# frequency domain phase shift

While going through FFT I came to know that, shifting of phase in frequency domain results on shifting of time in time domain.

According to this, in wireless communications, from the transmitter side having 2 antennas,signal s1 transmitted from Ant 1 and phase shifted version of sig2 in frequency domain transmitted from Ant2, sig2 will become a delayed version of sig1. And when both the antennas transmitting sig1 and sig2 at the same time how sig2 will become a delayed version or a multipath component to sig 1.

• What's your question exactly? How to achieve a time shift of sig2? – Deve May 17 '13 at 11:10

Suppose one signal is $a_1(t) = \cos(\omega t)$ and the other is $a_2(t) = \cos(\omega t - \theta)$. Note the phase shift $\theta$ between the two signals, and note also that we can write $$a_2(t) = \cos(\omega t - \theta) = \cos\left(\omega\left(t-\frac{\theta}{\omega}\right)\right) = \cos(\omega (t - t_0)) = a_1(t-t_0)$$ where $t_0 = \frac{\theta}{\omega}$ showing that $a_2(t)$ is a delayed version of $a_1(t)$. Suppose that $a_1(t)$ and $a_2(t)$ are transmitted from the two separate antennas simultaneously, that is, no additional time delay between them, and reach the receiver over paths of delays $\tau_1$ and $\tau_2$ respectively and gains $A_1$ and $A_2$ respectively. Then the received signals are \begin{align} A_1a_1(t-\tau_1) &= A_1\cos(\omega (t -\tau_1)) = A_1\cos(\omega \hat{t})\\ A_2a_2(t-\tau_2) &= A_2\cos(\omega (t -t_0 - \tau_2))\\ &= A_2\cos(\omega (\hat{t} +\tau_1 -t_0-\tau_2))\\ &= A_2\cos(\omega (\hat{t} - (t_0 + \tau_2-\tau_1))\\ &= A_2\cos(\omega \hat{t} -\hat{\theta}) \end{align} where $\hat{t} = t-\tau_1$ is the time as measured by the receiver clock. Thus we see that the second signal is delayed with respect to the former by $t_0+\tau_2-\tau_1$, that is, the delay introduced at the transmitter plus the difference in the delays between the two paths, and this can be expressed as a phase shift of $\hat{\theta}= \frac{t_0+\tau_2-\tau_1}{\omega}$.
• If $\tau_1 = \tau_2$ so that the two paths are of identical lengths, the second signal looks like a multipath signal with respect to the first with a delay of $t_0$ (equivalently the same phase shift $\theta$ as existed at the transmitter).
• If $t_0 = 0$ so that effectively only one signal is transmitted, but is received over paths of different lengths, the second signal is a multipath signal with respect to the first with a delay of $\tau_2-\tau_1$ and it too can be expressed as a phase shifted version of the first signal.