Solution manual of linear

For periodic signals, determine the fundamental period.. 2.6 Determine whether the following signals are energy or power, or neither and calculate the corresponding energy or power in th

Chapter 2

2.1 Consider the signals displayed in Figure P2.1.Show that each of these signals can be

expressed as the sum of rectangularΠ and/or triangular ( )( )t Λ t pulses

− 1−

c x t3( )=x2(− + t 2)

d x t4( )=sinc 2( ) ( )t Π t/ 2

Solution:

2.4 Determine whether the following signals are periodic For periodic signals,

determine the fundamental period

T can be written as ratio of

integers In the present case,

1 2

2 2 4

T T

×

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -0.4

-0.2 0 0.2 0.4 0.6 0.8 1

-1.5 -1 -0.5 0 0.5 1 1.5 2

-0.2 0 0.2 0.4 0.6 0.8 1

-1.5 -1 -0.5 0 0.5 1 1.5 2

Therefore, x t is periodic with fundamental period 1( ) T such that o

T can be written as ratio of integers

In the present case,

1 2

T can be written as ratio of

integers In the present case,

1 2

1 5 5

T T

2.6 Determine whether the following signals are energy or power, or neither and

calculate the corresponding energy or power in the signal:

/ 2

1lim 9 cos (4 ) 4.5

To prove nonlinearity, let x t( )=αx t1( )+βx t2( ) The response of the system

To prove nonlinearity, let x t( )=αx t1( )+βx t2( ) The response of the system to ( )

x t is

the system is nonlinear

To show that the system is time-invariant, let x1(t)=x(t−t o)be the system input The corresponding output is

To prove nonlinearity, let x t( )=αx t1( )+βx t2( ) The response of the system

To prove nonlinearity, let x t( )=αx t1( )+βx t2( ) The response of the system

for all and α β

To show that the system is time-invariant, let x1(t)=x(t−t o)be the system input The corresponding output is

, 2 42

0 0

1 3

3 0

3

1 3

2.11 An LTI system has the impulse response h t( )=e−0.5(t−2)u t( − 2)

a Is the system casual?

Solution:

Yes The system is casual because h t( )=0 for t< 0

b Is the system stable?

Solution:

The system is stable because

0.5 0.5

The system is not causal but stable

2.12 Write down the exponential Fourier series coefficients of the signal

Yes as indicated by the presence of DC termC0 =10

2.14 Write down the complex exponential Fourier series for each of the periodic signals shown in Figure P2.4 Use odd or even symmetry whenever possible

Solution:

a 3,T o = f o =1 / 3

1 1

2

1

2

2

Since x t is an even function of time, 3( )

n n

, odd

n A

n n

1 0

cos(2 / 3) 1

n n

ππ

o

T

j nf t n

2.15 For the rectangular pulse train in Figure 2.23, compute the Fourier coefficients of the new periodic

signal y(t) given by

∞

=−∞

= ∑ where

j tf n

o T x n

T C

f

j n

o T x n

T C

( ) j nf t o

n n

∞

=−∞

= ∑ where

1

The normalized average power for a periodic signal x (t) is given by

2sinc sin 2 sinc

21

1( )

− +

− ℑ

( ) ( )

/ 2 2sinc 2/ 4 4sinc 4

2

1

2 sin c sinc

From Table 2.2,

1( )

e e πt2

Solution:

()(t y t X f Y f

X f

j π f

=+Determine the FT V f of the following signals: ( )

a What is the impulse response h(t)?

−

=

= −

2.22 The periodic input ( )x t to an LTI system is displayed in Figure P2.7 The frequency

response function of the system is given by

2( )

H f

j πf

=+

Figure P2.7

a Write the complex exponential FS of input ( )x t

In Exercise 2.15(a), we showed that time shifting introduces a linear phase shift

in the FS coefficients; theirmagnitudes are not changed The phase shift is equal

c Compute the complex exponential FS of the output ( )y t

Determine the output signal in each of the following cases:

a x t( )=5sin 400( πt)+2 cos 1200( π πt− / 2)−cos 2200( π πt+ / 4)

-30 -25 -20 -15 -10 -5 0

5, 1000 Hz( )

0, 1000 Hz0.0025 , 1000 Hz( )

Since H(200) =5and H(200)= −π / 2, the output of the system for an input

5sin(400πt) can now be expressed using (2.117) as 25sin(400πt −π/2)

Next H(600) =5and (600)H = −3 / 2π =π / 2, the output of the system for

an input 2 cos 1200( π πt− / 2)can now be expressed using (2.117) as

10 cos 1200π πt− / 2+π/ 2 =10 cos 1200πt

Now H(1100) =0.So the LP filter doesn’t passcos 2200( π πt+ / 4) The

output of the LP filter is therefore given by

ππ

Now

0.0025

2000sinc 2000 0.001252000

Since H(400) =5and H(400)= − , the output of the system for an input π

5 cos 800 tπ is 25 cos 800 t( π π− ) The response of the system to ( )δ t is ( )h t

The impulse response of the ideal LP filter is 10000sinc 2000⎡⎣ (t−0.00125)⎤⎦ Combining

( ) 25 cos 800 2 ( ) 25cos 800 20 10 sinc 2000 0.00125

( )H f = −0.0005πf Therefore,

(925) 0.0005 0.0005 925 0.462(950) 0.0005 0.0005 950 0.475

ℑ

− ℑ

− ℑ

2.26 The signal 2e u t−2t ( )is input to an ideal LP filter with passband edge frequency

equal to 5 Hz Find the energy density spectrum of the output of the filter Calculate

the energy of the input signal and the output signal

Solution:

4 2

The energy density spectrum of the output y(t), Y f( )2, is related to the energy

density spectrum of the input x(t)

2 2

2.27 Calculate and sketch the power spectral density of the following signals Calculate

the normalized average power of the signal in each case

/ 2 / 2

/ 2

1( ) lim ( ) ( )

1 lim 2 cos 1000 2 cos 1000

T T

T

T T

is zero because it integrates a sinusoidal function over a period in each case

Similarly, it can be shown that all other cross-correlation terms are zero

Therefore,

1( ) ( ) ( ) 2 cos 1000 cos 1850

0 0.2 0.4 0.6 0.8 1

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1