Resolver's Output Sine and Cosine signals

[Amr Wael]

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

 

[KlausST]

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.

V23401-T2071-B101 TE Connectivity AMP Connectors | Sensors, Transducers | DigiKey

Order today, ships today. V23401-T2071-B101 – Sensor Angle from TE Connectivity AMP Connectors. Pricing and Availability on millions of electronic components from Digi-Key Electronics.

www.digikey.com

--- Updated Mar 15, 2022 ---

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.

V23401-T2071-B101 TE Connectivity AMP Connectors | Sensors, Transducers | DigiKey

Order today, ships today. V23401-T2071-B101 – Sensor Angle from TE Connectivity AMP Connectors. Pricing and Availability on millions of electronic components from Digi-Key Electronics.

www.digikey.com

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?

 

[KlausST]

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".
or read the datasheet

Klaus