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Suppose I have a discrete-time signal vector, for example, x(n)=[1,a1,a2,…,aN]. The signal is then transmitted by using the single carrier pulses, constituting a single-carrier communication over a multipath channel. Hence, at the receiver side, the received signal experiences also the zeros of the z-transform of the channel impulse response. However, due to a frequency offset caused by the wireless multipath channel, the received signal is affected. Upon reception, the signal exhibits a CFO effect.

1- Could you please help me how to interpret this carrier frequency offset effect on the zeros of the z-transform of the received signal?

2- How a carrier frequency offset (CFO) causes a movement of all zeros of the received signal’s z−transform, from their original positions on the z-domain?

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The carrier offset moves the signal relative to the zeros of the channel as a shift in frequency assuming that the shift happened before the channel. (If the shift happens after the channel such as due to offsets in our local oscillator, then the channel zeros do not change relative to the signal).

Assuming it is the former case (a shift before the channel) and if we clarify a positive carrier offset to mean that the signal has been shifted higher in frequency, then if we hold the signal as a reference, the zeros of the channel shift in frequency in the opposite direction, relative to the signal. So the carrier offset is just a rotation of the Z-plane. Given a frequency offset $\omega_\Delta$ in normalized angular frequency radians/sample, the rotation will be $-\omega_\Delta$ radians.

CFO effect on zeros

For example consider if we had a channel zero directly at the frequency of the carrier. On the z-plane this would be located at $z=1$, which is DC as the baseband equivalent of the passband signal. If we had a carrier offset of +1 KHz for the signal, meaning the carrier was shifted 1 KHz higher, and if the received signal was sampled at 100 KHz, then the zero will now be at -1KHz relative to the original carrier frequency, or $-\pi/50$ radians/sample. (The graphic above shows a significantly larger rotation than this.)

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