# Difference between delayed down-converted signal and a down-converted delayed signal

I am having trouble understand what I think should be a pretty simple concept. Conceptually I can understand that a signal that has been down-converted, and then delayed in time is different than a signal that has been delayed in time and then down-converted.

I am trying to compensate for this difference on an FPGA (I am down-converting the signal, delaying in time, and then up-converting)

The way I see it, the delayed (and attenuated signal) looks like: $A e^{i(\omega_{\text{c}}-\omega_{\text{lo}})(t-\tau_1)} m(t-\tau_1)$

That is different from a down-converted delayed signal, so I need a phase correction constant: $\phi = e^{i\omega_{\text{lo}}\tau_1}$

Which gives: $A e^{i(\omega_{\text{c}}-\omega_{\text{lo}})t - i\omega_{\text{c}}\tau_1} m(t-\tau_1)$

So the correction is a function of the LO and the delay itself, right? I have my sampled signal in I and Q, but I need to multiply it by the correction before up-converting, and that is where I am lost.

Let your original signal be (leaving out any constant multiplicative factor):

$$s(t)=e^{i\omega_ct}m(t)$$

Then the down-converted signal is

$$\tilde{s}(t)=e^{i(\omega_c-\omega_{lo})t}m(t)$$

and its delayed version is

$$\tilde{s}(t-\tau)=e^{i(\omega_c-\omega_{lo})(t-\tau)}m(t-\tau)$$

whereas the delayed and down-converted signal is

$$s(t-\tau)e^{-i\omega_{lo}t}=e^{i\omega_c(t-\tau)}e^{-i\omega_{lo}t}m(t-\tau)= \tilde{s}(t-\tau)e^{-i\omega_{lo}\tau}$$

So the correction factor $e^{-i\omega_{lo}\tau}$ indeed depends on both $\omega_{lo}$ and the delay $\tau$.

• OK, thank you. So where do I got from there? If I have an LO of 2437MHz say, and a delay of 15ns, I am strugging to see what the I/Q adjustment would be. I = sin(2437e6*15e-9) and Q = cos(2437e6*15e-9)? – toozie21 Jul 1 '14 at 17:02
• @toozie21: If I understand your problem correctly, you need to apply the complex correction factor to the complex (combined) IQ signal, i.e. $(I+iQ)\cdot C$, where $C$ is the correction factor. By taking the real and imaginary parts of this corrected signal you get the new I and Q components. – Matt L. Jul 1 '14 at 17:23
• I think you are on track with what I am wanting. So in this case you are thinking that I would have a C of 36.55 and I would multiply that by I and iQ? Am I understanding your suggesting right? – toozie21 Jul 1 '14 at 17:29
• @toozie21: $C=e^{-i\omega_{lo}\tau}$ is the complex correction factor, so it can't be 36.55, but it is of the form $x+iy$ with $x=\cos\omega_{lo}\tau$ and $y=-\sin\omega_{lo}\tau$. – Matt L. Jul 1 '14 at 18:02