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Posted: **Wed Apr 27, 2022 5:50 pm**

: P1 A Standard Wired Telephone Circuit Has A Usable Frequency Range Extending From 600 Hz To 3000 Hz With A Signal To No 1 (115.25 KiB) Viewed 33 times

: P1 A Standard Wired Telephone Circuit Has A Usable Frequency Range Extending From 600 Hz To 3000 Hz With A Signal To No 2 (83.95 KiB) Viewed 33 times

: P1 A Standard Wired Telephone Circuit Has A Usable Frequency Range Extending From 600 Hz To 3000 Hz With A Signal To No 3 (86.09 KiB) Viewed 33 times

I already have a solution for the first two questions, but I want to solve the rest of the branches if possible. The previous solution is attached to the pictures. If the rest is solved, I will give a like to the owner of the solution
P1. A standard wired telephone circuit has a usable frequency range extending from 600 Hz to 3000 Hz with a signal-to-noise ratio of about 30 dB. According to the channel capacity expression, what is the maximum data rate that can be achieved with this system? P2. Evaluate the Shannon channel capacity theorem for an IS-54 cellular phone system. The channel bandwidth is 30 kHz. Assume a receive signal power of -60 dBm and no/2= le-18 W/Hz. P3. A transmitter requires the IM3 products to be -65dB below the main tones when the total output power is 20 dBm. Calculate the output IP3 of this transmitter. P4. The output 1P3 of the final amplifier in a transmitter is 33 dBm. To reduce the distortion of the transmitted waveforms, this amplifier should operate at a 6 dB back-off mode. What is the peak output signal power that can be transmitted with this transmitter? Noted that Back-off mode means to reduce the gain of the amplifier by the value of back-off (6dB). P5. Calculate the OIP3 of the cascaded stages shown below. How would the performance be affected if the order of stages is reversed? G=20 dB OIP3 =22dBm Conversion Loss56 dB TIP3 = 13dBm OIP3 = ? Amplifier Mixer
Pi) Now, from given Channel capacity theorem. Acc to the c3 Blog2 ( (1+ Р. No B ) Bitt/sec c.=R6 B = 3000 600 H ६ B=2400 Hz H3 = 30dB Now, log (so) so 2103 = NO 1000 P = So= 1000 NO 1000 NO c= 2400 log2 (1+ (it ) NO (2400) : +) = 2400 log2 (it 10 = 1206 bps 24 Rb = c = 1.206 kbps is maximum data rate that can be achieved with the system,
channel As per given P2)=) According to the shanon capacity theorem. Р lit ) C = Blog, NOB Р c = 30x103 log2 fit No (30 X103) It رکیا N10 Pdig =-60 dBm 10 20910 Pmw = -60 dBm Pmw = 10-6 P=10 bmw P = 109 w Noo = 10 W/HZ 10-9 2. 104 60 c = 30x103 log, (i+ 1+ 2x10-18 X30X163 = 30x103 dog, (at ( = 30 x103 log2 (0+105) Co 420.743 kbps For Is 54 cellular phone. 1+ 6