Signals systems and transforms 5th edition phillips solutions manual

Signals Systems and Transforms 5th Edition Phillips SOLUTIONS MANUAL Full clear download (no formatting errors) at: http://testbanklive.com/download/signals-systems-and-transforms-5th-edition-phillipssolutions-manual/

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Chapter 2 solutions 2.1 (a)

(i)

.r(-t/3)

-9 -8 - 7 -G -5 -4 -3 -2 -1

1

2

3

4

5 6

n

7 8

(ii)

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'T

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2..

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t

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(1,c'.) ,x(�t-�) -=?Cli:-\ =l L

=3/:::-t,,

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t

=

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2-

(-2,(.:/')

s

X(3>+ t: ') ::.. ;t:, { ?:' 'J �) 'C"' = t + 3

1.:u-) X (J- t ') ==? T-=

;t-�

�t

= 'Z:""- 3

:;> t- = -1: +- 2X(2-�)

r

-1

2

4

-= 'r-a+2

.3 2

,2

-'i

-�

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z,

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= 'X. £ r"\

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3

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t= (-1-s, =-

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t

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= �(i-') =? T-:-3+t .:ry t z; ,Z--- 3

..---------'A-3 x:...

( 3 + t: )

2.--

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-q

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(e) X.L-t-)== ult)-u(--t) -=/t..(-t)=u(-t.'J-4(-t') .!1 -=- -[u.lt"> -u.l-t-)] t 1: (:t) X.G-t) : , 1...(t) ls ddd. j

I

=-

...l -

1

r

·- .

=- 5&-o.-.3t

-= �( 3[t+�j)

��t + ��

I 1,

t

� X{-t ') � 5 Cds 3 (-t')

t: {&; =- � (- t) .: ,

I

I

J

!

1'.'. ( t ) = -:t C - t ) :. 'Jl {t ) l :; e treY1 · '. _[ •

l+"i

'X.(t\

-=-·- U.(t-1\ T IA.(-t-i)

")(. (-t) � - it (-t-1) � { - t)

= -'Xli-)

)

-f-

u.(t-1)

r,� O

�

(+) i 5

0

Jd

1

•

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(e) x0(0)=-Xo(-O)= -x0(0). The only number with a=-a is azO so this implies Xo(O)=O. x(0)9ee(O)+x0(0)9ee(O) .

â&#x20AC;˘

(a) Let z(t) be the sum of two even functions x1(t) and x2(t). To show that .:(t) is even. we need to show that z (t) = z(-t) for all t. This ·is easy to show, since z( t) = xi ( t) + 1·2( t) and .:( -t) = l'i( -t) + :i.•2 (-t) (since to get z(-t) we just plug in -t everywhere fort, which amounts to just plugging in -t in 1·1(t) and 1·2(t)). Xow since 1:i(t) and 1·2(t) are even. by definition xi(t) = x1(-t) and :r2(t) = 1·2(-t) so x1(t) + :r2(t) = x1(-t) + x2(-t) so z(t) = z(-t). {b) Let x1(t) and 1·2(t) he two odd functions. Then 1·1(-t) x2(t)) which shows that 1'1 (t) + J:2(t) is odd.

+ x2(-t) =

-x1(t)

+ (-x2(t)) = -(l'1(t) +

(c) Let z(t) = x1(t) + 1)2(t) as in part a, where now x1(-t) = x1(t) and xz(-t) = -1·2(t). \Ve need to show that z(t) 'F z(-t): z(t) f -z(-t). Consider that z(-t) = xi (-t) + :1·2(-t) = 1·1 (t) - ;1.·2(t). In order to have .:(t) be even: we would therefore need to have 1·1 (t) + 1·2(t) = x1 (t) -1·2(t) for all t . which is equivalent to having 1:2(t) = -1:2(t) for all t, which is not possible for nonzero l'.2(t). Similarly, in order to have :(t) be odd. we would need to have z(t) = -z(t) � xi (t) + 1·2(t) = 1·2(t) - xi (t). which is not possible for nonzero x1(t). So the sum of an even and odd function must be neither even nor odd. (d) Let z(t) = x1(t)x2(t) where :t1(t.) = J:1(-t) and x2(t) x1(t)x2(t) = z(t) which shows that -�(t) is even. (e) Let .:(t) = 1:i(t)x2(t). where X[(t) .:(-t) = :ri(-t)x2(-t) = (-1·i(t)) (-x2(t))

= 1·1(-t)x:?(-t) =

=

Then z(-t) = 1·1(-t.)1·2(-t) =

-xi(-t) and x2(t} = -x2(-t). Clearly .:(t) is even because 1·i(t)1·2(t) = .:(t)� which is the definition of evenness.

(t) = -1·i(-t) and x2(t) = 1·2(-t). Clearly :;(t) is odd because (-x1 (t)) 1·2(t) = -xJ(t).r2(t) = -:(t). which is the definition of oddness.

(f) Let :;(t) = l'1(t):r2(t), where .:(-t)

=

= 1·2(-t).

x1

-x.L+\ i 2.-

�c)

C-t)

=- "l{-t-> - r..C-t"> 2..

x.e : : %l-t \) + 7C (-t) 2:...

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•

�

I

-

-+

l

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f

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(a) sin(t) = sin(t + n2rr) for any integer n, so 7sin(3t) = 7sin(3t + n2;r) = 7sin {3(t +n2:r)); therefore l'(t) is periodic with fundamental period To=� and fundamental frequency wo = = 3.

t

7-)

(b) sin(8(t + + �O) = sin(8t + 21r + 30) = sin(8t + 30). wo = 8 and To = 7 = ! . (c) eJt = cos(t) + j sin(t) is periodic with fundamental period 2rr: so e121 is periodic with fundamental period � = 1r. and fundamental frequency wo = 2.

J

I �-i:+ �Gt JV _ � T -=.

TI - 2 - II J 2-

(c\) �

=

Tc;={(l'T\: .. To

c_e')

�'tr -:::.... Ti 5 -t 12,

= .1]: -:= 5J /LCt_:ft o- o-/ ol% '2..

111fe�

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� =- z

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J

T0;:

�:

�

= 1f7'5( �; �&-G,

•

•

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:: �

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-- C� '\ ld1 (e')

i(..

ot -1- Al-wl5t

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C--+:) =- � 3-t

+ -� �-6

��n-'-t +

>G(-+') =

n-t) : :

� 47f-l:

+ CKL(St)

�41Ft -!-�$Ji,-t + <2,a,�1Tt

�C'-t)-=: �(Si:-

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J

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+3D0)

+

e a-2-t: +�L3rrr-&')

:hl)-r-.

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(a..)

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:=

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+ Y-u..+(

d.e-,

vi

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-=- �-=- 2

T, T

Y..1 (:t \ .,,� �d.Lc

T "2-

•

Cc..J-=:, =. 7

7,

To :::: 5 T, -=- :2. 7Y

•

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;J_

5.· -

w��i5

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(

{.,u I::

t.<J({z_ �·�

a:

"!:·�.+

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Tb =-)Vhf,= -I 44-

:..�

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(a) For :q (t) + 1·2(t) to he periodic we need some number T such that x1(t + T) + l'2(t + T) = l'1(t) + 1·2(t.) for all t. This can only be true if 1·1 (t+T) = 1·1 (t) and 1·2(t.+T) = 1·2(t)i which can only be true if T = A·1T1 and T = A·2T2 (T is an integer multiple of hoth the periods). So we need there to he some integers l· 1 and k,2 such that A·1T1 = �·2T2 � � = (b) Put in its most reduced form;; by canceling any common terms in the numerator and denominator;

I

i7·

fi

J

1

then To = nT2 = mTr.

1 f Problem 2. I 4

I j

(a) >> syms t >> xa:::::5*exp(-t/2); >> ezplot(xa), grid (c) >> symst >> xc=5*exp(t/2); >> ezplot(xc),grid

(e)

>> symst >> xe=5*(1-exp(-2*t)); >> ezplot(xe), grid (g) >> symst >> xg=5*exp{-2*t)*2*sin(2*t); >> ezplot(xg),grid

(b) >> syms t >> xb=5*exp(-2*t); >> ezplot(xb),grid

(d) >> syms t >> xd=S*(l-exp(-t/2)); >> ezplot(xd), grid (f) >> symst >> xf=5*2*sin(2*t); >> ezplot(xf),grid (h) >>syms t >> xhz:c5*exp(-0.5*t)*2*sin(2*t); >> ezplot(xh),grid

-l

l. l

J

--- -

---

---------1

t

I

i

Problem 2.15 (b)

(a) t

cos(O+(,) = Re{eJ(o+;>} =

sin(O+(,) = Im{e1<o+;l} = lm{e.fSel'}

Re{e19e1;}

= Re{(cos8+ jsinB)(cos(, + jsin(,)}

= Im{(cosB+ jsin8)(cos(, + jsinf

= Re{cos8cos(, + jsin8cos(,

= Im{(cosBcos(,+ jsinBcos,p

e

+ jcosBsin(,-sinBsin(,}

+ j cos sin t/J - sine sin t/J}

= cos Bsin (, + sin B cos tp

= cosBcos(,-sinBsin(, (c)

cosBcos(,

. eii+e-1;} {eJ(o+;)+e1(0-;)} = Re ---2 2

= Re { e19

= Re{eJ(o+;) + e1(o-;)} = cos(B+tp) + cos(B-(,) 2

2

Ptev BLt rY\ · ?{ ')

2

�,

2

I fu

X(_-t-, =- � � ( ;;t t)-t'1(

c-n = � (e8

•

\

t-

-i

+-

d

"f-Jl� e�2t �(�+J�)e-a2,t

= =

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.L ? 1-,

�(zt)

\/lb'/ti,,' t�l 2

�-t

t»:

l

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t�) =-D, 3� Y-ud. (

t V?;. �

= -f o , � 2.. ya

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� (f-) ::

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-==

x c+) ::

:z.f.

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z.

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+

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rio:

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J)

,

--------------

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...

'Prob[e'N\ �,[,� ,(Co11t/Y1ue.:!)

(b)

4- � �

=- 4 &r� (4rrt)

1LCt')

ft,

=�ce_,

d,'frT-t:

+e

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2...

z,

=(i-��J) e":" -1--(_;i1-3{Jj

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·=::

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.

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15

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2.

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(a:-t-hJ�'-(-t-- c.) J-1: =-±-- �{%..-C0 5;c-1:-!r'cJJ-1: -�

-= it_

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?Ro B l6'M ':? , I 9

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1

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: · � l + J -== m:Jy..fc .,.sJ

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- t;)a

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l 6�

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=�

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(c) Recall the rules about integrating delta functions: t�(t) is nonzero only at t = 0. so r(t)6(t) = x(0)6(t). and '5(t.)dt = 1, so .r(t)o(t)dt = :r(0)6(t)dt = x(O) 00 6(t)dt = J'(O). \Ve can time-shift the delta function: J(t- to) is nonzero only at t = to, so J'(t)'5(t-to) = :r(to)o(t - to) and :r(t)'5(t- to)dt = �·(to).

i)

J:.:

1:_:

J�

1�

1::

1� cos(2t)'5(t)dt = cos(2 · 0) 1:: J(t)dt = 1.

ii) o(t -

f) is a time-shifted version of <>(t)i and is nonzero only at t = f

1

: ,:;

-,,

So:

oo. sin(2 · .ii. )o(t - �)dt sin(2t)'5(t - -)dt = -I J-ex· 4 -C\: 4 - sin(f)J"' J(t-f)dt=sin(t)=l ...

-00 ...

-

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c) ;; (t - %)

=-J" '�c��1 s(t-TT/&Jcu-

t:

-�

�

=-�(:Z) Jb(t-717t,)Ji

�%; = o.scc

-cl'

-=

- -VO

5�{ t-%)1 £-Ct - �)dt

�

.,tJ

(Lil')

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(tr)

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,.,.

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·

1 11

+ -

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(a)u.(2t

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(C.)

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2)

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r,t (

I

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=

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cc1 > u

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= u.

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U(-3-t-+4') =

l

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::: Ct-1., c 1

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u (t-2) 1

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1:(t)

9 8

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1 2 3 4 5 1 2 3 4 5 6

'(b)

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r '

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rft-2

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-E_tl) Ll(t -i) J

Gt(t-2) -�-'3 ')'u(t-3"')]

!=+f-

i-:

. W-,< t �

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::

1-, l+, -== p:t-�-r;,!f'-3rf:+_);5 +3-:f::-g-=.

J

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I

lb) 't2-l-t J i�

periodic.. wi t:L, � .c,.a

1l:z. (-t ') -=-

e: X,

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=- £.

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YI --De

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f

j

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