( ) D. An information signal x( t) = 5cos( 1000πt) LSSB modulates a carrier with amplitude A c

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1 An inormation signal x( t) 5cos( 1000πt) LSSB modulates a carrier with amplitude A c 1. This signal is transmitted through a channel with 30 db loss. It is demodulated using a synchronous demodulator. Noise with power spectral density G n ( ) / adds to the signal at the receiver input. Find the S / N, in db. For SSB, ( S / N ) D and W 500. Thereore ( S / N ) D W, S A 2 c S x R L / or db

2 A signal x( t) 2cos( 1000πt) + cos( 2000πt) USSB modulates a carrier with amplitude A c 10. The modulated carrier is transmitted through a channel with a loss o 70 db. Noise o power spectral density ( ) / adds to the signal at the receiver input. (a) Find the G n signal-to-noise ratio S / N. (b) Suppose a bandpass ilter with passband between 400 and 1100 Hz is added to the output o the receiver. Find the improvement in S / N For SSB, ( S / N ) D in db. W, A 2 c S x 4L 10 2 ( 2 2 / / 2) , and W Thereore ( S / N ) 6 D or db. I we bandpass ilter the output signal between 400 and 1100 Hz we reduce the noise bandwidth by a actor o 700/1000 or 0.7. That increases the signal-to-noise ratio by the reciprocal o that actor making it or db, an increase o db.

3 SB modulates a carrier o amplitude A c 3. The power spectral is shown in the igure below. White noise o spectral density ( ) / is added at the receiver input. The channel loss. A signal x t density o x t G n is 35 db. Find the signal-to-noise ratio S / N The received signal is x c A c L x( t)cos ω ct ( t) A c L x( t)cos ( ω ct). Its signal power is 2 A 2 c L x2 ( t) cos 2 ω c t. x2 t 1/2 is the area under its power spectral density which is 10 m. So the received signal power is m and m or DSB, ( S / N ) D W m m or db. m G x 10 ( ) m

4 A message x( t), with W 5kHz, DSB modulates a carrier o requency c 1MHz and amplitude A c 1. The channel loss is 20 db. Nonwhite noise with power spectral density as shown in the igure below adds to the signal prior to detection with a synchronous demodulator. Find the signal-to-noise ratio S / N 1 W. assuming that the power o the signal is 10 9 G n ( ) 2MHz

5 The received signal is x c A c v t L x t + n i ( t) ( t) A c L x( t)cos ( ω t c ). The signal plus noise is cos ω t c n q t sin ω c t The synchronously demodulated signal plus noise is y( t) y D ( t) A c 2 L x( t) + n t i 2 G n G ni ( ) 10 9 Λ( / ) G ni ( ) G nq ( ) 10 9 Λ( ( ) / )u + c N D n i t 2 N D 1/ 2 2 G ni 1 / 4 W d 1 / 4 W W Λ S D A 2 c 4L x2 ( t) ( / 2 10 ) 6 / u + c +G n ( c )u( c ) ( / )u( c ) ( / )u( + c ) +Λ( ( 10 6 ) / )u c G n + c + Λ 10 6 W Λ W +Λ( 10 ) 6 d 1/ W S / N S D N D 1,000 or 30 db d

6 An inormation signal x( t) cos( 2000πt) requency modulates a carrier o requency 1MHz. A c 2 and Δ 100Hz/V. The channel loss is 90 db. The receiver noise temperature is T 2000K. (a) Find the signal-to-noise ratio S / N in db. (b) I the output signal is passed through a bandpass ilter with passband rom 900 to 1100 Hz ind the improvement in S / N in db. (c) Repeat parts (a) and (b) or AM with synchronous demodulation and compare results.

7 (a) A 2 c 2L and 2000 kt kt 0 ( T / T 0 ) W 1000 D and B 2 D +1 T W B T >> B T so the approximations or high signal-to-noise ratio apply. S ( S / N) D 3D 2 x W 3 ( 0.1 ) 2 1/ or db (b) I the output signal is passed through a bandpass ilter with passband rom 900 to 1100 Hz the noise bandwidth is reduced by a actor o 200/1000 or 0.2. This increases the signal-to-noise ratio by a actor o 1/0.2 or 5, making it or db, an increase o 6.99 db.

8 (c) For AM modulation (with µ1) ( S / N ) D A c 2 S x W ( 1+ S x ) / 2L / 2 1+ S x ( S / N ) D / / ( 1+1/ 2) or db I the output signal is passed through a bandpass ilter with passband rom 900 to 1100 Hz the noise bandwidth is reduced by a actor o 200/1000 or 0.2. This increases the signal-to-noise ratio by a actor o 1/0.2 or 5, making it or db, an increase o 6.99 db.

9 A signal x( t) with power spectral density G x ( ) KΠ( / 8000) is transmitted over a telephone channel with transer unction H c ( ) 4. The noise power spectral density at the receiver j To compensate or the channel distortion, the input is G n receiver ilter transer unction is chosen to be j + H D ( ) Π( / 8000). The receiver's ( S / N ) D is required to be at least 35 db. Determine the minimum required value o K and the corresponding transmitted power S T and the received power.

10 G xc G yd G yd ( ) G x ( ) H c ( ) 2 KΠ( / 8000) ( ) G xc ( ) H D ( ) 2 KΠ( / 8000) ( ) KΠ( / 8000) S D KΠ( / 8000) 42 d 16K d 0.008K 7 G n Π 2 / 8000 ( ) N D d / ( S / N ) D S D N D 35 db is a S / N 0.008K 7500K o 3,162.3 K 3,162.3 / S T Π( / 8000) 2 d tan 1 ( 1) tan d tan 1 /

11 A message x( t) is a zero-mean Gaussian random signal with W 40 khz and σ x 0.3. It AM modulates a carrier o amplitude A c 1 (with µ 1). The channel loss is 70 db. The noise temperature o the receiver is T 10,000K. I the demodulation is synchronous, ind the signal-to-noise ratio at the destination. A 2 c 2L 1+ S x kt ( S / N ) D S x W 1+ S x ( S / N ) D or db

Solution of ECE 342 Test 3 S12

Solution of ECE 342 Test 3 S12 Solution of ECE 34 Test 3 S1 1 A random power signal has a mean of three and a standard deviation of five Find its numerical total average signal power Signal Power P = 3 + 5 = 34 A random energy signal