HW4 Solution

HW4 Solution - P631[a First we tind the The'venin...

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Unformatted text preview: P631 [a] First, we tind the The'venin equivalent tor the source and resistances. The open—circuit voltage is given by , , s ,2”: 1H: if—t “i” “0‘” V5~1e+o+e In terms at phasors, this becomes: 9? V.=V—t 1 ' 5R5+§+afi (l Zeroing the source, we find the The’venin resistance: he: 1 a ' us, + 1313? + £1} Thus, the original circuit has the equivalent: 12; "' ‘1; C View —. The transfer tunction tor this circuit is: 'h" l m = _ 2 m ruins} ( j 1 where, it; = Emgfl, Using Equation {1] to substitute tor if; in Equation {2] and rearranging, we have: " ' “out at 1 mic-'— v, ‘e+s+s,"1+,i[f,ogi (33' {In} Evaluating tor the circuit components given, we have: a = 1.5 m r; =106.1Hz 0.5 H[f'}=1+flfffl3 A Mai-RAB program 1'0 plof The transfer—function magnitude is: fover‘szflflfll: 3; Hmag= ubsfiUfiJEl + [*fwer'fbjj; plu fifover‘be-Imag} axisflfl 3 U {15]} The rasul’ring ploi‘ is: W ‘15 0.4 0.3 0.2 [1.1 D 9.5 1 1.5 2 2.5 3 15%; F625 The Transfer funcTian is given by EquaTian 6.9 in The TexT: Hf =—i l" 1+ gigs} The given inpuT signal is x ( ~ vi“ {1‘} = 4 + E sinllflflflsf + 3U: flI+ 5 caslfilfl x103 sf] which has campanenTs wiTh frequencies af El, 500, and 15,300 Hz. EvaluaTing The Transfer funcTian far These frequencies yields: ~ 1 Ha =—=1 {T 1+ yiqeaa] H590) = mans — 45= Hl:lfl“:l= U.U333Lfi—88.UQ° Applying The apprapriaTe value af The Transfer funcTian Ta each campanenT of The inpuT signal yields The aquuT: le:f:l= 4+ 1.414cusjaaam—1as=]+a.1assmlaax 1a3s_ seas} P6.EI5 This is The firsT-arder high—pass filTer analyzed in SecTian 5.5 in The TexT. The Transfer funcTian is ac 3|:ch Hifl: fl =—{r——1+J.fié} l where IE=EHR€=IUDDHL The inpuT signal is given by v...” {3‘} = 5 cas{2flflsf:}+ 5 cas{EDDUrrf ] This signal has campanenTs aT f =1DU Hz and f = lflflfl Hz. The Transfer— funcTian values aT These frequencies are: _ “30.1 _ + HLIGDJ— 1+Jfll — 0.0995£34.29 Hflflflflj = “'l 1_ = fi.?fl?1£45° 1+Jr1 Applying These Transfer—funcTian values Ta The respecTive campanenTs yields: vmlf} = 0.4915 caleUflrzf + a4.29=]+ 3.536casl:2UUUIrf + 451' P727 (:1) F:(A+3)Z (b) F=A+3+ET€J (c) F=AB+(8C)+D P7.36 In This circui’r, The ou‘i‘pu’r is high only if switch A is open (A low) and if ei’rher of the oTher‘ Two swiTches is open. Thus, we can write D:Z(C_'+B) P7.43 I=ZBC+ABE+AB€ 2:»:(357) I=(A+8+6')(A+B+5)(A+§+€)(Z+B+6)(Z+8+5) = ZM(O,1*2,4,5) P7.47 Applying DeMorgan's Laws To The outpuT of The circuit, we have 04+ B)(C+D) : (A+8)+(C'+D) Thus, The circuiT is A Ana. 5 C CH? ...
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This homework help was uploaded on 04/12/2008 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Tech.

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HW4 Solution - P631[a First we tind the The'venin...

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