This impedance is a constant (50 Ω) with respect to attenuation– impedance does not change when attenuation is changed. The Z-(arrows) pointing toward the attenuator indicates that the impedance seen looking into the attenuator with a load Z on the opposite end is Z, Z=50 Ω for our case. The Z-(arrows) pointing away from the attenuator in the figure below indicate this. The T and Π attenuators must be connected to a Z source and Z load impedance. Find the attenuation in dB.Įxample: Find the voltage attenuation ratio (K= (V I / V O)) for a 20 dB attenuator. Find the attenuation in dB.Įxample: Find the voltage attenuation ratio (K= (V I / V O)) for a 10 dB attenuator.Įxample: Power into an attenuator is 100 milliwatts, the power out is 1 milliwatt. That is, the source and load resistance need to be equal.Įxample: Power into an attenuator is 10 Watts, the power out is 1 Watt. Once again, the voltage ratio form of equation is only applicable where the two corresponding resistors are equal. We will use the latter form, since we need the voltage ratio. The two most often used forms of the decibel equation are: The power change in decibels in terms of power ratio is:Īssuming that the load R I at P I is the same as the load resistor R O at P O (R I = R O), the decibels may be derived from the voltage ratio (V I / V O) or current ratio (I I / I O): A gain of -3 dB is the same as an attenuation of +3 dB, corresponding to half the original power level. An attenuation of 3 dB corresponds to cutting power in half, while a gain of 3 db corresponds to a doubling of the power level. \]Ī change of 1 dB in sound level is barely perceptible to a listener, while 2 db is readily perceptible.
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