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Impulse response; conditions for distortionless transmission

Short answer

Impulse response h(t) is the output of an LTI system to a unit impulse. Distortionless transmission requires constant amplitude response |H(f)|=K, linear phase theta(f)=-2 pi f td (constant delay), and a linear system, so output is K x(t-td).

What is impulse response? What are the conditions for distortionless transmission? Explain with necessary diagrams.

Impulse response

The impulse response h(t)h(t) of a linear time-invariant (LTI) system is its output when the input is a unit impulse δ(t)\delta(t) (zero initial conditions). It completely describes the system: for any input x(t)x(t), the output is the convolution y(t)=x(t)∗h(t)y(t) = x(t) * h(t), and its Fourier transform is the transfer function H(f)H(f).

Conditions for distortionless transmission

Transmission is distortionless if the output is an exact copy of the input, differing only by a constant gain KK and a constant delay tdt_d:

y(t)=K x(t−td)  ⇒  H(f)=K e−j2πftd,h(t)=K δ(t−td)y(t) = K\,x(t - t_d) \;\Rightarrow\; H(f) = K\,e^{-j2\pi f t_d}, \quad h(t) = K\,\delta(t - t_d)

So, over the whole band of the signal:

  1. Flat (constant) amplitude response: ∣H(f)∣=K|H(f)| = K. Every frequency component gets the same gain. Otherwise amplitude distortion occurs, e.g. high frequencies cut more than low ones, making sharp edges rounded.
  2. Linear phase response: θ(f)=−2πftd\theta(f) = -2\pi f t_d (or −2πftd±nπ-2\pi f t_d \pm n\pi). The phase shift is proportional to frequency, so every component has the same delay tdt_d (constant group delay τg=−12πdθdf=td\tau_g = -\frac{1}{2\pi}\frac{d\theta}{df} = t_d). Otherwise phase (delay) distortion occurs, which spreads pulses and causes ISI in data transmission.
  3. The system must be linear; a non-linear system creates harmonics and intermodulation (non-linear distortion).
 Input x(t)           Output y(t) = K x(t - td)
   |  __                 |        __
   | /  \                |       /  \   (K times)
   |/    \               |      /    \
 --+------\---> t      --+-----/------\---> t
                         |<-td->|
 |H(f)|                 theta(f)
   |                       |
 K +---------------        |\
   |                       | \  slope -2*pi*td
   |                     --+--\-----------> f
 --+---------------> f     |   \
  -B       0      B        |    \

The required band is only the signal bandwidth ∣f∣≤B|f| \le B; outside it, H(f)H(f) may be anything. Real channels are not ideal, so equalizers are used to make the overall response flat in amplitude and linear in phase.

✓ Answer checked by Claude Opus 5.5 (high effort)

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