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How can we generate exponentially growing sinusoidal signal (further describes below equation and ) using a simple analogue circuit (e.g. op-amps, transistors, or oscillator)?

Mathematically, \$v(t)=C_0e^{rt}\cos(\frac{2\pi}{T} t ),\$ where \$ r>0,\$ and \$ t<n_0 T\$

or, visually something like this (amplitude can be very small)

enter image description here

I already read this question, but I am looking for generating an oscillating signal.

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    \$\begingroup\$ Build an op-amp oscillator that's held in a starting state by analog switches, and that has no AGC to limit oscillation. Switch it to the run state for 5ms, and then switch it back. \$\endgroup\$ – TimWescott Aug 4 '20 at 17:44
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    \$\begingroup\$ Or just look over this link as that's pretty much how some oscillators start up. \$\endgroup\$ – jonk Aug 4 '20 at 17:52
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    \$\begingroup\$ Generate an exponentially rising signal and use an oscillator and multiplier to produce an exponentially rising sine wave. Trigger the exponential growth at a zero cross of the sine wave. Next question is how to generate an exponentially rising signal. \$\endgroup\$ – Andy aka Aug 4 '20 at 18:23
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    \$\begingroup\$ super-regenerative radio receivers do this, however they generally switch off more gently, since reciprocating energy in L&C is substantial and takes some time to dissipate. eix.co.uk/Articles/Radio/Welcome.htm \$\endgroup\$ – glen_geek Aug 4 '20 at 19:06
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    \$\begingroup\$ Why do you want to do this? \$\endgroup\$ – Bruce Abbott Aug 4 '20 at 22:53
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Emitter-coupled Franklin oscillator with switched power supply. oscillator

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The following oscillator should work with precision op amps (low offset voltage), as long as the values of \$r\$ and \$T\$ do not need to be very precise.

Oscillator schematic

Defining \$\omega=2\pi/T\$, I get \$r=\frac{R_4/R_3}{2 R_2 C_2}\$ and \$\omega^2=\frac{(R_4/R_3) (R_5/R_6)+(R_4/R_6)+(R_5/R_6)}{R_1 C_1 R_2 C_2}-r^2\$. SPICE simulation seems to give the expected behavior using mostly ideal op amps with the switch switching at \$t=1\$ ms:

.subckt sw_spdt 1 2 3
R1 1 2 R={time < 1ms ? 10Meg : 1m}
R2 2 3 R={time < 1ms ? 1m : 10Meg}
R3 1 3 100Meg
.ends

A plot of V(/OUT) is shown below:

Simulation plot

The initial voltage can be chosen to charge C0, and flipping the switch starts the oscillator. After a few periods, the op amps will start to behave differently and probably approach equilibrium again (0V). The output voltage can be divided if a smaller voltage is desired, and another switch could be used to ground the output at some time \$t=n_0 T\$.

Explanation

The objective function \$v(t)=V_0 e^{r t}\cos(\omega t)\$ is the unique (nontrivial) real solution to the IVP

\$v(t)=-\left(v''(t)-2 r v'(t)\right)/(r^2+\omega^2),\,v(0)=V_0.\$

Op amps are used to take the appropriate derivatives and connected in a feedback loop. Switching the voltage pulls the system out of equilibrium and starts the oscillator.

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Bias a WEIN_BRIDGE oscillator with charge on each of the two capacitors.

Then enable the opamp output.

A CA3080 has this Enable ability.

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Since you need to impose Initial Voltages on the 2 capacitors, this becomes a Switched Capacitor design. Plan on using two analog multiplexors per capacitor, or 4 total multiplexors.

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    \$\begingroup\$ This is not a complete explanation and doesn't even have links to external information \$\endgroup\$ – HackerBoss Aug 4 '20 at 19:43

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