## 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 + d A}.sin(wct)

where d A 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 + d A}.sin(wct) = –0.5{A + d A}cos(wc + w0)t + 0.5{A + d A}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.d A}cos(wc – w0)t

Sum term:

> 0.5.d Acos(wc + w0)t

The ratio of the amplitude of the wanted to unwanted terms is thus:

> Amplitude ratio (image supression) = {A + 0.5.d A} / 0.5.d A

Now the question states that the amplitude error in the local oscillator is 0.1 dB, hence:

> 20.log10{A / (A + d A)} = 0.1 dB

therefore

> A / (A + d A) = 0.987, and d A = 0.0132A

The amplitude ratio of wanted to unwanted signal is thus 150.5:1, or a relative power level of approximately 21 dB