---
title: "Optimum detector impulse response; Hilbert transform"
subject: "Communication Systems"
chapter: 9
key: "semester_6/communication_systems:optimum-detector-impulse-response-hilbert"
source: https://www.bctengineer.com/q/semester_6/communication_systems/optimum-detector-impulse-response-hilbert
site: BCT Engineering
---

# Optimum detector impulse response; Hilbert transform

Communication Systems · Chapter 9: Noise in Communication Systems

- Asked in: 2080 Chaitra, 2+6+2 marks
- Video: [What is a Matched Filter?](https://www.youtube.com/watch?v=Ci-EjiMJo3I) — Iain Explains Signals, Systems, and Digital Comms
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/optimum-detector-impulse-response-hilbert
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9

## Question

Define optimum detector and find the impulse response of optimum detector in the presence of additive white noise. Briefly explain Hilbert Transform.

## Answer

#### Optimum detector

An **optimum detector** is a receiver (filter followed by sampler and threshold device) that makes the decision with the **minimum probability of error**. For a known pulse in additive white Gaussian noise, this is achieved by the filter that **maximises the output signal-to-noise ratio at the sampling instant**; this optimum filter is the **matched filter**.

#### Impulse response of the optimum detector

**Set-up.** The received signal is $x(t)=s(t)+w(t)$, $0\le t\le T$, where $s(t)$ is a known pulse and $w(t)$ is white noise of two-sided PSD $N_0/2$. It passes through an LTI filter $h(t)\leftrightarrow H(f)$ and is sampled at $t=T$:

```text
 s(t)+w(t)    +--------+   y(t)    sample     y(T)
 ------------>|  h(t)  |---------->--/ ---->  decision
              +--------+           at t=T
```

Output $y(t)=s_o(t)+n_o(t)$ with

$$
s_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df
$$

We want $H(f)$ that maximises the output peak signal-to-noise ratio

$$
\eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}
$$

**Schwarz inequality:** $\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df$, with equality only when $\phi_1(f)=k\,\phi_2^*(f)$. Put $\phi_1=H(f)$ and $\phi_2=S(f)e^{j2\pi fT}$:

$$
\eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0}
$$

(using Rayleigh's theorem, $E$ = pulse energy). The maximum is reached when

$$
H_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT}
$$

**Impulse response.** Take the inverse transform. For real $s(t)$, $S^*(f)=S(-f)$, which is the transform of $s(-t)$; the factor $e^{-j2\pi fT}$ is a delay of $T$:

$$
\begin{aligned}
h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\
&= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\
&= k\,s(T-t)
\end{aligned}
$$

So the optimum (matched) filter impulse response is the input pulse **reversed in time and delayed by $T$**. Choosing the sampling instant $T$ equal to the pulse duration makes $h(t)$ causal.

```text
   s(t)                      h(t) = s(T - t)
    |\                          /|
    | \                        / |
    |  \                      /  |
 ---+---+---> t          ---+---+---> t
    0   T                   0   T
```

**Results:** maximum SNR $=2E/N_0$ depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it passes strongly the frequencies where the signal is strong.

#### Hilbert transform

**Hilbert transform.** The Hilbert transform $\hat g(t)$ of $g(t)$ is obtained by shifting the phase of every frequency component by $-90^\circ$ (positive frequencies) and $+90^\circ$ (negative frequencies), without changing amplitudes:

$$
\hat g(t)=g(t)*\frac{1}{\pi t}=\frac1\pi\int_{-\infty}^{\infty}\frac{g(\tau)}{t-\tau}d\tau,\qquad \hat G(f)=-j\,\operatorname{sgn}(f)\,G(f)
$$

Example: the Hilbert transform of $\cos\omega_0t$ is $\sin\omega_0t$. Properties: $g$ and $\hat g$ have the same amplitude spectrum and energy, and are orthogonal. Uses: SSB generation by the phase-shift method $s(t)=m(t)\cos\omega_ct\mp\hat m(t)\sin\omega_ct$, analytic signals $g_+(t)=g(t)+j\hat g(t)$, and band-pass signal representation.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What criterion defines an optimum detector for a known pulse in AWGN?
   A. Maximum bandwidth of the receiving filter
   B. Minimum probability of error, achieved by maximising output SNR at the sampling instant
   C. Minimum filter delay with unity gain
   D. Maximum noise power at the output
   Answer: B. The optimum detector minimises error probability, which for a known pulse in white Gaussian noise means maximising the output SNR at the sampling instant. (Key point: Optimum detector goal)

2. What transfer function maximises the output SNR?
   A. H(f) = k S*(f) e^(-j2 pi f T)
   B. H(f) = k S(f) e^(j2 pi f T)
   C. H(f) = k / S(f)
   D. H(f) = k |S(f)|^2
   Answer: A. Equality in the Schwarz inequality needs H to be proportional to the conjugate of S(f)e^(j2 pi f T), i.e. k S*(f) e^(-j2 pi f T). (Key point: Optimum transfer function)

3. What is the impulse response of the optimum detector?
   A. h(t) = k s(t - T), the pulse simply delayed by T
   B. h(t) = k s(t), the pulse itself
   C. h(t) = k s(-t) with no delay
   D. h(t) = k s(T - t), the pulse reversed in time and delayed by T
   Answer: D. The conjugate gives time reversal s(-t) and the exponential gives a delay T, so h(t) = k s(T - t). (Key point: Impulse response result)

4. What does the Hilbert transform do to a signal's frequency components?
   A. Shifts positive frequencies by -90 degrees and negative by +90 degrees, keeping amplitudes
   B. Doubles the amplitude of all components
   C. Removes the negative frequencies
   D. Shifts all components by +90 degrees
   Answer: A. H(f) = -j sgn(f), a pure phase shift. (Key point: Hilbert transform effect)

## Related in Communication Systems

- [Optimum detector impulse response in additive white noise](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aoptimum-detector-impulse-response) — Communication Systems, chapter 9, asked 2×
- [Noise equivalent bandwidth; optimum detector impulse response](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aneb-optimum-detector) — Communication Systems, chapter 9
- [Matched filter; impulse response of optimum detector](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amatched-filter-optimum-detector) — Communication Systems, chapter 9
- [Optimum detector; matched filter impulse response](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amatched-filter-impulse-response) — Communication Systems, chapter 9, asked 3×