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Microphone to cable frequency?
A microphone with an impedance of 18 kΩ is connected to a cable of length 370 ft. The cable is specified as having a capacitance of 59 pF/ft and a resistance of 31 ohms/1000ft. Ignoring what is connected at the other end of the cable, at what frequency (in kHz) would you expect the signals to be attenuated by about 6 dB?
2 Antworten
- JamesLv 6vor 8 JahrenBeste Antwort
dB = 20 log (Vo / Vi)
-6 = 20 log (Vo / Vi)
-0.3 = log (Vo / Vi)
anti log both sides
.501 = (Vo / Vi)
the ratio of the voltage is .501
since we are ignoring the connection at the far end , assume it is open circuit,
the equivalent circuit is a voltage divider. and since the far end is open the resistance of the cable has no effect ( no voltage drop if no current)
Vo =[ ( -j /wC) / ( 18 x 10^3 + ( -j /wC) ) ] x Vi
Vo/ Vi =[ ( -j /wC) / ( 18 x 10^3 + ( -j /wC) ) ]
.501 =[ ( -j /wC) / ( 18 x 10^3 + ( -j /wC) )
solve the above for w(omega)
freq = w/ (2 pi)
- inkavesvanitcLv 4vor 4 Jahren
Henre is sturdy, the fifty 8 is a lots better suitable vocal mic. i'm uncertain approximately in actuality making use of an XLR to a million/4" cable, that would desire to paintings. i might use an on the spot container to alter the mic from severe impedance to low for the amp.
