Hallo,
Die Überschrift sagt's ja schon, meine UHF Messoftware spukt Messwerte in db/√Hz aus ( so steht es am Plot dran ), die Software zum auswerten der Daten erwartet die Werte in dbm.
Ich stehe da mal wieder auf dem Schlauch die Daten korrekt um zu rechnen, und wäre dankbar wenn mir jemand helfen könnte.
Gruß Lars
db/√hz zu dbm
db/√hz zu dbm
wer anderen eine Bratwurst brät, braucht ein Bratwurst Bratgerät.
-
- Beiträge: 209
- Registriert: Di Sep 28, 2010 9:56 pm
- Wohnort: Ulm, Germany
Re: db/√hz zu dbm
Lars,
das hier schon gesehen:
https://answers.yahoo.com/question/inde ... 914AAfz5hR
oder:
http://www.edaboard.com/thread280079.html
Speziell der Eintrag #4:
Usually the noise is quantified in terms of power-density with respect to the bandwidth, that is (using "dB" values) dBm/Hz. It can also be expressed in voltage as V/sqrt(Hz) or current as A/sqrt(Hz).
Then, if you have, for instance, a constant power density (that is power over 1Hz bandwith) of -70dBm/sqrt(Hz) over a BW=5kHz, then the total noise power over that bandwidth will be Pn=-70+10*Log(5000)=-33 dBm. If you have, instead voltage or current units, to convert to power you need to know the resistance on which it is applied and use P=V^/R or P=R*I^2 (and, if you want a log form, convert to dBm).
In you post you wrote -70 dB, but dB is not a measurement unit; it represent instead the ratio between two quantities expressed in log-form.
In the picture the title is power(dB) that is wrong: to express power in "dB" you have to use dBm (power referred to 1mW), dBW (power referred to 1W), an so on. The y-axis is in dBV that is a voltage, not a power.
das hier schon gesehen:
https://answers.yahoo.com/question/inde ... 914AAfz5hR
oder:
http://www.edaboard.com/thread280079.html
Speziell der Eintrag #4:
Usually the noise is quantified in terms of power-density with respect to the bandwidth, that is (using "dB" values) dBm/Hz. It can also be expressed in voltage as V/sqrt(Hz) or current as A/sqrt(Hz).
Then, if you have, for instance, a constant power density (that is power over 1Hz bandwith) of -70dBm/sqrt(Hz) over a BW=5kHz, then the total noise power over that bandwidth will be Pn=-70+10*Log(5000)=-33 dBm. If you have, instead voltage or current units, to convert to power you need to know the resistance on which it is applied and use P=V^/R or P=R*I^2 (and, if you want a log form, convert to dBm).
In you post you wrote -70 dB, but dB is not a measurement unit; it represent instead the ratio between two quantities expressed in log-form.
In the picture the title is power(dB) that is wrong: to express power in "dB" you have to use dBm (power referred to 1mW), dBW (power referred to 1W), an so on. The y-axis is in dBV that is a voltage, not a power.
Re: db/√hz zu dbm
Das hatte ich nicht gefunden...
Ok, BW ist die dann Bandbreite meiner gesamten Messung ?
Also, wenn ich einen Scan von 400 MHz bis 500 MHz alle 25 kHz mache, und bei 300mhz z.b. -29db/√hz angezeigt bekomme, sieht die Formel z.b. So aus aus ?:
Pn=-29+10*Log(10000000)=-22 dBm
Gruß
Ok, BW ist die dann Bandbreite meiner gesamten Messung ?
Also, wenn ich einen Scan von 400 MHz bis 500 MHz alle 25 kHz mache, und bei 300mhz z.b. -29db/√hz angezeigt bekomme, sieht die Formel z.b. So aus aus ?:
Pn=-29+10*Log(10000000)=-22 dBm
Gruß
wer anderen eine Bratwurst brät, braucht ein Bratwurst Bratgerät.