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Impulse response, transfer function and distortionless transmission

Short answer

The impulse response h(t) is the output for a unit impulse δ(t); the transfer function H(f) is the Fourier transform of h(t), giving amplitude and phase. For distortionless transmission, H(f) = K e^{-j2π f t_d}, so |H(f)| must be constant K and phase θ(f) linear, θ(f) = -2π f t_d. The amplitude response is a flat line.

Define impulse response and transfer function of a system. Derive the transfer function for a system that provides distortionless transmission and draw its amplitude response as well.

Impulse response and transfer function

  • Impulse response h(t)h(t): the output of an LTI system when the input is a unit impulse δ(t)\delta(t), with zero initial conditions. For any input, y(t)=x(t)∗h(t)=∫x(λ)h(t−λ) dλy(t) = x(t) * h(t) = \int x(\lambda)h(t-\lambda)\,d\lambda.
  • Transfer function H(f)H(f): the Fourier transform of the impulse response, H(f)=∫h(t)e−j2πft dt=Y(f)/X(f)H(f) = \int h(t)e^{-j2\pi ft}\,dt = Y(f)/X(f). It gives the gain ∣H(f)∣|H(f)| and phase shift θ(f)\theta(f) the system applies to each frequency: H(f)=∣H(f)∣ejθ(f)H(f) = |H(f)|e^{j\theta(f)}.

Transfer function for distortionless transmission

Transmission is distortionless when the output is an exact replica of the input except for a constant gain KK and a constant delay tdt_d:

y(t)=K x(t−td)y(t) = K\,x(t - t_d)

Take the Fourier transform of both sides, using the time-shift property x(t−td)↔X(f)e−j2πftdx(t - t_d) \leftrightarrow X(f)e^{-j2\pi f t_d}:

Y(f)=K X(f) e−j2πftdY(f) = K\,X(f)\,e^{-j2\pi f t_d}

Therefore

H(f)=Y(f)X(f)=K e−j2πftdH(f) = \frac{Y(f)}{X(f)} = K\,e^{-j2\pi f t_d}

and the impulse response is h(t)=K δ(t−td)h(t) = K\,\delta(t - t_d).

Comparing with H(f)=∣H(f)∣ejθ(f)H(f) = |H(f)|e^{j\theta(f)}, the two conditions are:

  1. Amplitude response constant: ∣H(f)∣=K|H(f)| = K for all frequencies in the signal band.
  2. Phase response linear in frequency: θ(f)=−2πftd\theta(f) = -2\pi f t_d, i.e. constant time delay td=−12πdθdft_d = -\frac{1}{2\pi}\frac{d\theta}{df} for every component. (A phase of −2πftd±nπ-2\pi f t_d \pm n\pi is also acceptable; ±π\pm\pi only inverts the sign.)

Amplitude and phase response

 |H(f)|
    |
  K +-------------------------
    |
    +-------------------------> f
   -B            0           B

 theta(f)
    |\
    | \   slope = -2*pi*td
 ---+--\----------------------> f
    |   \
    |    \

The amplitude response is a horizontal line at height KK and the phase response is a straight line through the origin with negative slope −2πtd-2\pi t_d. In practice these need to hold only over the bandwidth −B≤f≤B-B \le f \le B of the signal. If ∣H(f)∣|H(f)| varies, amplitude distortion results; if θ(f)\theta(f) is not linear, phase (delay) distortion results.

✓ Answer checked by Claude Opus 5.5 (high effort)

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