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PI M
Pin
= ( (i) )2
Pd
P
I (i)
PI where
is the twotone thirdorder 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 [105] 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.
 Spring '08
 Staff
 Frequency, Volt

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