁ = 1/(2π × 3.162×10⁻⁶) ≈ 50.33 kHz.

Step 2, the out-of-phase mode. The effective capacitance is C + 2Cc = 10 nF + 2 nF = 12 nF. Then L(C + 2Cc) = 1.2×10⁻¹¹ s², with a square root of 3.464×10⁻⁶ s, so f₂ = 1/(2π × 3.464×10⁻⁶) ≈ 45.94 kHz.

Step 3, the splitting and beat period. The gap is f₁ − f₂ ≈ 4.39 kHz. The beat period is therefore 1/4390 ≈ 228 µs, and energy transfers fully from A to B in about 114 µs.

Step 4, the coupling coefficient. Using the measurement formula, k = (50.33² − 45.94²)/(50.33² + 45.94²) = (2533 − 2111)/(2533 + 2111) ≈ 0.0909. The direct formula Cc/(C + Cc) = 1/11 = 0.0909 agrees, which confirms the derivation.

Step 5, the resolution check. The critical coupling is 1/Q = 1/60 ≈ 0.017. Since k = 0.091 is more than five times larger, the product kQ is about 5.5 and the two peaks are clearly resolved.

Worked Example 2: Inductively Coupled Loops

Problem. Two identical series LC loops each have L = 2 mH and C = 0.5 µF. The mutual inductance between the coils is M = 0.4 mH. Find the two mode frequencies and k.

Solution. The coupling coefficient is k = M/L = 0.4/2 = 0.2. For in-phase currents the effective inductance is L + M = 2.4 mH, so f = 1/(2π√(2.4×10⁻³ × 0.5×10⁻⁶)) = 1/(2π × 3.464×10⁻⁵) ≈ 4.59 kHz. For opposing currents the effective inductance is L − M = 1.6 mH, so f = 1/(2π × 2.828×10⁻⁵) ≈ 5.63 kHz. Here the in-phase mode is the lower one, the reverse of the capacitive case, and the measurement formula gives (5.63² − 4.59²)/(5.63² + 4.59²) = 0.2, matching M/L.

Common Errors and How to Troubleshoot Them

Component mismatch is the main enemy. Suppose C1 and C2 differ by 5% while the coupling ratio is Cc/C = 0.1. The detuning between the tanks is then comparable to the coupling, and the modes stop being clean symmetric and antisymmetric patterns. The oscillation tends to stay in one tank, the peaks become unequal, and the measured gap no longer matches theory. This is why Step 1 insists on matching, and why ordinary ceramic capacitors with wide tolerance are a poor choice here.

Inductor losses come next. A real inductor has winding resistance and self-capacitance, which lower Q and slightly shift the resonance. If only one broad peak appears, the usual cause is that k is smaller than 1/Q. Raise CC, or use inductors with lower resistance, until kQ is comfortably above 1.

Loading and grounding deserve attention too. A 10x probe presents a few picofarads and a high resistance, which is harmless against 10 nF but would be noticeable in a pF-scale circuit. Always clip both probe grounds to the same ground point close to the circuit, and keep the generator connection through the large Rs. If the observed peak positions drift when probes are moved, lead capacitance is the likely cause.

Finally, read peaks carefully. With a sine sweep, slow down near each peak, use scope cursors or a frequency counter on the generator, and take the frequency that gives maximum amplitude rather than estimating from the knob.

The Cross Coupled Oscillator: A Different Circuit

The phrase cross coupled oscillator does not describe two oscillators joined to each other. It describes a single oscillator whose gain element is a pair of transistors, with the output of each connected to the input of the other. The most common form in communication hardware is the cross coupled LC oscillator: two NMOS transistors share a grounded source, the drain of M1 connects to the gate of M2 and the drain of M2 connects to the gate of M1, and an inductor with a capacitor (often a varactor, so the frequency can be tuned) sits between the two drains.

Why the Cross Coupled Pair Sustains Oscillation

Every real tank loses energy in its resistance. This loss is modelled as a parallel resistance Rp across the tank, and without a source of energy the oscillation dies. The cross-coupled pair supplies that energy back each cycle by acting as a negative resistance.

The result can be derived in a few lines. Apply a differential test voltage v across the two drains, so that the drain of M1 sits at +v/2 and the drain of M2 at −v/2. The gate of M1 is tied to the drain of M2, so its gate voltage is −v/2, and its drain current is gm × (−v/2) = −gm·v/2. The current drawn from the test source is therefore negative whenever v is positive, which is the signature of a negative resistance. Dividing, the differential resistance looking into the pair is

R_pair = −2/gm

Oscillation can start when the negative resistance magnitude is smaller than the tank loss resistance, which means 2/gm < Rp, or gm > 2/Rp. Designers keep a safety margin above this, since temperature and process changes alter gm and Rp. The amplitude then grows until the transistors begin to switch and limit the swing, which fixes the final amplitude. The frequency is set by the tank to first order, f ≈ 1/(2π√(L_diff·C_total)), where L_diff is the inductance connected between the two drains and C_total is the capacitance across it, including the transistors' parasitic capacitance.

Worked Example 3: Start-Up Condition

Problem. A cross coupled LC oscillator uses a differential inductance of L = 10 µH and a total capacitance of C = 100 pF. The inductor has a quality factor of 20 at the operating frequency. Find the oscillation frequency and the minimum gm for start-up.

Step 1, the frequency. LC = 10×10⁻⁶ × 100×10⁻¹² = 10⁻¹⁵ s², whose square root is 3.162×10⁻⁸ s. Then f₀ = 1/(2π × 3.162×10⁻⁸) ≈ 5.03 MHz.

Step 2, the tank resistance. At resonance the reactance of the inductor is ω₀L = 2π × 5.03×10⁶ × 10×10⁻⁶ ≈ 316 Ω. For a parallel tank, Rp = Q × ω₀L = 20 × 316 ≈ 6.32 kΩ.

Step 3, the start-up condition. gm > 2/Rp = 2/6320 ≈ 0.32 mS. A practical design would choose gm comfortably larger than this value, so that oscillation starts reliably despite variation.

The Other Cross Coupled Circuit: The Astable Multivibrator

Students also meet cross-coupling in the classic BJT astable multivibrator, where two transistors are cross-connected through capacitors. This is a relaxation oscillator, not a sinusoidal one, and its output is a square wave. With a supply that fully switches each transistor, each half period is about 0.693RC, so the total period is T ≈ 0.693(R1C1 + R2C2). For R = 10 kΩ and C = 0.1 µF in both halves, T = 0.693 × 2 × 10⁴ × 10⁻⁷ ≈ 1.39 ms, or about 721 Hz. If a question mentions a cross-coupled circuit with RC timing and a square-wave output, this is the circuit it refers to.

Comparing the Three Circuits

Where These Ideas Are Used

Coupled resonant circuits are the basis of double-tuned transformers in radio intermediate-frequency stages, where the coupling is adjusted so the passband is flat and wide rather than sharply peaked. Resonant wireless charging relies on the same physics between a transmitter and a receiver coil, and the frequency splitting seen in this experiment is exactly what engineers watch for when the two coils are brought too close. Two cross coupled LC oscillators can themselves be coupled to produce quadrature outputs, which is how many radio receivers generate the 90-degree-shifted signals they need.

The cross coupled LC pair is a common core of voltage-controlled oscillators inside phase-locked loops, which in turn generate the local-oscillator frequencies in radios and radar. Organisations such as ISRO, DRDO laboratories and BEL develop communication and radar hardware where this kind of frequency generation is routine.

Exam, Viva and Career Context

Practical examiners tend to ask about the physical meaning of the result rather than the arithmetic. Typical viva questions include why the frequencies split, which mode is higher for capacitive and for inductive coupling, what happens to the gap as coupling increases, and how to obtain k from measured frequencies. The answers are all in the sections above.

For written exams, mutual inductance and coupled circuits belong to standard network theory, while oscillators and the Barkhausen criterion belong to analog electronics. The relative importance of these topics differs between GATE ECE, GATE EE and instrumentation papers, and between SSC JE and RRB JE papers for different branches. The syllabus and weightage change over time, and they could not be verified for this article, so the official notification for the current year is the reference to follow.

On the career side, LC oscillator and PLL design is a recognised specialisation within RF and analog IC design, and the coupled-resonator physics from this experiment carries over directly to filter and wireless power work. Salary figures are deliberately not quoted here, since they could not be verified from an authoritative source.

Conclusion

The coupled oscillator experiment compresses a lot of physics into a few capacitors and inductors. Two identical tanks give two frequencies, the gap between them measures the coupling, and the beats on the oscilloscope show energy moving between the tanks in real time. The reliability of the result rests on component matching and on keeping kQ comfortably above one, and both can be checked before any signal is applied.

The cross coupled oscillator then builds on a related intuition from the opposite direction. Rather than letting two resonators trade energy, it uses a transistor pair to cancel the loss of one resonator. Keeping the three circuits apart, the coupled tanks, the cross coupled LC oscillator and the cross-coupled multivibrator, prevents the most common confusion in lab reports and interviews.

The best way to cement this is to build the circuit. Pick the matched components, record the table for three values of Cc, and compare the measured coupling coefficient with Cc/(C + Cc) in your own lab notebook. Students heading toward SSC JE, RRB JE, GATE or PSU interviews will find that a result they have measured themselves is far easier to explain than one they have only read about.

FAQ

Q: What is a coupled oscillator? 

A coupled oscillator is a pair or group of oscillators connected so that energy can pass between them. In electronics the oscillators are usually LC circuits and the link is a capacitor or mutual inductance. The coupled system has several normal-mode frequencies instead of one.

Q: Why does coupling split the resonant frequency in a coupled oscillator experiment? 

When two identical tanks are linked, the system can oscillate in two distinct patterns, in phase and out of phase. The coupling element acts differently in each pattern, so each has its own effective inductance or capacitance and therefore its own frequency. In the capacitive circuit above, only the out-of-phase mode sees the extra 2Cc, so it is the lower one.

Q: How is a cross coupled oscillator different from a coupled oscillator? 

A coupled oscillator is two separate oscillators linked by a coupling element, and it is studied for its normal modes. A cross coupled oscillator is one oscillator whose active device is a transistor pair connected gate to drain in a crossed arrangement, which supplies the negative resistance needed to sustain oscillation.

Q: How do I calculate the coupling coefficient from measured frequencies? 

Measure the higher and lower mode frequencies, then use k = (f_high² − f_low²)/(f_high² + f_low²). For the capacitive circuit it equals Cc/(C + Cc), and for the inductive circuit it equals M/L, so the measurement can be compared directly with the component values.

Q: Why does only one peak appear in my experiment? 

There are three common causes. The coupling may be weaker than 1/Q, so the peaks overlap. The two tanks may be mismatched, so one tank dominates. Or the source resistance may be too low and is loading the tank. Increase CC, match the parts better, or raise Rs.

Q: What condition must a cross coupled oscillator satisfy to start? 

The negative resistance of the pair must outweigh the tank loss. For a cross coupled NMOS pair this means gm > 2/Rp, where Rp is the differential parallel resistance of the tank. Designs include some margin above this minimum.