Classification of Receivers MCQ Quiz in தமிழ் - Objective Question with Answer for Classification of Receivers - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

The image frequency for a super-heterodyne receiver is given by:

fsi = fs + 2IIF ----(1)

fsi = Image frequency

fs = Tuned frequency of the signal

IIF = Intermediate frequency

fLO = Local Oscillator frequency, which is calculated as:

I_IF = fLO -  fs  ----(2)

This is explained with the help of the following spectrum analysis:

Calculation:

Given:

fs = 600 kHz   

fLO = 1000 kHz

From equation (2) I_IF can be calculated as:

I_IF = 1000 - 600 

I_IF = 400 kHz

Also, the image frequency can be calculated using equation (1):

fsi = 600 + (2 × 400)

fsi = 1400 kHz

• The image frequency is an undesired input frequency which is demodulated by the superheterodyne receivers along with the desired incoming signal. This results in two stations being received at the same time, thus producing interference.
• This is mainly because of poor front-end selectivity of the RF stage, i.e. due to insufficient adjacent channel rejection by the front-end RF stage.
• ∴ One of the main functions of the RF amplifiers in a super-heterodyne receiver is to have sufficient bandwidth for the rejection of the image frequency.

Image Frequency:

The concept is understood with the help of the following diagram:

Image frequency is given by fsi = fs + 2If

Where fsi = image frequency

fs = incoming signal

If = intermediate frequency

Key Points

 The image rejection should be achieved before IF stage because once it enters into IF amplifier it becomes impossible to remove it from wanted signals.

So, image channel selectivity depends upon pre-selector and RF amplifiers only.

Mistake Points
The IF amplifier helps in the rejection of adjacent channel frequency and not image frequency.

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Detailed Block Diagram:

• A superheterodyne receiver changes the RF frequency to a lower IF frequency. This IF frequency will be amplified and demodulated to get a video signal.
• Generally, Mixer will do the down Conversion in superheterodyne Receiver i.e.

           IF = fL – fs.

• Superheterodyne radio used to be the undoubted radio receiver technique of choice. It was almost universally used. However, nowadays with software-defined radios taking over the superheterodyne is used less widely. Superheterodyne was used in every form of radio from domestic broadcast radios to walkie-talkies, television sets, through to hi-fi tuners and professional communications radios, satellite base stations, and much more.

• Its flexibility and capabilities have meant that it was adopted for very many uses from broadcast reception, uses as a test receiver for EMI / EMC testing, two-way radio communications, reception for scientific applications, satellite signal reception, and many others.

• Air-Ground Radio is simply a means of channelling information, within the station's area of operation, about an aircraft's position and intentions, through a single reporting point so superheterodyne radios are not used in this as besides the wanted signal, it receives also a signal at the so-called image frequency. The IF needs to be selected in such a way that the image frequency can be easily rejected by a simple RF tunable circuit. Hence option (4) is the correct answer.

Concept:

The image frequency for a super-heterodyne receiver is given by:

fsi = fs + 2IIF

fsi = Image frequency

fs = Tuned frequency of the signal

IIF = Intermediate frequency

fLO = Local Oscillator frequency, which is calculated as:

fLO = f_s + IF

This is explained with the help of the following spectrum analysis:

Calculation:

With fIF = 450 kHz and  f_s = 1000 kHz, the oscillator frequency will be:

fLO = fs + IF = 1000k + 450k

fLO = 1450 kHz

Also, the image frequency will be:

fsi = fs + 2IIF = 1000k + 2 × 450k

f_si = 1900 kHz

Explanation:

When a radio receiver is tuned to a specific frequency, it uses a local oscillator to convert the desired radio frequency signal to an intermediate frequency (IF) for easier processing. The frequency of the local oscillator is chosen such that the difference between the local oscillator frequency and the desired signal frequency equals the intermediate frequency.

Given:

• Tuned Frequency (f_signal): 560 kHz
• Local Oscillator Frequency (f_LO): 1,000 kHz

The intermediate frequency (IF) can be calculated using the formula:

IF = |f_LO - f_signal|

Substituting the given values:

IF = |1,000 kHz - 560 kHz| = 440 kHz

This means that the intermediate frequency is 440 kHz. However, due to the nature of the mixing process in the radio receiver, another signal can also produce the same intermediate frequency. This other signal will be at a frequency such that the absolute difference between its frequency and the local oscillator frequency also equals the intermediate frequency.

Let f_other be the frequency of the other signal. Then,

|f_LO - f_other| = IF

Substituting the values:

|1,000 kHz - f_other| = 440 kHz

Solving for f_other:

f_other = 1,000 kHz - 440 kHz = 560 kHz

or

f_other = 1,000 kHz + 440 kHz = 1,440 kHz

Since 560 kHz is the frequency of the desired signal, the other signal must be at 1,440 kHz.

Thus, the frequency of the other station is 1,440 kHz.

Given, the range of carrier frequency, 0.535 MHz < f_c < 1.605 MHz

The required intermediate frequency, f_IF = 0.455 MHz

Since the intermediate frequency is lower than the input carrier frequency range, so we have

f_LO - fc  = f_IF

Therefore, we have local oscillator frequency,

f_LO = f_c + f_IF

For the given range, 0.535 < f_c < 1.605 MHz, we have the range of local oscillator frequency

(0.535 + 0.455) < f_LO < (1.605 + 0.455)

0.99 MHz < f_LO < 2.06 MHz

Hence option (3) is the correct answer.

Concept:

Image frequency is given by:

fsi = fs + 2IIF  if f_LO > f_s

fsi = fs - 2IIF  if fLO > fs

fsi = Image frequency

fs = Tuned frequency of signal

IIF = Intermediate frequency

f_LO = Local Oscillator frequency

Calculation:

Given IF = 15 MHz

Local oscillator frequency f_LO = 3.5 GHz

\({{\rm{f}}_{\rm{s}}} - {{\rm{f}}_{\rm{L}}} = {\rm{IF}}\)

∴ f_s = 3.5 + 0.015 = 3.515 GHz

fLO < fs

Now, the image Frequency will now be:

\({{\rm{f}}_{{\rm{si}}}} = {{\rm{f}}_{\rm{s}}} - 2{\rm{IF}} = 3485~{\rm{MHz}}\)

Concept: 

The relation between image frequency and intermediate frequency is given by

\({{\rm{F}}_{{\rm{si}}}}{\rm{\;}} = {\rm{\;}}{{\rm{f}}_{\rm{s}}}{\rm{\;}} + {\rm{\;}}2{\rm{IF}}\)

Application: As \({{\rm{F}}_{{\rm{si}}}}{\rm{\;}} = {\rm{\;}}{{\rm{f}}_{\rm{s}}}{\rm{\;}} + {\rm{\;}}2{\rm{IF}}\)

IF the IF is to be chosen such that image frequency is outside the band 58 MHz - 68 MHz then

58 MHz + 2 IF ≥ 68 MHz

⇒ IF ≥ 5 MHz

Concept:

The image frequency is an undesired input frequency which is demodulated by superheterodyne receivers along with the desired incoming signal. This results in two stations being received at the same time, thus producing interference.

Image frequency is given by fsi = fs + 2If

Where fsi = image frequency

fs = incoming signal

If = intermediate frequency

Solution: fsi = fs + 2If