Why the amplitude of the 90 shifted phase signal when use the analytic signal block is not equal to the original signal?
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I could manage to shift the phase of signal by 90 by using analytic signal block. But, the problem is that its amplitude is not equal to the original one. Can anyone please explain it?
8 Comments
Walter Roberson
on 8 Jun 2020
How are you measuring the amplitude?
saleh shlimet
on 8 Jun 2020
saleh shlimet
on 8 Jun 2020
Walter Roberson
on 8 Jun 2020
I see what you mean, but at the moment I have no idea why that is happening. :(
saleh shlimet
on 8 Jun 2020
Walter Roberson
on 8 Jun 2020
I wonder if it would be practical to implement in terms of laplace transform?
saleh shlimet
on 8 Jun 2020
Walter Roberson
on 8 Jun 2020
Sorry, I do not know.
Answers (1)
Jyotsna Talluri
on 16 Jun 2020
0 votes
You can try variable phase shift block in simulink.
6 Comments
saleh shlimet
on 18 Jun 2020
Edited: saleh shlimet
on 18 Jun 2020
Jyotsna Talluri
on 18 Jun 2020
What is the input signal you are considering?From the figure you attached,I see it is a sine wave and you can specify the phase,you want to shift in the block parameters itself
saleh shlimet
on 18 Jun 2020
Jyotsna Talluri
on 20 Jun 2020
The other option I could think of is transport delay block where we can specify the time delay corresponding to a phase shift of 90 degrees as per the equation
phase shift (in deg) = 360*time delay/period of signal.
Refer to the below link
saleh shlimet
on 20 Jun 2020
Walter Roberson
on 20 Jun 2020
Suppose you leave the program running for two weeks. Then a frequency as low as one cycle per week could be detected, mathematically. That frequency has to have its phase shifted by 90 degrees, just like all the other frequencies do.Is it mathematically possible for something like a hilbert transform to properly predict when to make the appropriate output adjustments before the system has seen one cycle of the wave? Nyquist would say NO.
I would argue that it is therefore not possible to do what you want in a real-time system, unless
- You put a lower-frequency boundary for detection (upper frequency is according to sampling frequency); and
- You have a transport delay corresponding to one cycle of that low frequency.
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