# Signal Generator, Oscilloscope and Capacitor

• User1265
In summary, the capacitor produces a square wave signal with a time constant of 0.45 * 10^-6 seconds and a time period of 10^-6 seconds.
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.

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Hi,
User1265 said:
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:
User1265 said:
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:
User1265 said:
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

Last edited:
User1265
BvU said:
Not correct.

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

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

User1265
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.

BvU

## 1. What is a signal generator and how does it work?

A signal generator is a device that produces electronic signals of various frequencies and waveforms. It works by converting a direct current (DC) input into an alternating current (AC) output, which can then be adjusted to produce a specific frequency and waveform.

## 2. What is an oscilloscope and how is it used?

An oscilloscope is a measuring instrument used to display and analyze electronic signals in the form of a graph. It works by taking in a voltage signal and displaying it on a screen in the form of a waveform, allowing users to measure and analyze various characteristics of the signal, such as amplitude, frequency, and phase.

## 3. How does a capacitor work in a circuit?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. When connected in a circuit, it blocks the flow of direct current (DC) while allowing alternating current (AC) to pass through. It can also be used to smooth out fluctuations in a circuit's voltage or to store a charge for later use.

## 4. What is the difference between an analog and digital oscilloscope?

An analog oscilloscope displays signals in the form of a continuous waveform on a cathode ray tube (CRT) screen. A digital oscilloscope, on the other hand, converts the incoming signal into a series of digital values, which are then displayed on a digital screen. Digital oscilloscopes often have more advanced features and are easier to use, but analog oscilloscopes are still preferred by some for their ability to display fast-changing signals more accurately.

## 5. How do I choose the right signal generator, oscilloscope, and capacitor for my project?

When choosing a signal generator, oscilloscope, and capacitor for a project, it is important to consider the required frequency range, accuracy, and features needed. The signal generator should have a frequency range that covers the required frequency for your project. The oscilloscope should have a bandwidth and sampling rate that can accurately capture and display the signals you are working with. And the capacitor should have the appropriate capacitance value and voltage rating for your circuit. It is also important to consider the quality and reliability of the equipment, as well as any budget constraints.

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