hwset12solution

# If all powers are expressed in dbm then i pi m pd 2pin

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Unformatted text preview: ationship: PI M Pin = ( (i) )2 Pd P I (i) PI where is the two-tone third-order intercept input power. If all powers are expressed in dBm, then: (i) PI M − Pd = 2(Pin − PI ). (i) Solving for PI : (i) PI 1 = Pin + (Pd − PI M ). 2 In this problem, Pin = −20 dBm, Pd = −3 dBm, PI M = −30 dBm; thus: (i) PI 1 = −20 + (−3 + 30) = −6.5 dBm. 2 (b) G = Pd − Pin = −3 − (−20) = 17 dB. 1 [10-5] Let us first calculate the equivalent noise temperature of the receiver, Ter NF Ter T0 8dB 10 log 1 Ter T0 10 NF 10 1 1539.8 K Now we can calculate the equivalent noise temperature of the preamplifier plus the receiver. Te Ter Ga Tea 200 1539.8 12 10 297.2 10 and Top = Ts + Te = 1297.2 K Since SNRout S in kTop Bn the input power required to give 15dB SNRout is S in kTop Bn SNRout (1.38 10 2.83 10 23 14 ) 1297.2 (50 10 3 ) 10 15 10 (W ) or in dBm S in 10 log 2.83 10 10 3 14 105.5 ( dBm ) (b) Due to the gain in front of the receiver, the input intercept point of the cascade(preamplifier + receiver) is reduced by Ga = 12dB. Thus ) PI(,iTotal 6.5 dBm 12dB 5.5 dBm The dynamic range in dB is DR 2 (i )...
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## This homework help was uploaded on 03/13/2014 for the course ECE 453 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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