## Example 1.5

#### A vector modulator is fed with a perfect quadrature sinewave at the input, but there is a small phase error of 5o between the notional quadrature inputs of the carrier signal. What will be the ratio in dB between the sum and difference outputs of the vector modulator as a result of this phase error?

### Solution

#### Let us write the input to the vector modulator as:

> cos(w0t), and sin(w0t)

and the carrier inputs as:

> cos(wct), and sin(wct + f)

where f is the phase error. Now:

> sin(wct + f) = sinwct cosf – coswct sinf

and for small phase errors this can be approximated to:

> sin(wct + f) = sinwct.cosf

The mixer outputs then become:

> cos(w0t).cos(wct) = 0.5cos(wc + w0)t + 0.5cos(wc – w0)t

and

> sin(w0t).sin(wct + f) = –0.5cos(wc + w0)t.cos f + 0.5cos(wc – w0)t.cos f

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:

> 0.5{1 + cos f}cos(wc – w0)t

Sum term:

> 0.5{1 – cos f}cos(wc + w0)t

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

> Amplitude ratio (image supression) = {1 + cosf} / {1 – cosf}

For a phase error of 5o, the amplitude ratio of wanted to unwanted signals is thus 525:1, or a relative power level of approximately 27 dB