hwk5_sol

hwk5_sol - EE 40 FALL 2008 PROF CHANG-HASNAIN HOMEWORK 3...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS 1. A voltage v(t) is applied across a 1 Ω resistor (a) v (t) = Acos(4πt) find average power PA dissipated by resistor: PA = (1/(one period) oneperiod v (t) ∗ i(t)dt One period – From t = 0 to 0.5 0.5 0 .5 (t PA = (1/0.5) 0 v (t) ∗ i(t)dt = (1/0.5) 0 v (t) ∗ vR ) dt = (1/0.5) 0.5 (1/0.5) 0 v (t)2 dt = (1/0.5) 0 A2 ∗ cos2 (4πt)dt = (1/0.5) (1/0.5)A2 ( t+sin(2∗4πt)/(2∗4∗π) )|0.5 0 2 PA = A2 /2 0.5 0.5 0 0.5 v (t)2 dt 1 0 = A2 ∗ 1+cos(2∗4πt) dt 2 = (b) v (t) = B sin(4πt) find average power PB dissipated by resistor. Same analysis as before: 0.5 0.5 (1/0.5) 0 B 2 ∗ sin2 (4πt)dt = (1/0.5) 0 B 2 ∗ 1−cos(2∗4πt) dt = 2 (1/0.5)B 2 ( t−sin(2∗4πt)/(2∗4∗π) )|0.5 0 2 PB = B2 /2 (c) v (t) = Acos(4πt) + B sin(4πt) find average power P1 dissipated by resistor. Same analysis as before: 0.5 0 (1/0.5) 0 (A2 ∗ cos2 (4πt) + B 2 sin2 (4πt))dt = (1/0.5)int0.5 (A2 ∗ 1+cos(2∗4πt) + B 2 ∗ 2 1−cos(2∗4πt) )dt = 2 P1 = A2 /2 + B2 /2 (d) P1 = PA + PB The same calculation can be done by finding the RMS voltage and squaring it as such: 2 Vrms = Vamp / (2) Pavg = Vrms /R = Vamp /(2 ∗ R) This is left as an exercise to the student to show that they are equal 1 ! ! 2 43 ! - %4% 6 ! 5"1 %,% 7! B A= > ! - ?@,A ."= $ .1 ! > !4% ! # %89% Hambley, 5.27! ." %:*+;,% # ! #"7 2. v1 (t) = 10 cos(ωt + 30◦ ) ! = 3*< % ! $-/5 !i1m (t)! ! $.1 = (2) ∗ i1rms = (2) ∗ 5 = 7.071 ! 6 6 2 EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS % ◦ " ! -& $ .1 ! %4%V1 = 10∠30 % % I1 = 7.071∠(30 + 20)◦ 4:3, ! ! #"2#2 %4% # i1 (t) = 7.071 cos(ωt + 50◦ ) % ! B 2 '= ( ! 25 ?@,'!= ) -5 ! ( ! D 23 ! # * D 2:3, ! !"5!2 ! 8 " ! 25&-5 ! & ! $ 8999 ;((((= $ !2 8 : 899 )2 ! !"5!2&15 !% 8 ! < 89 " % < 9'8 ! 8 C2 '= ( ! !" (((((((((((((((((((((($5!2 ?@,'!= ) 15 ( ! 9'98 % < 9 ! (((((((((((((((((((((($ 899 # ! 7"1/*#"(%2"($-/6$0/!,()/!%($#(3"!45(%2"($-)"./01"($#()*!"+,(!"#$#%$&"' 3. Hambley, 5.458 E)%*:)%0CB)F%=G)%)<H:),,[email protected]% & $ B999 ;((((= $ % 1 ,CFA!= >Vs1 ?@,A!= )<-5 ! " % < 9'B9 != ) 215 ! > % )8 : 899 % 8 ! B9 > ) 1 ?@,A = 10∠0◦ −j/ω C = −j/(500 8 [email protected]):=CF0%=@%HG*,@:,%J)%@&=*CF% ∗◦uµ F) = -j1000 % (((((((((((((((((((((($I = Vs ! = 10∠0 ◦ ! ! 9'98−j/wc 1&215 ! % R% % ) 1& $ K5 ) 1&-5< 9'8C1000−j!1000µF ) = 7.071∠45 mA ◦ ◦ Vr = I ∗ R = 7 $ .∠45 ∗ 1000 7.071∠ %(((((((((($ L 1 ) (($--52 )( #"155.071$--52 ) L ' "<=!>?@ # 45 V (((((((((( ." V '>CB ∗ (−j/ωC!) =9'.071∠45◦1>' 51000 = 7.071∠ − 45◦ V > = I L ' E>'8D " 7 AABC # ∗ −j % c 6GM,N%J)%G*B)% I leads ()/!%($#(0"6/%$&"5(%2"($-)"./01"($#(1/)/1$%$&"' 7"1/*#"(%2"($-/6$0/!,Vs by 45◦ 1 ,CFA!= > ) 1 ?@,A!= ) -5 ! > ) 1 ?@,A!= ) 215 ! > ! 5 % ! ! % &$ $ I ' < !H 89(9 ! 8999 ' < 8999 '# ( % ( 21/ $ @[email protected](AC ! (-G ! ! ! 'H $ !' < !H "& $ @[email protected]( ' AC ! (F 'I $ I& $ @[email protected](AC ! (F &(+"/.#('# (J,(AC ! ! &$ '# ! ! ! I ' < !H ! 89(9 ! $ 8999 ' < B999 $ [email protected](>?'A?! (-G EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS 3 4. Hambley, 5.47 Is = 0.5∠0◦ jωL = j * 100 ! 1 = 100j"%&! * % ' #1 # #!$1 V)s = Is ∗('/200+1/(#100) = 0.5∠0◦ ∗ 1/200+1#!"!"! )44.72∠63.44◦ = 1 j &$*#*#" $ #"! /(j 100) "' ◦ Ir = Vs /R = 44.72∠63.44 /200 = 0.2236∠63.44◦ Il +,-.//01)'23)345,6./3-')7,875,'9).83:) 0.4472∠ − 26.56◦ = Vs /(jωL) = 44.72∠63.44◦ /(100j ) = V leads Is by 63.44◦ & & ! ! ! ! & ! *& & & (. # ($""( ! ! * ' # (. * %(( $ * , *(( # &&$'%"#)$&& ! (- # ' - # ($%%)#"#)$&& ! (+ # ' ,!+ # ($&&'%" ! %#$"#! '!1234.!(. !/0!#)$&& ! ! & ) 5. Hambley, (. # *(("( ! !89 5.92 # * $%&'( ;-<38)=>3-?7,875,')7=-<,',=-91)@3)2.63) ' # (. # Zero out the current source makes the capacitor, inductor and resistor all in series & * *(( $ * Z! =#−j 3Ω + j 3Ω + %D*"&! # #&"%C$AB! )H) giving: % ! , %(( & =7 # E ! % F 4Ω = 4Ω t' ) !.G $ Under%#$"#! !< open-circuit conditions – The capacitor is removed (no current can flow through # :$;&&I,'2)'23)9=5873)J38=3<1)@3)/==K)G.7K),-'=)'23)'38L,-./9).-<)933) "! i). The voltage is just the current flowing through the inductor and resistor: ( %# "#! % (- # ' - V):;$&&+ '! # ∗ $F % 0!8910! 36! &)V # = (4 " j 3) $ 2∠ %◦ F= % ∠ # .87◦ t t It is#substantially&& ! !89 calculate In = Vt but the direct approach is given as: (= # ' % ,!=M3N'1)'23)M=8'=-)75883-'),9) & &&$'%"#)$ easier to Z & 4+3j ◦ ! = 2.5∠36.87◦ 4+3 ' j & & In) = 2∠0"-∗ # j −3# *$""%C$AB! ) This gives the ( following Thevenin and Norton equivalent circuits: '!13>.!(. !/0!%#$"#! ! ' (. # *("( ! !89 *& ! & & ' # (. * * - $ * ,!+ $ ,!= * * *((( ! , ($((" $ , ($((" # *( !% # *("( ! !< (- # ' - # *("( ! !89 ! #(+ # ' ,!+ # "(" ! ;( !89 $%&')#% ,!= & # "(";( ! !89 (= # ' +=8)L.N,L5L)>[email protected])'8.-9O381)'23),L>3<.-73)=O)'23)/=.<)92=5/<)G3)'23) 7=L>/3N)7=-F5P.'3)=O)'23)Q2R63-,-),L>3<.-73:) ! ) [email protected]!BC! E+ %D & [email protected]!DE82.!13F>2F!D63G!D62!.BAFH2!HAFF2GDL!!56E.! ) ) ( /=.< # #& $ F " # T/=.< $ F '!S /=.< ( # # E.!7B..E/12!/2H3A.2!HAFF2GD!EG!D62!H373HED3GH2!/313GH2.!D62!HAFF2GD!EG!D62! U3'',-P)83./)>.8'9)345./:) ) 4 EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS Part (a) Frequency = 10 GHz ω = 2 ∗ pi ∗ 10 ∗ 109 = 6.283E 10 New impedances are: ZR = 40Ω ZL = j ∗ ω ∗ L = j ∗ 6.283E 10 ∗ 1nH = j 62.8Ω ZC = 1/(j ∗ ω ∗ C ) = 1/(j ∗ 6.283E 10 ∗ 1pF ) = −j 15.9Ω Current source = 6∠0◦ Repeating the above calculations give: ! Zt = ZR +!ZL + ZC = 61.6∠49.5◦ Ω = 40 + 46.9jΩ Vt = (! R + ZL ) ∗%&&'()*+!,-.!/0',#+.12)/)3)0*!&4)*5)&'.6!7.!-#/.8! 6∠0◦ = 240 + 376.8jΩ = 446.7∠57.5◦ "#$%% Z "#$! ! =A !=@ # =A " = != # =A " =A & 6. Hambley, 6.33 ? !; " $ 0:, $ $ ! $A @ &)* =@ # =A # < A!;> @ # < A!;> !=@ # =A " @ # < !; ;9 " (a) Derive an expression for the transfer function H(f ) = Vout /Vin : ! From voltage divider: 7-.4.!;9 $ !=@ # =A " !A!> " ! R1 + R2 2 2/ H(f ) = Vout /Vin = R1 +R2+R2 = R1 +R2 +j 2πf L = 1+jR2 /(L/(+R+)R2 ) = R1+(Rf1/fR2 ) ! L 2πf R1 j( B) So: "B$! C/#':#,)*+!;04!,-.!50D&0*.*,!/#':.3!+)/.*6!7.!-#/.8! fB = (R1 + R2 )/(2πL) ! ; 1 $ [email protected] = 10mH (b) Given: R1 = 1k Ω, R2 = 9 k Ω, LJK!H?I : fB = (R1 + R2 )/(2πL) = 31.83kHz ! ! ! GFE ? !; " $ @ # < !; ;9 " 0.5 B ! The transfer function is sketched as: %!3H.,5-!0;!,-.!,4#*3;.4!;:*5,)0*!D#+*),:2.!)38! ! H(f ) = 1+j (f /f ) ! ! "#$%'! point (fB , −3dB ) or (fB , 0.35) should be marked The L04!,[email protected]?I!3)+*#'6!,-.!,4#*3;.4!;:*5,)0*!D#+*),:2.!)3!! ! & GF E % A @ @ ! ! ? "; $ $ 0:, $ $ GFAE $ $ A &)* A% A @ # "; N ;B $ @ # "@G M N ;B $ A ! ! EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS 5 7. Hambley, 6.45 Cascading the filters gives: H(f ) = H1 (f )H2 (f ) |H(f )|dB = | H1 (f ) |dB + | H2 (f ) |dB For the above two equations to be valid, H1 (f ) must be the transfer function of the first filter WITH the second filter attached!! Part (a) – Cascading two first-order filters |H (f = f1 )| ≈ −28.5dB Finding |H (f = 20 ∗ f1 )| where f1 is much greater than the half power frequency of both filters. This allows for the approximation: H(f ) = 1 1+j ff ≈ 1 j ff B B Increasing f from f1 to f1 ∗ 20 leads to |H(f )|dB | changing by: 20 ∗ log( 201∗f1 / 1f1 ) = 20 ∗ log 1/20 = −26dB j Both first order filters go down by -26dB. Thus, from above the cascaded filters have the overall magnitude: Change in |H(f )|dB = | H1 (f ) |dB + | H2 (f ) |dB = −26dB + −26db fB j/ f b |H(f = 20 * f1 )|dB = −28.5dB + −52dB = 80.5dB % % ! 6 % ! 8. Hambley, 6.55 EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS % "#$%%! C % E ". C:: # D ". # ! C $ E ". C::# D ". # ! C $ ". C::# C $ ". C::# 9 9 ! C! % % % !"/$F% D ". # 2; ! 9: '+1"C# ! : 5% !",%8"($,%#$% % 9 & ()0*(-". C:: # 5% !",%($?&8*+*#0%;+2,%8'+*$%(),B% % % 9C6 % EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS 7 1+j ( Repeat for: H(f ) = 1+j (ff /10) /1000) √2 1 +(f /10)2 |H(f )| = √ 2 2 1 +(f /1000) !"#$%!&'( '#" '!" &#" &!" %#" %!" $#" $!" #" !" $" $!" $!!" $!!!" $!!!!" $!!!!!" ()*+,(-." !"#$%& #!!" +!" *!" )!" (!" '!" &!" %!" $!" #!" !" #" #!" #!!" #!!!" #!!!!" #!!!!!" ,-./0" 8 EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS 9. Hambley, 6.60 There are two general methods to solve this problem. The first is to use a Thevenin Equivalent Circuit and work out the transfer function. The other approach is to use voltage divider directly. The first has slightly easier math, but is less intuitive. ! I! attached both solutions: "#$#%! "#$%&'!()!*#+,!&-)!.-/0)+#+!)12#034)+&!*5$!&-)!%52$6)!3+,!&-)!$)%#%&3+6)%7!! .-)!.-/0)+#+!$)%#%&3+6)!#%! ! : <& " " 899!! ! ! : <: # : <; ! 3+,!&-)!.-/0)+#+!054&3=)!#%! ! &" & <: # <; <; &#+ " 97:&#+ ! .-2%'!3+!)12#034)+&!*5$!&-)!5$#=#+34!6#$62#&!#%>! ! .-#%!#%!3!45(?3%%!*#4&)$!-30#+=!3!&$3+%*)$!*2+6&#5+!=#0)[email protected]!B123&#5+!C7D! E(#&-!6-3+=)%!#+!+5&3&#5+F>! &52& : " ! ! & : # H $* *G % & (-)$)!*G " : $;!<& L % " :KC7D!IJ 7! M%#+=!&-)!*36&!&-3&! & " 97:&#+ '!()!-30)!! I $* % " & .-)!G5,)!?45&%!3$)>! Approach Two: &52& 97 : " ! &#+ : # H $* *G % Vout Voltage divider: Vin ! = 1 jωC ∗R1 ∗R2 /(R1 +R2 )+1 1 1/R2 +jωC R1 + 1/R 1 jωC ;:K 2+ = 1 jωC ∗R1 +R1 /R2 +1 = = 1 j (f /fB )+1 , EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS 9 This can be shown to be equivalent to above. These give the following Bode Plots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ambley, 6.67 Using voltage divider we arrive at: H(f ) = fB = R/(2πL) = 1 MHz Vout Vin = jωL R+jωL = j 2πf L/R 1+j 2πf L/R = jf /fB 1+j (f /fB ) !"#$%!&'( $% !&$% !#$% !($% !'$% !"$$% !"&$% !"#$% "% $% % % % % $% $$ $$ % $, "$ "$ $$ "$ $$ $( *+ "$ *+ $$ "$ $$ $$ *+ $' % -./01-23% ") !"#$%& #!!" +!" *!" )!" (!" '!" &!" %!" $!" #!" !" #" !" " " " " !" !! !! " !) #! #! !! #! !! !( -. #! -. #, !! #, !! #! !! -. !* " ") ") /0123" #, ...
View Full Document

This note was uploaded on 10/24/2009 for the course EE 40 taught by Professor Chang-hasnain during the Fall '07 term at Berkeley.

Ask a homework question - tutors are online