The correlator is just a filtering operation. So in the case of passband pulse-amplitude modulation (PAM), such as QPSK or QAM, the (noiseless) received analytic signal before demodulation and sampling can be written as

$$r(t)=e^{j((\omega_c+\Delta\omega) t+\phi))}\sum_{k}A_kp(t-kT)\tag{1}$$

where $\Delta\omega$ is a frequency offset, $\phi$ is a phase offset, $\omega_c$ is the carrier frequency as estimated by the receiver, $A_k$ are the complex data symbols, and $T$ is the symbol period. The pulse $p(t)$ represents the combination of the transmit filter, linear distortions on the channel, and the receive filter (correlator).

Demodulation with $e^{-j\omega_ct}$ and sampling at the symbol rate $1/T$ gives

$$\tilde{r}[n]=e^{j(\Delta\omega nT +\phi)}\sum_kA_kp((n-k)T)\tag{2}$$

If for simplicity we assume that the combined pulse $p(t)$ satisfies the Nyquist criterion $p(kT)=\delta[k]$ (i.e., there is no intersymbol interference), then $(2)$ simplifies to

$$\tilde{r}[n]=e^{j(\Delta\omega nT +\phi)}A_n\tag{3}$$

which should make it clear that at time $t=nT$ the received symbol is a rotated version of the original symbol, where the rotation has a constant component $\phi$ (the phase offset), and a time-dependent component $\Delta\omega nT$ caused by the frequency offset. Consequently, if $\Delta\omega\neq 0$, the constellation rotates.

The correlator is just a filtering operation. So in the case of passband pulse-amplitude modulation (PAM), such as QPSK or QAM, the (noiseless) received analytic signal before demodulation and sampling can be written as

$$r(t)=e^{j((\omega_c+\Delta\omega) t+\phi))}\sum_{k}A_kp(t-kT)\tag{1}$$

where $\Delta\omega$ is a frequency offset, $\phi$ is a phase offset, $\omega_c$ is the carrier frequency as estimated by the receiver, $A_k$ are the complex data symbols, and $T$ is the symbol period. The pulse $p(t)$ represents the combination of the transmit filter, linear distortions on the channel, and the receive filter (correlator).

Demodulation with $e^{-j\omega_ct}$ and sampling at the symbol rate $1/T$ gives

$$\tilde{r}[n]=e^{j(\Delta\omega nT +\phi)}\sum_kA_kp((n-k)T)\tag{2}$$

If for simplicity we assume that the combined pulse $p(t)$ satisfies the Nyquist criterion $p(kT)=\delta[k]$ (i.e., there is no inter-symbol interference), then $(2)$ simplifies to

$$\tilde{r}[n]=e^{j(\Delta\omega nT +\phi)}A_n\tag{3}$$

which should make it clear that at time $t=nT$ the received symbol is a rotated version of the original symbol, where the rotation has a constant component $\phi$ (the phase offset), and a time-dependent component $\Delta\omega nT$ caused by the frequency offset. Consequently, if $\Delta\omega\neq 0$, the constellation rotates.

The correlator is just a filtering operation. So in the case of passband pulse-amplitude modulation (PAM), such as QPSK or QAM, the (noiseless) received analytic signal before demodulation and sampling can be written as

$$r(t)=e^{j((\omega_c+\Delta\omega) t+\phi))}\sum_{k}A_kp(t-kT)\tag{1}$$

where $\Delta\omega$ is a frequency offset, $\phi$ is a phase offset, $\omega_c$ is the carrier frequency as estimated by the receiver, $A_k$ are the complex data symbols, and $T$ is the symbol rate. The pulse $p(t)$ represents the combination of the transmit filter, linear distortions on the channel, and the receive filter (correlator).

Demodulation with $e^{-j\omega_ct}$ and sampling at the symbol rate $nT$ gives

$$\tilde{r}[n]=e^{j(\Delta\omega nT +\phi)}\sum_kA_kp((n-k)T)\tag{2}$$

If for simplicity we assume that the combined pulse $p(t)$ satisfies the Nyquist criterion $p(kT)=\delta[k]$ (i.e., there is no intersymbol interference), then $(2)$ simplifies to

$$\tilde{r}[n]=e^{j(\Delta\omega nT +\phi)}A_n\tag{3}$$

which should make it clear that at time $t=nT$ the received symbol is a rotated version of the original symbol, where the rotation has a constant component $\phi$ (the phase offset), and a time-dependent component $\Delta\omega nT$ caused by the frequency offset. Consequently, if $\Delta\omega\neq 0$, the constellation rotates.

The correlator is just a filtering operation. So in the case of passband pulse-amplitude modulation (PAM), such as QPSK or QAM, the (noiseless) received analytic signal before demodulation and sampling can be written as

$$r(t)=e^{j((\omega_c+\Delta\omega) t+\phi))}\sum_{k}A_kp(t-kT)\tag{1}$$

where $\Delta\omega$ is a frequency offset, $\phi$ is a phase offset, $\omega_c$ is the carrier frequency as estimated by the receiver, $A_k$ are the complex data symbols, and $T$ is the symbol period. The pulse $p(t)$ represents the combination of the transmit filter, linear distortions on the channel, and the receive filter (correlator).

Demodulation with $e^{-j\omega_ct}$ and sampling at the symbol rate $1/T$ gives

$$\tilde{r}[n]=e^{j(\Delta\omega nT +\phi)}\sum_kA_kp((n-k)T)\tag{2}$$

If for simplicity we assume that the combined pulse $p(t)$ satisfies the Nyquist criterion $p(kT)=\delta[k]$ (i.e., there is no intersymbol interference), then $(2)$ simplifies to

$$\tilde{r}[n]=e^{j(\Delta\omega nT +\phi)}A_n\tag{3}$$

which should make it clear that at time $t=nT$ the received symbol is a rotated version of the original symbol, where the rotation has a constant component $\phi$ (the phase offset), and a time-dependent component $\Delta\omega nT$ caused by the frequency offset. Consequently, if $\Delta\omega\neq 0$, the constellation rotates.