Signals and Noise - Report Example

Physics Assignment: Signal and Noise Radio signal frequency lies in the range of few Hz to 300GHz .The process of communication by radio signal involves transmitter and receiver linked by free space. The generated signal by microphone is processed to the required frequency and power before released into the free space. The receiver on the other end has the ability to detect transmitted weak signal in the free space .It processes the detected signal to the required strength before retrieval of important information. Block diagram of communication channel 1 .Information source and input transducer: At this stage, there is a transducer that convert physical quantity e.g. sound into electrical quantity (signal).The electrical signal from the transducer is fed into transmitter. 2. Transmitter: This section has the following stages; RF Amplifier: The generated signal at the transducer is so weak that it needs to be amplified. The signal is amplified to the required power before passed through the modulator. Modulator; Modulation is the process of combining the signal and frequency carrier for transmission. Stage 3(Local oscillator) generates steady sine wave that is mixed with the signal in the process of modulation. The signal can be amplitude modulated or frequency modulated. The frequency of steady sine wave from the oscillator determines the frequency of transmission. 3 Frequency Generator: This is an oscillator that generate pure and steady sine wave .The steady sine wave is combined with the signal from microphone in the process of modulation. It determines the frequency of transmitted signal. RF Power Amplifier: At this stage the modulated signal is amplified into the required power for transmission. This is done to ensure that the transmitted signal reaches the receiver even though there is power loss in the free space. : 3. Channel. The channel in this case is free space. The signal is transmitted into the free space where it is detected by the antenna of the receiver. 4 Receiver: This stage has the ability to receive the weak signal from the free space and process the signal. It generally consists of a pre-selector and amplifier stage .The pre-selector is a broad tuned band pass filter with adjustable center frequency that is tuned to the desired frequency. Main purpose of the pre-selector is to provide enough initial band limiting to prevent a specific unwanted radio frequency. It also detects band limits and amplify received RF signal 5 Output transducer: This is a device that converts the electrical quantity into the required physical quantity. It can television screen .speaker etc QUESTION 2 Electromagnetic waves are categorized based on the frequencies ranges or wavelengths ranges. Categories of electromagnetic waves Radiofrequency Waves: Has the frequency range of 106 to 910 Hz .The short wave radio frequencies in the high frequency band are reflected by layer of ionosphere at a considerable altitude and travel great distances. Microwaves: The frequency range is from 109 to about 1012Hz.It is quite useful for studying physical optics , radar applications ,microwave ovens and studying the origin of universe . INFRARED WAVES: IT lies just beneath red light of the visible light range in the electromagnetic spectrum. It has frequencies from about 1012 to 4.3*1014Hz . It is divided into four regions Near IR (780-3000nm), intermediate IR (3000-6000nm), far IR (6000 -15000nm) and extreme IR (15000nm-1.0mm) .It is used in remote controls. Visible lights: It occupies a very small portion of the electromagnetic spectrum. Its frequency runs from 4*1014 to7*1014Hz and wavelength in the range of 400-780nm. ULTRAVIOLET LIGHT: Its range lies between x-rays and visible lights. It is produced by hot bodies and special lamps. The wavelengths lie between 4 to 400nm. X- Rays: This frequencies ranges from 2.4*1016 to5*1019Hz and have extremely short wavelengths. The photon energy of x-rays is within the range (100eV to 0.2MeV). GAMMA Rays: They are the highest energy in the range of 104eV to 1019Ev.Its frequency is above 1020Hz.The rays are transmitted by bodies undergoing transition within atomic nucleus. QUESTION 3 The signal spectra help to determine features of the signal that cannot be obviously seen in time domain. The representation of waveform by summation of infinite number of points each with amplitude and phase is what is called signal spectrum. The analysis involves changing the waveform from time domain to frequency domain. From the waveform given it is realized that the frequency is 0.018 seconds and amplitude is 405V. Converting the frequency into Hertz it becomes; =55.556Hz Fourier transform is used to convert time domain into frequency domain for better analysis. Discrete Fourier transform can be used in this case. QUESTIO 4(a) The cascade of the receiver given in the figure has devices that cause both causes loss and gain to the input signal. When voltage is given, the power gain and loss is calculated by the formula given below G=20Log Where VOUT=Voltage output to the device Vin=Voltage input to the device Calculating gain for the RF stage, we have G= =20log100 =40 Gain for the mixer stage G= =20log0.5 =-6.02059 In this case it is loss because the value is less than 1 Gain of the Filter stage G= =20log0.632. =-3.985658 This is loss Gain for the IF amp stage G= =20log993.670886 =59.94485 NOTE: The values of power loss and gain calculated above are in decibels. (b) The overall power gain in the cascaded receiver is given by; GT=Addition of the gain of the stages in the cascade in decibels GT=40+ (-6.2059) + (-3.985658)+59.94485 =89.7532933(in decibels) (c)The formula of the power gain is given by GT= Output power=89.7532933 (2*10-9) =1.79506586*10-7 =0.179506586µW QUESTION 4(b) The three cascaded amplifiers are connected by cables which contribute to losses in the system. The overall gain of the cascade is reduced due to these losses. The total gain of the cascade connected in series is given by the product of the gain of the amplifier stages present in the cascade .The reciprocal of losses is considered a gain. Losses in the connecting cables varies depending on the diameter and length of the when material of the same kind is used. Losses contributed by the cables are calculated as shown below Loss contributed by cable 1 = (0.1*1) =0.1Db =1.023292992W Loss contributed by cable 2= (0.4*1) = 0.4Db = 1.096478196W Loss contributed by cable 3= (2*1) =2Db =1.584893192W In conversion of various losses into gain we have The gain contributed by cable 1 is given as Gain= = =0.97723729 Gain contributed by cable 2 is given as Gain= = =0.912010839 Gain contributed by cable 3 is given as Gain= = =0.630957344 The overall gain contributed by the cables is 0.912010839*0.977223729*0.630957344=0.562341364 The overall gain for the cascade is given by product of the three amplifier gain and cable gain =100*31.6227766*10*0.562341364 =17,782.83162W The relationship between power input, power output and the overall gain is given by G = Re- arranging the equation by making power output the subject of the formula, we have Power output =power input* Total Gain =17782.79522*0.316227766 =5623.4136W Converting this value into Db we have =10log (5623.4136) =37.50000Db QUESTION 5(a) Noise is unwanted signal in the system that tends to degrade the desired signal in the system. Noise can be in the signal before fed into the system or contributed by the system. There are several types of noise that form part of the output but was not initially at the input. These are contributed by the devices. The amplifying system contributes noise to the input signal but still there is need for amplifier in communication system. The various types of noise signal contributed by active devices are white noise, thermal noise, flicker noise and shot noise. In the figure given in the question, there are active components that contribute to noise in the output of the system. Resistors in the circuit: Due to random motion of electrons in the resistors there is fluctuation of voltage across these devices. The field effect transistor (FET): This is an active device which degrades the signal input to some level. The noise varies with the variation in bandwidth of the signal. (i) The resistors form a source of noise in the system. It is made of metal conductor which contains free electrons. Due to thermal agitation these electrons are moving about continuously in the conductor causing collision with atoms and continuous exchange of energy take place. This accounts for the resistance property of the conductor. The random motion of electrons in the conductor produces voltage fluctuation across the conductor for the mean square noise voltage. Transistor in the figure contributes for white noise in the system. It is due to the fact that electrical current is not continuous but discrete flow of electrons. (ii) Flicker noise: This is a random fluctuation of electrical signal. In the figure it is found in the transistor used in the amplifier. Thermal noise: This is contributed by the resistors in the circuit. QUESTION 5(b) Factors used to quantify the cumulative effect of noise are Noise figure and signal noisre ratio.Noise figure is ameasure of degradation of the signal to noise ratio.Sigal to noise ratio is a ratio of signal to noise in the communication system. QUESTION 6 (a) Noise figure is a measure of degradation of signal to noise ratio and is given by the formula below F= = = Where K=1.38*10-23 , B=Bandwidth=1.0*106 T =290 The input noise power is given by the formula below Input noise power=FGKT………………1 Conversion of Db into Watts yield, 13Db=10logx X=19.9526W Taking the amplifier gain to be one and making substitution in the equation 1 above, we have Noise input power=1909526*290*1*(1.38*10-23)*(1*106) =7.985039*10-14W QUESTION 6(b) The question can be represented diagrammatically as shown below input output Amplifier The Noise Figure and Gain for the amplifier are 9Db and 15 Db respectively The overall noise Figure for the cascade is given by the formula below F =F1+………………………..1 Conversion of various values given in decibels into power yields G1=15Db =31.6227766W F1=9Db=7.9432823W F2=20Db=100W Making substitution to equation 1 above we have FT= 7.09432823+ =11.07393617 Works cited Misra, Devendra . Radio-frequency and microwave communication circuits : analysis and design. Hoboken, N.J: Wiley-Interscience, 2004. Print. Yadav, Abhishek . Digital Communication. n/a: Firewall Media, 2009. Print. Al-Azzawi, Abdul . Photonics: Principles and Practices Optical Science and Engineering. n/a: CRC Press, 2006. Print. MLA formatting by BibMe.org. Read More