telecom-Jacarri.doc

a.The amplitude and phase responses of a filter network are shown in Figure 1. Estimate (i)the cut-off frequency, (ii)the roll-off in dB per decade and (iii)the phase response at the cut-off frequency.1. (i) Cut off frequency, -3dB point = 1kHz from the graph (1 mark) 2. (ii) Roll-off: choose linear portion of graph say between decade 104 (-20db) and 105(-40dB). Therefore, roll-off is -20dB per decade. (2 marks) minus sign is important 3. (iii) Phase response is 45 degrees 2 mark b. With reference to figure 1, how would you categorise the filter? Give an example of where this type of filter might be used.This filter could be used to block all RF frequencies from a speech circuit. (2.5 marks)Determine the Q factor of a resonant circuit with a centre frequency of 200MHz and a bandwidth of 150 kHz. Briefly explain why this Q factor may be difficult to achieve with inductors and capacitors.Q=(fo/BW)= 200x106/150 x 103 = 1333 (3 marks)In practice, inductors and capacitors have resistive losses which limits the Q factor. In particular the component Q of a lumped inductor is much less than 500. (2 marks)Sketch the frequency spectrum for the complex waveform;v(t) = 2 Sin (1 x 106)t + 4 Sin (3.14 x 106)t + 8 Sin (6.28 x 106)tNote that the frequency scale must be shown with units of Hertz.Sinusoids are: (dividing radian frequency by 2pi). 2 marks for calculations, 3 marks for labeled sketch.159kHz,2V500kHz,4V1MHz, 8VAn receiver antenna has a power output of 36dBm into a load resistance of 75ohms. Calculate the rms voltage across the load.dBm=10log(P/(1x10-3) (2 mark)preferred method;36=30+3+3 which is a power ratio of 103 x2x2 = 4x103Therefore Pout= 1mW x 4 x103 = 4W (2 marks) v=SQRT(PR) = SQRT(4x75) = 17.32v(rms) (1 mark)其他解法:dBm=10log(P/(1x10-3)36=10log(P*1000)3.6=log(1000*Vrms2/R)103.6=1000* Vrms2/75Vrms=17.27vDraw a circuit diagram for a single transistor tuned amplifier that will provide maximum gain at a single frequency.The circuit should use an NPN transistor, with appropriate biasing components and an output load resistor. The values of the biasing components are not required.Multimode propagationThis type of propagation allows for beams to enter the fibre at different anglesSinglemode FibreThe core diameter is narrow, less than the wavelength of the ray travelling down itOnly one ray or mode can travel down the fibreDistance-bandwidth product is 100GHz-kmUNIT1-5.51.计算基准频率,谐波例1.The Fourier series for a square wave is shown. What is the power in dBm in the 89 harmonic if it is terminated in a 93 Ohm resistor?Vpeak=(4/) * (1/89)=0.0143vVrms=Vpeak / 2=0.01vP=(Vrms)2/R=1.1wdBm=10log(P/10-3)=-29.58例2.What frequency in Hertz is the 3rd harmonic of the following signal1.5sin(775329.8t)=2f775329.8=2*3.14*ff=123460.159HZso 3rd harmonic frequency=3*f=370380.48HZ2.满足波形的最小带宽:(有BPSK调制的频率不变,没有的话频率为原来的一半)例1.The waveform in the figure shows a 6kbit/s bit pattern of alternating ones and zeros. A pulse of +1Volt represents binary 1 and a pulse of 1Volt represents Binary 0.Calculate the minimum channel bandwidth in kHz that is required to transmit the signal without errors.Answer:3 kHz例2.This 3kbit/s data source is modulated using BPSK. Calculate the minimum bandwidth of this transmission in kHz. Answer:3 kHz3.PSK图形其带宽BW=fo4.画OFDM信号的频幅特性图,3个载波,并解释为何干扰小.There is little interference between sub channels because each sub channel is orthogonal to its adjacent (next, beside it) sub channel.5.说明QPSK,16QAM,64QAM的特点与应用QPSK, 16QAM and 64QAM modulation schemes are used in OFDM. Modulation schemes in adjacent channels can be different to ensure minimum interference. OTN,SDH,PDH的特点1.OTN Stronger Forward Error Correction More Levels of Tandem Connection Monitoring (TCM) Transparent Transport of Client Signals Switching Scalability 2.SDHSDH is a timing hierarchyAllows higher and more flexible bit-rates than PDHUses a different frame structure and synchronization Hierarchical synchronization More accurate synchronization reduces slippage3.PDHIn plesiochronous operation the nodes are synchronized on the E1 (T1)-level. Both mutual synchronization and master-slave synchronization modes possible. High precision clock means here a clock with a stability range of about 10-11 . 10-13, implying use of atomic clocks, possibly UTC or GPS coordinated. 4.PCM在SDH中的应用I.e. both E1 and T1 PCM multiplexes are compatible with SDH. 解释TDM通信系统的复用方式画3通道的TDM通信系统描述PCM编码过程8个分立的语音信号,每个最大4KHZ,传输于7BIT的.PCM帧结构说明,bit含义量化信噪比SQR(以上答案见下(question4)Question 4(a) Explain what is meant by multiplexing in a communications system. Draw a block diagram of a three channel TDM communications system (Either PCM or PAM).multiplexing is a technique where multiple information sources (speech or data) can share the same physical channel resource (cable or free space) One method is Time division multiplexing (TDM), another method is Frequency Division Multiplexing(FDM).(b) List and briefly explain the essential processes in a pulse code modulator (PCM) system.Since any digital representation of a continuous analogue signal must be pulses, the first step in the analogue to digital conversion process is (1) sampling of the analogue signal. The sample is then (2) quantized and then assigned a digital value according to its amplitude. (3)The digital number is then transmitted as digital pulses.Diagram also acceptable:(c) Eight separate speech signals, each with a maximum frequency component of 4 kHz are to be transmitted using a 7 bit PCM system. Calculate the lowest possible bit rate for the PCM encoder.Fm = 4kHzFs(min) = 2fm = 8kHz Bit rate per channel = fs *(no. of bits per sample) = 8000 * 7 = 56000bit/s Lowest possible bit rate = no. of channels * bit rate per channel= 56000 * 8 = 448 kbit/s (d) With the aid of a diagram and a brief explanation, describe the European Standard ITU primary rate PCM frame, mentioning its frame duration, its composition, its no. of bits and for which purposes are its time slots used.Channel 0 is used for frame synchronization Channel 16 is used for signallingIn this system, 30 voice channels are multiplexed, using 8000 Hz sampling and 8 bit coding. Framing and signalling are allocated 8 grouped bits each per frame. The total number of bits in each frame can be worked out as follows:30 * 8-bit voice time slots = 240 bits1 * 8- bit signalling time slot = 8 bits1 * 8- bit framing time slot = 8 bitstotal number of bits in one frame = 256 bitsThere are 8000 frames/second and therefore the system bit rate is 8000 * 256 = 2.042 Mb/sec.(e) A PCM system employing linear quantisation has a certain Signal to Quantisation Noise Ratio (SQR). Explain what SQR means and calculate the number of bits per sample for an SQR of 48dB.SQR is the ratio of signal power to quantization noise power. For a linear quantiser: SQR (dB) = 1.76 +6nN = (SQR-1.76)/6 = (48 -1.76)/6 = 7.7 Round to the next highest integer gives 8 bits1.AM的优缺点disadvantages:Power: We have seen from the previous lecture that in conventional AM, transmitted carrier power constitutes two-thirds or more of the total transmitted power. The carrier contains no information so this means that AM is not power efficient. Note that conventional AM is referred to as Full Wave transmission or Double Sideband Full Carrier (DSBFC).Bandwidth: The bandwidth of conventional AM is twice the maximum frequency component of the modulating signal . This is because the sidebands are duplicated (Upper sideband and Lower sideband) All the information can be conveyed by the use of one side- band only, resulting in considerable power savings.2.FM的优缺点advantages:the signal to noise ratio at the output of a FM detector is better than an AM detector. An FM detector in a radio receiver will only demodulate the strongest signal at a particular frequency. This means that if there is a weaker signal operating on the same frequency (say from a different geographical area) is will not interfere with the main signal. Variations in signal strength do not change the quality of the received audio (unless the signal strength drops below a minimum threshold) 代码的起始,校验等In serial data transmission, the following bits are received. Decode each character111000100010110100000101111The communications settings are: 7 data bits (ASCII), one parity bit, 1 start bit, 2 stop bits. Note that the left side is transmitted first to the line. Redraw the data pattern and indicate (i) positions of idle bits, start bits, stop bits and parity bits and (ii) the two actual ASCII characters transmitted.0100 0100 = 44Hex = D (2 marks) 0100 0001 = 41Hex = A (2 marks)Parity is even (2 marks)IF的内容,目的The IF is the difference between the wanted signal frequency and the local oscillator (LO) frequency.fLO = fs + fIF例.If the Intermediate Frequency in a Superhet radio receiver is 3262910Hz and the Local Oscillator is set to 22011686Hz what frequency will be received?fs=flo-fif=22011686-3262910 =18748776HZTRF射频通信AF(Audio Frequency)内容,目的PAM的解释Strictly speaking, PAM is an analogue system but it is the basis for digitizing speech, music and video signals. The basic idea is that regular pulses are used to sample the amplitude of the signal. The amplitude of each pulse represents the instantaneous amplitude of the signal. 尼奎斯特定律BW=1/2fdata告诉RF,阻抗,天线阻抗,设计网络The output impedance of an RF amplifier is (50 + j0)ohm. The amplifier must transfer maximum power into an aerial of impedance Z = (50 - j42)ohm at a frequency of 606MHz. A suitable antenna-matching network will be designed using one reactive component. What is the value of the component? Include (write) the correct units in the answer.(l是电感,H是单位亨)jl=j42l=422 *f*l=42l=42/(2*3.14*606*106)l=1.103e-8 H4G中MIMO的多控复用好处 Motivation: current wireless systems MIMO exploits the space dimension to improve wireless systems capacity, range and reliability MIMO-OFDM the corner stone of future broadband wireless access天线800MHZ,两个路径,计算相位差A fixed antenna receives a 800 MHz radio transmission via two indirect paths. Path 1 is 950.95m and Path 2 is 950.05m Calculate the phase difference between the two signals at the receiving antenna.Calculate wavelength:360 degrees = 0.375mDifference between path 1 and path 2 = 950.95 950.05 =0.9mCalculate fraction of a wavelength between two signals 0.9 =