# Resolver's Output Sine and Cosine signals

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#### Amr Wael

##### Member level 4
Hi all
I have been studying resolver of Motors lately , I understand the basics that there is a reference signal that is used to excite the resolver and then the resolver produces a sine and cosine signals whose amplitudes are proportional to the angle of the motor.
My question is why do we need both sine and cosine signals to compute the angle ? Isn't the Sine only enough for us to compute it as the cosine is always just a phase shift of the sine or is there another reason?
Thank you very much in advance
Amr

Hi,

Afaik the amplitudes are constant.
But the instantaneous values can be used to calculate the angle.

A single value can´t be uses, because a sine (cosine) have two times the same value for a 360° rotation. So you can´t calculate at which position you are.

Also with a single value you have no reference what is 100%.
With dual signals you can calculate the 100% (of the amplitude) value by amplitude = sqrt(sin^2 + cos^2)

example:
0.53V could be 32° or 147° of a 1.00V amplitude sine.
0.53V could be 20.7° or 159.3° of a 1.50V amplitude sine.
***
use two random values where you know they are sine and cos of same amplitude:
0.45V sine, 0.965V cos
--> calculate amplitude: = sqrt(0.45^2 + 0.965^2) = 1.065V
knowing that both values are positive you know you are in the first quadrant i.e. 0..90°
now you have several options for the angle: line tan(phi) = sin(phi)/ cos(phi): tan(phi) = (0.45/0.965) --> 25°
or sin(phi) = val/amplitude = 0.45/1.065 --> 25°
or cos(phi) = val/amplitude = 0.965/1.065 --> 25°

Klaus

• Amr Wael

Hi,

Afaik the amplitudes are constant.
But the instantaneous values can be used to calculate the angle.

A single value can´t be uses, because a sine (cosine) have two times the same value for a 360° rotation. So you can´t calculate at which position you are.

Also with a single value you have no reference what is 100%.
With dual signals you can calculate the 100% (of the amplitude) value by amplitude = sqrt(sin^2 + cos^2)

example:
0.53V could be 32° or 147° of a 1.00V amplitude sine.
0.53V could be 20.7° or 159.3° of a 1.50V amplitude sine.
***
use two random values where you know they are sine and cos of same amplitude:
0.45V sine, 0.965V cos
--> calculate amplitude: = sqrt(0.45^2 + 0.965^2) = 1.065V
knowing that both values are positive you know you are in the first quadrant i.e. 0..90°
now you have several options for the angle: line tan(phi) = sin(phi)/ cos(phi): tan(phi) = (0.45/0.965) --> 25°
or sin(phi) = val/amplitude = 0.45/1.065 --> 25°
or cos(phi) = val/amplitude = 0.965/1.065 --> 25°

Klaus
Thank you that's very helpful. Now I understand how to calculate the angle from the output sine and cosine signals.
I have 1 more question please. One step backward What's the relation between the reference voltage and the output sine and cosine waves? I thought that the reference voltage is the full scale of the output sine or cosine but if it's not , can the full scale for a specific resolver be calculated from the data sheet or something? I guess it's related to the number of turns of the primary and secondary coils of the internal transformer inside the resolver?
I am planning to use this resolver.
--- Updated ---

Thank you that's very helpful. Now I understand how to calculate the angle from the output sine and cosine signals.
I have 1 more question please. One step backward What's the relation between the reference voltage and the output sine and cosine waves? I thought that the reference voltage is the full scale of the output sine or cosine but if it's not , can the full scale for a specific resolver be calculated from the data sheet or something? I guess it's related to the number of turns of the primary and secondary coils of the internal transformer inside the resolver?
I am planning to use this resolver.
I think rT or transformation ratio = 0.5 which means that half the reference voltage will be maximum at the output sinusoidal voltage or am I wrong?

Hi,
What's the relation between the reference voltage and the output sine and cosine waves?
I don´t know. I had to read the datasheet. But it´s better you do this first.

can the fu--> pll scale for a specific resolver be calculated from the data sheet or something?
post#2: --> "calculate amplitude".

Klaus

• Amr Wael

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