hwk5_sol

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

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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= > ! - [email protected],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 [email protected],'!= ) -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 [email protected],'!= ) 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 [email protected],A!= )<-5 ! " % < 9'B9 != ) 215 ! > % )8 : 899 % 8 ! B9 > ) 1 [email protected],A = 10∠0◦ −j/ω C = −j/(500 8 [email protected]):=CF0%[email protected]%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 ' "<=!>[email protected] # 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 [email protected],A!= ) -5 ! > ) 1 [email protected],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 \$ A'[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 [email protected] # =A " = != # =A " =A & 6. Hambley, 6.33 ? !; " \$ 0:, \$ \$ ! \$A @ &)* [email protected] # =A # < A!;> @ # < A!;> [email protected] # =A " @ # < !; ;9 " (a) Derive an expression for the transfer function H(f ) = Vout /Vin : ! From voltage divider: 7-.4.!;9 \$ [email protected] # =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 ! "#\$#%! \$ "#\$#&\$ ! !"#\$%&'()*+',#'\$"&-"./((\$01\$%&2)#'\$&(\$("+34\$&4\$5&-6'#\$789:\$&4\$)"#\$;++<8\$\$ !"#\$"/2%*.+3#'\$%'#=6#4>?\$&(\$-&@#4\$;?\$%E ! 9 CDB!01 A8 \$ !"#\$(2+.#\$+%\$)"#\$"&-"*%'#=6#4>?\$/(?F.)+)#\$&(\$G#'+8\$\$!"#\$(2+.#\$+%\$)"#\$2+3* %'#=6#4>?\$/(?F.)+)#\$&(\$BH\$,EC,#>/,#8\$\$!"#\$/(?F.)+)#(\$F##)\$/)\$)"#\$"/2%* .+3#'\$%'#=6#4>?\$%E 8\$ ! "#\$#'(! I..2?&4-\$)"#[email protected]+2)/-#*,&@&(&+4\$.'&4>&.2#J\$3#\$"/@#K\$ ) L "% # ! +6) \$ \$ )&4 ! \$ \$ \$ \$ \$ \$ \$ \$\$\$\$\$\$\$\$ ! H89 \$\$\$\$\$\$\$\$ ! \$\$\$\$\$\$\$\$ ! \$\$\$\$\$\$\$\$ 09 \$ 0B \$ 9 M B!%1 \$ 0B "09 \$ 0B # ! 9 \$ 9 M B!%1 "09 \$ 0B # M B!%1 "09 \$ 0B # \$ 09 \$ 0B 9 \$ M B!%1 "09 \$ 0B # 0B M "% %E # \$ 09 \$ 0B 9 \$ M "% %E # 0B M "% %E # ! 9 \$ M "% %E # 0B 10 EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 3 SOLUTIONS 10. Hambley, 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" #, ...
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