Superheterodyne (SHD) Receivers MCQ Quiz in বাংলা - Objective Question with Answer for Superheterodyne (SHD) Receivers - বিনামূল্যে ডাউনলোড করুন [PDF]

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.

\add

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

Selectivity:

• The ability of a radio receiver to respond only to the radio signal it is tuned to and reject other signals nearby is termed as Selectivity.
• Selectivity is the ability of a receiver to reject the unwanted frequency signal.
• This function is performed by the tuned circuits ahead of the detector stage.

Sensitivity:

• Sensitivity is the ability to amplify weak signals.
• Radio receivers should have reasonably high sensitivity so that it may have a good response to the desired signal.
• It should not have excessively high sensitivity, otherwise, it will pick up all the undesired noise signals.

Fidelity:

• The fidelity of a receiver is the ability to reproduce all the modulating frequencies equally, i.e. the fundamental frequency and the harmonics of the fundamental frequency.
• The radio receiver should have high fidelity or accuracy without introducing any distortion.
• Ex- In an AM broadcast the maximum audio frequency is 5 kHz. Hence the receiver with high fidelity must produce the entire frequency up to 5 kHz.

SNR:

The signal to noise ratio (SNR) is defined as the ratio of the signal power to the noise power, i.e.

\(SNR=\frac{{Signal\;power}}{{Noise\;power}}\)

In dB, the SNR is expressed as:

SNR(dB) = 10log10(SNR)

Analysis:

From the given receiver diagram the receiver signal ranges 5.8 – 25.2 MHz

f_s = 5.8 – 25.2 MHz

Intermediate frequency = 2 MHz

Local frequency will be f_LO = f_IF + f_S

The range of local carrier frequency will be from 5.8 + 2 to 25.2 + 2 MHz, i.e.

7.8 MHz – 27.2 MHz

f_LO(max) = 27.2 MHz and f_LO(min) = 7.8 MHz

Now, the frequency generated by the local oscillator is given by:

\(f_L=\frac{1}{2\pi \sqrt{LC}}\), i.e.

\(f_L\propto\frac{1}{\sqrt C}\)

The frequency generated will be minimum for maximum value of capacitance, i.e.

\(f_{Lmin}\propto\frac{1}{\sqrt C_{max}}\)   ---(1)

Similarly:

\(f_{Lmax}\propto\frac{1}{\sqrt C_{min}}\)   ---(2)

Dividing (2) from (1), and squaring both sides, we get:

\((\frac{f_{Lmax}}{f_{Lmin}})^2=\frac{C_{max}}{C_{min}}\)

\(\frac{C_{max}}{{C_{min}}}= {\left( {\frac{{27.2}}{{7.8}}} \right)^2}\)