Question on reference voltage of the instrumentation amplifier (INA)

Hello.
I need to use only one battery to supply the INA with +4.5V as reference.
the image shows that the INA has +9V(Vdd) and +4.5V(Vss) and the input signal is 10mV.
the gain is 101 and the output voltage should be 1.01V centered at +4.5V, But the result is just dc voltage at +4.5V (red line) and not sinusoidal. the question is: How to get a correct sinusoidal output centered at +4.5V ??

Mo,The problem you are having is because you have set up the INA with the negative supply at 4.5V instead of the reference. This means that the input (which is referenced to ground) is outside the power supply of the INA.So to start with, you need to swap the connections to pins 4 and 5.
You may still have a problem because the input will be at the negative supply so it still might not work. To get around this you should reference the input to 4.5V (including the - pin)

First modify the connections as Elizabeth says.As an alternative you could use a device like the ICL7660 to generate a negative voltage and power the IC’s negative supply from this negative voltage.  Remember to keep you input signal referenced to ground.
Another alternative would be to buffer the 4.5V voltage derived from the resistive divider with a unity gain buffer,then usethis reference to feed the IA and the return of it the input signal (so it is not connected to ground). This of course works fine in theory, but you may find the sensor return is connected to its body and then earthed by its mechanical assembly.

So far in the two answers so far, there is no real reason for using an IA. A simple op-amp could be used if you use a negative supply or use some level shifting. An IA measures the differential of the input so another option would be after correcting the supply, to “float” the negative of the sensor (again, provided the body is isolated) and connect it via a large value resistor >~100K to the 4.5V supply. Again a buffer on the 4.5V would improve the performance. 
You may want to look at the TI TLE2426 to generate the 4.5V. It includes your resistor divider and an op-amp buffer Treating the in;it as differential in this way will significantly improve the CMRR if you use a wire pair for connection.

Thanks, but I didn't understand what do mean by "you should reference the input to 4.5V (including the - pin)", I modified the circuit by ground pin4 and connect +4.5v to pin5 "ref pin" and the output is still not correct and it's dc line at about 4V.

I really want to supply INA with only one battery but in the datasheet of INA: the supply range is (±2.25 to ±18V) , Is that mean I should supply it with + and - voltage or not?

I'm working on project of design EMG circuit and I don't know what range of supply I should have and I know that related to the gain but I mean How INA will work if the supply voltage is small (e.g. ±3V)