Chapter 1.pdf

Chapter 1: Problem SolutionsReview of Signals and Systems

Signals

à Problem 1.1

a) x@nD = -0.5 ∆@n + 1D + ∆@nD + 0.5 ∆@n - 1D + ∆@n - 2D - 0.8 ∆@n - 3D

b) x@nD = -0.5 ∆@n + 5D + ∆@n + 4D + 0.5 ∆@n + 3D + ∆@n + 2D - 0.8 ∆@n + 1D

à Problem 1.2

a) I = e-1

b) I = e-1

c) I = 0 since the interval of integration does not include the point t = -1, where the impulse is centered.

d) I = 1

e) I = cos2 H0.1 ΠL

f) I = e

g) let Λ = - 12

t, then t = -2 Λ and dt = -2 dΛ. Substitute in the integral to obtain

I = Ù+¥

-¥

H-2 ΛL2 ∆ IΛ +

12

M H-2 dΛL =

= Ù-¥

+¥

8 Λ2 ∆ IΛ + 12

M dΛ = 2

h) let Λ = 3 t, then t = Λ 3 and dt = I 13

M dΛ. Then the integral becomes

I = Ù-¥

+¥

eΛ3 ∆ HΛ - 1L I 13

M dΛ = e13

3

à Problem 1.3

Amplitude A = 2,

Period T0 = 0.01 sec , then frequency F0 = 1T0

= 100 Hz = 0.1 kHz

Phase Α = 0

Then the signal can be written as x HtL = 2 cos H200 ΠtL.

à Problem 1.4

a) Frequency F0 = 1 T0 = 1 I310-3M = 13

103 Hz. Then

x HtL = 2.5 cos I 23

1000 Πt + 150M

b) Digital Frequency Ω0 = 2 ΠF0 Fs = 2 Π 6 = Π 3 rad. Therefore the sampled sinusoid becomes

x@nD = 2.5 cos I Π

3 n + 150M

à Problem 1.5

All sinusoids are distinct. In continuous time there is no ambiguity between

frequency and signal.

à Problem 1.6

First bring all frequencies within the interval -Π to Π. This yields

x2@nD = 2 cos H1.5 Πn - 0.1 Π - 2 ΠnL = 2 cos H-0.5 Πn - 0.1 ΠL= 2 cos H0.5 Πn + 0.1 ΠL

and also

x3@nD = 2 cos H1.5 Πn + 0.1 Π - 2 ΠnL = 2 cos H-0.5 Πn + 0.1 ΠL= 2 cos H0.5 Πn - 0.1 ΠL

Therefore we can see that x1@nD = x2@nD and x3@nD = x4@nD.

A different way of solving this problem is graphically. The frequency

plots for all four signals are shown next. The key point is to understand

that in discrete time, all frequencies within the interval -Π, Π yield complex Exponentials with dis-tinct values:

2 CurrentValue[FileName]

x1@nD Frequency Representation


CurrentValue[FileName] 3



Again you see that x1@nD and x2@nD have the same representation, and the same for x3@nD and x4@nD.

à Problem 1.7

The sinusoid x@nD = 3 cos H1.9 Πn + 0.2 ΠL has the same samples, since

3 cos H0.1 Πn - 0.2 ΠL = 3 cos H-0.1 Πn + 2 Πn + 0.2 ΠL

à Problem 1.8


à

Problem 1.8

a) since 150 = 15 * Π 180 = Π 12, we can write

x HtL = 32

IejΠ12 ej100Πt + e-jΠ12 e-j100ΠtM

b) x HtL = e-j0 .1 Π ej10Πt + ej0 .1 Π e-j10Πt

c) x HtL = -2.5 je-j0 .2 Π ej20Πt + 2.5 jej0 .2 Π

e-j20Πt

d) x HtL = 5 jej1000Πt- 5 je-j1000Πt

e) x HtL = 1.5 je-j0 .2 Πt ej200Πt - 1.5 jej0 .2 Πt

e-j200Πt

à Problem 1.9

a) since 2 + j = 2.2361 ej0 .4636 , we can write x HtL = 4.4722 cos H100 Πt + 0.4636L

b) since 1 + 2 ej0 .1 Π = 2.9672 ej0 .2098, then x HtL = 5.9344 cos H10 Πt + 0.2098L

c) the signal can be written as xHtL = 12

sinH5 Πt + 0.1 ΠL J,

2 e j Π

4 ej10Πt +,

2 e- j Π

4 e-j10ΠtN and therefore

x HtL = 1

,2

sin H5 Πt + 0.1 ΠL cos I10 Πt + Π

4M

à Problem 1.10

a) x HtL = j20Πx HtL

b) Ù x HtL dt = 1

j20Π x HtL

c) x Ht - 0.1L = e-j2Π x HtL = x HtL

à Problem 1.11

In order to solve this problem we have to decompose the signals into complex

exponentials.

a) x HtL = 2 cos H100 Πt + 0.2 ΠL = ej0 .2 Π ej100Πt + e-j0 .2 Π e-j100Πt


b) x HtL = 32

e-j0 .1 Π ej100Πt + 32

ej0 .1 Π e-j100Πt

- 32 j

ej0 .2 Π ej110Πt + 32 j

e-j0 .2 Π e-j110Πt

This becomes

x HtL = 32

e-j0 .1 Π ej100Πt + 32

ej0 .1 Π e-j100Πt

32

ej0 .7 Π ej110Πt + 32

e-j0 .7 Π e-j110Πt

c) x HtL = e-j0 .2 Π ej100Πt + ej0 .2 Π e-j100Πt

- 32 j

ej0 .1 Π ej100Πt + 32 j

e-j0 .1 Π e-j100Πt


c) x HtL = e-j0 .2 Π ej100Πt + ej0 .2 Π e-j100Πt

- 32 j

ej0 .1 Π ej100Πt + 32 j

e-j0 .1 Π e-j100Πt

Combining terms we obtain

x HtL = 0.9071 ej1 .1800 ej100t + 0.9071 e-j1 .1800 e-j100t

d) x HtL =,25 ej

Π

4 ej1000Πt +,5 ej2 .67795 e-j1100Πt

e) x HtL = ej0t + 12

ej10Πt + 12

e-j10Πt - 12 j

ej20Πt + 12 j

e-j20Πt

This becomes

x HtL = ej0t + 12

ej10Πt + 12

e-j10Πt 12

ej Π

2 ej20Πt + 12

e-j Π

2 e-j20Πt


x HtL = ej0t + 12

ej10Πt + 12

e-j10Πt 12

ej Π

2 ej20Πt + 12

e-j Π

2 e-j20Πt

à Problem 1.12

Referring to the figure above:

a) Ω0 = Π 2

b) Ω0 = 2 Π 3

c) Ω0 = Π

d) Ω0 = 0.75 Π, since 1000 Π 800 = 1.25 Π > Π. We want the frequency to be between -Π and Π , we use the fact that this sinusoid has the same samples as Ω0 = 2 Π - 1.25 Π = 0.75 Π.


d) Ω0 = 0.75 Π, since 1000 Π 800 = 1.25 Π > Π. We want the frequency to be between -Π and Π , we use the fact that this sinusoid has the same samples as Ω0 = 2 Π - 1.25 Π = 0.75 Π.

e) Ω0 = 0, since 1000 Π 500 = 2 Π and therefore Ω0 = 2 Π - 2 Π = 0.

à Problem 1.13

a) Since Ω0 = 0.2 Π = 2 ΠF0 2000 we solve for the frequency as F0 = 200 Hz.

b) Sinusoids with frequencies F0 + Fs = 2.2 kHz, or Fs - F0 = 1.8 kHz have the same sample values.

à Problem 1.14

a) Since x HtL = ej100Πt + e-j100Πt +

32

ej Π

2 ej120Πt + 32

e-j Π

2 e-j120Πt + 2 ej150Πt + 2 e-j150Πt

we obtain the frequency plot shown below.

b) The maximum frequency is FMAX = 75 Hz. Therefore the sampling frequency has to be such that Fs > 150 Hz.

à Problem 1.15

The bandwidth is 4 kHz and therefore the sampling frequency has to be at least Fs = 8 kHz. This means that we need 8000 samples for every second of data, and therefore 168000 = 128 kbytes sec. For one minute of data we need at least 60120 = 7200 kbytes = 7.2 Mbytes to store the file.


Systems

à Problem 1.16

If a property is not specified it is assumed Linear, Time Invariant, Causal,

BIBO Stable.

b) Non Linear. In fact if you apply superposition:

x1@nD ® S ® y1@nD = 2 x1@nD - 1x2@nD ® S ® y2@nD = 2 x2@nD - 1

and therefore

x@nD = x1@nD + x2@nD ® S ® y@nD = 2 Hx1@nD + x2@nDL - 1

which shows that the output is different from y1@nD + y2@nD = 2 Hx1@nD + x2@nDL - 2.

c) Time Varying, due to the time varying coefficient "t". Non BIBO Stable, since x HtL = u HtL , a Bounded Input , yields y HtL = tu HtL, a Non Bounded Output;

d) Time Varying,due to the coefficient e-t;

e) Non Linear. If x HtL = x1 HtL + x2 HtL then the output is y HtL = È x1 HtL + x2 HtL È different from y1 HtL + y2 HtL = È x1 HtL È + È x2 HtL È;

f) Non Causal, since the output at any time t depends on a future input at time t + 1;

h) Non Causal since the output at time n depends on a future input at time n + 1;

i) Non Linear (due to the term x2). Time Varying (due to the time varying coefficient n). Not BIBO Stable, since y@nD = nu@nD + ... , ie Non Bounded, when the input x@nD = u@nD, a Bounded input.

à Problem 1.17

a) y@nD = Sk=-¥

+¥ u@kD 0.5n-k

u@n - kD.

If n ³ 0:

y@nD = Sk=0

n 0.5n-k

= Sk=0

n 0.5k =

1-0.5n+1

1-0.5= 2 - 0.5n

If n < 0 then u@kD u@n - kD = 0 for all k, and therefore y@nD = 0.



Therefore y@nD = I2 - 0.5nM u@nD.

b) y@nD = Sk=-¥

+¥ u@k - 2D 0.5n-k

u@n - kD.

If n ³ 2:

y@nD = Sk=2

n 0.5n-k

= Sk=0

n 0.5k - 0.5n - 0.5n-1

=

= 1-0.5

n+1

1-0.5- 0.5n - 0.5n-1

= 2 I1 - 0.5n-2M

If n < 2 then u@k - 2D u@n - kD = 2 for all k, and therefore y@nD = 0.

Therefore y@nD = I2 - 0.5n-2M u@n - 2D.

c) y@nD = Sk=-¥

+¥ 0.8k u@kD 0.5n-k

u@n - kD.

If n ³ 0:

y@nD = Sk=0

n 0.8k 0.5n-k

= 0.5n Sk=0

n 1.6k = 0.5n

1-1.6n+1

1-1.6= 1.250.5n - 20.8n


Therefore y@nD = I1.250.5n - 20.8nM u@nD.

d) y@nD = Sk=-¥

+¥ 1.2k u@-kD 0.5n-k

u@n - kD

If n < 0 then u@-kD u@n - kD = u@n - kD. This implies

y@nD = Sk=-¥

n 1.2k 0.5n-k

= 0.5n Sk=-¥

n 2.4k =

0.5n Sk=-n

+¥ 2.4-k

=

0.5n K Sk=0

+¥ 2.4-k

- Sk=0

-n-1 2.4-kO =

0.5n J 1

1-2.4-1 -

1-2.4n

1-2.4-1 N =

= 1.71431.2n

If n ³ 0 then u@-kD u@n - kD = u@-kD. This implies

y@nD = Sk=-¥

0 1.2k 0.5n-k

= 0.5n Sk=-¥

0 2.4k =

= 0.5n Sk=0

+¥ 2.4-k

= 1.71430.5n

Finally we can write the answer y@nD = 1.71431.2n u@-n - 1D + 1.71430.5n u@nD.

e) y@nD = Sk=-¥

+¥ ej0 .2 Πn-k 0.5k u@kD = K S

k=0

+¥ 0.5k e-j0 .2 ΠkO ej0 .2 Πn

Applying the geometric series we obtain


e) y@nD = Sk=-¥

+¥ ej0 .2 Πn-k 0.5k u@kD = K S

k=0

+¥ 0.5k e-j0 .2 ΠkO ej0 .2 Πn

Applying the geometric series we obtain

y@nD = 1

1-0.5 e-j0 .2 Π ej0 .2 Πn = 1.5059 ej H0.2 Πn-0.4585L

f) y@nD = Sk=-¥

+¥ ej0 .2 Πn-k u@n - kD 0.5k u@kD = K S

k=0

n 0.5k e-j0 .2 ΠkO ej0 .2 Πn if n ³ 0.

Then, applying the geometric series we obtain

y@nD =ikjjj 1-I0.5 e-j0 .2 Π M

n+1

1-0.5 e-j0 .2 Π

yzzz ej0 .2 Πn =

= 1.5059 Iej H0.2 Πn-0.4585L - e-j0 .2 Π 0.5n+1M

and y@nD = 0 when n < 0. Therefore

y@nD = 1.5059 Iej H0.2 Πn-0.4585L - e-j0 .2 Π 0.5n+1M u@nD

g) Since x@nD = ej0 .2 Πn + e-j0 .2 Πn, using the solution to problem e) above we obtain

y@nD = 1.5059Iej H0.2 Πn-0.4585 + e-j H0.2 Πn-0.4585M =

= 3.0118 cos H0.2 Πn - 0.4585L

h) Since x@nD = ej0 .2 Πn u@nD + e-j0 .2 Πn u@nD, using the solution to problem f) above we obtain

y@nD = 3.0118 Icos H0.2 Πn - 0.4585L - cos H0.2 ΠL 0.5n+1M u@nD

= 3.0118 Icos H0.2 Πn - 0.4585L - 1.2183 0.5nM u@nD

à Problem 1.18

a) h@nD = ∆@nD + ∆@n - 1D + ∆@n - 2D

b) y@nD = u@nD + u@n - 1D + u@n - 2D

c) y@nD = ej0 .5 Πn + ej0 .5 Πn Hn-1L + ej0 .5 Πn Hn-2L. Therefore, since e-j0 .5 Π = -j, we can write

y@nD = H1 - j - 1L ej0 .5 Πn = eHj0 .5 Πn-0.5 ΠL

d) y@nD = cos H0.5 Πn - 0.5 ΠL

à Problem 1.19


à

Problem 1.19

To determine the impulse response ∆@nD substitute x@nD = ∆@nD:

a) h@nD = 3 ∆@nD + 2 ∆@n - 1D - ∆@n - 2D

b) h@nD has to satisfy the recursion h@nD = h@n - 1D + ∆@nD. Therefore:

if n < 0 then h@nD = 0;

if n = 0 then h@0D = h@-1D + 1 = 1, since h@-1D = 0;

if n > 0 then h@nD = h@n - 1D that is to say h@nD is a constant, which must be equal to h@0D = 1.

Putting everything together you can verify that h@nD = u@nD;

c) h@nD = 2 ∆@n - 2D

For the next three problems recall that any signal x@nD can be written as

x@nD = Sk=-¥

+¥ x@kD ∆@n - kD.

d) y@nD = Sk=0

+¥ 0.5k ∆@n - kD = 0.5n u@nD

e) y@nD = Sk=-¥

+¥ 0.5ÈkÈ

∆@n - kD = 0.5ÈnÈ

f) h@nD = Sk=-¥

+¥ 0.5k-2

u@k - 5D ∆@n - k - 2D = Sk=-¥

+¥ 0.5k u@k - 3D ∆@n - kD. Therefore the

impulse response is h@nD = 0.5n u@n - 3D.

à Problem 1.20

We have to check whether or not S-¥

+¥Ë h@nD Ë < +¥. Recall that the geometric series S

n=0

+¥ an < ¥ if

and only if È a È < 1.

a) Sn=0

+¥ 0.5n < +¥ , then BIBO Stable;

b) Sn=0

+¥ 0.5n + S

n=-¥

-1 0.5-n

= Sn=0

+¥ 0.5n + S

n=1

+¥ 0.5-n

< +¥ , then BIBO Stable;

c) Sn=0

+¥ 0.5n + S

n=-¥

-1 0.5n = S

n=0

+¥ 0.5n + S

n=1

+¥ 0.5-n

= +¥ then it is NOT BIBO Stable

d) Sn=-¥

0 0.5n = S

n=0

+¥ 2n = +¥ then NOT BIBO Stable

e) Sn=1

+¥ 1n

= +¥, then NOT BIBO Stable

f) Sn=1

+¥ 1n2

< +¥, then BIBO Stable


f) Sn=1

+¥ 1n2

< +¥, then BIBO Stable

g) Sn=0

+¥ 1 = +¥, then NOT BIBO Stable

h) Sn=-¥

+¥Ë ej0 .2 Πn Ë = S

n=-¥

+¥ 1 = +¥, then NOT BIBO Stable

i) Sn=0

+¥Ì ej0 .2 Πn Ì = S

n=0

+¥ 1 = +¥then NOT BIBO Stable

j) Sn=0

+¥Ì 0.8n ej0 .2 Πn Ì = S

n=0

+¥ 0.8n < +¥ then BIBO Stable

k) Sn=-¥

+¥Ë 0.8ÈnÈ

cos H0.2 ΠnL Ì £ Sn=0

+¥ 0.8n + S

n=-¥

-1 0.8-n

= Sn=0

+¥ 0.8n + S

n=1

+¥ 0.8n < +¥ then

BIBO Stable

l) Sn=-¥

0 2n = S

n=0

+¥ 0.5n < +¥ then BIBO Stable

m) Sn=0

+¥ 2n = +¥ then NOT BIBO Stable

z- Transforms

à Problem 1.21

a) X HzL = Sn=0

+¥ 0.5n z-n =

11-0.5 z-1 =

zz-0.5

, ROC É 0.5 z-1 É < 1, ie È z È > 0.5;

b) X HzL = Sn=-¥

-1 0.5n z-n = S

n=1

+¥ 0.5-n

zn = 1

1-2 z- 1 = -

zz-0.5

=, ROC É 0.5-1 z É < 1, ie

È z È < 0.5;

c) Sn=-¥

0 0.5n z-n = S

n=0

+¥ 0.5-n

zn = 1

1-2 z, ROC È z È < 0.5;

d) X HzL = Sn=0

+¥ ej0 .4 Πn z-n =

11-ej0 .4 Π z-1 =

zz-ej0 .4 Π

, ROC É ej0 .4 Π z-1 É < 1, ie È z È > 1;

e) X HzL = Sn=-¥

-1 ej0 .4 Πn z-n + S

n=0

+¥ ej0 .4 Πn z-n = S

n=1

+¥ e-j0 .4 Πn zn + S

n=0

+¥ ej0 .4 Πn z-n . The first

series converges when È z È < 1 and the second when È z È > 1. Therefore there is no common ROC and X HzL for this signal DOES NOT exist!


e) X HzL = Sn=-¥

-1 ej0 .4 Πn z-n + S

n=0

+¥ ej0 .4 Πn z-n = S

n=1

+¥ e-j0 .4 Πn zn + S

n=0

+¥ ej0 .4 Πn z-n . The first

series converges when È z È < 1 and the second when È z È > 1. Therefore there is no common ROC and X HzL for this signal DOES NOT exist!

f) X HzL = Sn=-¥

-1 ej0 .4 Πn z-n = S

n=1

+¥ e-j0 .4 Πn zn =

11-e-j0 .4 Π z

- 1 = - z

z-ej0 .4 Π, ROC

É e-j0 .4 Π z É < 1, ie È z È < 1;

g) X HzL = Sn=-¥

-1 0.5-n

z-n + Sn=0

+¥ 0.5n z-n = S

n=1

+¥ 0.5n zn + S

n=0

+¥ 0.5n z-n. See the two series:

Sn=1

+¥ 0.5n zn =

11-0.5 z

- 1 = - zz-2

, when È z È < 2;

Sn=0

+¥ 0.5n z-n =

11-0.5 z-1 =

zz-0.5

, when ÈzÈ>0.5.

Therefore the ROC has to be the intersection of the ROC’s, which yields

X HzL = z

z-0.5-

zz-2

, ROC: 0.5 < È z È < 2

h) X HzL = 1 + z-1 + z-2, ROC: 0 < È z È;

i) X HzL = z + z2 + z3, ROC È z È < +¥

à Problem 1.22

Recall

Z 8an u@nD< = zz-a

ROC: È z È > È a È

Z 8an u@-n - 1D< = - zz-a

ROC: È z È < È a È

a) x@nD = 2 0.5n-1 u@n - 1D, then X HzL = 2 z-1

zz-0.5

= 2

z-0.5

b) x@nD = 0.5n u@-n - 1D + ∆@nD, then X HzL = -z

z-0.5+ 1 =

-0.5z-0.5

d) x@nD = I0.5ej0 .1 ΠMn u@nD, then X HzL =

zz-0.5 ej0 .1 Π

e) x@nD = H-0.5Ln u@nD, then X HzL =

zz+0.5

f) x@nD = 12

I0.5 ej0 .1 ΠMn u@nD +

12

I0.5 ej0 .1 ΠMn u@nD, then

X HzL = 12

z

z-0.5 ej0 .1 Π+

12

z

z-0.5 e-j0 .1 Π. Combining terms w obtain

X HzL = z2-cos H0.1 ΠL z

z2-cos H0.1 ΠL+0.25

g) x@nD = 0.5n u@nD + 0.5-n u@-n - 1D, then X HzL =

zz-0.5

- zz-2

= -1.5 z

z2-2.5 z+1, ROC:

0.5 < È z È < 2

h) Using the result in g) above, X HzL = z-1 -1.5 z

z2-2.5 z+1=

-1.5z2-2.5 z+1


h) Using the result in g) above, X HzL = z-1 -1.5 z

z2-2.5 z+1=

-1.5z2-2.5 z+1

à Problem 1.23

a) X HzLz

= 2z

+ 0.5774 ej2 .6180

z-ej2 .0944 + 0.5774 e-j2 .6180

z-e-j2 .0944 therefore

X HzL = 1 + 0.5774 ej2 .6100 z

z - ej2 .0944+ 0.5774 e-j2 .6100

z

z - e-j2 .0944

Since È z È > 1 all terms are causal. Therefore

x@nD = ∆@nD + 0.5774 ej H2.0944 n+2.6100L u@nD + 0.5774 e-j H2.0944 n+2.6100L u@nD

Combining terms we obtain

x@nD = ∆@nD + 1.0548 cos H2.0944 n + 2.6100L u@nD

b) Same X HzL as in the previous problem, but different ROC. All terms are noncausal,

and therefore

x@nD =

∆@nD - 0.5774 ej H2.0944 n+2.6100L u@-n - 1D - 0.5774 e-j H2.0944 n+2.6100L u@-n - 1D

and, after combining terms

x@nD = ∆@nD 11.0548 cos H2.0944 n + 2.6100L u@-n - 1D

c) X HzLz

= 0.5z

- 2z-1

+ 1.5z-2

, therefore X HzL = 0.5 - 2 zz-1

+ 1.5 zz-2

. Form the ROC all terms are causal. Therefore

x@nD = 0.5 ∆@nD - 2 u@nD + 1.52n u@nD

d) Same X HzL with different ROC. In this case all terms are anticausal,

x@nD = 0.5 ∆@nD + 2 u@-n - 1D - 1.52n u@-n - 1D

e) Same X HzL with different ROC. In this case zz-1

yields a causal term, and zz-2

yields an anticausal term. Therefore

x@nD = 0.5 ∆@nD - 2 u@nD - 1.52n u@-n - 1D

f) X HzLz

= 1

z2+1=

0.5 jz+j

- 0.5 jz-j

therefore X HzL = 0.5 j zz+j

- 0.5 j zz-j

. From the ROC

È z È > 1 all terms are causal, therefore

x@nD = 0.5 jej Π

2 n

u@nD - 0.5 je-j Π

2 n

u@nD

Combining terms:

x@nD = cos I Π

2 n +

Π

2M u@nD

g) Same X HzL but all anticausal:

x@nD = -0.5 jej Π

2 n

u@-n - 1D + 0.5 je-j Π

2 n

u@-n - 1D

Combine terms:


Combine terms:

x@nD = cos I Π

2 n -

Π

2M u@-n - 1D

à Problem 1.24

a) H HzL = 1

1-0.5 z-1 = z

z-0.5 Since it is causal, the impulse response is h@nD = 0.5n u@nD

b) H HzL = 1

1-0.5 z=

-2z-2

. Since H HzLz

= -2

z Hz-2L = 1z

- 1z-2

then H HzL = 1 - zz-2

. The impulse response is anticausal, form the difference equation,

therefore the impulse response is h@nD = ∆@nD + 2n u@-n - 1D.

c) H HzL = 1+z-1+z-2

1-z-2 = z2+z+1z2-1

. Therefore H HzLz

= z2+z+1

z Iz2-1M=

-1z

+ 1.5z-1

+ 0.5z+1

which leads to the

decomposition H HzL = -1 + 1.5 zz-1

+ 0.5 zz+1

. Since the difference equation is causal, then the impulse response is

causal, which leads to h@nD = -∆@nD + 1.5 u@nD + 0.5 H-1Ln u@nD.

d) H HzL = 1

1-z-1-z-2 = z2

z2-z-1, then

H HzLz

= z

z2-z-1=

0.2764z+0.6180

+ 0.7236z-1.6180

and therefore

H HzL = 0.2764 z

z+0.6180+ 0.7236

zz-1.6180

. The difference equation is causal and therefore the impulse response is

h@nD = 0.2764 H-0.6180Ln u@nD + 0.72361.6180n u@nD

à Problem 1.25

a) The system is causal with transfer function H HzL = z

z-0.5, with ROC È z È > 0.5. The unit

circle is in the ROC and therefore the system is BIBO stable.

The impulse response is h@nD = 0.5n u@nD.

b) Now H HzL = -2 z

1-2 z=

zz-0.5

, same as before. But since it is clearly anticausal (the output depends

from future values only) the ROC is È z È < 0.5 and the impulse response h@nD = -0.5n u@-n - 1D. The system is NOT BIBO Stable, since the unit circle is not within the

ROC.

c) If the input is x@nD = u@nD, its z-Transform is X HzL = zz-1

with ROC È z È > 1.

When the system is causal then

Y HzL = z

z-0.5

zz-1

= - z

z-0.5+ 2

zz-1

, ROC È z È > 1

therefore y@nD = -0.5n u@nD + 2 u@nD;

When the system is anticausal, the z-Transform is the same, but with

different ROC given by the intersection of È z È > 1 and È z È < 0.5, which is empty. Therefore in this case the output y@nD does not have a z-Transform.


When the system is anticausal, the z-Transform is the same, but with

different ROC given by the intersection of È z È > 1 and È z È < 0.5, which is empty. Therefore in this case the output y@nD does not have a z-Transform.

à Problem 1.26

a) The difference equation can be implemented in a number of ways, so we have

not enough information to assess causality or the ROC of the transfer

function.

b) The transfer function is H HzL = z-1

1-2.5 z-1+z-2 = z

z2-2.5 z+1=

zHz-2L Hz-0.5L . The possible ROC’s

are the following:

È z È > 2, the system is causal , and NOT BIBO stable, since the unit circle is not

in the ROC;

0.5 < È z È < 2, the system is non causal, but BIBO Stable, since the unit circle is in

the ROC;

È z È < 0.5, the system is anticausal, NOT BIBO Stable since the unit circle is not

in the ROC.

c) The step response has transfer function Y HzL =

z2

Hz-2L Hz-0.5L Hz-1L = 0.6667 z

z-0.5- 2

zz-1

+ 1.333 zz-2

When the system is causal the ROC is È z È > 2, then y@nD = 0.66670 .5n u@nD - 2 u@nD + 1.3332n u@nD;

When the system is non casual, the ROC is 1 < È z È < 2, then

y@nD = 0.66670 .5n u@nD - 2 u@nD - 1.3332n u@-n - 1D;

When the system is anticausal, the ROC of the transfer function (0.5 < È z È) and the ROC of the input ( È z È > 1) have empty intersection, and therefore the output has no z-Transform.

Frequency Response

à Problem 1.27

The transfer function of the system is

H HzL = 1

1-0.5 z-1 = z

z-0.5

Therefore the system is stable (pole at z = 0.5 inside the unit circle) and its frequency response is


Therefore the system is stable (pole at z = 0.5 inside the unit circle) and its frequency response is

H HΩL = ejΩ

ejΩ-0.5

a) the input is a complex exponential with frequency Ω = 0.2 Π and therefore the corresponding output is y@nD = H H0.2 ΠL ej0 .2 Πn. Substituting for the frequency response H HΩL we obtain y@nD = 1.5059 e-j4585 ej0 .2 Πn = 1.5059 ej H0.2 Πn-0.4585L.

b) Since x@nD = ej0 .1 Π ej0 .5 Πn + e-j0 .1 Π e-j0 .5 Πn the response is y@nD = H H0.5 ΠL ej0 .1 Π ej0 .5 Πn + H H-0.5 ΠL e-j0 .1 Π e-j0 .5 Πn. Substituting for H HΩL this yields

y@nD = 0.8944 e-j0 .4636 ej0 .1 Π ej0 .5 Πn + 0.8944 ej0 .4636 e-j0 .1 Π e-j0 .5 Πn

= 0.4472 cos H0.5 Πn - 0.1495L

c) Since H H0L = 1

1-0.5= 2 and H H0.3 ΠL = 1.2289 e-j5202 then

y@nD = 12 + 51.2289 cos H0.3 Πn - 0.5 Π - 0.5202L which yields y@nD = 2 + 6.1443 cos H0.3 Πn - 2.0910L;

d) H H0L = 2 and H HΠL = 0.6667, then y@nD = 2 + 0.6667 H-1Ln;

e) H H0.2 ΠL = 1.5059 e-j4585, therefore y@nD = 1.5059 sin H0.2 Πn - 0.4585L + 1.5059 cos H0.2 Πn - 0.4585L;

à Problem 1.28

The system has transfer function H HzL = Z 8h@nD< = Z 90.8n u@nD + 1.25n u@-n - 1D=. This yields

H HzL = z

z-0.8-

zz-1.25

, ROC: 0.8 < È z È < 1.25

The unit circle is within the ROC therefore the system is stable and it

has a Frequency Response H HΩL = ejΩ

ejΩ-0.8-

ejΩ

ejΩ-1.25.

a) H H0.2 ΠL = 1.0417, then y@nD = 1.0417 ej0 .2 Πn;

b) H H0.5 ΠL = 0.2195, then y@nD = 0.4390 cos H0.5 Πn + 0.1 ΠL;

c) H H0L = 9, H H0.3 ΠL = 0.5146, then y@nD = 9 + 2.5731 cos H0.3 Πn - 0.5 ΠL;

d) H H0L = 9, H HΠL = 0.1111, therefore y@nD = 9 + 0.1111 H-1Ln;

e) H H0.2L = 1.0417, then y@nD = 1.0417 sin H0.2 ΠnL + 1.0417 cos H0.2 ΠnL

à Problem 1.29

The system ha transfer function H HzL = -2 z1-2 z

= z

z-0.5 and it is anticausal.Therefore the ROC is

given by È z È < 0.5 and the unit circle is NOT in the ROC. As a consequence the system is

UNSTABLE and the steady state frequency response does not exist.

à Problem 1.30


à

Problem 1.30

The transfer function is H HzL = z

1+z-1+z-2 , the system is causal and the poles are on the unit circle. Therefore

ROC: È z È > 1, the unit circle is NOT in the ROC and the system is NOT stable. Again

the frequency response does not exist!

à Problem 1.31

a) From the graph H H0.1 ΠL = 6.5 e-j. Therefore the output is

y@nD = 13 cos H0.1 Πn - 1L

b) From the graph we obtain H H0L = 8, H H0.5 ΠL = 1.8 ejΠ and H HΠL = 0. Therefore the output is

y@nD = 16 + 1.8 cos H0.5 Πn + ΠL

Time, z-, and Frequency Domains

à Problem 1.32

a) First we have to determine the digital frequency of the disturbance,

as Ω0 = 2 ΠF0 Fs = 0.4 Π. Therefore the filter must have at least two zeros at z1,2 = e±j0 .4 Π, and its transfer function becomes

H HzL = b0 + b1 z-1 + b2 z-2 = b0 z2+b1 z+b2

z2

= b0 Iz-ej0 .4 ΠM Iz-e-j0 .4 ΠM

z2= b0 I1 - 0.6180 z-1 + z-2M

The poles are both z = 0.

b) With the constraint that H H0L = 1, ie that H HzL = 1 when z = 1 we obtain an equation to solve for the coefficient b0. Therefore

H H1L = b0 H1 - 0.6180 + 1L = 1

which yields b0 = 0.7236. Therefore the difference equation of the filter becomes

y@nD = 0.7236 x@nD - 0.4472 x@n - 1D + 0.7236 x@nD


à Problem 1.33

If we keep the same zeros at z1,2 = e±j0 .4 Π , we add two poles close to the zeros as, say, p1,2 = 0.95 e±j0 .4 Π, then the frequency response becomes more selective. In this way the

Transfer Function becomes

H HzL = b0 Hz-z1L Hz-z2LHz-p1L Hz-p2L = b0

z2-0.6180 z+1z2-0.5871 z+0.9025

In order to have the Frequency Response with H H0L = 1, we need to solve for the coefficient b0 from the equation

b0 1-0.6180+1

1-0.5871+0.9025= 1

which yields b0 = 0.9518. Finally the difference equation for the filter becomes

y@nD = 0.5871 y@n - 1D - 0.9025 y@n - 2D +0.9518 x@nD - 0.5883 x@n - 1D + 0.9518 x@n - 2D

à Problem 1.34

Since x@nD is periodic with period N, by definition x@nD = x@n - ND for all n, and therefore x@nD - x@n - ND = 0 for all n. Therefore in this case the difference equation becomes

y@nD = -a1 y@n - 1D - a2 y@n - 2D + 0

and if the system is stable y@nD ® 0 from any initial condition.

a) When N = 4 the filter has Transfer Function

H HzL = b0 z4-1

z2+a1 z+a2

The zeros must be the solutions of the equation z4 = 1, ie z4 = ejk2Π with k integer. This yields z = ejkΠ2, for k = 0, 1, 2, 3, Therefore the four zeros are z = 1, j, -1, -j

b) Choose the poles close to the zeros, inside the unit circle, at z = Ρ, jΡ, -Ρ, -jΡ, with 0 < Ρ < 1, close to one.

à Problem 1.35

The digital frequencies of the disturbance are Ω1 = 2 Π100 1000 = 0.2 Π, Ω2 = 2 Π150 1000 = 0.3 Π, Ω3 = 2 Π200 1000 = 0.4 Π.

a) zeros at z1,2 = e±j0 .2 Π, z3,4 = e±j0 .3 Π, z5,6 = e±j0 .4 Π, and poles at p1,2 = Ρe±j0 .2 Π, p3,4 = Ρe±j0 .3 Π, p5,6 = Ρe±j0 .4 Π with Ρ again positive, close to one. Say Ρ = 0.9.

b) From the zeros and the poles the transfer function becomes

H HzL = b0 z6-3.4116 z5+6.6287 z4-7.9988 z3+6.6287 z2-3.4116 z+1

z6-3.0705 z5+5.3692 z4-5.8312 z3+4.3491 z2-2.0145 z+0.5314


H HzL = b0 z6-3.4116 z5+6.6287 z4-7.9988 z3+6.6287 z2-3.4116 z+1

z6-3.0705 z5+5.3692 z4-5.8312 z3+4.3491 z2-2.0145 z+0.5314

If we want the DC gain to be one, ie H HΩL = 1 when Ω = 0, then b0 = 0.7664;

c) The difference equation then becomes

y@nD = 3.0705 y@n - 1D - 5.3692 y@n - 2D + 5.8312 y@n - 3D -4.3491 y@n - 4D + 2.0145 y@n - 5D - 0.5314 y@n - 6D +

+ 0.7664 Hx@nD - 3.4116 x@n - 1D + 6.6287 x@n - 2D -7.9988 x@n - 3D + 6.6287 x@n - 4D - 3.4116 x@n - 5D + x@n - 6DL

à Problem 1.36

The digital frequency of the signal we want to enhance is Ω0 = 2 Π1.5 5 = 0.6 Π radians.

a) The poles are p1,2 = Ρe±j0 .6 Π with 0 < Ρ < 1, close to one. We can choose any zeros we like, say z1,2 = ±1 to attenuate the low frequencies (Ω = 0) and the high frequencies (Ω = Π).

b) Choose (say) Ρ = 0.9, and the transfer function becomes

H HzL = b0 z2-1

z2-0.5562 z+0.81

We can choose b0 to satisfy any normalization we like. For example, if we want the

frequency response at frequency Ω0 to have unit magnitude, we impose È H H0.6 ΠL È = 1 which yields b0 = 0.0950.

c) The difference equation becomes

y@nD = 0.5562 y@n - 1D - 0.81 y@n - 2D + 0.0950 x@nD - 0.0950 x@n - 2D

à Problem 1.37

a) The filter has zeros at z1,2 = ±j, poles at p1,2 = ±j0 .9. Its frequency response is as shown;

b) Notch Filter;

c) y@nD = -0.81 y@n - 2D + x@nD + x@n - 2D.


Fourier Analysis of Discrete Time Signals

à Problem 1.38

a) Period N = 3, therefore X@kD = Sn=0

2 x@nD e-j

2 Π

3 kn, for k = 0, 1, 2. Just substitute for one

period as x@0D = 1, x@1D = 2, x@2D = 3 to obtain X@0D = 6.000, X@1D = -1.5 + j0 .8660, X@2D = -1.5 - j0 .8660.

b) Period N = 2, then X@kD = Sn=0

1 x@nD e-j

2 Π

2 kn = x@0D + H-1Lk

x@1D, for k = 0, 1. Therefore

X@0D = 1 - 1 = 0, X@1D = 1 - H-1L = 2.

c) Period N = 4, then X@kD = Sn=0

3 x@nD e-j

2 Π

4 kn = S

n=0

3 x@nD H-jLkn, for k = 0, 1, 2, 3.

Substitute the numerical values of the sequence to obtain X@0D = X@2D = 0, X@1D = X@3D = 2 or, in vector form,

X = @0, 2, 0, 2D

d) The signal x@nD = É cos I Π

4 nM É is shown below.

Plot of x@nD = É cos I Π

4 nM É

The period can be seen by inspection as N = 4 and therefore the expansion is of the form

x@nD = 14

Sk=0

3 X@kD e-j

Π

2 kn, with the DFS being

X@kD = Sn=0

3 x@nD jkn, for k = 0, 1, 2, 3

since ej Π

2 = j. Therefore


since ej Π

2 = j. Therefore

X = @2.4142, 1, -0.4142, 1D

à Problem 1.39

The transfer function of the system is H HzL = 0.2 zz-0.8

, by inspection. The system is causal and stable therefore we can define

the frequency response as

H HΩL = 0.2 ejΩ

ejΩ-0.8

a) The DFS of the signal is X = @6, -1.5 + j0 .866, -1.5 - j0 .866D. If we call y@nD the output signal, it is going to be periodic with the same period N = 3 and DFS given by Y@kD = H Ik

2 Π

NM X@kD, for k = 0, 1, 2. By substitution we find Y@0D =

0.21-0.8

H6L = 6,

Y@1D = 0.2 e

j 2 Π

3

ej

2 Π

3 -0.8 H-1.5 + j0 .866L = -0.123 + j0 .1846 and

Y@2D = 0.2 e

j2 2 Π

3

ej2

2 Π

3 -0.8 H-1.5 - j0 .866L = -0.123 - j0 .1846. In vector form:

Y = @6, -0.123 + j0 .1846, -0.123 - j0 .1846D

b) Period N = 2, therefore Y@kD = H Ik 2 Π

2M X@kD for k = 0, 1. This yields

Y@kD = 0.2 H-1Lk

H-1Lk-0.8 X@kD for k = 0, 1. Substitute the numerical values of X@kD to obtain

Y = @0, 0.5D

c) Period N = 4, therefore Y@kD = H Ik 2 Π

4M X@kD for k = 0, 1, 2, 3. This yields

Y@kD = 0.2 H-jLk

H-jLk-0.8 X@kD. Substitute for X@kD to obtain

Y = @0, 0.2439 + j0 .1951, 0, 0.2439 - j0 .1951D

à Problem 1.40

a) The length of the sequence is N = 4. Therefore the DFT becomes X@kD = Sn=0

3 x@nD e-j

2 Π

4 kn for

k = 0, 1, 2, 3. Substitute for the sequence to obtain

X = @0, 0, 4, 0D

b) Length N = 4, then again X@kD = Sn=0

3 x@nD e-j

2 Π

4 kn for k = 0, ..., 3 to obtain

X = @4, 0, 0, 0D

c) Length N = 3, then X@kD = Sn=0

2 x@kD = e-j

2 Π

3 kn for n = 0, 1, 2. This yields


c) Length N = 3, then X@kD = Sn=0

2 x@kD = e-j

2 Π

3 kn for n = 0, 1, 2. This yields

X = @6.0000, -1.5000 + j0 .8660, -1.5000 - j0 .8660D

d) Length N = 4, then again X@kD = Sn=0

3 x@nD e-j

2 Π


X = @0, 2, 0, 2D

e) Length N = 4, then again X@kD = Sn=0

3 x@nD e-j

2 Π


X = @1 + j2 .4142, 1 - j2 .4142, 1 - j0 .4142 i, 1 + j0 .4142D

f) Length N = 4, then again X@kD = Sn=0

3 x@nD e-j

2 Π


X = @-4.0961, 4.1904 - j2 .4938, 2.7687, 4.1904 + j2 .4938D

g) Length N = 8, then X@kD = Sn=0

7 x@nD e-j

2 Π


X = @0, 7.0534 + j9 .7082, 0, 0, 0, 0, 0, 7.0534 - j9 .7082D

à Problem 1.41

Recall the definition X HΩL = DTFT 8x@nD< = Sn=-¥

+¥ x@nD e-jΩn. Then:

a) X HΩL = Sn=-¥

+¥ 0.8ÈnÈ

e-jΩn = Sn=-¥

-1 0.8-n

e-jΩn + Sn=0

+¥ 0.8n e-jΩn . After some simple

manipulations:

X HΩL = Sn=0

+¥ 0.8n ejΩn - 1 + S

n=0

+¥ 0.8n e-jΩn =

11-0.8 ejΩ

- 1 + 1

1-0.8 e-jΩ=

0.361-1.6 cos HΩL+0.64

An alternative way would be to use the z-Transforms. Since, in this case

x@nD = 1.25n u@-n - 1D + 0.8n u@nD

the z-Transform is X HzL = - z

z-1.25+

zz-0.8

, ROC : 0.8 < È z È < 1.25

the unit circle È z È = 1 is within the ROC. Therefore just substitute z = ejΩ to obtain:

X HΩL = - ejΩ

ejΩ-1.25+

ejΩ

ejΩ-0.8

and the result follows with some algebra.

b) X HΩL = Sn=0

+¥ 0.5n e-jΩn =

11-0.5 e-jΩ

= ejΩ

ejΩ-0.5

c) x@nD = 1 when 0 £ n £ 4 and x@nD = 0 otherwise. Therefore


c) x@nD = 1 when 0 £ n £ 4 and x@nD = 0 otherwise. Therefore

X HΩL = Sn=0

4 e-jΩn =

1-e-j5Ω

1-e-j Ω= e-j1 .5 Ω

sin H2.5 ΩLsin HΩL

magnitude È X HΩL È

d) x@nD = 1.5 ej0 .2 Π ej0 .1 Πn + 1.5 e-j0 .2 Π e-j0 .1 Πn.

Recall that DTFT 9ejΩ0 n= = 2 Π∆ HΩ - Ω0L and therefore we obtain

X HΩL = 3 Πej0 .2 Π ∆ HΩ - 0.1 ΠL + 3 Πe-j0 .2 Π ∆ HΩ + 0.1 ΠL


X HΩL magnitude and phase

e) x@nD = -0.5 jej0 .5 Πn+ 0.5 je-j0 .5 Πn

+ 0.5 ej0 .25 Πn + 0.5 e-j0 .25 Πn. Therefore the DTFT becomes

X HΩL = Πe-j0 .5 Π ∆ HΩ - 0.5 ΠL + Πej0 .5 Π ∆ HΩ + 0.5 ΠL ++Π ∆ HΩ - 0.25 ΠL + Π∆ HΩ + 0.25 ΠL



f) x@nD = 2 I-0.5 jej0 .5 Πn+ 0.5 je-j0 .5 ΠnM

2=

= 2 I-0.25 ejΠn - 0.25 e-jΠn + 20.25M =

= 1 - ejΠn

Therefore X HΩL = 2 Π∆ HΩL - 2 Π∆ HΩ - ΠL =

= 2 Π∆ HΩL + Πe-jΠ ∆ HΩ - ΠL + ΠejΠ ∆ HΩ + ΠL

Notice that we split the term 2 Π∆ HΩ - ΠL = Πe-jΠ ∆ HΩ - ΠL + ΠejΠ ∆ HΩ + Π) to preserve the symmetry of the DTFT.



g) x@nD = H-1Ln= e-jΠn = 0.5 ejΠn + 0.5 e-jΠn therefore

X HΩL = Π∆ HΩ - ΠL + Π ∆ HΩ + ΠL

again we split it to preserve the symmetry of the DTFT.



Fourier Transform

à Problem 1.42

You know that

FT 8rect HtL< = sinc HFL


-3 -2 -1 1 2 3t

-0.5

0.5

1

1.5

2rectHtL

-6 -4 -2 2 4 6F

-0.5

0.5

1

1.5

2sincHFL

Apply the properties of the Fourier Transform:

a) FT 8rect H2 tL< = 12

sinc I F2

M

-3 -2 -1 1 2 3t

-0.5

0.5

1

1.5

2rectH2tL

-6 -4 -2 2 4 6F

-0.5

0.5

1

1.5

20.5sincHF 2L

b) FT 8rect H2 t - 5L< = FT 8rect H2 Ht - 2.5LL< = e-j2ΠF 2.5 FT 8rect H2 tL< which yields

FT 8rect H2 t - 5L< = e-j5ΠF 12

sinc I F2

M

c) FT 8rect H5 tL cos H20 ΠtL< = FT 9rect H5 tL 12

Iej2Π 10 t + e-j 2 Π 10 tM=. Since

FT 8rect H5 tL< = 15

sinc I F5

M then

FT 8rect H5 tL cos H20 ΠtL< = 110

sinc HF - 10L + 110

sinc HF + 10L

-0.4-0.2 0.2 0.4t

-1.5

-1

-0.5

0.5

1

1.5xHtL

-15-10 -5 5 10 15F

-0.1

-0.05

0.05

0.1

0.15

0.2XHFL

d) FT 8sinc H0.1 tL<. By the duality property, FT 8sinc HtL< = rect H-FL. Since the "rect" function is symmetric, we can write FT 8sinc HtL< = rect HFL. Therefore


d) FT 8sinc H0.1 tL<. By the duality property, FT 8sinc HtL< = rect H-FL. Since the "rect" function is symmetric, we can write FT 8sinc HtL< = rect HFL. Therefore

FT 8sinc H0.1 tL< = 10 rect H10 FL

-20 -10 10 20t

-0.5-0.25

0.250.5

0.751

1.251.5

sincH0.1tL

-0.4-0.2 0.2 0.4F

2468

101214

10 rectH10FL

e) FT 8sinc H0.1 tL cos H20 Πt + 0.1 ΠL< =

ej0 .1 Π FT 9ej2Π10t sinc H0.1 ΠtL= + e-j0 .1 Π FT 9e-j2Π10t sinc H0.1 ΠtL=. Therefore

X HFL = ej0 .1 Π 0.1 rect H10 HF - 10LL + e-j0 .1 Π 0.1 rect H10 HF + 10LL

f) Since FT 8rect HatL< = 1

ÈaÈ sinc I Fa

M , then IFT 9sinc I Fa

M= = È a È rect HatL.

Therefore let a = 1 10 and then

IFT 8sinc H10 FL< = 0.1 rect H0.1 tL

g) IFT 8rect H10 FL< = 0.1 sinc H0.1 tL for the same reasons as in f);

h) Recall that FT : Sn=-¥

+¥ ∆ Ht - nT0L> = F0 S

k=-¥

+¥ ∆ HF - kF0L, where F0 = 1 T0. Also recall that

Sn=-¥

+¥ x HnT0L ∆ Ht - nT0L = x HtL S

n=-¥

+¥ ∆ Ht - nT0L.

With this in mind we can write

Sn=-¥

+¥ sinc H0.2 nL ∆ Ht - 0.1 nL = sinc H2 tL S

n=-¥

+¥ ∆ Ht - 0.1 nL, and therefore

FT : Sn=-¥

+¥ sinc H0.2 nL ∆ Ht - 0.1 nL> = FT 8sinc H2 tL< * FT : S

n=-¥

+¥ ∆ Ht - 0.1 nL>

= 0.5 rect H0.5 FL * 10 Sk-¥

+¥ ∆ HF - 10 kL = 5 S

k-¥

+¥ rect H0.5 HF - 10 kLL

An alternative way to arrive at the same answer is by using the "rep" and

"comb" operators. Recall that

FT 8combT0 x HtL< = F0 repF0 X HFL

In our case T0 = 0.1, F0 = 1 T0 = 10, x HtL = sinc H2 tL, X HFL = FT 8sinc H2 tL< = 0.5 rect H0.5 FL. Therefore


In our case T0 = 0.1, F0 = 1 T0 = 10, x HtL = sinc H2 tL, X HFL = FT 8sinc H2 tL< = 0.5 rect H0.5 FL. Therefore

FT 8comb0.1 sinc H2 tL< = 10 rep10 0.5 rect H0.5 FL

i) Again apply duality to obtain

FT 9repT0 x HtL= = F0 combF0 X HFL

Therefore FT : Sn=-¥

+¥ rect Ht - 2 nL> = FT 8comb2 rect HtL< = 0.5 rep0.5 sinc HFL, and

therefore

FT : Sn=-¥

+¥ rect Ht - 2 nL> = 0.5 S

k=-¥

+¥ sinc H0.5 kL ∆ HF - 0.5 kL

j) By the same arguments as in the previous problem, we can write

FT : Sn=-¥

+¥ rect Ht - nL> = FT 8comb1 rect HtL< = rep1 sinc HFL

and therefore

FT : Sn=-¥

+¥ rect Ht - nL> = S

k=-¥

+¥ sinc HkL ∆ HF - kL

But now recall that sinc HkL = 0 for all k ¹ 0. Also it is easy to see that Sn=-¥

+¥ rect Ht - nL = 1

for all t. Therefore

FT 81< = ∆ HFL

since all other terms sinc HkL ∆ HF - kL , for k ¹ 0, in the summation, are zero.

à Problem 1.43

Let a signal x HtL have Fourier Transform as shown.


Sketch the following:

a) Sn=-¥

+¥ x H0.01 nL ∆ Ht - 0.01 nL = x HtL S

n=-¥

+¥ ∆ Ht - 0.01 nL = comb0.01 x HtL. There-

fore its Fourier Transform is given by

FT : Sn=-¥

+¥ x H0.01 nL ∆ Ht - 0.01 nL> = 100 rep100 X HFL

shown below

b) Sn=-¥

+¥ x Ht - 0.01 nL = rep0.01 x HtL. Therefore its Fourier Transform is 100 comb100 X HFL

shown below


c) FT 8x HtL cos H100 ΠtL< = 12

X IF - 50 + 12

X HF + 50L

d) Since cos2 H100 ΠtL = I 12

ej100Πt + 12

e-j100ΠtM2

= 14

Iej200Πt + 2 + e-j200ΠtM then

FT 9x HtL cos2 H100 ΠtL= = 12

X HFL + 14

X HF - 100L + 14

X HF + 100L

shown below.

e) From the properties, FT 8x H5 tL< = 15

X I F5

M shown below:


f) From the properties again, FT 8x H0.1 tL< = 10 X H10 FL, as shown:


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