# Output voltage from LC tank oscillates only for 1 period

Status
Not open for further replies.

#### eecs4ever

##### Full Member level 3 So suppose i have a LC tank with a cap of size 1 F . and a inductor of size 1 H.

they are place in parallel.

I feed in a pulse of current through an ideal current source.

I've drawn the picture in the

The output voltage osscillates for exactly 1 period. (I've simulated it in spice).
The voltage goes close to 0 after time 2*pi.

why doesn't the osscillation continue?

#### v_c Re: LC tank, voltage

First note that the natural frequency of this circuit is $\omega_0=\frac{1}{\sqrt{LC}}= 2\pi f_0=1$. So $f_0=\frac{1}{2\pi}$ and the period is $T_0=\frac{1}{f_0}=2\pi$.

Let us start with decomposing the input current. The current $i(t)$ can be written as

$i(t) = u(t) - u(t- 2\pi)$

this is a positive amplitude step function and a shifted, negative amplitude step function.

Use superpostion and take each component of the current at a time. The voltage response to $u(t)$ is simpy $v_1(t)=\sin(\omega_0 t) u(t)$. The response to $-u(t- 2 \pi)$ is the same as the previous response but it is zero until $t= 2\pi$ and it is a negative version. So the response to $-u(t - 2 \pi)$ is $v_2(t) = -\sin( \omega (t-2\pi)) u(t-2 \pi)$. To total response is the sum $v(t)=v_1(t) + v_2(t) = \sin(\omega_0 t) u(t) -\sin( \omega (t-2\pi)) u(t-2 \pi)$. This can be simplified to $v(t) = \sin(\omega_0 t) [ u(t) - u(t-2 \pi) ]$. This just looks like a sinewave that has been multiplied by the pulse shape of the current waveform.

So what happens is the shifted and inverted sine wave due to the second step function effectively cancels the sine wave from the first step function. But the cancellation does not occur until $t=2 \pi$.

Does this help validate what you saw in the simulation.

Best regards,
$v_c$

Last edited by a moderator:
• pllram

Points: 2

### pllram

Points: 2

#### eecs4ever

##### Full Member level 3 Re: LC tank, voltage

How do you know that the response to u(t)

is v(t) = sin(wo t) u(t) ?

#### v_c Re: LC tank, voltage

It is a classic second order differential equation. There are no losses in the circuit so there will be sustained oscillations.
Mark voltages $v_C$ and $v_L$ for the inductor and capacitor from the top node to ground and mark currents $i_C$ and $i_L$ for the inductor and capacitor flowing from the top node to ground.
Just write the differential equations for the circuit.

$i(t) = C \frac{d v_C}{dt} + i_L$ ----(1)
$v_L = L \frac{d i_L}{dt}$ -----(2)
note also that $v_C=L \frac{d i_L}{dt}$ (since $v_C=v_L$). Now taking derivative on both sides of $v_C=L \frac{d i_L}{dt}$ gives $\frac{d v_C}{dt} = L \frac{d^2 i_L}{dt^2}$. Now substitute this into (1) and you will get
$i(t) = LC \frac{d^2 i_L}{dt^2} + i_L$

This is a classic 2nd order differential equation. Given that $i(t)=I_m u(t)$ is a step function of amplitude $I_m$ and that initial inductor current and capacitor voltage is zero $i_L(0)=0$, $v_C(0)=0$, the solution is

$i_L(t)= I_m - I_m \cos( \frac{t}{\sqrt{LC}}) = I_m - I_m \cos(\omega_0 t)$
$v_C(t)= I_m Z_0 \sin( \frac{t}{\sqrt{LC}}) = I_m Z_0 \sin(\omega_0 t)$

where $Z_0=\sqrt{\frac{L}{C}}$ is called characteristic impedance of the tank circuit. In the example that you had $I_m=1$, $Z_0=1$ and $\omega_0=1$, so your solutions were

$i_L(t)= 1 - \cos(t)$
$v_C(t)= \sin(t)$

Best regards,
$v_c$

Added after 1 hours 45 minutes:

Sorry to reply to my own post but I had to correct some typos and clean up the equations and put them in TeX mode. I think they should be OK now.

$v_C$

Last edited by a moderator:
• pllram

Points: 2

### pllram

Points: 2

#### flatulent Re: LC tank, voltage

Just for the fun of it, try different driving pulse widths. If the width is very narrow the circuit should ring indefinitely because the Q is infinite.

#### v_c Re: LC tank, voltage

Even with a delta function it will ring indefinitely. With different pulse widths you can see any portion of the sine wave you like.

#### flatulent Re: LC tank, voltage

You can look at it from the conservation of energy view. During the first half cycle the current source is putting energy into the LC. (Current is flowing against a positive voltage load.) During the second half cycle the current source is absorbing energy from the load. (Current is flowing against a negative voltage load.) Since the two halves of the sine wave are equal in amplitude and shape these two energy flows are equal in magnitude and opposite in direction. At the end there is no energy left in the LC.

#### v_c Re: LC tank, voltage

Yes, it is quite interesting. The voltage and current evaluated at $t=2 \pi$ both evaluate to zero so the energy stored in each is also zero.

This system is just like a forced mass and spring (with zero friction). Once you give it a little push it should oscillate forever where the energy will slosh back and forth between the potential energy and the kinetic energy. The pulsed current case is a bit harder to imagine with a mass-spring -- you have to imagine a forcing function that will make only one oscillation and then stop.

Best regards,
v_c

Last edited by a moderator:

Status
Not open for further replies.