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This module examines signal power, looking at instantaneous and average power. It uses orthogonality properties to derive a simple expression for average power. It also defines and displays a power spectrum.

An interesting question you could ask about a signal is its average power . A signal's instantaneous power is defined to be its square, as if it were a voltage or current passing through a 1Ωresistor. The average power is the average of the instantaneous power over some time interval. For a periodicsignal, the natural time interval is clearly its period; for nonperiodic signals, a better choice would be entire time ortime from onset. For a periodic signal, the average power is the square of the root-mean-squared (rms) value. We define therms value of a periodic signal to be

rms s 2 1 T t 0 T s t 2
and thus its average power is rms s 2 .
power s rms s 2 1 T t 0 T s t 2

What is the rms value of the half-wave rectified sinusoid?

A half-wave rectified sinusoid has half the average power of the original sine wave since it is zero half the time. A sine wave's average power equals A 2 2 , making the rms value of the half-wave rectified signal A 2 .

To find the average power in the frequency domain, we need to substitute the spectral representation of the signal into thisexpression.

power s 1 T t 0 T a 0 k 1 a k 2 k t T k 1 b k 2 k t T 2
The square inside the integral will contain all possible pairwise products. However, the orthogonality properties say that most of these crossterms integrate to zero. The survivors leave a rather simple expression for thepower we seek.
power s a 0 2 1 2 k 1 a k 2 b k 2

Power spectrum of a half-wave rectified sinusoid

Power spectrum of a half-wave rectified sinusoid.

It could well be that computing this sum is easier than integrating the signal's square. Furthermore, the contributionof each term in the Fourier series toward representing the signal can be measured by its contribution to the signal'saverage power. Thus, the power contained in a signal at its k th harmonic is a k 2 b k 2 2 . The power spectrum P s k , such as shown in , plots each harmonic's contribution to the total power.

In stereophonic systems, deviation of a sine wave from the ideal is measured by the total harmonic distortion, whichequals the total power in the harmonics higher than the first compared to power in the fundamental. Find anexpression for the total harmonic distortion for any periodic signal. Is this calculation most easily performedin the time or frequency domain?

Total harmonic distortion equals k 2 a k 2 b k 2 a 1 2 b 1 2 . Clearly, this quantity is most easily computed in the frequency domain. However, the numerator equals the thesquare of the signal's rms value minus the power in the average and the power in the first harmonic.

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Source:  OpenStax, Signal and information processing for sonar. OpenStax CNX. Dec 04, 2007 Download for free at http://cnx.org/content/col10422/1.5
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