[PDF] ECEN 314: Signals and Systems - Homework 2 Solutions I Reading

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ECEN 314: Signals and Systems - Homework 2 Solutions •Date Assigned: Monday June 4, 2018 •Date Due: Wednesday, June 13, 2018I Reading Exercise

Chapter 1 - sections 1.1, 1.2 and 1.3

II Problems

1. (1.3) Determine the value of P

1and E1for each of the following signals

(a)x(t) =e2tu(t) (b)x(t) =ej(2t+4 (d)x[n] = (12 )nu[n] (f)x[n] = cos4 n

Solution :

a) E 1=Z 1 0 e4tdt= 14 e4t 1 0 =14 P 1= 0 b) E 1=Z 1 1 jx2(t)j2dt=Z 1 1 dt= [t]11=1 P

1= limT!112TZ

T

Tdt= limT!12T2T= 1

d) E 1=1X n=1jx2[n]j2=1X n=0 14 n =114 1114
=43 P 1= 0 1 f) E 1=1X n=1cos 24
n =1

Period = 8

P

1=P8=18

7 X 0cos 2(4 n) =18 4 =12

2. (1.4) Letx[n] be a signal withx[n] = 0 forn <2 andn >4. For each signal given below,

determine the values of n for which it is guaranteed to be zero (c)x[n] (e)x[n2]

Solution :

c

The signalx[n] is

ipped. The ipped signal will be zero forn <4 andn >2 e

The signalx[n] is

ipped and the ipped signal is shifted by 2 to the left. This new signal will be zero forn <6 andn >0

3. (1.5) Letx(t) be a signal withx(t) = 0 fort <3. For each signal given below, determine the

values of t for which it is guaranteed to be zero. (b)x(1t) +x(2t) (c)x(1t)x(2t)

Solution :

b From (a), we know thatx(1t) is zero fort >2. Similarly,x(2t) is zero fort >1.

Therefore,x(1t) +x(2t) will be zero fort >1

2 c x(1t) is zero for 1t <3 =)t >2.Similarly,x(2t) is zero fort >1. Hence x(1t)x(2t) is 0 fort >2

4. (1.10) Determine the fundamental period of the signal x(t) = 2cos(10t+ 1)sin(4t1)

Solution :

Period of rst term in RHS =

210
=5

Period of the second term in RHS =2

Therefore, the overall signal is periodic with a period which is the least common multiple of the periodic of the rst and second terms. This is equal to

5. (1.11) Determine the fundamental period of the signalx[n] = 1 +ej4n=7ej2n=5

Solution :

Period of the rst term in the RHS = 1

Period of the second term in the RHS =m24=7

= 7(whenm= 2)

Period of the third term in the RHS =m22=5

= 5 (whenm= 1) Therefore, the overall signalx[n] is periodic with a period which is the least common multiple of the periods of the three terms inx[n]. This is equal to 35.

6. (1.25) Determine whether or not each of the following continuous-time signals is periodic.If

the signal is periodic, determine its fundamental period (b) expj(t1) (c) (cos(2t=3))2 (d)Evfcos(4t)u(t)g (e)Evfsin(4t)u(t)g

Solution :

b

Periodic, period = 2

c

Periodic, period ==2

3 d

Periodic, period = 1=2

e

Not periodic

7. (1.26) Determine whether or not each of the following discrete-time signals is periodic. If the

signal is periodic, determine its fundamental period. (b)cos(n=8) (c)cos(8 n2)

Solution :

b

Not periodic

c

Periodic, period = 8

8. (1.34) In this problem, we explore several of the properties of even and odd signals.

(a) Show that if x[n] is an odd signal, then 1 X n=1x[n] = 0 (b) Show that ifx1[n] is an odd signal andx2[n] is an even signal, thenx1[n]x2[n] is an odd signal. (c) Letx[n] be an arbitrary signal with even and odd parts denoted by x e[n] =Evfx[n]g x o[n] =Odfx[n]g

Show that

1 X n=1x

2[n] =1X

n=1x

2e[n] +1X

n=1x 2o[n] 4 (d) Although parts (a)-(c) have been stated in terms of discrete-time signals, the analogous properties are also valid in continuous time. To demonstrate this, show that Z 1 n=1x2(t)dt=Z 1 n=1x2e(t)dt+Z 1 n=1x2o(t)dt wherexe(t) andxo(t) are, respectively, the even and odd parts ofx(t).

Solution :

a

Consider

1X n=1x[n] =x[0] +1X n=1x[n] +x[n] If x[n] is odd,x[n] +x[n] = 0. Therefore, the sum equals zero. b

Lety[n] =x1[n]x2[n]. Then,

y[n] =x1[n]x2[n] =x1[n]x2[n] =y[n] c 1 X n=1x

2[n] =1X

n=1x e[n] +xo[n]2=1X n=1x

2e[n] +1X

n=1x

2o[n] + 21X

n=1x e[n]xo[n]

From b, we knowxe[n]xo[n] is an odd signal. So,

2 1X n=1x e[n]xo[n] = 0

Therefore,

1X n=1x

2[n] =1X

n=1x

2e[n] +1X

n=1x 2o[n] 5 d Z 1 1 x2(t)dt=Z 1 1 x2e(t)dt+Z 1 1 x2o(t)dt+ 2Z 1 1 x e(t)xo(t)dt

Again,xe(t)xo(t) is odd, so:Z1

1 x e(t)xo(t) = 0

Therefore,

Z1 1 x2(t)dt=Z 1 1 x2e(t)dt+Z 1 1 x2o(t)dt

9. (1.36) Letx(t) be the continuous-time complex exponential signal

x(t) =ejw0t with fundamental frequencyw0and fundamental periodTo=2w

0. Consider the discrete-time

signal obtained by taking equally spaced samples of x(t)-that is, x[n] =x(nT) =ejw0nT (a) Show that x[n] is periodic if and only ifT=T0is a rational number-that is, if and only if some multiple of the sampling interval exactly equals a multiple of the period ofx(t). (b) Suppose that x[n] is periodic-that is, that TT 0=pq where p and q are integers. What are the fundamental period and fundamental frequency ofx[n]? Express the fundamental frequency as a fraction ofw0T. (c) Again assuming thatT=T0satisesTT 0=pq , determine precisely how many periods of x(t) are needed to obtain the samples that form a single period of x[n].

Solution :

a Ifx[n] is periodicej!0(n+N)T=ej!0nT, where!0= 2=T0. This implies that2T

0NT= 2k.

So TT 0=kN = a rational number. b IfT=T0=p=qthenx[n] =ej2n(p=q). The fundamental period isq=gcd(p;q) and the funda- mental frequency is 2q gcd(p;q) =2p pq gcd(p;q) =!0p gcd(p;q) =!0Tp gcd(p;q) 6 c p=gcd(p;q) periods ofx(t) are needed 7quotesdbs_dbs9.pdfusesText_15

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