$C o n s i d e r t h e b l o c k d i a g r a m o f a f r e q u e n c y S y n t h e s i z e r c o n s i s t i n g o f a p h a s e l o c k e d l o o p (P L L) a s s h o w n b e l o w :$

$i f t h e r e f e r e n c e i n p u t f r e q u e n c y (f_{i n}) t o t h e a b o v e s y n t h e s i z e r i s 0.48 M H z a n d t h e f r e q u e n c y d i v i d e r i n t h e f e e d b a c k p a t h h a s N = 50, t h e n t h e s t e a d y s t a t e o u t p u t f r e q u e n c y (f_{o u t}) w i l l b e . . . . M H z .$
24

Open in App

Solution

The correct option is A 24
$f o r t h e g i v e n f r e q u e n c y s y n t h e s i z e r, i n t h e s t e a d y s t a t e, \frac{f_{o u t}}{N} = f_{n} f_{o u t} = N f_{n} = 50 \times 0.48 = 24 M H z$

Suggest Corrections

Similar questions

Q. Consider the block diagram as shown in the figure below :

The input signal x(t) is an AM wave of bandwidth 10 kHz and carrier frequency that may lie anywhere in the range of 0.535 MHz to 1.605 MHz. It is required to translate this signal to a frequency band centered at a fixed intermediate frequency (IF) of 0.455 MHz. The range of tuning that must be provided in the local oscillator to achieve this requirement is

Q. A ramp input applied to an unity feedback system results in

5 %

steady state error. The type number and zero frequency gain of the system are respectively

Q. The three stage Johnson counter as shown in figure below is clocked at a constant frequency of

15 M H z

from the starting state of

Q_{2} Q_{1} Q_{0} = 101

. The frequency of output

Q_{2} Q_{1} Q_{0}

will be _____

M H z .

Q. In half-wave rectification, what is the output frequency if the input frequency is

50 H z

What is the output frequency of a full-wave rectifier for the same input frequency.

A frequency modulated signal with a carrier frequency of 10 MHz and a modulation index of 20 rad is to be generated using an indirect FM modulator shown in the figure below.

If the input message signal is a 15 kHZ sinusoid, then the maximum frequency deviation required at the output of the NBFM generator will be equal to kHz.

Join BYJU'S Learning Program

The correct option is A 24for the given frequency synthesizer, in the steady state, foutN=fn fout=Nfn=50×0.48=24MHz

The correct option is A 24
$f o r t h e g i v e n f r e q u e n c y s y n t h e s i z e r, i n t h e s t e a d y s t a t e, \frac{f_{o u t}}{N} = f_{n} f_{o u t} = N f_{n} = 50 \times 0.48 = 24 M H z$