If the input signal is a(t)cos(ωt+Θ)+b(t)sin(ωt+Θ), then the I/Q output should be:
I --> a(t)cosΘ + b(t)sinΘ
Q--> -a(t)sinΘ + b(t)cosΘ
If the input signal is a(t)cos(ωt+Θ) to the upper branch and b(t)sin(ωt+Θ) to the lower branch, then the I/Q output should be:
I --> a(t)cosΘ
Q--> b(t)cosΘ
This graph is from Razavi RF Microelectronics, P129.
Yep!!
The i/p should have been written as
a(t)cosΘcos(ωt) + b(t)sineΘsine(ωt) Which is mathematically correct
But here we are only interested in frequncy domain analysis, we need to
show frequency translation hence in the diagram i/p is shown as
a(t)cos(ωt+Θ) + b(t)sine(ωt+Θ) which is not mathematically correct
Yep!!
The i/p should have been written as
a(t)cosΘcos(ωt) + b(t)sineΘsine(ωt) Which is mathematically correct
But here we are only interested in frequncy domain analysis, we need to
show frequency translation hence in the diagram i/p is shown as
a(t)cos(ωt+Θ) + b(t)sine(ωt+Θ) which is not mathematically correct
I do think there is something wrong with the original diagram. It shows two sigals linearly adding at the input, but they are of the same frequency! So there is no difference in these two signals in the frequency domain, so using a mixer and filter is no longer going to seperate the two signals at the outputs.
If the input signal were comprised of two signals at different frequencies (spaced farther apart than 2 X lowpass filter cuttoff), then the block diagram starts to make some sense.
Really Razavi says about digital modulation (p.128). For this reason the picture is not correct. Output of digital modulator in common case is described as a(t)cosΘcos(ωt) - b(t)sineΘsine(ωt) (thanks to nandgates for formular). If we can neglect amplitude mismatch, a(t) = b(t), then input signal is a(t)cos(ωt+Θ). And no complex signals.
Regards.