ACE Notes - Uddhav PDF

# UNIT-1 ## Communication ### Block diagram - Information Source - Input Transducer - Transmitter - Channel - Receiver - Destination - Noise ### Speed of Radio Signal - Velocity of light = $3 \times 10^{8}$ m/s - Wavelength of Radio Wave = $\lambda = \frac{c}{f}$ ### Question 9.1 A Radio-station...

# UNIT-1 ## Communication ### Block diagram - Information Source - Input Transducer - Transmitter - Channel - Receiver - Destination - Noise ### Speed of Radio Signal - Velocity of light = $3 \times 10^{8}$ m/s - Wavelength of Radio Wave = $\lambda = \frac{c}{f}$ ### Question 9.1 A Radio-station transmit signal of 30MHz what is the wavelength of Radio signal. $ \lambda=\frac{c}{f}=\frac{3 \times10^{8}}{30\times 10^{6}} =10^{-2}=0.01 \times 10^{0}$ $\lambda = 10m$ #### Question 9.2 A signal send by Radio Transmitter to a satellite has a frequency of 10 GHz, What is its wavelength. $\lambda=\frac{c}{f}=\frac{3\times10^{8}}{10\times10^{9}}=\frac{3 \times 10^{-1}}{10} =0.03m$ $\lambda= 3cm$ ## Radio frequency Spectrum | Frequency Range | Name of The Band | | ------------- | ------------- | | 3 to 30kHz | Very Low | | 30kHz to 300kHz | Low | | 300kHz to 3000KHz (3MH2) | Moderate (Medium freq) | | 3 MHz to 30MHz | High | | 30MHz to 300MHz | Very High | | 300MHz to 3000MHz (3GHz) | Ultra High | | 3GHz to 30GHz | Super High | | 30 GHz to 300 GHZ | Extreme High | ## Baseband signal Transmission & Bandwidth requirements - **Telephone** - Baseband Signal - Analogue Information - Telephone - **Voice of analogue baseband signal** - Baseband Signal - Analogue Communication - TV set ## Video as analogue baseband signal - Telephone - Baseband Signal - Digital Information - ADC (Parallel) - Analogue Information - DAC - Telephone - Analogue Information ## Voice as analogue baseband signal converted to digital form - Computer - Baseband Signal - Digital Information - Computer ## Computer data as digital baseband signal ## Bandwidth requirement of same baseband signal - speech, music, & video ### Note - Bandwidth is important for long distance communication - bandwidth requirements, to avoid mixing of signals in communication ## Types of Transmission media - Guided - Wired - physically required - Unguided - Wireless - No need ## Modulation Techniques - Signals - Information Signal - Low frequency - Baseband - Carrier Signal - High Frequency - Modulation signal - General Equation of carrier wave can be written as - X = A sin (wt + β) - X = Instantaneous value of voltage and current - A = Max Amplitude - w = angular freq (2πf) - β = Phase angle - If anyone parameter is varied in accordance with Instantaneous "modulating" signal, then process is called modulation process - the process of changing Amplitude, frequency, phase of the carrier wave in accordance with Instantaneous value of modulating value signal is called as modulation - There are three modulation Techniques - Amplitude modulation - frequency modulation - Phase modulation - we can perform one by one modulation by keeping another two techniques as constant and definition is the same. ## Simplex and Duplex System - Simplex System - Keyboard - Monitor - Walkie Talkie - Duplex system - Half duplex - Full duplex - Telephone - Simultaneous from both side ## Modes of communication - Point - to - Point communication - Phone - one-to-one - Broodcast communication - TV - Transmitte one person & receivers are many. ## Necessity of Modulation - Length of Antenna - Interference - Poor Transmission ## Length of Antenna - Modulation allows us to transmit data over long distance effectively by utilizing specific frequency ranges overcoming limitations of antenna lengh. ## Interference - Modulation helps differentiate signals from different sources. - Reducing the chances of interference - Enabling multiple devices to share the same frequency band - Modulation enchances signal quality and resilence mitigating issues like poor transmission due to noise, fading and distortion in the communication channel. ## Mathematical Amplitude Modulation Expression - Modulating voltage and carrier voltage represented mathematically as - $V_m = V_m sin W_mt$ - (1) - $V_c = V_c sin W_ct$ - (2) - $φ$ is ignored because it is constant in amplitude modulation. - The amplitude of the carrier ($V_c$) will change in accordance with instantaneous value of Modulating voltage ($V_m$). - Let A = amplitude of Resultant amplitude Modulated wave! - ie.. $A = V_c + V_m$ - (3) - Put value of equation (1) in (3) - $∴ A = V_c + V_m.sin w.mt$ - $A = V_c (1+ \frac{V_m}{V_c} sin w_m.mt)$ - (4) - Modulation index of a amplitude modulated wave is - $m = \frac{V_m}{V_c}$ - (5) - $A = V_c(1 + m.sin w_m.mt)$ - $m$ suffix everywhere, take note. (Amplitude Modulation) - The instantaneous value of resulting "e_Am" wave is denoted by e_Am. - $e_Am = A sin W_ct$ - (5) - Now put the value of 'A' i.e. equation (4) in ( 5) - $e_Am = V_c(1+m.sin w_m.mt).sin W_ct$ - $∴ e_Am = V_c(sin W_ct + m.sin w_m.mt.sin W_ct)$ - $W.K. t. 'sin A .sin B =\frac{1}{2} [cos(A-B)-cos (A+B)]'$ - Now , put formula - $∴ e_Am = V_csin W_ct + \frac{V_m}{2}[cos(W_ct-w_mt)-cos (W_ct + w_mt)]'$ - $e_Am = V_csin. W_ct + \frac{V_m}{2}.cos(W_ct-w_mt)-\frac{V_m}{2}.cos [W_ct + w_mt].$ - $∴ e_Am = V_csin. W_ct + \frac{Vim}{2}.cos(Wc-wm)t-\frac{V_m}{2}cos(Wc+wm)t.$ - The equation of Amplitude modulated wave suggest that it contains three frequencies - i) Original camier frequency - ($W_c = 2πf_c$) - ii) Upper onde band freq. (USB) → ($W_c+W_m = 2πf_c + 2πf_m$) ↳ Sum of camer + modulating signal freq. ↳ Also denoted by (fc + fm). - iii) Lower side Band freq. → ($W_c-W_m$) ↳ Difference beton modulating sig. freq. and cammer freq. ↳ Denoted by (fc-fm). ### Question 9.1 An Audio signal of 500Hz is used to modulate a camier of 500KHz. Determine i) Bandwidth ii) Side Band freq. - Audio ongenal - Modulated signal - $f_m$ - Camier signal - Lanier freq. - $f_c$ #### i) - $Man - Min = B.W$ - ie.. $2f_m$ - $BW = 2 \times 500 Hz$ - $BW = 1000Hz$ #### ii) - $Side Band freq. $ - $LSB= f_c - f_m= 500KHz - 500Hz = 500KHz-0.5KHz$ - $LSB= 499.5KHz$ - $USB= f_c + f_m $= 500KHz + 500 Hz $= 500.5KHz$ ## 9.2 An audio signal is given by em = 48sin3600t its amplitude modulades a camer wave given by fc = 120sin 62800t calculate and ) Audio frequency 2) Camer frequency 3) % of Modulation. ### Note - audio signal = modulating signd - $e_m=V_m$ - $e_c= V_c$ - camera signal ### Given - $e_m = 48sin3600t$ - (1) - $V_m = V_m sin W_mt$. - (2) - By comparing (1) & (2) - $W_mE = 3600t$ - $W_m = 3600$ - Audio freq. - $f = \frac{W_m}{2π} = \frac{3600}{2\times3.14}$ - $f= 572.95 Hz$ ### i) - $f_c = 120sin 62800t$ - (3) - $V_c = V_csin W_ct$ - ( 4) - By comparing $ (3) + (4) $ - $W_ct = 62800t$ - $W_c = 62800$ - Camer freq. - $f= \frac{W_c}{2π} = \frac{62800}{2π}$ - $f = 9994.930 Hz $ or $f= 9.994k세$ ### iii) - Modulation Indea - $m = \frac{V_m}{V_c} = \frac{48}{120} = 0.4$ - % Modulation - $% Modulation = 0.4 \times 100 = 40%$ ## 9.2 A spectrum analysis of an em wave gives the following information 50vot @ 750 kHz 20 V @ 748 KHz @752 kHz 20 v then find 1) Camer freq. 2) freq. at side band 3) % of Modulation 4) B.W. - $V_o=50v$ - $m_v_c=200$ ### i) - $Cammer freq = f_c= 750kHz$ ### ii) - $freq. at side bands$ - $USB= f_c+ f_m = 752 KHz$ - $LSB = f_c - f_m = 748KHz$ ### iii) - % of modulation - $Modulation = \frac{m_v_c}{V_c}=\frac{2 \times 20}{50}=0.8$ - $m% = 80$ ### iv) - $B.W = f_{max}-f_{min}$ - $=(f_c+f_m)-(f_c-f_m)$ - $=f_c+f_m-f_c+f_m$ - $= 2f_m $ - $=(752-748) KHz$ - $B.W = 4KHz$ ## 9.2 Calculate the modulation factor and % of modulation factor for the following figure. - $E_{max} = 77v$ - $E_{min} = 37v$ ### i) - $Modulation factor = \frac{E_{max}-E_{min}}{E_{max}+E_{min}}=\frac{77-37}{77+37}=\frac{40v}{114v}$ - $M.F = 0.3508$ ### ii) - $% Modulation factor = 100 \times 0.3508$ - $% M.F = 35.08%$ ## 9.3 An amplitude modulated camier wave is viewed on an oscilloscope and has created voltage of 55V P-P the bottom point of the wave measures 10V p-p. find 1) What is the M.F. 2)% Modulation 3) See what is P-P unmodulated camier voltage. - $f_{max} = 55v$ - $ f_{min} = 10v$ ### i) - $Modulation factor = M.f =\frac{E_{max}-E_{min}}{E_{max}+E_{min}}=\frac{55 - 10}{55 + 10} = \frac{45v}{65v}$ - $M.F. = 0.692$ ### ii) - % Modulation factor = $100 \times 0.692$ - $%M.F = 69.2%$ ### iii) - $Unmodulated P-P camier voltage$ = average of 2 voltages i.e.. $E_{max}$ & $E_{min}$ - $U.C.V = \frac{E_{max}+ E_{min}}{2}= \frac{55+10}{2} = \frac{65}{2}$ - $U.Vc = 32.5 v$ ## Types of Modulation ### Different Conventions - **Analogue to Analogue Conversion** - AM - FM - PM - **Analogue to Digital Conversion** - PCM - DM - AOM - **Digital to Digital Conversion** - Line coding - Block coding - **Digital to Analogue Conversion** - ASK - FSK - PSK - **Analogue to Analogue Modulation** - **Also known as Source Coding Modulation** - **Digital Continuous Wave Signal** - **Pulse Analogue Modulation** - PAM - PWM - PTM - PPM - PFM # Modulation - **Analogue Modulation** - AM - DSBFC - DSBSC - SSB - LSB - VSB - Angle Modulation - Analogue Pulse Modulation - PAM - PWM - PPM - PFM - **Digital Modulation** - Digital Pulse Modulation - PCM - DM - Digital Continuous Wave Modulation - ASK - FSK - PSK ## Sampling Theorem - Ideal sampling - Natural Sampling - Flat Top sampling - most easy to recognize ### Natural sampling - High speed switch is turned on for a small period of time when sampling is applied. ### Ideal sampling: - Pulses of a Analogue signal are sampled and it is difficult to implement. ### Flat Top sampling - Sample level is held using capacitor, based - switching (ON/OFF), this is widely used sampling method. - The proces of sampling is called PAM. (Pulse amplitude modulation). ## Nyguist Theorem - $BW = f_{max} - f_{min}$ - $BW = f_{max} - 0$ - $BW = f_{max}$ - $Low Pass signal, Nyquist rate = 2f_{max}$ - $BW=f_{max}-f_{min}$ - $∴ BW= f_{max}$ - $Band Pass filter, Nyquist rate = 2f_{max}$ - $f_s = f$ - $I/p Signal$ - $Recovered sine wave at f_s = f (Recovery gives distorted O/P)$ - $I/P Signal$ - $X_1$ - $X_2$ - $X_3$ - $X_4$ - $X_1$ - $X_2$ - $X_3$ - $X_4$ - $Recovered sine wave at f_s = 2f (Recovery gives good O/P)$ - $I/P Signal$ - $X_1$ - $X_2$ - $X_3$ - $X_4$ - $X_5$ - $X_6$ - $X_7$ - $X_8$ - $X_9$ - $Recovered destre sine wave if f_s = 4f (Recovery gives best O/P)$ ## Introduction to PAM ### Pulse Amplitude Modulation - Double Polarity - Single Polarity - PAM System ### Note - Analogue/Amplitude Modulation - Pulse Modulation - Analogue - Digital - Low/High - Modulating - High/Low - Signal Freq - Information Signal - Low frequency modulating signal or low freg. baseband signal. - Cammer pulse Train - Time axis - double polarity PAM sig - t - x(t) - t - cammer - t - y(t) - t - single polarity pulse train ## Generation of double polarity PAM - x(t) - Modulating signal - Low pass filter (P) - Multiplier XW/FET (+) - cammer (Pube train) Generator - y(t) - Generate PAM signal ## Demodulation of PAM - PAM signal - LPF - Modulating signal - y(t) ## Pulse width Modulation - Pulse Time Modulation - PWM - Pulse position modulation - x(t) - pulse width - low frequency modulating signal - time axis - high frequency Cammer - time axis - y(t) - pulse width modulated signal - time axis ### Note - Amplitude Main - PAM - I/P - cammer wave - sine wave - PWM - I/P - pulse train - Cammer wave -> width of pulse train. - **Def**: - The width of cammer wave (pulses) is changed in accordance with instantaneous value of modulating signal is called as pulse width modulation - Similarly for PAM and AM ## Generation of Pulse width Modulation - $S(t)$ - modulating signal - $D_A$, $R_A$ - Cammer sig - $R_B$, Discharge - Supply - $IC 555$ - Trigger - $Control Vtg, GND$ - Supply - $O/P PWM supply y(+)$ - $s(t)$ - $Cammer signal s(t)$ - $S(t)C_A$ - $V_{in}, R_A$ - '' Point waveform - These sharp negative going pulses are applied to the trigger I/P of IC555 ## Demodulation of PWM - $PWM signal J/P$ - PWM to PAM Converting Block - LPF - O/P modulating signal ## Practical diagram of PWM demodulation - $PWM signal$ - $Schmitt Triggered Ckt as a pulse genr$ - $Ramp generator Ckt$ - $Adder$ - $Level Shifter Block$ - $Delay Ckt$ - $React-fied Block$ - $LPF$ - $Recovered original information O/P$ ## Waveforms of pwm - PWM J/P signal with noise - Time - O/P of Schmitt Trigger (Regenerated PWM) - Time - Ramp generated O/P - Reference pulse or generated O/P or synchronisation pulse - Time - Adder O/P - Time - Level Shifter O/P - $pwm$ - Rectified output O/P - Time - Recovered original Mod n signal - Time ## Pulse Position Modulation - Modulation signal - Cammier wave signal - Time - PWM - Time - PPM - Time ## Generation of PPM - Modulating signal - $JC555 Monostable Multivibrator$ $JC555$, monastable multivibrator - PPM. - Camier signal - Same diagram of pwm is repeated in ppm. ## Demodulation - $PPM$ - $Schmitt Triggered Ckt on pulse generator$ - $Pulse generation for synchronisation$ - R - Flip - S - Flop - Pulse - O/P Modl of signal ## Multiplexing - Many signals - Signal1 - Signal2 - Signals - $Multi-plexing$ - $Transimission line (signals)$ - $De'MUX$ - Signal 1 - Signal 2 - Signal n - $Multiplexing$ - FDM - Frequency Division Multiplexing - TDM - Time Division Multiplexing - Synchronized - Asynchronized ## Frequency division Multiplexing (FDM) - $Signal 1 J/P$ - $f_{n1} cammer$ - $Modulation$ - $CH_1$ - $Signal 2 J/P$ - $f_{n2} cammer$ - $Modulation$ - $CH_2$ - $Signal n J/P$ - $f_{n} cammer$ - $Modulator$ - $CH_n$ - $Multiplexing sig O/P$ - $Linear common transmission adder$ - $Line forall.$ ## Power Relation in AM -1) Unmodulated camier ($f_c$)= $P_c$ - 2) Lower SideBand ($f_{LSB} = f_c-f_m = P_{LSB}$ - 3) Upper side band ($f_{USB} = f_c + f_m = P_{USB}$ - $∴ P_T = P_c + P_{LSB}+ P_{USB}$ - (1) - $P_T = \frac{E_c^2}{R}+\frac{E_{LSB}^2}{R}+\frac{E_{USB}^2}{R}$ - (2) - $Now : E_c, E_{LSB}, E_{USB}$ are the RMS value of the 'camier', Lower side band, Upper side band respectively. - R = is the characteristic Impedance of Antenna in which total Power is dissipated. - $∴ cammer paver = P_c = \frac{E_c^2}{R} = \frac{(f_c/√2)^2}{R}$ - $∴ R_c=\frac{E_c^2}{2R}$ ### 2) Power in side bonds = $P_{SB} = P_{LSB} = P_{USB} = \frac{F_{SB}^2}{R}= \frac{E_{SB}^2}{R}$ - $∴ P_{SB} = \frac{E_{SB}^2}{R}$ - **The peak amplitude of each side bond i's F= $\frac{m_v_c}{2} = m_v_c E $** - $ ∴ P_{SB} = (\frac{m_v_c}{2})^2 \frac{1}{R} = (\frac{m_v_cE}{2})^2 \frac{1}{R}$ - $= \frac{m^2V_c^2}{8} \frac{1}{R} => \frac{m^2V^2}{8R}$ - $= \frac{m^2V_c^2}{8R}$ '' $V = E$ '' - $∴ P_{LSB} = P_{USB} = \frac{m^2E_c^2}{8R}$ - $∴ P_c = \frac{E_c^2}{2R}$--- BY cammer power formula. - $Similarly P_{LSB} = P_{USB} = \frac{m^2.P_c}{4}$ - Now putting values in equation (2) - $P_T = P_c + \frac{m^2}{4} P_c +\frac{m^2}{4}P_c$ - $P_T = P_c(1+\frac{2m^2}{4})$ - $∴ P_T = P_c(1+\frac{m^2}{2})$. - $∴ \frac{P_T}{P_c} = (1+\frac{m^2}{2})$. - $∴ \frac{P_T}{P_c}-1=m^2$ - $∴ m = √2( \frac{P_T}{P_c} -1)$ - $∴ n = Transmission efficiency of AM wave$ - $n = \frac{Total SB Power }{Total Transmitted Power}$ - $n = \frac{ P_{LSB} + P_{USB} }{P_T}$ - $n=\frac{\frac{E_{LSB}^2}{R}+\frac{E_{USB}^2}{R}}{P_T}$ - By Simplified - $n = \frac{ \frac{m^2P_c}{4} + \frac{m^2}{4} P_c }{ P_c ( 1 + \frac{m^2}{2} ) } $ - $= \frac{ P_c ( \frac{m^2}{2} ) }{P_c(1 + \frac{m^2}{2} ) }$ - $= \frac{ \frac{m^2}{2} }{ (1+\frac{m^2}{2} ) }$ - $= \frac{ 2m^2 }{ (2+m^2) } = \frac{m^2}{(2+m^2) } $ - $n = \frac{m^2}{ (2+m^2) }$ - $∴ In % we can write $ - $% n = n \times 100 = \frac{ m^2 }{ 2+m^2} \times 100$ ## AM Power in term of current - $P_T = I_t^2 \times R$ - (1) - $P_c = I_c^2 \times R$ - (2) - $∴ \frac{P_T}{P_c} = \frac{I_t^2}{I_c^2}$ - by using (1) & (2) - $∴ \frac{P_T}{P_c} = (\frac{I_t}{I_c})^2 = (1+ \frac{m^2}{2})^2$ - $∴ (\frac{I_t}{I_c})^2 = (1+\frac{m^2}{2})^2$ - $∴ I_t = I_c(1 + \frac{m^2}{2})$ - $∴ I_t = √I_c^2(1 + \frac{m^2}{2})$ - $∴ I_t = I_c √(1 + \frac{m^2}{2})$ - $∴ m^2 = 2(\frac{I_t}{I_c})^2-1$ - Modulation Index in form of 'I' - $Now : In the terms of 'm'$ - $m = √2(\frac{I_t}{I_c})^2-1 $ ## Modulation Index in the from of Voltage - $m = √2(\frac{E_T}{E_c})^2-1$ - $∴ m = √2(\frac{E_T}{E_c})^2-1$ ## Power relation in AM ### Question 9.1 A 12KW carrier wave is amplitude modulated at 80% depth of modulation by a sinusoidal modulating signal. calculate 1) side band power, 2) Total power 3) Transfer efficiency of AM wave ### Given - $P_c = 12KW = 12 \times 10^3 W$ - $m = 0.8$ - $P_{LSB}=P_{USB}=\frac{m^2}{4} \times P_c = \frac{(0.8)^2}{4} \times 12\times10^3 = 0.64 \times 12\times 10^3 =0.16\times 12RW=1.92KWW \times2$ - $P_{SB} = 3.84KW$ ### ii) - $P_T = P_c(1+\frac{m^2}{2}) = 12K(1+\frac{0.64}{2})=12K(1.32)$ - $P_T = 15.84KW$ ### iii) - $Efficiency of Transver$ - $n = \frac{m^2}{ (2+m^2) } = \frac{(0.8)^2}{(2+0.8)^2}=\frac{0.64}{2.84}=0.242 $ - $n= 0.242$ - $% n = 24.2%$ ### Question 9.2 Transmitter transmitt 18kw of power without modulation and 18kw of power after amplitude modulation what i's the modulation Inden and to modulation. ### Given - $P_c = 16kW = 16 \times 10^3 W$ - Before modulation - $P_T = 18KW = 18 \times 10^3 W$ - After modulation, ### Formula, - $P_T = (1 + \frac{m^2}{2})P_c$ - $m = √2( \frac{P_T}{P_c} -1) = √2( \frac{18}{16} -1)$ - $m = √2( \frac{2}{16}) = √ \frac{2\times2}{16} $ - $m = \frac{1}{2}$ - $% m = 50.0%$ ### Question 9.3 A Total Antenna cument of an Am transmitter is 6A with the modulation index is 0.6 calculate the antenna cument when only the carrier is set. ### Given - $I_T = 6A$ - $m=0.6$ - $∴ I_c = I_T √(1 + \frac{m^2}{2}) = 6√(1 + \frac{0.6^2}{2}) $ - $∴ I_c = 6 √1.18$ - $∴ I_c = 6 \times 1.086$ - $∴ I_c = 6.1810A$ - $∴ I_c = 5.523A$ ### Question 9.4 The Rms Antenna current of an AM Transmitter increases by 20% over its unmodulated value when sinusoidal modulation by 4KHz signal is applied calculate modulation index. ### Given - $I_T= 4.8 KHz = 4.8 \times 10^3 Hz$ - $I_c = 4 KHz = 4 \times 10^3 Hz$ - $m = √2(\frac{I_T}{I_c})^2 -1 = √2(\frac{4.8}{4})^2-1 = √2(1.44)-1 = √2.88-1 = √1.88$ - $M=1.37$ ## 9.4 Calculate Total modulation index if camer wave i's amplitude modulated by 3 modulation signal with modulation Indices (0.7, 0.35, 0.45) ### Formula - $M=√{m_1^2+m_2^2+m_3^2}$ - $m =√{(0.7)^2 + (0.35)^2+ (0.45)^2}$ - $M = 0.9027$ ## 9.5 AM Transmission radiates 10.1KW with the camer un modulated and a 11.9k0o when the conter is sinusoidally modulated another calculate modulation index, if camier sine wave cones-ponding to 35% modulation, is transmitted simultaneury * Determine the total radiated power. ### Given - $P_T = 11.9kW = 11.9 \times 10^3 W$ - $P_c= 10.1kW = 10.1 \times 10^3 W$ ### i) - $m_1 = ? $ - $m_1 = √2(\frac{P_T}{P_c}-1) = √2(\frac{11.9}{10.1}-1) = √1.356 = 1.164 = 0.599$ ### ii) - $m_2 = 0.35$ ### iii) - $m = √{m_1^2+m_2^2}=√{0.4789}$ - $M= 0.692$ ### iv) - $ P_T = P_c (1 + \frac{m^2}{2}) => P_T = 10.1K(1+0.4788)^2$ - $Radiated Power = P_T = 10.1K (1.2394)$ - $P_T = 12.394KW$ ## 9.7 Broadcast AM Transmitter radiates 55kw of camer power If the modulation index is 0.80, what will be the radiated power. ### Given: - $P_c = 55kW = 55\times10^3 W$ - $m = 0.80$ ### To find: - $Rodiated power = P_T$ - $P_T = P_c (1 + \frac{m^2}{2} ) = 55K ( 1 + \frac{0.64}{2} ) = 55K (1.32)$ - $∴ P_T = 72.6 KW$ ## 9.8 When to modulation is 70% of an Am wave transmitter produces Itkw, what will be the camier power. ### Given: - $m = 0.7$ - $P_T = 11KW$ - $P_c = ?$ - $P_c = \frac{P_T}{(1 + \frac{m^2}{2}) } = \frac{11K}{(1+\frac{0.49}{2}) } $ - $∴ P_c = \frac{11K}{(1.245)} = \frac{11K}{1.245}$ - $P_c = 8.835KW$ - $∴ P_c = 0.1131 \times 10^3$ - $P_c = 8.841 KW$