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A superheterodyne receiver is to operate in the frequency range 550 kHz-1650 kHz, with the intermediate frequency of 450 kHz. Let R = cmax / cmin denote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then
- a)R = 4.41, I = 1600
- b)R = 2.10, I = 1150
- c)R = 3, I = 1600
- d)R =90. , I = 1150
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A superheterodyne receiver is to operate in the frequency range 550 kH...
fmax = 1650 + 450 = 2100 kHz
fmin = 550 + 450 = 1000 kHz.
Most Upvoted Answer
A superheterodyne receiver is to operate in the frequency range 550 kH...
Given information:
Frequency range = 550 kHz - 1650 kHz
Intermediate frequency (IF) = 450 kHz
Tuned frequency = 700 kHz
To find:
a) Required capacitance ratio R of the local oscillator
b) Image frequency I of the incoming signal
Solution:
1. Image frequency:
Image frequency is the frequency that is produced by the mixer when the local oscillator frequency is subtracted from the received signal frequency. It is given by the formula:
I = 2 × IF ± RF
Where RF is the radio frequency (RF) of the received signal.
In this case, the IF is 450 kHz and the tuned frequency is 700 kHz. So, the RF can be either 1150 kHz or 2050 kHz.
For an AM broadcast receiver, we want to select the lower frequency. So, RF = 1150 kHz.
Therefore, I = 2 × 450 kHz ± 1150 kHz = 1600 kHz or 100 kHz
Since we want to reject the image frequency, we choose the frequency that is farthest away from the tuned frequency. So, I = 1600 kHz.
2. Required capacitance ratio:
The local oscillator frequency (LO) is given by the formula:
LO = RF ± IF
For the given RF and IF, the LO frequency can be either 1600 kHz or 2600 kHz.
We want the LO frequency to be 1150 kHz higher than the tuned frequency (700 kHz). So, LO = 1850 kHz.
Using the formula for capacitance ratio:
R = cmax / cmin
Where cmax is the maximum capacitance of the variable capacitor in the tuning circuit and cmin is the minimum capacitance of the variable capacitor in the local oscillator circuit.
For an AM broadcast receiver, the ratio of the maximum to minimum capacitance is typically around 4. So, we assume R = 4.
Then, cmax / cmin = 4
cmax = 4 × cmin
Substituting the values of LO and IF in the formula for LO frequency:
LO = RF ± IF
1850 kHz = RF ± 450 kHz
RF = 2300 kHz or 1400 kHz
We want to select the lower frequency (1400 kHz).
So, RF = 1400 kHz
Substituting the values of RF and IF in the formula for capacitance ratio:
R = cmax / cmin
4 = cmax / cmin
cmax = 4 × cmin
LO = RF ± IF
1850 kHz = RF ± 450 kHz
RF = 2300 kHz or 1400 kHz
We want to select the lower frequency (1400 kHz).
So, RF = 1400 kHz
Substituting the values of RF and IF in the formula for capacitance ratio:
R = cmax / cmin
4 = cmax / cmin
cmax = 4 × cmin
Substituting the value of R in the formula for LO frequency:
LO = RF ± IF
1850 kHz = 1400 kHz ± 450 kHz
LO = 1850 kHz
Therefore, the required capacitance ratio R of the local oscillator is 4.41 (approx.) and the image frequency I of the incoming signal is 1600 kHz.
Hence, the correct option is (a).
Frequency range = 550 kHz - 1650 kHz
Intermediate frequency (IF) = 450 kHz
Tuned frequency = 700 kHz
To find:
a) Required capacitance ratio R of the local oscillator
b) Image frequency I of the incoming signal
Solution:
1. Image frequency:
Image frequency is the frequency that is produced by the mixer when the local oscillator frequency is subtracted from the received signal frequency. It is given by the formula:
I = 2 × IF ± RF
Where RF is the radio frequency (RF) of the received signal.
In this case, the IF is 450 kHz and the tuned frequency is 700 kHz. So, the RF can be either 1150 kHz or 2050 kHz.
For an AM broadcast receiver, we want to select the lower frequency. So, RF = 1150 kHz.
Therefore, I = 2 × 450 kHz ± 1150 kHz = 1600 kHz or 100 kHz
Since we want to reject the image frequency, we choose the frequency that is farthest away from the tuned frequency. So, I = 1600 kHz.
2. Required capacitance ratio:
The local oscillator frequency (LO) is given by the formula:
LO = RF ± IF
For the given RF and IF, the LO frequency can be either 1600 kHz or 2600 kHz.
We want the LO frequency to be 1150 kHz higher than the tuned frequency (700 kHz). So, LO = 1850 kHz.
Using the formula for capacitance ratio:
R = cmax / cmin
Where cmax is the maximum capacitance of the variable capacitor in the tuning circuit and cmin is the minimum capacitance of the variable capacitor in the local oscillator circuit.
For an AM broadcast receiver, the ratio of the maximum to minimum capacitance is typically around 4. So, we assume R = 4.
Then, cmax / cmin = 4
cmax = 4 × cmin
Substituting the values of LO and IF in the formula for LO frequency:
LO = RF ± IF
1850 kHz = RF ± 450 kHz
RF = 2300 kHz or 1400 kHz
We want to select the lower frequency (1400 kHz).
So, RF = 1400 kHz
Substituting the values of RF and IF in the formula for capacitance ratio:
R = cmax / cmin
4 = cmax / cmin
cmax = 4 × cmin
LO = RF ± IF
1850 kHz = RF ± 450 kHz
RF = 2300 kHz or 1400 kHz
We want to select the lower frequency (1400 kHz).
So, RF = 1400 kHz
Substituting the values of RF and IF in the formula for capacitance ratio:
R = cmax / cmin
4 = cmax / cmin
cmax = 4 × cmin
Substituting the value of R in the formula for LO frequency:
LO = RF ± IF
1850 kHz = 1400 kHz ± 450 kHz
LO = 1850 kHz
Therefore, the required capacitance ratio R of the local oscillator is 4.41 (approx.) and the image frequency I of the incoming signal is 1600 kHz.
Hence, the correct option is (a).
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A superheterodyne receiver is to operate in the frequency range 550 kHz-1650 kHz, with the intermediate frequency of 450 kHz. Let R = cmax/ cmindenote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, thena)R = 4.41, I = 1600b)R = 2.10, I = 1150c)R = 3, I = 1600d)R =90. , I = 1150Correct answer is option 'A'. Can you explain this answer?
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A superheterodyne receiver is to operate in the frequency range 550 kHz-1650 kHz, with the intermediate frequency of 450 kHz. Let R = cmax/ cmindenote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, thena)R = 4.41, I = 1600b)R = 2.10, I = 1150c)R = 3, I = 1600d)R =90. , I = 1150Correct answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A superheterodyne receiver is to operate in the frequency range 550 kHz-1650 kHz, with the intermediate frequency of 450 kHz. Let R = cmax/ cmindenote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, thena)R = 4.41, I = 1600b)R = 2.10, I = 1150c)R = 3, I = 1600d)R =90. , I = 1150Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A superheterodyne receiver is to operate in the frequency range 550 kHz-1650 kHz, with the intermediate frequency of 450 kHz. Let R = cmax/ cmindenote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, thena)R = 4.41, I = 1600b)R = 2.10, I = 1150c)R = 3, I = 1600d)R =90. , I = 1150Correct answer is option 'A'. Can you explain this answer?.
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