Solutions to Chapter 6 Electrical Engineering Problems, Assignments of Electrical and Electronics Engineering

Download Solutions to Chapter 6 Electrical Engineering Problems and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! Homework CH6 Solutions CH6.1 E6.3 The input signal )30002cos(3)10002cos(21)( tttv ⋅π+⋅π+= has three components with frequencies of 0, 1000 Hz and 3000 Hz. For the dc component, we have 414)()0()( 1,out,1 =×=×= tvHtv in For the 1000-Hz component, we have: ooo 30602303)1000( in,2out,2 ∠=∠×∠== VV H )3010002cos(6)(out,1 o+⋅= ttv π For the 3000-Hz component: 0030)3000( in,3out,3 =∠×== oVV H 0)(out,3 =tv Thus, the output for all three components is )3010002cos(64)(out o+⋅π+= ttv P6.9* The phasors for the input and output are: oo 201 and 252 ∠=−∠= outin VV The transfer function for Hz 5000=f is ( ) o455.05000 ∠== inoutH VV CH6.2 E6.4 (do not turn in) Using the voltage-division principle, we have: fLjR R π2inout + ×= VV Then the transfer function is: BfjfRfLjfLjR RfH /1 1 /21 1 2 )( in out + = + = + == ππV V E6.5 From Equation 6.9, we have Hz 200)2/(1 == RCfB π , and from Equation 6.9, we have . /1 1)( in out Bfjf fH + == V V For the first component of the input, the frequency is 20 Hz, ,71.5995.0)( o−∠=fH o010in ∠=V , and o71.595.9)( inout −∠== VV fH Thus the first component of the output is )71.540cos(95.9)(out,1 o−= ttv π For the second component of the input, the frequency is 500 Hz, ,2.68371.0)( o−∠=fH o05in ∠=V , and o2.6886.1)( inout −∠== VV fH Thus the second component of the output is )2.6840cos(86.1)(out,2 o−= ttv π For the third component of the input, the frequency is 10 kHz, ,9.88020.0)( o−∠=fH o05in ∠=V , and o9.88100.0)( inout −∠== VV fH Thus the third component of the output is )9.88102cos(100.0)( 4out,2 o−×= ttv π Finally, the output with for all three components is: )9.88102cos(100.0 )2.6840cos(86.1)71.540cos(95.9)( 4 out o oo −×+ −+−= t tttv π ππ CH6.3 E6.7 (a) dB 15)(log20)( dB == fHfH 0.75 /2015)(log ==fH 623.510 )( 0.75 ==fH (b) dB 30)(log20)( dB == fHfH 1.5 /2030)(log ==fH 62.3110 )( 1.5 ==fH E6.9 (a) To find the frequency halfway between two frequencies on a logarithmic scale, we take the logarithm of each frequency, average the logarithms, and then take the antilogarithm. Thus 2.3161010 5.22/)]1000log()100[log( === +f Hz is half way between 100 Hz and 1000 Hz on a logarithmic scale. (b) To find the frequency halfway between two frequencies on a linear scale, we simply average the two frequencies. Thus (100 + 1000)/2 = 550 Hz is halfway between 100 and 1000 Hz on a linear scale. E6.10 To determine the number of decades between two frequencies we take the difference between the common (base-ten) logarithms of the two frequencies. Thus 20 Hz and 15 kHz are 875.2)20log()1015log( 3 =−× decades apart. P6.59 This is the first-order high-pass filter analyzed in Section 6.5 in the text. The transfer function is ( ) ( )( )B B in ffj ffjfH + == 1 out V V where Hz 1000 2 1 == RC fB π . The input signal is given by ( ) ( ) ( )tttvin ππ 2000cos5200cos5 += This signal has components at Hz. 1000 and Hz 100 == ff The transfer- function values at these frequencies are: ( ) ( ) o o 457071.0 11 11000 29.840995.0 1.01 1.0100 ∠= + = ∠= + = j jH j jH Applying these transfer-function values to the respective components yields: ( ) ( ) ( )oo 452000cos536.329.84200cos4975.0 +++= tttvout ππ CH6.6 P6.63* At the resonant frequency: o o o 9010 9010 01 −∠= ∠= ∠= C L R V V V CH6.7 P6.69* MHz 592.1 2 1 0 == LC f π 00.10 2 0 == Lf RQp π kHz 2.1590 == pQ fB CH6.8 P6.77 The circuit diagram of a second-order lowpass filter is: mH 592.1 2 0 == f RQL s π ( ) pF 15922 1 0 == fRQ C s π CH6.9 MHz 069.1 2 MHz 181.1 2 kHz 5.112 102 MHz 125.1 2 1 0 0 0 0 0 =−≅ =+≅ == == == Bff Bff Q fB R LfQ LC f L H s s π π