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Ideal band pass and low pass filters (DT)

Short answer

Ideal lowpass filter with cutoff ω_c has h_LP[n]=sin(ω_c n)/(π n), h[0]=ω_c/π. Ideal bandpass with passband ω_1 to ω_2 has h_BP[n]=[sin(ω_2 n)-sin(ω_1 n)]/(π n) = 2 cos(ω_0 n) sin(W n)/(π n). Both are non-causal and not absolutely summable, so unrealizable.

Explain the impulse response of ideal band pass and low pass filters for discrete time signal.

Ideal filters have unit gain in the passband and zero gain in the stopband. A DT frequency response is periodic with period 2π2\pi, so it is defined over −π≤ω≤π-\pi\le\omega\le\pi. The impulse response is found by the inverse DTFT, h[n]=12π∫−ππH(ejω)ejωndωh[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}H(e^{j\omega})e^{j\omega n}d\omega.

Ideal low pass filter

HLP(ejω)={1,∣ω∣≤ωc0,ωc<∣ω∣≤πH_{LP}(e^{j\omega})=\begin{cases}1,&|\omega|\le\omega_c\\0,&\omega_c<|\omega|\le\pi\end{cases} hLP[n]=12π∫−ωcωcejωndω=sin⁡(ωcn)πn,hLP[0]=ωcπh_{LP}[n]=\frac{1}{2\pi}\int_{-\omega_c}^{\omega_c}e^{j\omega n}d\omega=\frac{\sin(\omega_c n)}{\pi n},\qquad h_{LP}[0]=\frac{\omega_c}{\pi}

Ideal band pass filter

Passband ω1≤∣ω∣≤ω2\omega_1\le|\omega|\le\omega_2:

           H(e^jw)
   +--+              +--+
   |  |              |  |  1
 --+--+------+-------+--+---> w
 -w2 -w1     0      w1  w2

It equals an LPF with cut-off ω2\omega_2 minus an LPF with cut-off ω1\omega_1:

hBP[n]=sin⁡(ω2n)−sin⁡(ω1n)πn,hBP[0]=ω2−ω1πh_{BP}[n]=\frac{\sin(\omega_2 n)-\sin(\omega_1 n)}{\pi n},\qquad h_{BP}[0]=\frac{\omega_2-\omega_1}{\pi}

Using sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}, with centre ω0=ω1+ω22\omega_0=\frac{\omega_1+\omega_2}{2} and half-width W=ω2−ω12W=\frac{\omega_2-\omega_1}{2}:

hBP[n]=2cos⁡(ω0n)⋅sin⁡(Wn)πnh_{BP}[n]=2\cos(\omega_0 n)\cdot\frac{\sin(Wn)}{\pi n}

So the band pass impulse response is a low pass sinc (cut-off WW) modulated by 2cos⁡ω0n2\cos\omega_0 n, which shifts the passband to ±ω0\pm\omega_0.

Remarks

  • Both responses are sinc-shaped, non-zero for n<0n<0: the ideal filters are non-causal.
  • They decay only as 1/n1/n, so they are not absolutely summable: not BIBO stable, hence not realizable exactly. Practical filters truncate (window) and delay h[n]h[n].

✓ Answer checked by Claude Opus 5.5 (high effort)

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