Question E4D05:
What transmitter frequencies would create an intermodulation-product signal in a receiver tuned to 146.70 MHz when a nearby station transmits on 146.52 MHz?  

Choices:  
A: 146.34 MHz and 146.61 MHz  
B: 146.88 MHz and 146.34 MHz  
C: 146.10 MHz and 147.30 MHz  
D: 146.30 MHz and 146.90 MHz    

Correct Answer is  A: 146.34 MHz and 146.61 MHz  

To get a third-order intermodulation product at the receiver frequency, use:
- Third-order IMD formulas for two strong signals $f_1$ and $f_2$:
  $$f_{\text{IM3}} = 2f_1 - f_2 \quad \text{and} \quad f_{\text{IM3}} = 2f_2 - f_1$$

Given a nearby transmitter on $f_A = 146.52\ \text{MHz}$ and the receiver tuned to $146.70\ \text{MHz}$, solve for another offending signal $f_X$:

1) Use $2f_A - f_X = 146.70$  
$$f_X = 2(146.52) - 146.70 = 293.04 - 146.70 = 146.34\ \text{MHz}$$

2) Use $2f_X - f_A = 146.70$  
$$f_X = \frac{146.70 + 146.52}{2} = \frac{293.22}{2} = 146.61\ \text{MHz}$$

Therefore, 146.34 MHz and 146.61 MHz would create an intermod product at 146.70 MHz with 146.52 MHz present. That matches choice A.

Why the other choices are incorrect:
- B: 146.88 MHz and 146.34 MHz  
  - 146.34 MHz works (as above).  
  - 146.88 MHz does not:  
    $$2(146.52) - 146.88 = 146.16,\quad 2(146.88) - 146.52 = 147.24$$
- C: 146.10 MHz and 147.30 MHz  
  - 146.10 MHz:  
    $$2(146.52) - 146.10 = 146.94,\quad 2(146.10) - 146.52 = 145.68$$
  - 147.30 MHz:  
    $$2(146.52) - 147.30 = 145.74,\quad 2(147.30) - 146.52 = 148.08$$
- D: 146.30 MHz and 146.90 MHz  
  - 146.30 MHz:  
    $$2(146.52) - 146.30 = 146.74,\quad 2(146.30) - 146.52 = 146.08$$
  - 146.90 MHz:  
    $$2(146.52) - 146.90 = 146.14,\quad 2(146.90) - 146.52 = 147.28$$

Rule of thumb:
- Third-order IMD products appear symmetrically spaced: if $f_2$ is $\Delta f$ below $f_1$, then $2f_1 - f_2 = f_1 + \Delta f$ (same spacing above $f_1$). Here, $146.34$ is $0.18$ MHz below $146.52$, so the product appears $0.18$ MHz above $146.52$, i.e., at $146.70$ MHz.