could you tell me how to get value of voltage from this oscilloscope I do n't know how to get it.

 

Measure the height of each waveform on the screen. For yours one is about one division and the other is about 3 divisions. Multiply those values by the volts/div for that channel. If the X5 MAG is pulled for either channel, then multiply that channel by 5.

This assumes both variable volts/div controls are in the cal position.

 

The higher amplitude sine wave looks to be about 2.5 divisions peak to peak. The right hand picture show that the gain is set to 20 mV per division. So that makes it 2.5 x 20 mV = 50 mV. You then have to consider that most scope probes attenuate the signal by a factor of 10. So if you are using 1/10 probes then the voltage at the probe tip would be 500 mV. If you want to convert the peak to peak value to RMS then you divide it by 2 the square root of 2. So you divide by 2 x 1.414 = 2,828

Les.

 

So the sin graph would be 2.8*20mV = 56mVp-p and another is 1*5V = 5Vp-p
and If I want to calculate Voltage gain (it's a common emitter circuit) it would be (5Vp-p)/(56mVp-p) = 89.29, is that wrong?

 

Yes, I forgot about X10 probes. But from the pictures you can't tell which waveform belongs to which channel.

 

I used channel 1 for the sine wave(input signal) and channel 2 for another graph(output signal).
My work is almost done thanks a lot.

 

Make sure the VARIABLE knob on the INPUT controls are set to the calibrated position CAL.

Also note that your output signal is badly distorted. Looks like the bias on your BJT common-emitter amplifier is set too low.