Wavetek 132 Spezifikationen PDF herunterladen (Seite 86)

4

4

4

s(t)

=

.5cos2ir

10000

cosß(—

sin2*1000

/

+

sin2*3000/

+

sin2*

5000/)

it

3rc

5*

4

4

4

.5sin2*

10000

sinß(—

sin2*

1000

/

+

sin2*3000/

+

sin2rc5000/)

ir

3*

5*

4

4

4

Sj

=

cos

ß(—

sin2u

1000/

+

sin

2%

3000/

+

sin

2*

5000/)

%

3%

5%

4

4

4

S

0

=

sin

ß(—

sin

2%

1000/

+

sin2*3000/

+

sin2it5000/)

s

(/)

=S

T

+

/

S

n

compemr-

J

I

J

Q

3

b)

Q:

What

are

positive

and

negative

frequency

deviations,

f

and

f".

What

is

the

peak

frequency

deviation?

What

is

the

modualtion

index

(ß)?

Measure

2Af

on

the

spectrum

analyzer

for

this

value

of

ß?

A:

Measured

values

off

1

and

f:

f

=

(1.720

xlO'

3

-

1.58

xlO"

3

)"

1

=

7142

Hz

f

=

(1.250

xlO"

3

-

1.03

xlO"

3

)"

1

=

10.526

«

10

kHz

2Af

=

f

-

f

=

3383

Hz

Af

=

1692

Hz

ß

=

Af/4

=

1692/1000

=1.692

Hz

measured

value

of

2Af

=

10980

-

7568

=

3412

Hz

Q:

Using

Carsons

rule,

what

is

the

bandwidth

for

this

signal?

A:

B

=

2Af+2f

m

=

3

412

+

2(1000)

=

5412

Hz

Lab

3

Solutions

Page

6

76

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