- 2075 Bhadra (CS I) · 2+6 marks
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 is obtained by shifting the phase of every frequency component by (positive frequencies) and (negative frequencies), without changing the amplitudes:
Example: the Hilbert transform of is . 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 , gain and delay has
Its impulse response is the inverse Fourier transform:
where .
h(t) peak 2BK at t = td
| *
| * *
. . . . | . * * . . . .
----------------+----+-+----+-----------> t
(non-zero 0 td
for t < 0)
Causality check: a system is causal only if for all . The sinc function extends from to ; it is zero only at the discrete points , . For example at , in general. However large the delay is made, the oscillating tails for never vanish completely.
So for , 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 over a whole band.) Practical filters (Butterworth, Chebyshev) only approximate the ideal response; a large delay makes the truncated sinc a good approximation.
✓ Answer checked by Claude Opus 5.5 (high effort)
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