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eee206_2004_sols - Solutions to EEE206/274_Spring Semester...

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Unformatted text preview: Solutions to EEE206/274_Spring Semester 2003—2004 B.Chamhers l.(a) The type of diode detector and its mode of operation is dependent on the working point on the diode’s I-V characteristic. Square-Law detector works in the non—linear region of the characteristic. Good for small signal input but demodulated signal may have distortion — OK for radar receiver — detection of weak signals in the noise. Envelope detector works on linear part of the characteristic, hence needs large signal input or signal + dc. bias. Demodulated signal has much less distortion if m is small, hence OK for sound broadcasting. (b) and (c) Students to reproduce from lecture notes (at) m = 0.6, mm = 2 11: 5x103. This gives RC g 4.244 x 10'5. Suggest R = 3K9 Q, C = 0.01 uF. Then RC = 3.9 x 10'5 This is OK and uses preferred values. (e) Analysis assumes small linear capacitor discharge between carrier peaks. This requires f; > 100 fm. Hence the hidden dependence of equation 1 on fc. If this is not fiJIfilled the discharge becomes sawtooth-like with increased distortion. (f) When in approaches 1, parts of the signal envelope approach 0 and the voltage is then below the diode turn-on point. Hence distortion of the demodulated signal. 2(a) TRF obtains signal gain using several stages at the input signal frequency, followed by demodulation. Difficult and expemsive to tune the amplifier stages and make them ‘track’. Superhet obtains most of its gain at the fixed IF frequency by using a tunable L0 and mixer. Possibility of image frequencies. 0)) Draw receiver diagram from notes. AGC makes use of do. signal from the envelope demodulator, whose value is proportional to signal strength, to control the gain of amplifier stages prior to demodulation. (0) Local oscillator frequencies 145 ~ 69 = 76 MHZ, 69 + 9 = 78 MHz 9 + .455 = 9.455 MHZ Images directly into the mixers Imagel= l45—2x69=7MHz Image2=69+2x9 =87MI-Iz Image 3 = 9 + 2x .455 = 9.91 MHZ Image into the 2'ml mixer Image 4 = 78 'l' 9.91 = 87.9] W2 Images into the l51 mixer Image 5 = 76 + 87.91 = 163.91 MHz Imag66=76+87 = 163 MHz s 3. C=Bl .1+— (a) 0s.( N) C = information carrying capacity of the channel in bits / sec B = bandwidth in Hz S = signal power in watts N = noise power in watts (b) Plot either log;(] + S/N) or SIN against B Consider a two-level signal, 6. g. square wave. Then this has good noise immunity, hence needs only small signal power, but needs wide bandwidth. If now make the signal multi-level, it becomes more "rounded" - hence needs less bandwidth but has less noise immunity and so needs increased signal power for a given SfN. (c) 32 vertical lines / frame, therefore 32 pixels per frame horizontally. Aspect ratio is 3 :2 (portrait) and so there are 48 pixels vertically per frame. Number of pixels / frame = 32 x 48 = 1536 12.5 frames a“ sec and so 12.5 x 1536 = 19200 pixels / sec Each pixel has 8 possible brightness levels = 3 bits Bits / sec = 3 x 19200 = 57600 SIN = 30 dB = 1000 (linear) B = C I log (1 + 1000) = 57600 I log; (1001) = 5760019967 = 5.779 kHz Actual bandwidth of AM DSB signal is twice this = 11.558 KHz (d) Can estimate bandwidth using Carson’s rule B = 2 fm (l + m) - fm is approximately the number of bits /sec = 5 7600 and m = 1. Hence B = 2 x 57600 x 2 = 230400 = 230.4 kHz. This is very large compared with AM DSB but exchanges BW for SIN. 4(a) Advantages: can handle mixed (i.e. analogue and digital) data, multiplexing, error detection and correction, regeneration Disadvantages: wider bandwidth (b) Bookwork, covered in lectures. PCM performance will be governed by sampling rate, signal to quantisation noise ratio, i.e bits/sample. (c) (i) Quantisation interval for 8 bitslsample = 5f256 V (ii) 10000000 = 5x1281256 = 2.5 V, 11100101 = 5x229/256 : 4.472 V (d) For NRZ pulses and s bits/sample, bandwidth = 2 x 6 x 103 x 8 = 96 kHz ...
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