Answer 1.10

Let us write the inputs to the vector modulator as:
cos(w0t), and sin(w0t)
and the carrier inputs as:
A.cos(wct), and {A + dA}.sin(wct)
where dA is the amplitude error. The mixer outputs then become:
cos(w0t).Acos(wct) = 0.5Acos(wc + w0)t + 0.5Acos(wc – w0)t
and
sin(w0t).{A + dA}.sin(wct) = –0.5{A + dA}cos(wc + w0)t + 0.5{A + dA}cos(wc – w0)t
At the output of the summing device we get a wanted term at the difference frequency and an unwanted term (usually referred to as the image) at the sum frequency as follows:

  Difference term:
{A + 0.5.dA}cos(wc – w0)t
  Sum term:
0.5.dAcos(wc + w0)t
The ratio of the amplitude of the wanted to unwanted terms is thus:
Amplitude ratio (image supression) = {A + 0.5.dA} / 0.5.dA
Now the question states that the amplitude error in the local oscillator is 0.1 dB, hence:
20.log10{A / (A + dA)} = 0.1 dB
therefore
A / (A + dA) = 0.987, and dA = 0.0132A
The amplitude ratio of wanted to unwanted signal is thus 150.5:1, or a relative power level of approximately 21 dB