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Hilbert transform; ideal LPF is non-causal

Short answer

Hilbert transform shifts all frequency components by -90 degrees for positive frequencies and +90 for negative, equivalent to convolution with 1/(pi t). An ideal low-pass filter has impulse response 2BK sinc[2B(t-td)], which is nonzero for t less than td, so the filter is non-causal.

Define Hilbert transformation. Show that impulse response of an ideal low pass filter is non-causal.

Hilbert transform

The Hilbert transform of a signal x(t)x(t) is obtained by shifting the phase of every frequency component by −90∘-90^\circ (positive frequencies) and +90∘+90^\circ (negative frequencies), without changing the amplitudes:

x^(t)=1π∫−∞∞x(τ)t−τ dτ=x(t)∗1πt\hat{x}(t) = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{x(\tau)}{t-\tau}\,d\tau = x(t) * \frac{1}{\pi t} X^(f)=−j sgn(f) X(f)\hat{X}(f) = -j\,\text{sgn}(f)\,X(f)

Example: the Hilbert transform of cos⁡2πfct\cos 2\pi f_c t is sin⁡2πfct\sin 2\pi f_c t. It is used in SSB generation (phase-shift method) and in analytic signal / pre-envelope representation.

Impulse response of an ideal LPF is non-causal

An ideal low-pass filter of bandwidth BB, gain KK and delay tdt_d has

H(f)={K e−j2πftd,∣f∣≤B0,∣f∣>BH(f) = \begin{cases} K\,e^{-j2\pi f t_d}, & |f| \le B \\ 0, & |f| > B \end{cases}

Its impulse response is the inverse Fourier transform:

h(t)=∫−BBK e−j2πftdej2πft df=K∫−BBej2πf(t−td) df=K ej2πB(t−td)−e−j2πB(t−td)j2π(t−td)=K sin⁡2πB(t−td)π(t−td)=2BK sinc[2B(t−td)]\begin{aligned} h(t) &= \int_{-B}^{B} K\,e^{-j2\pi f t_d}e^{j2\pi f t}\,df = K\int_{-B}^{B} e^{j2\pi f(t-t_d)}\,df \\ &= K\,\frac{e^{j2\pi B(t-t_d)} - e^{-j2\pi B(t-t_d)}}{j2\pi(t-t_d)} \\ &= K\,\frac{\sin 2\pi B(t-t_d)}{\pi(t-t_d)} = 2BK\,\text{sinc}\left[2B(t-t_d)\right] \end{aligned}

where sinc(x)=sin⁡(πx)/(πx)\text{sinc}(x) = \sin(\pi x)/(\pi x).

             h(t)  peak 2BK at t = td
               |       *
               |      * *
     . .  .  . |  .  *   *  .  .   . .
 ----------------+----+-+----+-----------> t
   (non-zero     0    td
    for t < 0)

Causality check: a system is causal only if h(t)=0h(t) = 0 for all t<0t < 0. The sinc function extends from −∞-\infty to +∞+\infty; it is zero only at the discrete points t=td+n/(2B)t = t_d + n/(2B), n≠0n \ne 0. For example at t=−14Bt = -\frac{1}{4B}, h(t)≠0h(t) \ne 0 in general. However large the delay tdt_d is made, the oscillating tails for t<0t < 0 never vanish completely.

So h(t)≠0h(t) \ne 0 for t<0t < 0, i.e. the filter would respond before the impulse is applied. Hence the ideal LPF is non-causal and physically unrealizable. (This also follows from the Paley–Wiener criterion: a causal filter cannot have ∣H(f)∣=0|H(f)| = 0 over a whole band.) Practical filters (Butterworth, Chebyshev) only approximate the ideal response; a large delay tdt_d makes the truncated sinc a good approximation.

✓ Answer checked by Claude Opus 5.5 (high effort)

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