The whole class got a zero on this exam (superheterodyne receiver)

[kring_c14]

this is our last quiz on communication.. fortunately, the whole class got zero on this quiz so the professor arranged a retake and made it a take home exam instead.. this is my last hope on passing the course.. so any help would be greatly appreciated..

1.A superheterodyne receiver having an RF amplifier and an IF of 450 kHz is tuned to 15Mhz. Calculate the Qs of the RF and mixer input tuned circuits, both being the same, if the receiver's image rejection is to be 120

2. WHen a superheterodyne receiver is tuned to 555 kHZ, its local oscillator provides the mixer with an input at 1010 kHZ. What is the image frequency? The antenna of this receiver is connected to the mixer via tuned circuit whose loaded Q is 40. What will be the rejection ratio for the calculated image frequency?


3.Calculate the image frequency rejection of a double conversion receiver which has a first IF of 2 MHz and a second IF of 200 kHz, an RF amplifier whose tuned circuit has a Q of 75 (the same as that of the mixer) and which is tuned to a 30 Mhz signal. The answer is to given in decibels.

4. Caculate the image rejection of the receiver having an RF amplifier and an IF of 450 kHz., if the Qs of the relevant coils are 65, at an incoming frequency of
a.) 1200kHz
b.) 20 Mhz


Formulas are:

f_i-image frequency
f_s-desired signal frequency
f_IF-intermediate frequency

f_i = f_s$$\stackrel{+}{-}$$ 2f_IF

IFRR-image freq. ratio
IFRR=$$\sqrt{1+Q^{2}p^{2}}$$

p=$$\frac{fi}{fs}$$-$$\frac{fs}{fi}$$


These are the only formulas on my notes.. i have difficulty identifying which frequency is which particularly the tuned frequency..is the tuned frequency the desired frequency, or intermediate frequency?

 

[kring_c14]

ive found the answers on these problems.. bur i can't still figure out the solution..help!

#1. 1465 khz
#2. a.)5870 b.) 33.8
#3. 93.6
#4. 51 db

 

[Askalany]

The tuned frequency is the desired frequency, the IF is a lower frequency than tuned one and higher than baseband frequency.

 

[vk6kro]

Sounds like there is some confusion about how a superheterodyne receiver actually works.

Suppose you wanted to receive a signal on 10 MHz. (There is a very nice, accurate time signal on exactly that frequency).

You first build a stable, sensitive amplifier on some much lower frequency. A common one is 0.455 MHz. It doesn't have to move in frequency, so it can be made very selective, too.

Now, to receive 10 MHz, you can provide an oscillator on 10.455 MHz and mix it with the incoming 10 MHz in a mixer. Or, you could provide an oscillator on 9.545 MHz and mix it with the incoming 10 MHz frequency signal. Both will give an output from the mixer of 0.455 MHz.

Notice, though that in either case, there are other possibilities.
What if you had a signal on 10.910 MHz? This will still mix with the first oscillator and give an output at 0.455 MHz.
A signal on 9.090 MHz will give the same result with the second oscillator.
These are called image frequencies.

If the oscillator frequency is not given, the question is ambiguous because it could be on either side of the signal frequency. If this happens, I would always assume it is on the high side of the signal frequency, but it is a free choice.