Signal Generator, Oscilloscope and Capacitor

[User1265]

Homework Statement

A signal generator, resistance 300Ω , produces a square-wave signal of frequency
1.0 MHz. The output of the generator is connected to an oscilloscope, with negligible
capacitance, by a coaxial cable of capacitance 1.5 × 10^-9F and negligible resistance.
Sketch the trace observed on the oscilloscope screen.
Give any necessary calculations.

Relevant Equations

time constant of Capacitor = RC

1/f = T

I started by working out the time constant of the Capacitor = RC = 300 * 1.5*10^-9 = 0.45 *10^-6
and the time period (t) of the square wave signal produced by the signal generator : 1/(1*10^6 )= 10^-6

Then I drawed the graph as follows:

https://www.physicsforums.com/data/attachments/234/234320-391937ca3e0804dfc6126d030d757447.jpg I sketched what I thought the circuit was for this Q below the graph. I unsure how the graph of the capacitor would look like for when the voltage applied by the singal generator is negative.

T /2 is roughly equal to one time constant so on the graph, at T/2 I showed the Chargge/Voltage/current of the capacitance at 63% roughly of the p.d applied across the capactior and the resistor.

However the worked solutions below show a different graph, along with some of the reasoning.
https://www.physicsforums.com/data/attachments/234/234321-6a47ec112bdd2f8099738f570ac71db3.jpg

1) I'm really confused on the reasoning given for the graph for the capacitor shown in the solutions, and thus I can't understand the graph. If it could be explained more clearly I would be grateful.

Also, Q1) Why does the graph of the capactior start at a negative value at t=0 , surely it would have no charge on its plates intially?

12) Am I correct that the capacitor, when the voltage is reversed by the signal generator, after T/2, the capactior reaches the same magntiude of charge (on its other plate this time) , but in a smaller time than the time constant - as in the reverse direction, the current reaching the capactior plate would be greater due to not having to go through the resistor.

 

[BvU]

Hi,

I sketched what I thought the circuit was for this Q below the graph. I unsure how the graph of the capacitor would look like for when the voltage applied by the singal generator is negative.

Same differential equation: the current flows the other way. Resistor is still sitting there ! So:

Am I correct that the capacitor, when the voltage is reversed by the signal generator, after T/2, the capactior reaches the same magntiude of charge (on its other plate this time) , but in a smaller time than the time constant - as in the reverse direction, the current reaching the capactior plate would be greater due to not having to go through the resistor.

Not correct.

Note that your solution -- if continued for a few more cycles -- and the worked solution are as good as equal:

Also, Q1) Why does the graph of the capactor start at a negative value at t=0 , surely it would have no charge on its plates intially?

Because that's what you get to see on the oscilloscope: the periodic part of the solution.

You are correct with your analysis (assuming the exercise actually mentions the initial state of the capacitor -- which it does not) but is a transient solution that lasts only for a few microseconds

 

[User1265]

Not correct.

Thanks for the clarification. May I ask why my thinking is not correct for this part?

 

[BvU]

If the voltage inside the signal generator is negative, the current still has to go through its internal resistance of 300 Ohm.

Consider the diagram of a short-circuit situation and then turn it upside down

 

[rude man]

You can get an exact output waveform expression with the following approach and fact:
1. Peak-to-trough time = trough-to-peak time = 1/2 the period of the input square wave.
2. Assuming say +/- 1V input square wave, write the expression for the output voltage starting at the (unknown) negative-most voltage and aiming for the positive-most voltage. During this time the excitation voltage is +1V.
After 1/2 period the output voltage is at peak and its magnitude is the same as that of the trough voltage.

That's it, you should be able to take it from there.