Your closedloop crossover frequency (when the magnitude of $P(s)\,C(s)$ is equal to one) lies at roughly 1 rad/s. This means that the feedback controller already causes the system to track reference signals that have a frequency content sufficiently below that. So adding feedforward that only approximates the inverse of the plant below the crossover frequency should not aid in the tracking performance, so that is why for both $\tau=1$ and $\tau=10$ the magnitude of the overall transfer functions both start to drop off around 1 rad/s, similar to if you wouldn't use any feedforward ($C_f(s)=0$).

However, for $\tau=1$ the magnitude of $C_f(s)\,P(s)$ near 1 rad/s is still close to 0 dB, but the phase at 1 rad/s has already dropped to -135°, so the actual feedforward is actually doing partially the opposite of what the ideal feedforward would do. For $\tau=10$ the magnitude of $C_f(s)\,P(s)$ near 1 rad/s is already close to -60 dB. So, even though the phase is already way lower compared when using $\tau=1$, the magnitude is so low that the feedback controller can counteract the little disturbance the feedforward term is causing near that frequency. So that is why $\tau=10$ performs better (less overshoot) than $\tau=1$. Though, it can be noted that for $\tau=10$ there is a 0.3 dB peak near 0.06 rad/s and an overshoot of 0.03 after 20 seconds, while the system with only feedback has no overshoot.

Your closedloop crossover frequency (when the magnitude of $P(s)\,C(s)$ is equal to one) lies at roughly 1 rad/s. This means that the feedback controller already causes the system to track reference signals that have a frequency content sufficiently below that. So adding feedforward that only approximates the inverse of the plant ($C_f(s)\,P(s)\approx1$) below the crossover frequency should not aid in the tracking performance, so that is why for both $\tau=1$ and $\tau=10$ the magnitude of the overall transfer functions both start to drop off around 1 rad/s, similar to if you wouldn't use any feedforward ($C_f(s)=0$).

However, for $\tau=1$ the magnitude of $C_f(s)\,P(s)$ near 1 rad/s is still close to 0 dB, but the phase at 1 rad/s has already dropped to -135°, so the actual feedforward is actually doing partially the opposite of what the ideal feedforward would do and thus steering away from better reference tracking. For $\tau=10$ the magnitude of $C_f(s)\,P(s)$ near 1 rad/s is already close to -60 dB. So, even though the phase is already way lower compared when using $\tau=1$, the magnitude is so low that the feedback controller can counteract the little disturbance the feedforward term is causing near that frequency. So that is why $\tau=10$ performs better (less overshoot) than $\tau=1$. Though, it can be noted that for $\tau=10$ there is a 0.3 dB peak near 0.06 rad/s and an overshoot of 0.03 after 20 seconds, while the system with only feedback and no feedforward has no overshoot.

Your closedloop crossover frequency (when the magnitude of $P(s)\,C(s)$ is equal to one) lies at roughly 1 rad/s. This means that the feedback controller already causes the system to track reference signals that have a frequency content sufficiently below that. So adding feedforward that only approximates the inverse of the plant below the crossover frequency should not aid in the tracking performance, so that is why for both $\tau=1$ and $\tau=10$ the magnitude of the overall transfer functions both start to drop off around 1 rad/s, similar to if you wouldn't use any feedforward ($C_f(s)=0$).

However, for $\tau=1$ the magnitude of $C_f(s)\,P(s)$ near 1 rad/s is still close to 0 dB, but the phase at 1 rad/s has already dropped to -135°, so the actual feedforward is actually doing partially the opposite of what the ideal feedforward would do. For $\tau=10$ the magnitude of $C_f(s)\,P(s)$ near 1 rad/s is already close to -60 dB. So, even though the phase is already way lower compared when using $\tau=1$, the magnitude is so low that the feedback controller can counteract the little disturbance the feedforward term is causing near that frequency. So that is why $\tau=10$ performs better than $\tau=1$.

Your closedloop crossover frequency (when the magnitude of $P(s)\,C(s)$ is equal to one) lies at roughly 1 rad/s. This means that the feedback controller already causes the system to track reference signals that have a frequency content sufficiently below that. So adding feedforward that only approximates the inverse of the plant below the crossover frequency should not aid in the tracking performance, so that is why for both $\tau=1$ and $\tau=10$ the magnitude of the overall transfer functions both start to drop off around 1 rad/s, similar to if you wouldn't use any feedforward ($C_f(s)=0$).

However, for $\tau=1$ the magnitude of $C_f(s)\,P(s)$ near 1 rad/s is still close to 0 dB, but the phase at 1 rad/s has already dropped to -135°, so the actual feedforward is actually doing partially the opposite of what the ideal feedforward would do. For $\tau=10$ the magnitude of $C_f(s)\,P(s)$ near 1 rad/s is already close to -60 dB. So, even though the phase is already way lower compared when using $\tau=1$, the magnitude is so low that the feedback controller can counteract the little disturbance the feedforward term is causing near that frequency. So that is why $\tau=10$ performs better (less overshoot) than $\tau=1$. Though, it can be noted that for $\tau=10$ there is a 0.3 dB peak near 0.06 rad/s and an overshoot of 0.03 after 20 seconds, while the system with only feedback has no overshoot.