benji_styler
New Member
Hi i have attempted answering this question:
The unwanted noise power is measured at the output of an amplifier by measuring the output power with the input shorted to 0v (ground). The measurement is 2mW. An input is then applied to produce full output power. The result is 26W.Calculate the signal to noise ration of the amplifier in db.
this is my answer si it correct?
Sn ratio = 20 log 10 ps/pn
=
20 log 10 26w/0.002w = 180641 db
I think i have miscalculated the end answer as i do not have a scientific calc to hand and am using the windows one...
Can anyone tell me if i have calculated it correctly?
Many thanks - wasnt sure where to post but it has been bothering me![Confused :confused: :confused:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
The unwanted noise power is measured at the output of an amplifier by measuring the output power with the input shorted to 0v (ground). The measurement is 2mW. An input is then applied to produce full output power. The result is 26W.Calculate the signal to noise ration of the amplifier in db.
this is my answer si it correct?
Sn ratio = 20 log 10 ps/pn
=
20 log 10 26w/0.002w = 180641 db
I think i have miscalculated the end answer as i do not have a scientific calc to hand and am using the windows one...
Can anyone tell me if i have calculated it correctly?
Many thanks - wasnt sure where to post but it has been bothering me