The PI controller.

\(G_c(s)=K_p+\dfrac{K_i}{s}=\dfrac{K_p s+K_i}{s}\)

It therefore contributes one pole at the origin and one finite zero at \(s=-K_i/K_p\). Everything in the question follows from that.

E — "PI adds a pole at s = 0, hence the type increases by 1." TRUE. The system type is by definition the number of poles at the origin in the open-loop transfer function, so inserting 1/s raises it by one.

A — "PI increases the type by 1, therefore improves the steady-state error by one order." TRUE. Steady-state error depends on the static error constants:

\(e_{ss}(\text{step})=\dfrac{1}{1+K_p},\qquad e_{ss}(\text{ramp})=\dfrac{1}{K_v}\)

Raising the type from 0 to 1 makes \(K_p \to \infty\), so the step error becomes zero instead of finite; a Type-1 plant becomes Type-2 and tracks a ramp with zero error. This elimination of offset is the whole purpose of integral action.

B — "PI reduces the rise time." FALSE. Integral action is sluggish: it accumulates error over time, so the closed-loop response becomes slower and generally more oscillatory. The rise time increases and the overshoot tends to grow. (It is derivative action, PD, that speeds up the transient.)

C — "PI increases the bandwidth." FALSE. The 1/s term boosts low-frequency gain while contributing −90° of phase; this is lag-type compensation, which reduces the gain-crossover frequency and hence narrows the bandwidth. Slower response and reduced bandwidth are two statements of the same fact.

D — "PI is a high-pass filter." FALSE. Its magnitude is largest at low frequency (infinite at d.c.) and falls to K_p at high frequency — that is low-pass / lag behaviour. This also gives PI the useful side effect of attenuating high-frequency measurement noise.

Summary of the trade-off. PI buys zero steady-state error at the price of slower response, less bandwidth and reduced relative stability — which is why PID adds derivative action to recover the speed.

Hence, the correct answer is A and E only.