6.003 Homework #3 Solutions

Transcription

1 6.00 Homework # Solutios Problems. Complex umbers a. Evaluate the real ad imagiary parts of j j. π/ Real part = Imagiary part = 0 e Euler s formula says that j = e jπ/, so jπ/ j π/ j j = e = e. Thus the real part is e π/ ad the imagiary part is 0. b. Evaluate the real ad imagiary parts of ( j ). Real part = 4096 Imagiary part = 0 The magitude of = j is, ad its phase agle is 60. So the magitude of is = 4096 ad its phase is 60, which is a multiple of 60, ad therefore the same as 0. So = Its real part is 4096 ad its imagiary part is 0. c. Express the real part of e 5jθ i terms of si θ ad cos θ. Real part = cos 5 θ 0 cos θ si θ + 5 cos θ si 4 θ From Euler s formula, e 5jθ = cos 5θ + j si 5θ. Its real part is cos 5θ. But we eed to write it i terms of si θ ad cos θ. So we better start with So e jθ = cos θ + j si θ. e 5jθ = e jθ 5 = (cos θ + j si θ.) 5. The real part of the expasio is the odd-umbered terms, sice those have oly eve powers of j. I order, they are cos 5 θ, 0 cos θ si θ, 5 cos θ si 4 θ. So the real part, which is also cos 5θ, is cos 5 θ 0 cos θ si θ + 5 cos θ si 4 θ.

2 6.00 Homework # Solutios / Fall 0. Yi-Yag Determie the system fuctioal Y for the followig system X A X + B C + Y where A, B, ad C represet the system fuctioals for the boxed subsystems. Y = X A + B + BC Let W represet the iput to subsystem B. The Black s equatio ca be used to show that W X = + BC. The output Y is the sum of AW ad BW. So the result is Y X = A + B + BC.

3 6.00 Homework # Solutios / Fall 0. Z trasforms Determie the Z trasform (icludig the regio of covergece) for each of the followig sigals: a. x [] = u[ ] X = 8 ( ) ROC: > X () = x [] = (/) =. Let l =. The = = 0 l+ ( /) X () = = = ( /) 8 ( ). l=0 The ROC is > /. 8 A alterative approach is to thik of x [] as times a versio of u[] that is delayed by. The Z trasform of u[] is. Delayig it by multiplies the trasform by ; ad scalig by 8 scales the trasform similarly. b. x [] = ( + ) u[] X = ROC: > Start with the defiitio of the Z trasform: 0 X() = x[] Differetiate both sides with respect to : dx() 0 0 = x[]( ) = x[] d This shows that Sice x[] dx() d u[], >.

4 6.00 Homework # Solutios / Fall 0 it follows that ( ) ( ) d u[], >. = ( d ) 4 X () = + =, ( ) > ( )

5 6.00 Homework # Solutios / Fall 0 ( c. x [] = ) 5 X = ( )( + ) ( ) ( ) ROC: < < x [] = ( ) u[] ( ) ( ) u[] + u[ ] ( ) ; >. ( ) u[ ] ; <. ( ) X () = ( ( )( + ) + = ) ( ) ( ) ( ; < <. ) d. x 4 [] 0 4 X 4 = + + ROC: 0 < < 0 X 4 () = x[] = x 4 [ ] + x 4 [] + x 4 [] = = + + ; 0 < <.

6 6.00 Homework # Solutios / Fall 0 4. Iverse Z trasforms Determie all possible sigals with Z trasforms of the followig forms. a. X () = Eter expressios (or umbers) i the followig table to discribe the possible sigals. Each row should correspod to a differet sigal. If there are fewer sigals tha rows, eter oe i the remaiig rows. < = = 0 = > x [] = or or oe oe oe oe oe This trasform has a pole at =, as does the uit-step sigal: 0 u[] = = =0 Multiplyig this trasform by represets a uit delay i time: 0 u[ ] = =0 = Thus there is a right-sided iverse trasform x R [] = u[ ] ; > ad a left-sided iverse trasform x L [] = u[ ] ; <. x L [] x R []

7 6.00 Homework # Solutios / Fall 0 b. X () = ( ) 7 < = = 0 = > x [] = 0 or or oe oe oe oe oe X () = ( ) The first term ( ) represets two uits of delay. The secod term correspods to a uit ramp startig at = 0. Because the ramp has a pole at, there are two regios of covergece. The right-sided regio correspods to a uit ramp startig at = (because of the two uits of delay). The left-sided regio correspods to a ramp with slope proceedig backwards i time from =. x L [] x R []

8 6.00 Homework # Solutios / Fall 0 c. X () = = = = 0 = = x [] = or or oe oe oe oe oe X () = ( ) j e jπ/ e jπ/ = ( j e jπ/ ) e jπ/ Both terms correspod to geometric sequeces. The base of the first is e jπ/, the base of the secod is e jπ/. The factor represets a uit of delay. Sice the poles are o the uit circle, there are two regios of covergece: greater tha ad less tha. For >, For <, ( e j( )π/ e j( )π/) u[ ] = si x R [] = j =, 5, 8,... = =, 6, otherwise ( e j( )π/ e j( )π/) u[ ] = si x L [] = j = 0,, 6, 9,... = =, 4, otherwise x L [] x R [] π( ) u[ ] π( ) u[ ]

9 6.00 Homework # Solutios / Fall 0 ( ) d. X 4 () = 9 = = = 0 = = x 4 [] = 0 0 or or oe oe oe oe oe oe oe oe oe oe X 4 () = + 4 = + This sum coverges for all. Therefore there is a sigle regio of covergece 0 < <. x 4 [] = δ[ ] δ[] + δ[ + ] x 4 [] 4 0 4

10 6.00 Homework # Solutios / Fall Poles The followig diagrams represet systems with poles (idicated by x s) but o eros. The scale for each diagram is idicated by the circle, which has uit radius. X () X () X () X 4 () a. Which (if ay) of X () through X 4 () could represet a system with the followig uit-sample respose? 0 Eter a subset of the umbers through 4 (separated by spaces) to represet X () through X 4 () i the aswer box below. If oe of X () through X 4 () apply, eter oe. ad/or ad/or ad/or 4 or oe: oe We ca write the desired sigal as x[] = a, 0 < a <. The the Z-trasform is give by: 0 0 X() = a + a = =0 = (a a ) ( a)( a ) ; a < < a. This Z-trasform has poles at = a, = a, ad a ero at = 0. Therefore the aswer is oe of the above.

11 6.00 Homework # Solutios / Fall 0 b. Which (if ay) of the systems could be stable? Hit: A system is stable iff the regio of covergece of its Z trasform icludes the uit circle. ad/or ad/or ad/or 4 or oe: System will be stable if the regio of covergece is outside the circle defied by the pole; system will be stable if the regio of covergece is the aulus defied by the two poles, ad system will be stable if the regio of covergece is outside the circles defied by the poles. System 4 caot be stable, sice oe of the poles is o the uit circle. c. Which (if ay) of systems could be causal? Hit: A liear, time-ivariat system is causal if its uit-sample respose is ero for t < 0. ad/or ad/or ad/or 4 or oe: 4 Ay of the pole diagrams could represet such a system if the regio of covergece is outside all of the circles defied by all of the poles. d. Which (if ay) of the systems could be both causal ad stable? ad/or ad/or ad/or 4 or oe: For stability, the ROC must iclude the uit circle. A causal system must be rightsided. A right-sided system has ad ROC of the form > a. Oly X () ad X () represet such systems. If we decompose a system fuctio usig partial fractios, the we ca cosider the uit-sample respose of the system as a sum of compoets that each correspod to oe of the poles of the system. If the cotributio of a pole is right-sided, the its Z trasform coverges for all with magitudes bigger tha that of the pole. To be stable, the magitude of that pole must be less tha. It follows that the regio of covergece icludes the uit circle. A similar argumet holds for left-sided cotributios.

12 6.00 Homework # Solutios / Fall 0 6. Periodic system Cosider this variat of the Fiboacci system: y[] = y[ ] y[ ] + x[] where x[] represets the iput ad y[] represets the output. a. Compute the uit-sample respose ad show that it is periodic. Eter the period i the box below. period = 6 The system fuctioal is H = Y X = R + R. Oe ca expad the system fuctioal usig sythetic divisio as follows: +R +0R R R 4 +0R 5 +R 6 + R + R R +R R R R R +R R R +R 4 R 5 R 4 +R 5 R 4 +R 5 R 6 The uit sample respose is the sequece h :,, 0,,, 0,,, 0,,, 0,... The form of this solutio suggests that +R 6 + H = + R R R 4 + R 6 + R 7 R 9 R = ( + R)( R )( + R 6 + R + R 8 + ) = ( + R)( R ) R 6 which is true because this expressio equals the origial system fuctioal: ( + R)( R ) R 6 = R + R. Equality ca be see by cross multiplicatio ad expasio ( + R)( R )( R + R ) = R 6. It is clear from H = ( + R)( R ) R 6 that the uit-sample respose is periodic ad that the period is 6.

13 6.00 Homework # Solutios / Fall 0 b. Eter the poles of the system i the box below (separated by spaces). jπ/ poles = e e jπ/ Substitute R = + i the system fuctioal to get ad fid the roots of the deomiator: = ± j = e ±jπ/. c. Decompose the system fuctioal ito partial fractios, ad use the result to determie a closed-form expressio for h[], the uit-sample respose. Eter your expressio i the box below. h[] = R + R = j si π si π( + )/ si π/ ( ) e jπ/ e jπ/ R e jπ/ e jπ/ R The correspodig uit-sample respose is ( h[] = j si π e jπ(+)/ e jπ(+)/) si π( + )/ = si π/

14 6.00 Homework # Solutios / Fall 0 Egieerig Desig Problems 7. Scalig time A system cotaiig oly adders, gais, ad delays was desiged with system fuctioal Y H = X which is a ratio of two polyomials i R. Whe this system was costructed, users were dissatisfied with its resposes. Egieers the desiged three ew systems, each based o a differet idea for how to modify H to improve the resposes. 4 System H : every delay elemet i H is replaced by a cascade of two delay elemets. System H : every delay elemet i H is replaced by a gai of followed by a delay. System H : every delay elemet i H is replaced by a cascade of three delay elemets. For each of the followig parts, evaluate the truth of the associated statemet ad eter yes if the statemet is always true or o otherwise. j5π/4 a. If H has a pole at = j =, the H has a pole at = e. Yes or No: Yes Explai. The poles of H are the roots of the deomiator of H R. But H = H R R. Thus the = (H R R ) R poles of H are the roots of the deomiator of H R It follows that the poles of H are the square roots of the poles of H. = H R. If H has a pole at = j the H must have poles at = ± j. The two square roots of j are e jπ/4 ad e j5π/4. Thus e j5π/4 is a pole of H.

15 6.00 Homework # Solutios / Fall 0 b. If H has a pole at = p the H has a pole at = p. 5 Yes or No: No Explai. The poles of H are the roots of the deomiator of H R. But H = H R R/. Thus the ( ) R poles of H are the roots of the deomiator of H R = H R R/ = H R. It follows that the poles of H are half those of H. If H has a pole at = p the H must have poles at = p/ (ot p). c. If H is stable the H is also stable (where a system is said to be stable if all of its poles are iside the uit circle). Yes or No: Yes Explai. The poles of H are the roots of the deomiator of H R. But H = H R R. Thus the = (H R R ) R poles of H are the roots of the deomiator of H R It follows that the poles of H are the cube roots of the poles of H. = H R. If H is stable, the the magitudes of all of its poles are less tha. It follows that the magitudes of all of the poles of H are also less tha sice the magitude of the cube root of a umber that is less tha is also less tha. Thus H must also be stable.

16 6.00 Homework # Solutios / Fall 0 8. Complex Sum Each diagram below shows the uit circle i the complex plae, with the origi labeled with a dot. Each diagram illustrates the sum 6 00 α S =. =0 Determie the diagram for which α = j. diagram = F C A B D E F G H The first two terms i the series represetatio of S are ad j. All of the curves start with a horiotal lie to the right that stops at the itersectio with the uit circle. Thus all of the curves correctly represet the first term. The secod term should have a legth that is somewhat shorter tha ad a agle of ta (0.5). We ca immediately

17 6.00 Homework # Solutios / Fall 0 discard curves C, D, ad E because their secod terms have egative imagiary parts, which is wrog. How quickly should the terms i the series coverge? The magitude of α is The magitude of the last term i the sum is the approximately < 0 9 which would ot be visible i the plots. Thus, the curve should appear to coverge to a limit, which elimiates curves A ad H. Furthermore, it says that the sum of 00 terms is very early equal to the ifiite sum: 0 S S = α = α = ( j) = 0. 0.j =0 = 0. e = 5 e jπ/4 jπ/4 Thus the fial values of B ad G are wrog, ad the aswer is F. 7

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