Detection of Periodic Signals in the Presence of Noise (by Cross-Correlation)

Signals and SystemsElectronics & ElectricalDigital Electronics

Detection of Periodic Signals in the Presence of Noise

The noise signal is an unwanted signal which has random amplitude variation. The noise signals are uncorrelated with any periodic signal.

Detection of the periodic signals masked by noise signals is of great importance in signal processing. It is mainly used in the detection of radar and sonar signals, the detection of periodic components in brain signals, in the detection of periodic components in sea wave analysis and in many other areas of geophysics etc. The solution of these problems can be easily provided by thecorrelation techniques. The cross-correlation function, therefore can be used to detect a periodic signal which masked by a noise signal.

Consider $\mathit{x}\mathrm{(\mathit{t})}$ is a periodic signal and $\mathit{n}\mathrm{(\mathit{t})}$ is the noise signal. Then, the cross-correlation function of the signals $\mathit{x}\mathrm{(\mathit{t})}$ and $\mathit{n}\mathrm{(\mathit{t})}$ is given by,

$$\mathit{R_{xn}}\mathrm{(\mathit{\tau})}\:\mathrm{=}\:\lim_{\mathit{T} \rightarrow \infty}\frac{\mathrm{1}}{\mathit{T}}\int_{-\mathit{T/\mathrm{2}}}^{\mathit{T/\mathrm{2}}}\mathit{x}\mathrm{(\mathit{t})}\mathit{n}\mathrm{(\mathit{t-\tau})}\mathit{dt}\:\mathrm{=}\:{0}; (for\: all \:\mathit{\tau})$$

Detection of Periodic Signal by Cross Correlation

The cross-correlation can be used to detect a periodic signal which is mixed with another periodic signal of the same frequency. The disadvantage of the detection by cross-correlation is that it is necessary to know beforehand the frequency of the signal to be detected.

Now, consider the periodic signal $\mathit{x}\mathrm{(\mathit{t})}$is mixed with a noise signal $\mathit{n}\mathrm{(\mathit{t})}$, then the received signal is given by,

$$\mathit{y}\mathrm{(\mathit{t})}\:\mathrm{=}\:\mathit{x}\mathrm{(\mathit{t})}\mathrm{+}\mathit{n}\mathrm{(\mathit{t})}$$

Also, consider$\mathit{z}\mathrm{(\mathit{t})}$is a locally generated signal of the same frequency as that of the periodic signal$\mathit{x}\mathrm{(\mathit{t})}$. Therefore, the cross correlation function of$\mathit{y}\mathrm{(\mathit{t})}$ and $\mathit{z}\mathrm{(\mathit{t})}$ is given by,

$$\mathit{R_{yz}}\mathrm{(\mathit{\tau})}\:\mathrm{=}\:\lim_{\mathit{T} \rightarrow \infty}\frac{\mathrm{1}}{\mathit{T}}\int_{-\mathit{T/\mathrm{2}}}^{\mathit{T/\mathrm{2}}}\mathrm{\mathit{y}\mathrm{(\mathit{t})}\mathit{z}\mathrm{(\mathit{t-\tau})}}\mathit{dt}$$ $$\Rightarrow\mathit{R_{yz}}\mathrm{(\mathit{\tau})}\:\mathrm{=}\:\lim_{\mathit{T} \rightarrow \infty}\frac{\mathrm{1}}{\mathit{T}}\int_{-\mathit{T/\mathrm{2}}}^{\mathit{T/\mathrm{2}}}\mathrm{[\mathit{x}\mathrm{(\mathit{t})}\mathrm{+}\mathit{n}\mathrm{(\mathit{t})}]}\mathit{z}\mathrm{(\mathit{t-\tau})}\mathit{dt}$$ $$\Rightarrow\mathit{R_{yz}}\mathrm{(\mathit{\tau})}\:\mathrm{=}\:\lim_{\mathit{T} \rightarrow \infty}\frac{\mathrm{1}}{\mathit{T}}\int_{-\mathit{T/\mathrm{2}}}^{\mathit{T/\mathrm{2}}}\mathrm{\mathit{x}\mathrm{(\mathit{t})}}\mathit{z}\mathrm{(\mathit{t-\tau})}\mathit{dt}\mathrm{+}\lim_{\mathit{T} \rightarrow \infty}\frac{\mathrm{1}}{\mathit{T}}\int_{-\mathit{T/\mathrm{2}}}^{\mathit{T/\mathrm{2}}}\mathrm{\mathit{n}\mathrm{(\mathit{t})}}\mathit{z}\mathrm{(\mathit{t-\tau})}\mathit{dt}$$ $$\therefore\mathit{R_{yz}}\mathrm{(\mathit{\tau})}\:\mathrm{=}\:\mathit{R_{xz}}\mathrm{(\mathit{\tau})}\mathrm{+}\mathit{R_{nz}}\mathrm{(\mathit{\tau})}$$

As the function $\mathit{z}\mathrm{(\mathit{t})}$ is a periodic function and it is uncorrelated with the noise signal $\mathit{n}\mathrm{(\mathit{t})}$. Thus, their cross-correlation function is equal to zero, i.e.,

$$\mathit{R_{nz}}\mathrm{(\mathit{\tau})}\:\mathrm{=}\:{0}$$ $$\therefore\mathit{R_{yz}}\mathrm{(\mathit{\tau})} \:\mathrm{=}\:\mathit{R_{xz}}\mathrm{(\mathit{\tau})}$$

Here, the signals $\mathit{x}\mathrm{(\mathit{t})}$ and $\mathit{z}\mathrm{(\mathit{t})}$ are signals of the same frequency. Therefore, the correlation function $\mathit{R_{xz}}\mathrm{(\mathit{\tau})}$ is also a periodic function of the same frequency. Hence, if the cross-correlation of the mixed signal $\mathit{y}\mathrm{(\mathit{t})}$ with $\mathit{z}\mathrm{(\mathit{t})}$ results a periodic signal, then the signal $\mathit{y}\mathrm{(\mathit{t})}$ must contain a periodic component of the same frequency as that of the signal $\mathit{z}\mathrm{(\mathit{t})}$. In this way, we can detect a periodic signal in the presence of noise using cross-correlation.

raja
Updated on 07-Jan-2022 11:24:48

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