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Reconstruction of signal from samples

Short answer

Reconstruction recovers a continuous signal from samples. The sampled spectrum contains the primary component and copies at multiples of ω_s. If sampling rate exceeds twice the signal bandwidth, an ideal low-pass filter recovers the signal exactly via sinc interpolation. Practical systems use zero-order hold (ZOH) or first-order hold (FOH) with transfer functions (1-e^{-Ts})/s and (Ts+1)/T((1-e^{-Ts})/s)² respectively.

Explain about reconstruction of original signal from sampled signal with the help of figure and necessary.

A sampled signal carries values only at t=kTt = kT. Reconstruction produces a continuous signal from these values, so that the output of a digital controller can drive an analog plant.

Spectrum of the sampled signal

The impulse-sampled signal has spectrum

X∗(jω)=1T∑n=−∞∞X(j(ω+nωs)),ωs=2πTX^*(j\omega) = \frac{1}{T}\sum_{n=-\infty}^{\infty} X(j(\omega + n\omega_s)), \quad \omega_s = \frac{2\pi}{T}

i.e. the original spectrum (primary component, n=0n=0) plus copies (complementary components) centred at ±ωs,±2ωs,…\pm\omega_s, \pm 2\omega_s, \dots

 |X*(jw)|
   _/\_      _/\_      _/\_
 --------+---------+---------> w
       -ws    0    ws
      copies  |  primary
         <-- ideal LPF -->
           -ws/2    ws/2

Condition for exact recovery

If x(t)x(t) is band-limited to ωm\omega_m and ωs>2ωm\omega_s > 2\omega_m (sampling theorem), the copies do not overlap. Then an ideal low-pass filter with gain TT and cut-off ωs/2\omega_s/2 keeps only the primary component and gives back x(t)x(t) exactly:

x(t)=∑kx(kT) sin⁡[ωs(t−kT)/2]ωs(t−kT)/2x(t) = \sum_{k} x(kT)\,\frac{\sin[\omega_s(t-kT)/2]}{\omega_s(t-kT)/2}

If ωs<2ωm\omega_s < 2\omega_m the copies overlap (aliasing) and recovery is impossible.

Practical reconstruction

The ideal filter is non-causal, so practical systems use hold circuits, which build the signal from past samples only:

  1. Zero order hold (ZOH): keeps x(kT)x(kT) constant until (k+1)T(k+1)T. Transfer function Gh0(s)=1−e−TssG_{h0}(s) = \frac{1-e^{-Ts}}{s}. Its magnitude T∣sin⁡(ωT/2)ωT/2∣T\left|\frac{\sin(\omega T/2)}{\omega T/2}\right| is a rough low-pass filter; it passes some high-frequency components (ripple) and adds a lag of ωT/2\omega T/2.
  2. First order hold (FOH): extrapolates with the slope of the last two samples. Gh1(s)=Ts+1T(1−e−Tss)2G_{h1}(s) = \frac{Ts+1}{T}\left(\frac{1-e^{-Ts}}{s}\right)^2.
 samples:   .   .   .
 ZOH:      |‾‾‾|‾‾‾|‾‾‾   staircase
 FOH:      /  /  /        ramps

Choosing a small TT (high sampling rate) and following the hold with a smoothing filter makes the reconstructed signal close to the original.

✓ Answer checked by Claude Opus 5.5 (high effort)

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