ELECTRICAL WIRING:RESIDENT.-TEXT (PB)
19th Edition
ISBN: 9781337116213
Author: MULLIN
Publisher: CENGAGE L
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Chapter 30, Problem 6R
What is the purpose of grounding and bonding? _____________________________
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PSD
A certain signal f(t) has the following PSD (assume 12 load):
| Sƒ(w) = π[e¯\w\ + 8(w − 2) + +8(w + 2)]
(a) What is the mean power in the bandwidth w≤ 1 rad/sec?
(b) What is the mean power in the bandwidth 0.99 to 1.01 rad/sec?
(c) What is the mean power in the bandwidth 1.99 to 2.01 rad/sec?
(d) What is the total mean power in (t)?
Pav=
+
2T
SfLw) dw
- SALW)
An AM modulation waveform signal:-
p(t)=(8+4 cos 1000πt + 4 cos 2000πt) cos 10000nt
(a) Sketch the amplitude spectrum of p(t).
(b) Find total power, sideband power and power efficiency.
(c) Find the average power containing of each sideband.
Can you rewrite the solution because it is
unclear?
AM
(+) = 8(1+ 0.5 cos 1000kt +0.5 ros 2000ks)
=
cos 10000 πt.
8 cos wat + 4 cos wit + 4 cos Wat coswet.
-Jet
jooort
J11000 t
= 4 e
jqooort jgoort
+4e
+ e
+e
j 12000rt.
12000 kt
+
e
+e
jooxt
igoo t
te
(w) = 8ES(W- 100007) + 8IS (W-10000)
USB
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ELECTRICAL WIRING:RESIDENT.-TEXT (PB)
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- Can you rewrite the solution because it is unclear? AM (+) = 8(1+0.5 cos 1000kt +0.5 ros 2000 thts) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. J4000 t j11000rt $14+) = 45 jqooort +4e + e + e j 12000rt. 12000 kt + e +e +e Le jsoort -; goon t te +e Dcw> = 885(W- 100007) + 8 IS (W-10000) - USBarrow_forwardCan you rewrite the solution because it is unclear? Q2 AM ①(+) = 8 (1+0.5 cos 1000πt +0.5 ros 2000kt) $4+) = 45 = *cos 10000 πt. 8 cos wat + 4 cosat + 4 cos Wat coswet. j1000016 +4e -j10000πt j11000Rt j gooort -j 9000 πt + e +e j sooort te +e J11000 t + e te j 12000rt. -J12000 kt + с = 8th S(W- 100007) + 8 IS (W-10000) <&(w) = USB -5-5 -4-5-4 b) Pc 2² = 64 PSB = 42 + 4 2 Pt Pc+ PSB = y = Pe c) Puss = PLSB = = 32 4² = 8 w 32+ 8 = × 100% = 140 (1)³×2×2 31 = 20% x 2 = 3w 302 USB 4.5 5 5.6 6 ms Ac = 4 mi = 0.5 mz Ac = 4 ५ M2 = =0.5arrow_forwardA. Draw the waveform for the following binary sequence using Bipolar RZ, Bipolar NRZ, and Manchester code. Data sequence= (00110100) B. In a binary PCM system, the output signal-to-quantization ratio is to be hold to a minimum of 50 dB. If the message is a single tone with fm-5 kHz. Determine: 1) The number of required levels, and the corresponding output signal-to-quantizing noise ratio. 2) Minimum required system bandwidth.arrow_forward
- Find Io using Mesh analysisarrow_forwardFM station of 100 MHz carrier frequency modulated by a 20 kHz sinusoid with an amplitude of 10 volt, so that the peak frequency deviation is 25 kHz determine: 1) The BW of the FM signal. 2) The approximated BW if the modulating signal amplitude is increased to 50 volt. 3) The approximated BW if the modulating signal frequency is increased by 70%. 4) The amplitude of the modulating signal if the BW is 65 kHz.arrow_forwardAn FDM is used to multiplex two groups of signals using AM-SSB, the first group contains 25 speech signals, each has maximum frequency of 4 kHz, the second group contains 15 music signals, each has maximum frequency of 10 kHz. A guard bandwidth of 500 Hz is used bety each two signals and before the first one. 1. Find the BWmultiplexing 2. Find the BWtransmission if the multiplexing signal is modulated using AM-DSB-LC.arrow_forward
- An FM signal with 75 kHz deviation, has an input signal-to-noise ratio of 18 dB, with a modulating frequency of 15 kHz. 1) Find SNRO at demodulator o/p. 2) Find SNRO at demodulator o/p if AM is used with m=0.3. 3) Compare the performance in case 1) and 2).. Hint: for single tone AM-DSB-LC, SNR₁ = (2m²) (4)arrow_forwardFind Va and Vb using Nodal analysisarrow_forwardCalculate the nodal voltage in the circuitarrow_forward
- Calculate the mesh currents, find Ia, Ib, Ic. Apply mesh analysisarrow_forwardFind Va and Vb using Nodal analysisarrow_forward4. A battery operated sensor transmits to a receiver that is plugged in to a power outlet. The device is continuously operated. The battery is a 3.6 V coin-cell battery with a 245mAHr capacity. The application requires a bit rate of 36 Mbps and an error rate of less than 10^-3. The channel has a center frequency of 2.4 GHz, a bandwidth of 10 MHz and a noise power spectral density of 10^-14 W/Hz. The maximum distance is 36 meters and the losses in the channel attenuates the signal by 0.25 dB/meter. Your company has two families of chips that you can use. An M-ary ASK and an M-ary QAM chip. The have very different power requirements as shown in the table below. The total current for the system is the current required to achieve the desired Eb/No PLUS the current identified below: Hokies PSK Chip Set Operating Current NOT Including the required Eb/No for the application Hokies QAM Chip Set Operating Current NOT Including the required Eb/No for the application Chip ID M-ary Voltage (volts)…arrow_forward
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