HW1 solutions – ECE351

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Problem 1.44 There are 2 methods to solve this problem. Method 1: From )cos()( φω += tAtx

Power = tdtAT

T2

0

1 ])cos([ φω∫ +

= tdtTA T

)(cos φω∫ +0

22

We know that

21

222 +=

)(cos)(cos θθ

Thus, Power = td

TAtdt

TA TT

∫∫ ++0

2

0

2

212

21 )][(cos φω

Note that 020

=+∫ tdtT

)][(cos φω ;

Power = 2

021

210

22

0

2 ATTAt

TA

T

=−⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛+ )( watts.

Method 2: From )cos()( φω += tAtx Average power

RVRIVIP

22 === where V, I, and R are Voltage(rms), Current (rms), and Resistance, respectively.

V(rms) of x(t) = AA 70702

.≈ where A is the amplitude of x(t).

Assume R = 1 ohm. That is the average power ( )21

222

2 AR

VPA

=== watts.

Problem 1.50

HW1 solutions – ECE351

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Problem 1.52

)()( tytx −−1

HW1 solutions – ECE351

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(g)

HW1 solutions – ECE351

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Problem 1.57 (a) from ( )nnx π15

8cos][ = For discrete-time signal to be periodic, we require that

][][ pNnxnx +=

Therefore,

( )( )( )p

p

Nn

Nnnnx

ππ

ππ

158

158

158

158

+=

+=

=

cos

)(coscos][

We know that )cos()(cos θπθ += k2 where k = integer Thus,

kN p ππ 2158 =

kkN p 4152

158

==ππ

The small integer k that results in Np = integer is k =4. That means this is periodic signal with fundamental period Np = 15 samples. (b) From ( )nnx π15

7cos][ =

We use the same procedure as in part (a). Therefore,

kN p ππ 2157 =

kkN p 7302

157

==ππ

The small integer k that results in Np = integer is k =7. That means this is periodic signal with fundamental period Np = 30 samples. (c) from ( ) ( )tttx 32 sincos)( += Let ( )ttx 21 cos)( = and ( )ttx 32 sin)( = The general expression for sinusoidal signals is

( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=== t

Ttfttx

11

1222 ππ coscoscos)( where T1 is fundamental period for x1(t).

Thus, tT

t1

122 π= π=1T .

For x2(t), we have

( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=== t

Ttfttx

22

1223 ππ sinsinsin)( where T2 is fundamental period for x2(t).

HW1 solutions – ECE351

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Thus, t

Tt

2

123 π= 3

22

π=T .

The fundamental period of the system x (t) is

21 TqTrTsum == where r and q are small positive integer. For r =2 and q = 3, we have π221 === TqTrTsum

That means this signal is periodic with fundamental period of π2 . (d) First, we plot the graph ∑

∞

∞−=

−−=k

k kttx )()()( 21 δ

Note that this signal can be considered as discrete-time signal. Thus, this is periodic signal with fundamental period of 4 seconds (as continuous-time signal) or 2 samples (as discrete-time signal). (e) From }][][{][ 23 knknnx

k

−+−= ∑∞

∞−=

δδ

Note that ∑

∞

∞−=

−k

kn ][ 3δ linearly increases when k (integer) increases. this term is periodic signal.

However, ][ 2knk

−∑∞

∞−=

δ exponentially increases when k (integer) increases. this cannot be periodic signal.

Thus, x[n] is not a periodic signal. (f) From )()(cos)( tuttx = That means x(t) = 0 when t < 0. On the other hand, )(cos)( ttx = when t > 0. Therefore, this is not a periodic signal.

HW1 solutions – ECE351

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(g) We plot the signal x(t) = v(t) + v(-t) as follows: Therefore, this is a periodic signal. From ( ) ( ) ⎟

⎠⎞

⎜⎝⎛== t

Ttft 122 ππ coscoscos

where T is fundamental period Thus, t

Tt 12π= π2=T .

(h) We plot the signal x(t) = v(t) + v(-t) as follows: Therefore, this is not a periodic signal. (i) From

( ) ( )nnnx ππ 51

31 cossin][ =

We know that { })sin()sin(cossin BAbABA −++=

21

Therefore, { })sin()sin(][ nnnx ππ 15

2158

21

+=

Let )sin(][ nnx π15

81 = and )sin(][ nnx π15

22 =

Use the concept as used in problem 1.57(a).

HW1 solutions – ECE351

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Assume N1 is fundamental period for x1[n] Thus,

kN ππ 21158 = ;where k = interger

kkN4

152

1581 ==ππ

The small integer k that results in N1 = integer is k =4. That means this is periodic signal with fundamental period N1 = 15 samples. Assume N2 is fundamental period for x2[n] Thus,

kN ππ 22152 = ;where k = interger

kkN 152

1522 ==ππ

The small integer k that results in N2 = integer is k =1. That means this is periodic signal with fundamental period N2 = 15 samples. The fundamental period of the system x[n] is

21 NqNrN sum == where r and q are small positive integer. For r =1 and q = 1, we have 1521 === NqNrN sum

That means the signal x[n] is periodic with fundamental period of 15 samples.