# How can I find the time period for two sine waves using the oscilloscope in Multisim?

The formula for calculating the phase difference requires a time delay and a time period which is given in the formula:

Φ = td/tp × 360

Where td is the time delay between the two waves at zero point and tp is the time period.

Since I know td already, I would like to know how tp is calculated using two sine waves. Do I need to just subtract the periods between the two waves, or there are other methods?

• What do you really want to find? In your title you ask to find the "Period", yet in the question you "want to calculate the phase difference". Those are two different things.
– Nedd
May 26 at 2:10
• Did you read through the question or just read the first line? I'll edit and make it more clear. May 26 at 6:40

Methods of phase measurement described by others are good.

• One must be careful when measuring zero-crossings of an oscilloscope display.
• One must be confident that wave shape is sinusoidal, containing no even-order harmonics.

Measuring time difference (or scale distance) at zero crossings of the two waveforms is valid only if zero-crossings are aligned on the same horizontal line. The line chosen is usually aligned with the oscilloscope's reticle. OP's example waveform is nicely aligned on the centre-line:
You can ensure that no DC offsets foul the zero-crossing measurement by looking at both positive-going zero crossing and negative-going zero crossing. Both time-spans should be equal. If not, the waveform contains a non-zero DC voltage, or else the oscilloscope has added a DC offset.
The other possible source of non-equal zero crossings on a waveform is presence of even-order harmonics: it is not a pure sine wave.

One should also check that BOTH oscilloscope channels are DC-coupled or else BOTH oscilloscope channels are AC-coupled. Especially at low frequency, AC-coupling adds a phase shift that DC-coupling does not. DC-coupling is probably safer.

Once you've verified that BOTH waveforms are aligned with equal-span positive & negative zero crossings, proceed to measure their time ratio or scale ratio. Since OP's waveform is conveniently aligned so period is exactly 5 division of reticle, the blue waveform's zero crossing is offset by 1.1 division...
So phase shift is $$\ {{1.1}\over{5}}\times 360 = 79.2 degrees\$$

One might change the time scale so that one cycle spans the entire display (10 divisions), enhancing time resolution.
One might also boost the gain of the blue waveform so that its amplitude is larger...this makes its zero-crossing less shallow making estimation of zero crossing more easily eye-balled.
One can even boost amplitude so that waveform tops & bottoms are off-screen - we're concerned with zero crossings, not amplitudes. Be careful with this...the oscilloscope's amplifiers can be driven into non-linear regions that add DC offsets that foul your zero-crossing alignment.

• I like how you added the image for clear visuals making the explanation more understandable. I really appreciate your help. May 26 at 13:46

You can only use this equation if the two waves are the same period (which you can verify by calculating the period of both of them and comparing). Phase difference between two waves of different periods/frequencies conceptually makes no sense, as the difference would be changing with time.

Calculating the minus of the two periods will only find the difference between them, which will be 0 for any situation where a phase difference conceptually makes any sense.

Choose two identical points on one of the waves, and measure the time between them. The easiest point is probably to use the positive-going 0 crossing point of the blue wave, as its conveniently close to a grid line for accuracy.

This is because the period is defined as the time it takes for one cycle to occur.

It looks like there is 5 'boxes' (called a 'Div' or 'division') between the first and second positive going crossing, and since the timebase is 200us per Div, the time period is 5 x 200us = 1ms.

So by my measurements, phase difference will be (220us/1000us)*360 = 79.2 degrees

• Thank you very much for the detailed explanation. I am new to this topic and appreciate your help May 26 at 13:44

I agree with Entropy.

I just wanted to add that you could've made it much easier on yourself, if you had placed cursor 2 (yellow) at the positive-going zero crossing of the blue wave (I marked it green).

You then would've gotten following readings:

Result:

$$\Φ = td/tp × 360° = (T2-T1)/tp × 360° = 220μs/1000μs × 360° = 79.2°\$$

NB: this is only true when both periods are the same (which is the case here).

• My eyes failed me on this one. I appreciate you pointing this out. Thanks May 26 at 13:48